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[0812.4763-English-]0812.4763.English[http://arxiv.org/PS_cache/arxiv/pdf/0812/0812.4763v6.pdf]

Quaternion Rhapsody

Aleks Kleyn Dedicated to I. M. Gelfand
Abstract.

In this paper I explore the set of quaternion algebras over field. Quaternion algebra E(C,1,1)E(C,-1,-1) is isomorphic to tensor product of complex field CC and quaternion algebra H=E(R,1,1)H=E(R,-1,-1). Considered the set of quaternion functions, which satisfy to equation similar to Cauchy-Riemann equation for a complex function.

Key words and phrases:
algebra, linear algebra, division ring, derivative

1. Preface

When I started my research in area of calculus over division ring, I paid attention to large volume of papers dedicated to regular functions of quaternion and analogue of the Cauchy-Riemann equation. I wanted to understand whether the Cauchy-Riemann equation appears in the frame of the theory that I explore. This was the reason why I considered division ring as vector space over center.

Introduction of basis simplifies some constructions and presents a bridge between the Gâteaux derivative and Jacobian matrix of map. This is exactly the place, where the Cauchy-Riemann equations should appear. The exploration of derivative of function of complex numbers reveals that the Cauchy-Riemann equation has algebraic origin and is related with statement that there exists RR-linear function over complex field, however this function is not CC-linear. For instance, conjugation of complex number is linear over real field, however it is not linear map over complex field.

In quaternion algebra HH, there exist only RR-linear map. The corollary of this statement is the ability represent every RR-linear map using quaternion and absence of the evident analogue of the Cauchy-Riemann equation in quaternion algebra. However, in numerous papers and books dedicated to calculus over quaternion algebra mathematicians explore sets of maps that have properties similar to properties of functions of complex variable. Some of authors do not restrict themselves to quaternion algebra and explore more general algebras.

In the paper [3], Gelfand explores the quaternion algebra over arbitrary field assuming that product depends on arbitrary parameters. We assume H=E(R,1,1)H=E(R,-1,-1). Using this paper, I decided to explore two cases that are important for me.

The algebra E(R,a,b)E(R,a,b) was interesting for me because I supposed to find parameters aa, bb such that the system of linear equations [2]-(LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation) is singular. It was important to understand what happens in this case. When I explored the structure of linear map over division ring it was not evident how non singularity of system of linear equations [2]-(LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation) have an influence on answer. However, the solution to this problem was not the one I had expected. It turned out that the system of linear equations is so simple that everybody can see that this system cannot be singular.

I have wrote that the Cauchy-Riemann equation is related with statement that complex field has real field as subfield. I assumed also that similar statement is possible in algebras with enough aggregate center. This is why I expected to see analogue of the Cauchy-Riemann equation in the quaternion algebra over complex field. In the course of solving the problem I realized that algebra E(C,1,1)E(C,-1,-1) is isomorphic to tensor product CHC\otimes H. Therefore linear functions of this algebra satisfy to the Cauchy-Riemann equation for CC-component of tensor product. Therefore I can tell the same about Jacobian matrix of arbitrary function. Natural extension of this topic is exploration of tensor product C(CH)C\otimes(C\otimes H). In particular, I put my attention is non associativity of tensor product.

In the book [2] on the base of which I wrote this paper, I explore the Gâteaux derivative of function over division ring. However in this paper I consider arbitrary algebras that not always are division rings. During the time that I explore the Gâteaux derivative, I realized that this subject can be generalized to more wide set of algebras. I will prepare the complete research later. I wrote this paper in order to explore conditions when the Cauchy-Riemann equations are possible.

2. Conventions

  1. (1)

    Function and mapping are synonyms. However according to tradition, correspondence between either rings or vector spaces is called mapping and a mapping of either real field or quaternion algebra is called function.

  2. (2)

    We can consider division ring DD as DD-vector space of dimension 11. According to this statement, we can explore not only homomorphisms of division ring D1D_{1} into division ring D2D_{2}, but also linear maps of division rings. This means that map is multiplicative over maximum possible field. In particular, linear map of division ring DD is multiplicative over center Z(D)Z(D). This statement does not contradict with definition of linear map of field because for field FF is true Z(F)=FZ(F)=F. When field FF is different from maximum possible, I explicit tell about this in text.

  3. (3)

    Let AA be free finite dimensional algebra. Considering expansion of element of algebra AA relative basis e¯¯\overline{\overline{e}}{} we use the same root letter to denote this element and its coordinates. However we do not use vector notation in algebra. In expression a2a^{2}, it is not clear whether this is component of expansion of element aa relative basis, or this is operation a2=aaa^{2}=aa. To make text clearer we use separate color for index of element of algebra. For instance,

    a=a𝒊e¯𝒊a=a^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}
  4. (4)

    If free finite dimensional algebra has unit, then we identify the vector of basis e¯𝟎\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} with unit of algebra.

  5. (5)

    Without a doubt, the reader may have questions, comments, objections. I will appreciate any response.

3. Linear Function of Complex Field

Theorem 3.1 (the Cauchy-Riemann equations).

Let us consider complex field CC as two-dimensional algebra over real field. Let

(3.1) e¯=C𝟎1e¯=C𝟏i\begin{array}[]{rr}\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=1&\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=i\end{array}

be the basis of algebra CC. Then in this basis product has form

(3.2) e¯=C𝟏2e¯C𝟎\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{2}=-\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0}}}

and structural constants have form

(3.3) CC=𝟎𝟎𝟎1CC=𝟎𝟏𝟏1CC=𝟏𝟎𝟏1CC=𝟏𝟏𝟎1\begin{array}[]{r@{}rr@{}r}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&1&C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1&C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-1\end{array}

Matrix of linear function

y𝒊=x𝒋f𝒋𝒊y^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}=x^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}

of complex field over real field satisfies relationship

(3.4) f𝟎𝟎\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =f𝟏𝟏\displaystyle=f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
(3.5) f𝟎𝟏\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =f𝟏𝟎\displaystyle=-f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
Proof.

Equations (3.2) and (3.3) follow from equation i2=1i^{2}=-1. Using equation [2]-(LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation) we get relationships

(3.6) f𝟎𝟎=f𝒌𝒓CCCC𝒌𝟎𝒑=𝒑𝒓𝟎f𝟎𝒓CCCC𝟎𝟎𝟎+𝟎𝒓𝟎f𝟏𝒓CCCC𝟏𝟎𝟏=𝟏𝒓𝟎f𝟎𝟎f𝟏𝟏f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}
(3.7) f𝟎𝟏=f𝒌𝒓CCCC𝒌𝟎𝒑=𝒑𝒓𝟏f𝟎𝒓CCCC𝟎𝟎𝟎+𝟎𝒓𝟏f𝟏𝒓CCCC𝟏𝟎𝟏=𝟏𝒓𝟏f𝟎𝟏+f𝟏𝟎f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}
(3.8) f𝟏𝟎=f𝒌𝒓CCCC𝒌𝟏𝒑=𝒑𝒓𝟎f𝟎𝒓CCCC𝟎𝟏𝟏+𝟏𝒓𝟎f𝟏𝒓CCCC𝟏𝟏𝟎=𝟎𝒓𝟎f𝟎𝟏f𝟏𝟎f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=-f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}-f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}
(3.9) f𝟏𝟏=f𝒌𝒓CCCC𝒌𝟏𝒑=𝒑𝒓𝟏f𝟎𝒓CCCC𝟎𝟏𝟏+𝟏𝒓𝟏f𝟏𝒓CCCC𝟏𝟏𝟎=𝟎𝒓𝟏f𝟎𝟎f𝟏𝟏f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0r}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}

(3.4) follows from equations (3.6) and (3.9). (3.5) follows from equations (3.7) and (3.8). ∎

Theorem 3.2 (the Cauchy-Riemann equations).

Since matrix

(y𝟎x𝟎y𝟎x𝟏y𝟏x𝟎y𝟏x𝟏)\begin{pmatrix}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\\ \vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\end{pmatrix}

is Jacobian matrix of map of complex variable

x=x𝟎+x𝟏iy=y𝟎(x𝟎,x𝟏)+y𝟏(x𝟎,x𝟏)ix=x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i\rightarrow y=y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}},x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})+y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}},x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})i

over real field, then

(3.10) y𝟏x𝟎=y𝟎x𝟏y𝟎x𝟎=y𝟏x𝟏\begin{array}[]{r@{}r}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}=&\displaystyle-\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\\ \vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}=&\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\end{array}
Proof.

The statement of theorem is corollary of theorem 3.1. ∎

Theorem 3.3.

Derivative of function of complex variable satisfyes to equation

(3.11) yx𝟎+iyx𝟏=0\frac{\partial y}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}+i\frac{\partial y}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}=0
Proof.

Equation

y𝟎x𝟎+iy𝟏x𝟎+iy𝟎x𝟏y𝟏x𝟏=0\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}+i\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}+i\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}=0

follows from equations (3.10). ∎

Equation (3.11) is equivalent to equation

(3.12) (1i)(y𝟎x𝟎y𝟎x𝟏y𝟏x𝟎y𝟏x𝟏)(1i)=0\begin{pmatrix}1&i\end{pmatrix}\begin{pmatrix}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\\ \vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}\end{pmatrix}\begin{pmatrix}1\\ i\end{pmatrix}=0

4. Quaternion Algebra

In this paper I explore the set of quaternion algebras defined in [3].

Definition 4.1.

Let FF be field. Extension field F(i,j,k)F(i,j,k) is called the quaternion algebra E(F,a,b){\color[rgb]{.4,0,.9}E(F,a,b)} over the field FF111I follow definition from [3]. if multiplication in algebra EE is defined according to rule

(4.1) ijkiakajjkbbikajbiab\begin{array}[]{c|ccc}&i&j&k\\ \hline\cr i&a&k&aj\\ j&-k&b&-bi\\ k&-aj&bi&-ab\\ \end{array}

where aa, bFb\in F, ab0ab\neq 0. ∎

Elements of the algebra E(F,a,b)E(F,a,b) have form

x=x𝟎+x𝟏i+x𝟐j+x𝟑kx=x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i+x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}j+x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}k

where x𝒊Fx^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\in F, 𝒊=𝟎\boldsymbol{{\color[rgb]{1,.3,.6}i}}=\boldsymbol{{\color[rgb]{1,.3,.6}0}}, 𝟏\boldsymbol{{\color[rgb]{1,.3,.6}1}}, 𝟐\boldsymbol{{\color[rgb]{1,.3,.6}2}}, 𝟑\boldsymbol{{\color[rgb]{1,.3,.6}3}}. Quaternion

x¯=x𝟎x𝟏ix𝟐jx𝟑k\overline{x}=x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i-x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}j-x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}k

is called conjugate to the quaternion xx. We define the norm of the quaternion xx using equation

(4.2) |x|2=xx¯=(x𝟎)2a(x𝟏)2b(x𝟐)2+ab(x𝟑)2|x|^{2}=x\overline{x}=(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-a(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-b(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}+ab(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}

From equation (4.2), it follows that E(F,a,b)E(F,a,b) is algebra with division only when a<0a<0, b<0b<0. In this case we can renorm basis such that a=1a=-1, b=1b=-1.

We use symbol E(F){\color[rgb]{.4,0,.9}E(F)} to denote the quaternion division algebra E(F,1,1)E(F,-1,-1) over the field FF. We will use notation H=E(R,1,1){\color[rgb]{.4,0,.9}H}=E(R,-1,-1). Multiplication in quaternion algebra E(F)E(F) is defined according to rule

(4.3) ijki1kjjk1ikji1\begin{array}[]{c|ccc}&i&j&k\\ \hline\cr i&-1&k&-j\\ j&-k&-1&i\\ k&j&-i&-1\\ \end{array}

In algebra E(F)E(F), the norm of the quaternion has form

(4.4) |x|2=xx¯=(x𝟎)2+(x𝟏)2+(x𝟐)2+(x𝟑)2|x|^{2}=x\overline{x}=(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}+(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}+(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}+(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}

In this case inverse element has form

(4.5) x1=|x|2x¯x^{-1}=|x|^{-2}\overline{x}

The inner automorphism of quaternion algebra HH222See [6], p.643.

(4.6) pqpq1q(ix+jy+kz)q1=ix+jy+kz\begin{matrix}p\rightarrow qpq^{-1}\\ \vphantom{\overset{\rightarrow}{1}^{1}}q(ix+jy+kz)q^{-1}=ix^{\prime}+jy^{\prime}+kz^{\prime}\end{matrix}

describes the rotation of the vector with coordinates xx, yy, zz. The norm of quaternion qq is irrelevant, although usually we assume |q|=1|q|=1. If qq is written as sum of scalar and vector

q=cosα+(ia+jb+kc)sinαa2+b2+c2=1q=\cos\alpha+(ia+jb+kc)\sin\alpha\ \ \ a^{2}+b^{2}+c^{2}=1

then (4.6) is a rotation of the vector (x,y,z)(x,y,z) about the vector (a,b,c)(a,b,c) through an angle 2α2\alpha.

5. Tower of Algebras

Let F1F_{1} be algebra over the field F2F_{2}. Let e¯¯12\overline{\overline{e}}{}_{12} be basis of algebra F1F_{1} over the field F2F_{2}. Let C12𝒊𝒋𝒌C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ij}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}k}}} be structural constants of algebra F1F_{1} over the field F2F_{2}.

Let F2F_{2} be algebra over the field F3F_{3}. Let e¯¯23\overline{\overline{e}}{}_{23} be basis of algebra F2F_{2} over the field F3F_{3}. Let C23𝒊𝒋𝒌C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ij}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}k}}} be structural constants of algebra F2F_{2} over the field F3F_{3}.

I will consider the algebra F1F_{1} as direct sum of algebras F2F_{2}. Each item of sum I identify with vector of basis e¯¯12\overline{\overline{e}}{}_{12} . Accordingly, I can consider algebra F1F_{1} as algebra over field F3F_{3}. Let e¯¯13\overline{\overline{e}}{}_{13} be basis of algebra F1F_{1} over the field F3F_{3}. Index of basis e¯¯13\overline{\overline{e}}{}_{13} consists from two indexes: index of fiber and index of vector of basis e¯¯23\overline{\overline{e}}{}_{23} in fiber.

I will identify vector of basis e¯12𝒊\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}} with unit in corresponding fiber. Then

(5.1) e¯=13𝒋𝒊e¯e¯23𝒋12𝒊\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}}=\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}

The product of vectors of basis e¯¯13\overline{\overline{e}}{}_{13} has form

(5.2) e¯e¯13𝒋𝒊=13𝒎𝒌e¯e¯23𝒋e¯12𝒊e¯23𝒎=12𝒌C23e¯𝒋𝒎𝒂C1223𝒂e¯𝒊𝒌𝒃12𝒃\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}mk}}}=\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}m}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}

Because C12𝒊𝒌𝒃F2C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\in F_{2}, then expansion C12𝒊𝒌𝒃C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}} relative to basis e¯¯23\overline{\overline{e}}{}_{23} has form

(5.3) C12=𝒊𝒌𝒃C12e¯𝒊𝒌𝒃𝒄23𝒄C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}=C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}c}}}

Let us substitute (5.3) into (5.2)

e¯e¯13𝒋𝒊13𝒎𝒌\displaystyle\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}mk}}} =C23e¯𝒋𝒎𝒂C1223𝒂e¯𝒊𝒌𝒃𝒄e¯23𝒄12𝒃\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}c}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}
(5.4) =C23C23𝒋𝒎𝒂C12𝒂𝒄𝒅e¯𝒊𝒌𝒃𝒄e¯23𝒅12𝒃\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ac}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}d}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}
=C23C23𝒋𝒎𝒂C12𝒂𝒄𝒅e¯𝒊𝒌𝒃𝒄13𝒅𝒃\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ac}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}

Therefore, we can define structural constants of algebra F1F_{1} over field F3F_{3}

(5.5) C13=𝒋𝒊𝒎𝒌𝒅𝒃C23C23𝒋𝒎𝒂C12𝒂𝒄𝒅𝒊𝒌𝒃𝒄C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}mk}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ac}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}

To verify construction, let us consider the product

e¯e¯12𝒊12𝒌\displaystyle\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}} =e¯e¯13𝟎𝒊13𝟎𝒌\displaystyle=\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0k}}}
=C13e¯𝟎𝒊𝟎𝒌𝒅𝒃13𝒅𝒃\displaystyle=C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}
(5.6) =C23C23𝟎𝟎𝒂C12𝒂𝒄𝒅e¯𝒊𝒌𝒃𝒄13𝒅𝒃\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}\ C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ac}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}
=C23C23𝟎𝟎𝟎C12𝟎𝒄𝒅e¯𝒊𝒌𝒃𝒄13𝒃𝒅\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\ \overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}bd}}}
=C23C12𝟎𝒄𝒅e¯𝒊𝒌𝒃𝒄13𝒃𝒅\displaystyle=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}\ \overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}bd}}}

On the other hand

e¯e¯12𝒊12𝒌\displaystyle\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}} =C12e¯𝒊𝒌𝒃12𝒃\displaystyle=C_{12\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}
(5.7) =C12e¯𝒊𝒌𝒃𝒅e¯23𝒅e¯23𝟎12𝒃\displaystyle=C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bd}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}d}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}
=C12C23𝒊𝒌𝒃𝒅e¯𝟎𝒅𝒄e¯23𝒄12𝒃\displaystyle=C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bd}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0d}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}c}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}c}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}
=C12C23𝒊𝒌𝒃𝒅e¯𝟎𝒅𝒄13𝒃𝒄\displaystyle=C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}bd}}}C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0d}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}c}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}

Expressions (5.6), (5.7) coincide.

Theorem 5.1.

If C12𝐢𝐤𝐛F3C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\in F_{3}, then we multiply components e¯12𝐢\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}, e¯23𝐤\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}} of vector e¯13𝐢𝐤\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ik}}} independently

(5.8) C13=𝒋𝒊𝒎𝒌𝒅𝒃C23C12𝒋𝒎𝒅𝒊𝒌𝒃C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}mk}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}db}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}C_{12\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}
Proof.

From equation (5.2), it follows

(5.9) e¯e¯13𝒋𝒊=13𝒎𝒌C23e¯𝒋𝒎𝒂C1223𝒂e¯𝒊𝒌𝒃=12𝒃C23C12𝒋𝒎𝒂e¯𝒊𝒌𝒃23𝒂𝒃\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ji}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}mk}}}=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}}=C_{23\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}a}}}C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ \overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ab}}}

Equation (5.8) follows from the equation (5.9). ∎

Theorem 5.2.

Let F1F_{1} be algebra over the field F2F_{2}. Let F2F_{2} be algebra over the field F3F_{3}. There exists equation

(5.10) a12𝒌=a13𝒊𝒌e¯23𝒊a_{12}^{\boldsymbol{{\color[rgb]{1,.3,.6}k}}}=a_{13}^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}

between coordinates a12𝐤a_{12}^{\boldsymbol{{\color[rgb]{1,.3,.6}k}}} of element aF1a\in F_{1} relative to basis e¯¯12\overline{\overline{e}}{}_{12} and coordinates a13𝐣𝐤a_{13}^{\boldsymbol{{\color[rgb]{1,.3,.6}jk}}} of element aF1a\in F_{1} relative to basis e¯¯13\overline{\overline{e}}{}_{13}

Proof.

Element aF1a\in F_{1} has expansion

(5.11) a=a12𝒌e¯12𝒌a=a_{12}^{\boldsymbol{{\color[rgb]{1,.3,.6}k}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}

relative basis e¯¯12\overline{\overline{e}}{}_{12} and expansion

(5.12) a=a13𝒊𝒌e¯13𝒊𝒌a=a_{13}^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}

relative basis e¯¯13\overline{\overline{e}}{}_{13} From equations (5.1), (5.12), it follows that

(5.13) a=a13𝒊𝒌e¯e¯23𝒊12𝒌a=a_{13}^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}

The equation (5.10) follows from equations (5.11),(5.13). ∎

Theorem 5.3 (The Cauchy-Riemann equations).

Let F1F_{1} be algebra over the field F2F_{2}. Let F2F_{2} be algebra over the field F3F_{3}. Matrix of F2F_{2}-linear mapping

f:F1F1f:F_{1}\rightarrow F_{1}

of F2F_{2}-algebra F1F_{1} satisfies relationship

(5.14) f𝒋𝒍𝒊𝒌C23=𝒎𝒑𝒋C23f𝒑𝒍𝒏𝒌𝒎𝒏𝒊f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mn}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}nk}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}
Proof.

Since ff is F2F_{2}-linear mapping, then

(5.15) f(ax)=af(x)aF2xF1\begin{matrix}f(ax)=af(x)&a\in F_{2}&x\in F_{1}\end{matrix}

Let

a=a𝒋e¯23𝒋x=x13𝒋𝒍e¯13𝒋𝒍\begin{matrix}a=a^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}&x=x_{13}^{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}\end{matrix}

Then

(5.16) f(ax)𝒊𝒌=(ax)𝒋𝒍f𝒋𝒍𝒊𝒌=C23a𝒎𝒎𝒑𝒋x𝒑𝒍f𝒋𝒍𝒊𝒌f(ax)^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}=(ax)^{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}m}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}

Similarly,

(5.17) (af(x))𝒊𝒌=C23a𝒎𝒎𝒏𝒊(f(x))𝒏𝒌=C23a𝒎𝒎𝒏𝒊f𝒑𝒍𝒏𝒌x𝒑𝒍(af(x))^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mn}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}m}}}(f(x))^{\boldsymbol{{\color[rgb]{1,.3,.6}nk}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mn}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}m}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}nk}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}

From equations (5.15), (5.16), (5.17) , it follows that

(5.18) f𝒋𝒍𝒊𝒌C23a𝒎𝒎𝒑𝒋x𝒑𝒍=C23a𝒎𝒎𝒏𝒊f𝒑𝒍𝒏𝒌x𝒑𝒍f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}jl}}}C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}m}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mn}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}m}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}nk}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}pl}}}

Since aF2a\in F_{2}, xF1x\in F_{1} are arbitrary, then the equation (5.14) follows from equation (5.18) ∎

Theorem 5.4 (The Cauchy-Riemann equations).

Matrix of F2F_{2}-linear mapping

f:F2F2f:F_{2}\rightarrow F_{2}

of F3F_{3}-algebra F2F_{2} satisfies relationship

(5.19) f𝒋𝒊C23=𝒎𝒑𝒋C23f𝒑𝒏𝒎𝒏𝒊f^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mn}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}n}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}
Proof.

Statement of the theorem is proved if we assume F1=F2F_{1}=F_{2} in the theorem 5.3. ∎

Theorem 5.5 (The Cauchy-Riemann equations).

Matrix of CC-linear mapping

f:CCf:C\rightarrow C

of RR-algebra CC satisfies relationship

(5.20) f𝟎𝟎=f𝟏𝟏f𝟏𝟎=f𝟎𝟏\begin{array}[]{r@{\,}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{array}
Proof.

For complex field CC, the equation (5.19) has form

(5.21) f𝟎𝟎CC+𝒎𝒑𝟎f𝟏𝟎CC𝒎𝒑𝟏=CCf𝒑𝟎𝒎𝟎𝟎+CCf𝒑𝟏𝒎𝟏𝟎f𝟎𝟏CC+𝒎𝒑𝟎f𝟏𝟏CC𝒎𝒑𝟏=CCf𝒑𝟎𝒎𝟎𝟏+CCf𝒑𝟏𝒎𝟏𝟏\begin{array}[]{r@{\,}l}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\end{array}

From equations (3.3), (5.21), it follows that

(5.22) {f𝟎𝟎CC𝟎𝟎𝟎=CCf𝟎𝟎𝟎𝟎𝟎f𝟎𝟎CC𝟏𝟏𝟎=CCf𝟏𝟏𝟏𝟏𝟎f𝟏𝟎CC𝟎𝟏𝟏=CCf𝟏𝟎𝟎𝟎𝟎f𝟏𝟎CC𝟏𝟎𝟏=CCf𝟎𝟏𝟏𝟏𝟎f𝟎𝟏CC𝟎𝟎𝟎=CCf𝟎𝟏𝟎𝟏𝟏f𝟎𝟏CC𝟏𝟏𝟎=CCf𝟏𝟎𝟏𝟎𝟏f𝟏𝟏CC𝟎𝟏𝟏=CCf𝟏𝟏𝟎𝟏𝟏f𝟏𝟏CC𝟏𝟎𝟏=CCf𝟎𝟎𝟏𝟎𝟏\left\{\begin{array}[]{r@{\,}l}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}&=C_{C\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{array}\right.

Equations (5.20) follow from equations (3.3), (5.22). ∎

I intentionally wrote the equation (5.22) in the intermediate form, to show how much more is tedious proof of the theorem 5.5 compared with proof of the analogous theorem 3.1. never used the statement that the product in the F1F_{1}-algebra F2F_{2} is commutative. Therefore, the theorem 5.4 can be applied to an arbitrary algebra.

Theorem 5.6 (The Cauchy-Riemann equations).

Matrix of HH-linear mapping

f:HHf:H\rightarrow H

of RR-algebra HH satisfies relationship

(5.23) f𝟎𝟎=f𝟏𝟏=f𝟐𝟐=f𝟑𝟑f𝟏𝟎=f𝟎𝟏=f𝟑𝟐=f𝟐𝟑f𝟐𝟎=f𝟑𝟏=f𝟎𝟐=f𝟏𝟑f𝟑𝟎=f𝟐𝟏=f𝟏𝟐=f𝟎𝟑\begin{array}[]{r@{\,}r@{\,}r@{\,}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{array}
Proof.

According to the theorem [2]-LABEL:0812.4763-English-theorem:_Quaternion_over_real_field, structural constants of quaternion algebra have form

(5.24) CH=𝟎𝟎𝟎1CH=𝟎𝟏𝟏1CH=𝟎𝟐𝟐1CH=𝟎𝟑𝟑1CH=𝟏𝟎𝟏1CH=𝟏𝟏𝟎1CH=𝟏𝟐𝟑1CH=𝟏𝟑𝟐1CH=𝟐𝟎𝟐1CH=𝟐𝟏𝟑1CH=𝟐𝟐𝟎1CH=𝟐𝟑𝟏1CH=𝟑𝟎𝟑1CH=𝟑𝟏𝟐1CH=𝟑𝟐𝟏1CH=𝟑𝟑𝟎1\begin{array}[]{r@{\,}r@{\ \ }r@{\,}r@{\ \ }r@{\,}r@{\ \ }r@{\,}r}C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1\\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&-1\\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&-1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1\\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-1&C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-1\end{array}

For quaternion algebra HH, the equation (5.19) has form

(5.25) f𝟎𝟎CH+𝒎𝒑𝟎f𝟏𝟎CH+𝒎𝒑𝟏f𝟐𝟎CH+𝒎𝒑𝟐f𝟑𝟎CH𝒎𝒑𝟑=CHf𝒑𝟎𝒎𝟎𝟎+CHf𝒑𝟏𝒎𝟏𝟎+CHf𝒑𝟐𝒎𝟐𝟎+CHf𝒑𝟑𝒎𝟑𝟎i=0f𝟎𝟏CH+𝒎𝒑𝟎f𝟏𝟏CH+𝒎𝒑𝟏f𝟐𝟏CH+𝒎𝒑𝟐f𝟑𝟏CH𝒎𝒑𝟑=CHf𝒑𝟎𝒎𝟎𝟏+CHf𝒑𝟏𝒎𝟏𝟏+CHf𝒑𝟐𝒎𝟐𝟏+CHf𝒑𝟑𝒎𝟑𝟏i=1f𝟎𝟐CH+𝒎𝒑𝟎f𝟏𝟐CH+𝒎𝒑𝟏f𝟐𝟐CH+𝒎𝒑𝟐f𝟑𝟐CH𝒎𝒑𝟑=CHf𝒑𝟎𝒎𝟎𝟐+CHf𝒑𝟏𝒎𝟏𝟐+CHf𝒑𝟐𝒎𝟐𝟐+CHf𝒑𝟑𝒎𝟑𝟐i=2f𝟎𝟑CH+𝒎𝒑𝟎f𝟏𝟑CH+𝒎𝒑𝟏f𝟐𝟑CH+𝒎𝒑𝟐f𝟑𝟑CH𝒎𝒑𝟑=CHf𝒑𝟎𝒎𝟎𝟑+CHf𝒑𝟏𝒎𝟏𝟑+CHf𝒑𝟐𝒎𝟐𝟑+CHf𝒑𝟑𝒎𝟑𝟑i=3\begin{array}[]{r@{\,}l@{\ \ \ }r}&f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}\\ =&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m2}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m3}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}&i=0\\ &f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}\\ =&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m2}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m3}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}&i=1\\ &f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}\\ =&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m2}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m3}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}&i=2\\ &f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}mp}}}\\ =&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m0}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m1}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m2}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}+C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}m3}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}&i=3\end{array}

From equations (5.24), (5.25), it follows that333I do not consider equations which are trivial or follow from the equations (5.26).

(5.26) {f𝟏𝟎CH=𝟏𝟎𝟏CHf𝟎𝟏𝟏𝟏𝟎i=0m=1p=0f𝟎𝟎CH=𝟏𝟏𝟎CHf𝟏𝟏𝟏𝟏𝟎i=0m=1p=1f𝟑𝟎CH=𝟏𝟐𝟑CHf𝟐𝟏𝟏𝟏𝟎i=0m=1p=2f𝟐𝟎CH=𝟏𝟑𝟐CHf𝟑𝟏𝟏𝟏𝟎i=0m=1p=3f𝟐𝟎CH=𝟐𝟎𝟐CHf𝟎𝟐𝟐𝟐𝟎i=0m=2p=0f𝟑𝟎CH=𝟐𝟏𝟑CHf𝟏𝟐𝟐𝟐𝟎i=0m=2p=1f𝟎𝟎CH=𝟐𝟐𝟎CHf𝟐𝟐𝟐𝟐𝟎i=0m=2p=2f𝟏𝟎CH=𝟐𝟑𝟏CHf𝟑𝟐𝟐𝟐𝟎i=0m=2p=3f𝟑𝟎CH=𝟑𝟎𝟑CHf𝟎𝟑𝟑𝟑𝟎i=0m=3p=0f𝟐𝟎CH=𝟑𝟏𝟐CHf𝟏𝟑𝟑𝟑𝟎i=0m=3p=1f𝟏𝟎CH=𝟑𝟐𝟏CHf𝟐𝟑𝟑𝟑𝟎i=0m=3p=2f𝟎𝟎CH=𝟑𝟑𝟎CHf𝟑𝟑𝟑𝟑𝟎i=0m=3p=3\left\{\begin{array}[]{r@{\,}l@{\ \ \ }r@{\ \ }r@{\ \ }r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&i=0&m=1&p=0\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&i=0&m=1&p=1\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&i=0&m=1&p=2\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&i=0&m=1&p=3\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&i=0&m=2&p=0\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&i=0&m=2&p=1\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&i=0&m=2&p=2\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&i=0&m=2&p=3\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&i=0&m=3&p=0\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&i=0&m=3&p=1\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&i=0&m=3&p=2\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}=&C_{H\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&i=0&m=3&p=3\end{array}\right.

Equations (5.23) follow from equations (5.24), (5.26). ∎

Theorem 5.7.

Let F1F_{1} be algebra over the field F2F_{2}. Let F2F_{2} be algebra over the field F3F_{3}. Let F2F_{2}-linear mapping

f:F1F1f:F_{1}\rightarrow F_{1}

have representation

(5.27) f(a)𝒊=f𝒋𝒊a𝒋f𝒋𝒊=f𝒋𝒊𝒌e¯23𝒌\begin{matrix}f(a)^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}&f^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}\end{matrix}

relative to basis e¯¯12\overline{\overline{e}}{}_{12} and have representation

(5.28) f(a)𝒌𝒊=f𝒍𝒋𝒌𝒊a𝒍𝒋f(a)^{\boldsymbol{{\color[rgb]{1,.3,.6}ki}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}ki}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}

relative to basis e¯¯13\overline{\overline{e}}{}_{13} . Coordinates f𝐢𝐥𝐣𝐤f^{\boldsymbol{{\color[rgb]{1,.3,.6}jk}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}il}}} of mapping ff relative to basis e¯¯13\overline{\overline{e}}{}_{13} have form

(5.29) f𝒍𝒋𝒑𝒊=C23f𝒋𝒊𝒌𝒌𝒍𝒑f^{\boldsymbol{{\color[rgb]{1,.3,.6}pi}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}kl}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
Proof.

Since

a𝒋=a𝒍𝒋e¯23𝒍a^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}=a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}l}}}

then, from the equation (5.27), it follows that

(5.30) f(a)𝒊=C23f𝒋𝒊𝒌𝒌𝒍𝒑a𝒍𝒋e¯23𝒑f(a)^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}kl}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}

From the equation (5.28), it follows that

(5.31) f(a)𝒊=f(a)𝒑𝒊e¯=23𝒑f𝒍𝒋𝒑𝒊a𝒍𝒋e¯23𝒑f(a)^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}=f(a)^{\boldsymbol{{\color[rgb]{1,.3,.6}pi}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}=f^{\boldsymbol{{\color[rgb]{1,.3,.6}pi}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}

From equations (5.30), (5.31), it follows that

(5.32) f𝒍𝒋𝒑𝒊a𝒍𝒋e¯=23𝒑C23f𝒋𝒊𝒌𝒌𝒍𝒑a𝒍𝒋e¯23𝒑f^{\boldsymbol{{\color[rgb]{1,.3,.6}pi}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}=C_{23\cdot}{}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}kl}}}f^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}_{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}}\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}

The equation (5.29) follows from the equation (5.32) because e¯¯23\overline{\overline{e}}{}_{23} is basis and coordinates a𝒍𝒋a^{\boldsymbol{{\color[rgb]{1,.3,.6}lj}}} of element aF1a\in F_{1} are arbitrary. ∎

6. Quaternion Algebra over Complex Field

In this section, I will consider quaternion algebra E(C,1,1)E(C,-1,-1), where CC is complex field.

Product in algebra E(C,1,1)E(C,-1,-1) is defined according to table

(6.1) e¯12𝟏e¯12𝟐e¯12𝟑e¯12𝟏1e¯12𝟑e¯12𝟐e¯12𝟐e¯12𝟑1e¯12𝟏e¯12𝟑e¯12𝟐e¯12𝟏1\begin{array}[]{c|rrr}&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ \hline\cr\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&-1&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&-\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\\ \overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&-\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&-1&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ \overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&-\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&-1\\ \end{array}

According to theorem [2]-LABEL:0812.4763-English-theorem:_Quaternion_over_real_field, structural constants of quaternion algebra have form

C12𝟎𝟎𝟎\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =\displaystyle= 1\displaystyle 1 C12𝟎𝟏𝟏\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =\displaystyle= 1\displaystyle 1 C12𝟎𝟐𝟐\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =\displaystyle= 1\displaystyle 1 C12𝟎𝟑𝟑\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =\displaystyle= 1\displaystyle 1
C12𝟏𝟎𝟏\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =\displaystyle= 1\displaystyle 1 C12𝟏𝟏𝟎\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =\displaystyle= 1\displaystyle-1 C12𝟏𝟐𝟑\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =\displaystyle= 1\displaystyle 1 C12𝟏𝟑𝟐\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =\displaystyle= 1\displaystyle-1
C12𝟐𝟎𝟐\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =\displaystyle= 1\displaystyle 1 C12𝟐𝟏𝟑\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =\displaystyle= 1\displaystyle-1 C12𝟐𝟐𝟎\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =\displaystyle= 1\displaystyle-1 C12𝟐𝟑𝟏\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =\displaystyle= 1\displaystyle 1
C12𝟑𝟎𝟑\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =\displaystyle= 1\displaystyle 1 C12𝟑𝟏𝟐\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =\displaystyle= 1\displaystyle 1 C12𝟑𝟐𝟏\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =\displaystyle= 1\displaystyle-1 C12𝟑𝟑𝟎\displaystyle C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =\displaystyle= 1\displaystyle-1

Let e¯=23𝒌e¯C𝒌\overline{e}{}_{23\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}=\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}. Product in algebra CC is defined according to rule (3.2). According to the theorem 3.1, structural constants of complex field over real field have form (3.3)

Therefore, algebra E(C,1,1)E(C,-1,-1) is isomorphic to tensor product CHC\otimes H. So we can select basis

e¯=13𝒊𝒋e¯C𝒊e¯12𝒋\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ij}}}=\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\otimes\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
(6.2) e¯=13𝟎𝟎11e¯=13𝟎𝟏1ie¯=13𝟎𝟐1je¯=13𝟎𝟑1ke¯=13𝟏𝟎i1e¯=13𝟏𝟏iie¯=13𝟏𝟐ije¯=13𝟏𝟑ik\begin{array}[]{rrrr}\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 00}}}=1\otimes 1&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 01}}}=1\otimes i&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 02}}}=1\otimes j&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 03}}}=1\otimes k\\ \overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 10}}}=i\otimes 1&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 11}}}=i\otimes i&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 12}}}=i\otimes j&\overline{e}{}_{13\cdot\boldsymbol{{\color[rgb]{1,.3,.6}\cdot 13}}}=i\otimes k\end{array}
Theorem 6.1.

Table of product in algebra E(C,1,1)E(C,-1,-1) over field RR has form

1i1j1ki1iiijik1i111k1jiii1ikij1j1k111iijiki1ii1k1j1i11ikijiii1i1iiijik111i1j1kiii1ikij1i111k1jijiki1ii1j1k111iikijiii11k1j1i11\begin{array}[]{l|rrrrrrrr}&1\otimes i&1\otimes j&1\otimes k&i\otimes 1&i\otimes i&i\otimes j&i\otimes k\\ \hline\cr 1\otimes i&-1\otimes 1&1\otimes k&-1\otimes j&i\otimes i&-i\otimes 1&i\otimes k&-i\otimes j\\ 1\otimes j&-1\otimes k&-1\otimes 1&1\otimes i&i\otimes j&-i\otimes k&-i\otimes 1&i\otimes i\\ 1\otimes k&1\otimes j&-1\otimes i&-1\otimes 1&i\otimes k&i\otimes j&-i\otimes i&-i\otimes 1\\ i\otimes 1&i\otimes i&i\otimes j&i\otimes k&-1\otimes 1&-1\otimes i&-1\otimes j&-1\otimes k\\ i\otimes i&-i\otimes 1&i\otimes k&-i\otimes j&-1\otimes i&1\otimes 1&-1\otimes k&1\otimes j\\ i\otimes j&-i\otimes k&-i\otimes 1&i\otimes i&-1\otimes j&1\otimes k&1\otimes 1&-1\otimes i\\ i\otimes k&i\otimes j&-i\otimes i&-i\otimes 1&-1\otimes k&-1\otimes j&1\otimes i&1\otimes 1\end{array}
Proof.

Table is written according to equation

(e¯C𝒊e¯)12𝒌(e¯C𝒋e¯)12𝒎=(e¯e¯C𝒊)C𝒋(e¯e¯12𝒌)12𝒎(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\otimes\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}})(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\otimes\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}m}}})=(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}}\ \overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}})\otimes(\overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}\ \overline{e}{}_{12\cdot\boldsymbol{{\color[rgb]{1,.3,.6}m}}})

and definition of basis (6.2). ∎

Theorem 6.2.

Structural constants of the algebra E(C,1,1)E(C,-1,-1) over field RR have form

C13=𝟎𝟎𝟎𝟎𝟎𝟎1C13=𝟎𝟎𝟎𝟏𝟎𝟏1C13=𝟎𝟎𝟎𝟐𝟎𝟐1C13=𝟎𝟎𝟎𝟑𝟎𝟑1C13=𝟎𝟏𝟎𝟎𝟎𝟏1C13=𝟎𝟏𝟎𝟏𝟎𝟎1C13=𝟎𝟏𝟎𝟐𝟎𝟑1C13=𝟎𝟏𝟎𝟑𝟎𝟐1C13=𝟎𝟐𝟎𝟎𝟎𝟐1C13=𝟎𝟐𝟎𝟏𝟎𝟑1C13=𝟎𝟐𝟎𝟐𝟎𝟎1C13=𝟎𝟐𝟎𝟑𝟎𝟏1C13=𝟎𝟑𝟎𝟎𝟎𝟑1C13=𝟎𝟑𝟎𝟏𝟎𝟐1C13=𝟎𝟑𝟎𝟐𝟎𝟏1C13=𝟎𝟑𝟎𝟑𝟎𝟎1\begin{array}[]{r@{}rr@{}rr@{}rr@{}r}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&1&C_{13\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&-1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&-1\end{array}
C13=𝟎𝟎𝟏𝟎𝟏𝟎1C13=𝟎𝟎𝟏𝟏𝟏𝟏1C13=𝟎𝟎𝟏𝟐𝟏𝟐1C13=𝟎𝟎𝟏𝟑𝟏𝟑1C13=𝟎𝟏𝟏𝟎𝟏𝟏1C13=𝟎𝟏𝟏𝟏𝟏𝟎1C13=𝟎𝟏𝟏𝟐𝟏𝟑1C13=𝟎𝟏𝟏𝟑𝟏𝟐1C13=𝟎𝟐𝟏𝟎𝟏𝟐1C13=𝟎𝟐𝟏𝟏𝟏𝟑1C13=𝟎𝟐𝟏𝟐𝟏𝟎1C13=𝟎𝟐𝟏𝟑𝟏𝟏1C13=𝟎𝟑𝟏𝟎𝟏𝟑1C13=𝟎𝟑𝟏𝟏𝟏𝟐1C13=𝟎𝟑𝟏𝟐𝟏𝟏1C13=𝟎𝟑𝟏𝟑𝟏𝟎1\begin{array}[]{r@{}rr@{}rr@{}rr@{}r}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&-1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1\end{array}
C13=𝟏𝟎𝟎𝟎𝟏𝟎1C13=𝟏𝟎𝟎𝟏𝟏𝟏1C13=𝟏𝟎𝟎𝟐𝟏𝟐1C13=𝟏𝟎𝟎𝟑𝟏𝟑1C13=𝟏𝟏𝟎𝟎𝟏𝟏1C13=𝟏𝟏𝟎𝟏𝟏𝟎1C13=𝟏𝟏𝟎𝟐𝟏𝟑1C13=𝟏𝟏𝟎𝟑𝟏𝟐1C13=𝟏𝟐𝟎𝟎𝟏𝟐1C13=𝟏𝟐𝟎𝟏𝟏𝟑1C13=𝟏𝟐𝟎𝟐𝟏𝟎1C13=𝟏𝟐𝟎𝟑𝟏𝟏1C13=𝟏𝟑𝟎𝟎𝟏𝟑1C13=𝟏𝟑𝟎𝟏𝟏𝟐1C13=𝟏𝟑𝟎𝟐𝟏𝟏1C13=𝟏𝟑𝟎𝟑𝟏𝟎1\begin{array}[]{r@{}rr@{}rr@{}rr@{}r}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&-1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&-1\end{array}
C13=𝟏𝟎𝟏𝟎𝟎𝟎1C13=𝟏𝟎𝟏𝟏𝟎𝟏1C13=𝟏𝟎𝟏𝟐𝟎𝟐1C13=𝟏𝟎𝟏𝟑𝟎𝟑1C13=𝟏𝟏𝟏𝟎𝟎𝟏1C13=𝟏𝟏𝟏𝟏𝟎𝟎1C13=𝟏𝟏𝟏𝟐𝟎𝟑1C13=𝟏𝟏𝟏𝟑𝟎𝟐1C13=𝟏𝟐𝟏𝟎𝟎𝟐1C13=𝟏𝟐𝟏𝟏𝟎𝟑1C13=𝟏𝟐𝟏𝟐𝟎𝟎1C13=𝟏𝟐𝟏𝟑𝟎𝟏1C13=𝟏𝟑𝟏𝟎𝟎𝟑1C13=𝟏𝟑𝟏𝟏𝟎𝟐1C13=𝟏𝟑𝟏𝟐𝟎𝟏1C13=𝟏𝟑𝟏𝟑𝟎𝟎1\begin{array}[]{r@{}rr@{}rr@{}rr@{}r}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&-1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&-1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&-1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&1&C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&1\end{array}
Proof.

We consider the statement of theorem either as corollary of the theorem 6.1, or as corollary of theorem 5.1. ∎

Theorem 6.3 (The Cauchy-Riemann equations).

Matrix of linear function

y𝒊𝒌=x𝒋𝒎f𝒋𝒎𝒊𝒌y^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}=x^{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}

of algebra E(C,1,1)E(C,-1,-1) satisfies relationship

(6.3) f𝟎𝒊𝟎𝒋=f𝟏𝒊𝟏𝒋f𝟎𝒊𝟏𝒋=f𝟏𝒊𝟎𝒋\begin{array}[]{r@{}r}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}\end{array}
Proof.

From equations [2]-(LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation), (5.8), (3.3) it follows

f𝟎𝒊𝟎𝒋\displaystyle f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}} =f𝒌𝒂𝒓𝒄C13C13𝒌𝒂𝟎𝒊𝒑𝒃𝒑𝒃𝒓𝒄𝟎𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}
=f𝒌𝒂𝒓𝒄CCC12𝒌𝟎𝒑CC𝒂𝒊𝒃C12𝒑𝒓𝟎𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟎𝒄CCC12𝟎𝟎𝟎CC𝒂𝒊𝒃C12𝟎𝟎𝟎𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
(6.4) +f𝟏𝒂𝟏𝒄CCC12𝟏𝟎𝟏CC𝒂𝒊𝒃C12𝟏𝟏𝟎𝒃𝒄𝒋\displaystyle+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟎𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋f𝟏𝒂𝟏𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=(f𝟎𝒂𝟎𝒄f𝟏𝒂𝟏𝒄)C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=(f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}})\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
f𝟎𝒊𝟏𝒋\displaystyle f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}} =f𝒌𝒂𝒓𝒄C13C13𝒌𝒂𝟎𝒊𝒑𝒃𝒑𝒃𝒓𝒄𝟏𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}
=f𝒌𝒂𝒓𝒄CCC12𝒌𝟎𝒑CC𝒂𝒊𝒃C12𝒑𝒓𝟏𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟏𝒄CCC12𝟎𝟎𝟎CC𝒂𝒊𝒃C12𝟎𝟏𝟏𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
(6.5) +f𝟏𝒂𝟎𝒄CCC12𝟏𝟎𝟏CC𝒂𝒊𝒃C12𝟏𝟎𝟏𝒃𝒄𝒋\displaystyle+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟏𝒄C12C12𝒂𝒊𝒃+𝒃𝒄𝒋f𝟏𝒂𝟎𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=(f𝟎𝒂𝟏𝒄+f𝟏𝒂𝟎𝒄)C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=(f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}})\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
f𝟏𝒊𝟎𝒋\displaystyle f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}} =f𝒌𝒂𝒓𝒄C13C13𝒌𝒂𝟏𝒊𝒑𝒃𝒑𝒃𝒓𝒄𝟎𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}
=f𝒌𝒂𝒓𝒄CCC12𝒌𝟏𝒑CC𝒂𝒊𝒃C12𝒑𝒓𝟎𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟏𝒄CCC12𝟎𝟏𝟏CC𝒂𝒊𝒃C12𝟏𝟏𝟎𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
(6.6) +f𝟏𝒂𝟎𝒄CCC12𝟏𝟏𝟎CC𝒂𝒊𝒃C12𝟎𝟎𝟎𝒃𝒄𝒋\displaystyle+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟏𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋f𝟏𝒂𝟎𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=(f𝟎𝒂𝟏𝒄+f𝟏𝒂𝟎𝒄)C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=-(f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}})\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
f𝟏𝒊𝟏𝒋\displaystyle f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}} =f𝒌𝒂𝒓𝒄C13C13𝒌𝒂𝟏𝒊𝒑𝒃𝒑𝒃𝒓𝒄𝟏𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}}\ C_{13\cdot}{}_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pb}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}
=f𝒌𝒂𝒓𝒄CCC12𝒌𝟏𝒑CC𝒂𝒊𝒃C12𝒑𝒓𝟏𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ka}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rc}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟎𝒄CCC12𝟎𝟏𝟏CC𝒂𝒊𝒃C12𝟏𝟎𝟏𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
(6.7) +f𝟏𝒂𝟏𝒄CCC12𝟏𝟏𝟎CC𝒂𝒊𝒃C12𝟎𝟏𝟏𝒃𝒄𝒋\displaystyle+f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=f𝟎𝒂𝟎𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋f𝟏𝒂𝟏𝒄C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}
=(f𝟎𝒂𝟎𝒄f𝟏𝒂𝟏𝒄)C12C12𝒂𝒊𝒃𝒃𝒄𝒋\displaystyle=(f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0c}}}-f^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1a}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1c}}})\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ai}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ C_{12\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}bc}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}

Equation (6.3) follows from comparison of equations (6.4) and (6.7), (6.5) and (6.6). ∎

Theorem 6.4 (The Cauchy-Riemann equations).

Since matrix

(y𝒊𝒋x𝒌𝒍)\left(\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ij}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}kl}}}}\right)

is Jacobian matrix of map in algebra E(C,1,1)E(C,-1,-1), then

y𝟏𝒋x𝟎𝒊\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}} =y𝟎𝒋x𝟏𝒊\displaystyle=-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}}
y𝟎𝒋x𝟎𝒊\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0j}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0i}}}} =y𝟏𝒋x𝟏𝒊\displaystyle=\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1j}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1i}}}}
Proof.

The statement of theorem is corollary of theorem 6.3. ∎

7. Algebra C(CH)C\otimes(C\otimes H)

Algebra C(CH)C\otimes(C\otimes H) is not quaternion algebra. However I consider this algebra here because this algebra is similar on algebra E(C,1,1)=CHE(C,-1,-1)=C\otimes H.

Algebra C(CH)C\otimes(C\otimes H) is interesting from other point of view also. When we consider this algebra it becomes evident that tensor product of noncommutative rings is nonassociative. Because CC is field then CCC\otimes C is isomorphic to CC, and therefore, (CC)H(C\otimes C)\otimes H is isomorphic CHC\otimes H. However, because HH is not algebra over complex field, then algebra C(CH)C\otimes(C\otimes H) is different from algebra (CC)H=CH(C\otimes C)\otimes H=C\otimes H.

Algebra C(CH)C\otimes(C\otimes H) has so large dimension, that it becomes unsuitable to consider product table and structural constants of this algebra. However it is easy to see that structure of this algebra is similar to structure of algebra considered in section 5.

We will represent basis of algebra C(CH)C\otimes(C\otimes H) as

(7.1) e¯=𝒌𝒋𝒊e¯C𝒌(e¯C𝒋e¯)H𝒊\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}kji}}}=\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}\otimes(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\otimes\overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}})

where e¯¯C\overline{\overline{e}}{}_{C} is basis of algebra CC and e¯¯H\overline{\overline{e}}{}_{H} is basis of algebra HH. Correspondingly, the product in algebra C(CH)C\otimes(C\otimes H) is defined componentwise

(7.2) a1(a2a3)b1(b2b2)=(a1b1)((a2b2)(a3b3))a_{1}\otimes(a_{2}\otimes a_{3})\ b_{1}\otimes(b_{2}\otimes b_{2})=(a_{1}b_{1})\otimes((a_{2}b_{2})\otimes(a_{3}b_{3}))
Theorem 7.1.

Structural constants of the algebra C(CH)C\otimes(C\otimes H) have form

B𝒑𝒋𝒊𝒒𝒎𝒌𝒓𝒅𝒃=CCCC𝒑𝒒𝒓CH𝒋𝒎𝒅𝒊𝒌𝒃B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pji}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}qmk}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rdb}}}=C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pq}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}r}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}

where CC𝐩𝐪𝐫C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pq}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}r}}} are structural constants of complex field, CH𝐢𝐤𝐛C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}} are structural constants of quaternion algebra.

Proof.

To prove the theorem it is enough to compare following equations

e¯e¯𝒑𝒋𝒊=𝒒𝒎𝒌\displaystyle\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pji}}}\ \overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}qmk}}}= B𝒑𝒋𝒊𝒒𝒎𝒌𝒓𝒅𝒃e¯𝒓𝒅𝒃\displaystyle B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}pji}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}qmk}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}rdb}}}\ \overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}rdb}}}
e¯e¯𝒑𝒋𝒊=𝒒𝒎𝒌\displaystyle\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pji}}}\ \overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}qmk}}}= e¯C𝒑(e¯C𝒋e¯)H𝒊e¯C𝒒(e¯C𝒎e¯)H𝒌)\displaystyle\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\otimes(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\otimes\overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}})\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q}}}\otimes(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}m}}}\otimes\overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}))
=\displaystyle= (e¯e¯C𝒑)C𝒒((e¯e¯C𝒋)C𝒎(e¯)H𝒊e¯)H𝒌)\displaystyle(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ \overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q}}})\otimes((\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}j}}}\ \overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}m}}})\otimes(\overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}i}}})\ \overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}k}}}))
=\displaystyle= CCCC𝒑𝒒𝒓CH𝒋𝒎𝒅e¯𝒊𝒌𝒃C𝒓(e¯C𝒅e¯)H𝒃\displaystyle C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pq}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}r}}}C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}r}}}\otimes(\overline{e}{}_{C\cdot\boldsymbol{{\color[rgb]{1,.3,.6}d}}}\otimes\overline{e}{}_{H\cdot\boldsymbol{{\color[rgb]{1,.3,.6}b}}})
=\displaystyle= CCCC𝒑𝒒𝒓CH𝒋𝒎𝒅e¯𝒊𝒌𝒃𝒓𝒅𝒃\displaystyle C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}pq}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}r}}}\ C_{C\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}jm}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}d}}}\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\ \overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}rdb}}}

Theorem 7.2.

Structural constants of the algebra C(CH)C\otimes(C\otimes H) have form

(7.3) B𝟎𝟎𝒊𝟎𝟎𝒌𝟎𝟎𝒃=CH𝒊𝒌𝒃B𝟎𝟎𝒊𝟎𝟏𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃B𝟎𝟏𝒊𝟎𝟎𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃B𝟎𝟏𝒊𝟎𝟏𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃B𝟎𝟎𝒊𝟏𝟎𝒌𝟏𝟎𝒃=CH𝒊𝒌𝒃B𝟎𝟎𝒊𝟏𝟏𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟎𝟏𝒊𝟏𝟎𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟎𝟏𝒊𝟏𝟏𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟎𝒊𝟎𝟎𝒌𝟏𝟎𝒃=CH𝒊𝒌𝒃B𝟏𝟎𝒊𝟎𝟏𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟏𝒊𝟎𝟎𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟏𝒊𝟎𝟏𝒌𝟏𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟎𝒊𝟏𝟎𝒌𝟎𝟎𝒃=CH𝒊𝒌𝒃B𝟏𝟎𝒊𝟏𝟏𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟏𝒊𝟏𝟎𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃B𝟏𝟏𝒊𝟏𝟏𝒌𝟎𝟏𝒃=CH𝒊𝒌𝒃\begin{array}[]{r@{}r|r@{}r}\displaystyle B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ -C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}&B_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11k}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01b}}}=&\ C_{H\cdot}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}ik}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}b}}}\end{array}
Proof.

The statement of theorem is corollary of theorems 7.1, 3.1. ∎

Theorem 7.3 (The Cauchy-Riemann equations).

Matrix of linear function

y𝒊𝒌𝒑=x𝒋𝒎𝒓f𝒋𝒎𝒓𝒊𝒌𝒑y^{\boldsymbol{{\color[rgb]{1,.3,.6}ikp}}}=x^{\boldsymbol{{\color[rgb]{1,.3,.6}jmr}}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}jmr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}ikp}}}

of algebra C(CH)C\otimes(C\otimes H) satisfies relationship

(7.4) f𝟎𝟎𝒊𝟎𝟎𝒋=f𝟏𝟎𝒊𝟏𝟎𝒋=f𝟎𝟏𝒊𝟎𝟏𝒋=f𝟏𝟏𝒊𝟏𝟏𝒋f𝟎𝟎𝒊𝟎𝟏𝒋=f𝟏𝟎𝒊𝟏𝟏𝒋=f𝟎𝟏𝒊𝟎𝟎𝒋=f𝟏𝟏𝒊𝟏𝟎𝒋f𝟎𝟎𝒊𝟏𝟎𝒋=f𝟏𝟎𝒊𝟎𝟎𝒋=f𝟎𝟏𝒊𝟏𝟏𝒋=f𝟏𝟏𝒊𝟎𝟏𝒋f𝟎𝟎𝒊𝟏𝟏𝒋=f𝟏𝟎𝒊𝟎𝟏𝒋=f𝟎𝟏𝒊𝟏𝟎𝒋=f𝟏𝟏𝒊𝟎𝟎𝒋\begin{array}[]{r@{}r@{}r@{}r}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}\end{array}
Proof.

Using equations [2]-(LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation), (7.3), we can repeat computing that we used to prove theorem 6.3. However, it is evident that this computing is different only by set of indexes. Thus, from theorem 6.3, it follows

(7.5) f𝟎𝒋𝒊𝟎𝒎𝒌=f𝟏𝒋𝒊𝟏𝒎𝒌f𝟎𝒋𝒊𝟏𝒎𝒌=f𝟏𝒋𝒊𝟎𝒎𝒌\begin{array}[]{r@{}r}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0ji}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0mk}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1ji}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1mk}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0ji}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1mk}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}1ji}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}0mk}}}\end{array}
(7.6) f𝒑𝟎𝒊𝒒𝟎𝒌=f𝒑𝟏𝒊𝒒𝟏𝒌f𝒑𝟎𝒊𝒒𝟏𝒌=f𝒑𝟏𝒊𝒒𝟎𝒌\begin{array}[]{r@{}r}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q0k}}}=&f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q1k}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p0i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q1k}}}=&-f_{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}p1i}}}^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}q0k}}}\end{array}

Equations (7.4) follow from equations (7.5), (7.6). ∎

Theorem 7.4 (The Cauchy-Riemann equations).

Since matrix

(y𝒊𝒋𝒑x𝒌𝒍𝒒)\left(\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}ijp}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}klq}}}}\right)

is Jacobian matrix of map in algebra C(CH)C\otimes(C\otimes H) , then

y𝟎𝟎𝒊x𝟎𝟎𝒋=y𝟏𝟎𝒊x𝟏𝟎𝒋=y𝟎𝟏𝒊x𝟎𝟏𝒋=y𝟏𝟏𝒊x𝟏𝟏𝒋y𝟎𝟎𝒊x𝟎𝟏𝒋=y𝟏𝟎𝒊x𝟏𝟏𝒋=y𝟎𝟏𝒊x𝟎𝟎𝒋=y𝟏𝟏𝒊x𝟏𝟎𝒋y𝟎𝟎𝒊x𝟏𝟎𝒋=y𝟏𝟎𝒊x𝟎𝟎𝒋=y𝟎𝟏𝒊x𝟏𝟏𝒋=y𝟏𝟏𝒊x𝟎𝟏𝒋y𝟎𝟎𝒊x𝟏𝟏𝒋=y𝟏𝟎𝒊x𝟎𝟏𝒋=y𝟎𝟏𝒊x𝟏𝟎𝒋=y𝟏𝟏𝒊x𝟎𝟎𝒋\begin{array}[]{r@{}r@{}r@{}r}\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01j}}}}=&\displaystyle-\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}01i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}10j}}}}=&\displaystyle\frac{\partial y^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}11i}}}}{\partial x^{\cdot\boldsymbol{{\color[rgb]{1,.3,.6}00j}}}}\end{array}
Proof.

The statement of theorem is corollary of theorem 7.3. ∎

8. Quaternion Algebra E(R,a,b)E(R,a,b)

Assume e¯=𝟎1\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=1, e¯=𝟏i\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=i, e¯=𝟐j\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=j, e¯=𝟑k\overline{e}{}_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=k. According to equation (4.1) structural constants of algebra E(R,a,b)E(R,a,b) have form

C𝟎𝟎𝟎=1C𝟎𝟏𝟏=1C𝟎𝟐𝟐=1C𝟎𝟑𝟑=1C𝟏𝟎𝟏=1C𝟏𝟏𝟎=aC𝟏𝟐𝟑=1C𝟏𝟑𝟐=aC𝟐𝟎𝟐=1C𝟐𝟏𝟑=1C𝟐𝟐𝟎=bC𝟐𝟑𝟏=bC𝟑𝟎𝟑=1C𝟑𝟏𝟐=aC𝟑𝟐𝟏=bC𝟑𝟑𝟎=ab\begin{array}[]{r@{}rr@{}rr@{}rr@{}r}C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&a&C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&a\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&-1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&b&C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-b\\ \vphantom{\overset{\rightarrow}{1}^{1}}C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&1&C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&-a&C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&b&C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-ab\end{array}
Theorem 8.1.

Standard components of linear function and coordinates of corresponding linear map over field RR satisfy relationship

(8.1) {f𝟎𝟎=f𝟎𝟎+af𝟏𝟏+bf𝟐𝟐abf𝟑𝟑f𝟏𝟏=f𝟎𝟎+af𝟏𝟏bf𝟐𝟐+abf𝟑𝟑f𝟐𝟐=f𝟎𝟎af𝟏𝟏+bf𝟐𝟐+abf𝟑𝟑f𝟑𝟑=f𝟎𝟎af𝟏𝟏bf𝟐𝟐abf𝟑𝟑\left\{\begin{array}[]{r@{}r@{}r@{}r@{}r}f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}-&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}+&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-&af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}+&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-&af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}-&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\end{array}\right.
(8.2) {f𝟎𝟏=f𝟎𝟏+f𝟏𝟎bf𝟐𝟑+bf𝟑𝟐f𝟏𝟎=af𝟎𝟏+af𝟏𝟎+abf𝟐𝟑abf𝟑𝟐f𝟐𝟑=f𝟎𝟏+f𝟏𝟎+bf𝟐𝟑+bf𝟑𝟐f𝟑𝟐=af𝟎𝟏+af𝟏𝟎abf𝟐𝟑abf𝟑𝟐\left\{\begin{array}[]{r@{}r@{}r@{}r@{}r@{}r}f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&&f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+&f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}-&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&&af^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}+&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&-&f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+&f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&-&af^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\end{array}\right.
(8.3) {f𝟎𝟐=f𝟎𝟐+af𝟏𝟑+f𝟐𝟎af𝟑𝟏f𝟏𝟑=f𝟎𝟐+af𝟏𝟑f𝟐𝟎+af𝟑𝟏f𝟐𝟎=bf𝟎𝟐abf𝟏𝟑+bf𝟐𝟎+abf𝟑𝟏f𝟑𝟏=bf𝟎𝟐abf𝟏𝟑bf𝟐𝟎abf𝟑𝟏\left\{\begin{array}[]{r@{}r@{}r@{}r@{}r}f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}+&f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}-&af^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}-&f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}+&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}-&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\end{array}\right.
(8.4) {f𝟎𝟑=f𝟎𝟑+f𝟏𝟐f𝟐𝟏+f𝟑𝟎f𝟏𝟐=af𝟎𝟑+af𝟏𝟐+af𝟐𝟏af𝟑𝟎f𝟐𝟏=bf𝟎𝟑+bf𝟏𝟐+bf𝟐𝟏+bf𝟑𝟎f𝟑𝟎=abf𝟎𝟑+abf𝟏𝟐abf𝟐𝟏abf𝟑𝟎\left\{\begin{array}[]{r@{}r@{}r@{}r@{}r@{}r}f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}=&&f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+&f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}-&f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}+&f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}=&&af^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}+&af^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}-&af^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}=&-&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}+&bf^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\\ \vphantom{\overset{\rightarrow}{1}^{1}}f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}=&-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}-&abf^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\end{array}\right.
(8.5) {f𝟎𝟎=14f𝟎𝟎+14f𝟏𝟏+14f𝟐𝟐+14f𝟑𝟑f𝟏𝟏=14af𝟎𝟎+14af𝟏𝟏14af𝟐𝟐14af𝟑𝟑f𝟐𝟐=14bf𝟎𝟎14bf𝟏𝟏+14bf𝟐𝟐14bf𝟑𝟑f𝟑𝟑=14abf𝟎𝟎+14abf𝟏𝟏+14abf𝟐𝟐14abf𝟑𝟑\left\{\begin{array}[]{r@{}c@{}r@{}c@{}r@{}c@{}r@{}c@{}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}=&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}=&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}=&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}=&-\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\end{array}\right.
(8.6) {f𝟏𝟎=14af𝟏𝟎+14f𝟎𝟏+14af𝟑𝟐+14f𝟐𝟑f𝟎𝟏=14af𝟏𝟎+14f𝟎𝟏14af𝟑𝟐14f𝟐𝟑f𝟑𝟐=14abf𝟏𝟎+14bf𝟎𝟏14abf𝟑𝟐+14bf𝟐𝟑f𝟐𝟑=14abf𝟏𝟎14bf𝟎𝟏14abf𝟑𝟐+14bf𝟐𝟑\left\{\begin{array}[]{r@{}c@{}c@{}r@{}c@{}r@{}c@{}r@{}c@{}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}=&&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}=&&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}=&-&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}=&&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\end{array}\right.
(8.7) {f𝟐𝟎=14bf𝟐𝟎14bf𝟑𝟏+14f𝟎𝟐14f𝟏𝟑f𝟑𝟏=14abf𝟐𝟎14abf𝟑𝟏14af𝟎𝟐+14af𝟏𝟑f𝟎𝟐=14bf𝟐𝟎+14bf𝟑𝟏+14f𝟎𝟐+14f𝟏𝟑f𝟏𝟑=14abf𝟐𝟎14abf𝟑𝟏+14af𝟎𝟐+14af𝟏𝟑\left\{\begin{array}[]{r@{}c@{}c@{}r@{}c@{}r@{}c@{}r@{}c@{}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}=&&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}=&&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}=&&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}=&-&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\end{array}\right.
(8.8) {f𝟑𝟎=14abf𝟑𝟎+14bf𝟐𝟏14af𝟏𝟐+14f𝟎𝟑f𝟐𝟏=14abf𝟑𝟎+14bf𝟐𝟏+14af𝟏𝟐14f𝟎𝟑f𝟏𝟐=14abf𝟑𝟎+14bf𝟐𝟏+14af𝟏𝟐+14f𝟎𝟑f𝟎𝟑=14abf𝟑𝟎14bf𝟐𝟏+14af𝟏𝟐+14f𝟎𝟑\left\{\begin{array}[]{r@{}c@{}c@{}r@{}c@{}r@{}c@{}r@{}c@{}r}f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}=&-&\displaystyle\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}=&-&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}=&&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}=&-&\displaystyle\vphantom{{\frac{1}{1}}^{1}}\frac{1}{4ab}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-&\displaystyle\frac{1}{4b}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+&\displaystyle\frac{1}{4a}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+&\displaystyle\frac{1}{4}&f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\end{array}\right.
Proof.

Using equation [2]-LABEL:0812.4763-English-eq:_linear_map_over_field,_division_ring,_relation we get relationships

f𝟎𝟎\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =f𝒌𝒓C𝒌𝟎𝒑C𝒑𝒓𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=f𝟎𝟎C𝟎𝟎𝟎C𝟎𝟎𝟎+f𝟏𝟏C𝟏𝟎𝟏C𝟏𝟏𝟎+f𝟐𝟐C𝟐𝟎𝟐C𝟐𝟐𝟎+f𝟑𝟑C𝟑𝟎𝟑C𝟑𝟑𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=f𝟎𝟎+af𝟏𝟏+bf𝟐𝟐abf𝟑𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}
f𝟎𝟏\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =f𝒌𝒓C𝒌𝟎𝒑C𝒑𝒓𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟏C𝟎𝟎𝟎C𝟎𝟏𝟏+f𝟏𝟎C𝟏𝟎𝟏C𝟏𝟎𝟏+f𝟐𝟑C𝟐𝟎𝟐C𝟐𝟑𝟏+f𝟑𝟐C𝟑𝟎𝟑C𝟑𝟐𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟏+f𝟏𝟎bf𝟐𝟑+bf𝟑𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}-bf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}
f𝟎𝟐\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =f𝒌𝒓C𝒌𝟎𝒑C𝒑𝒓𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟐C𝟎𝟎𝟎C𝟎𝟐𝟐+f𝟏𝟑C𝟏𝟎𝟏C𝟏𝟑𝟐+f𝟐𝟎C𝟐𝟎𝟐C𝟐𝟎𝟐+f𝟑𝟏C𝟑𝟎𝟑C𝟑𝟏𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟐+af𝟏𝟑+f𝟐𝟎af𝟑𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}-af^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}
f𝟎𝟑\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =f𝒌𝒓C𝒌𝟎𝒑C𝒑𝒓𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k0}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟑C𝟎𝟎𝟎C𝟎𝟑𝟑+f𝟏𝟐C𝟏𝟎𝟏C𝟏𝟐𝟑+f𝟐𝟏C𝟐𝟎𝟐C𝟐𝟏𝟑+f𝟑𝟎C𝟑𝟎𝟑C𝟑𝟎𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟑+f𝟏𝟐f𝟐𝟏+f𝟑𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}-f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}
f𝟏𝟎\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =f𝒌𝒓C𝒌𝟏𝒑C𝒑𝒓𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=f𝟎𝟏C𝟎𝟏𝟏C𝟏𝟏𝟎+f𝟏𝟎C𝟏𝟏𝟎C𝟎𝟎𝟎+f𝟐𝟑C𝟐𝟏𝟑C𝟑𝟑𝟎+f𝟑𝟐C𝟑𝟏𝟐C𝟐𝟐𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=af𝟎𝟏+af𝟏𝟎+abf𝟐𝟑abf𝟑𝟐\displaystyle=af^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}+abf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}
f𝟏𝟏\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =f𝒌𝒓C𝒌𝟏𝒑C𝒑𝒓𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟎C𝟎𝟏𝟏C𝟏𝟎𝟏+f𝟏𝟏C𝟏𝟏𝟎C𝟎𝟏𝟏+f𝟐𝟐C𝟐𝟏𝟑C𝟑𝟐𝟏+f𝟑𝟑C𝟑𝟏𝟐C𝟐𝟑𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟎+af𝟏𝟏bf𝟐𝟐+abf𝟑𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}-bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}+abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}
f𝟏𝟐\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =f𝒌𝒓C𝒌𝟏𝒑C𝒑𝒓𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟑C𝟎𝟏𝟏C𝟏𝟑𝟐+f𝟏𝟐C𝟏𝟏𝟎C𝟎𝟐𝟐+f𝟐𝟏C𝟐𝟏𝟑C𝟑𝟏𝟐+f𝟑𝟎C𝟑𝟏𝟐C𝟐𝟎𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=af𝟎𝟑+af𝟏𝟐+af𝟐𝟏af𝟑𝟎\displaystyle=af^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}-af^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}
f𝟏𝟑\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =f𝒌𝒓C𝒌𝟏𝒑C𝒑𝒓𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k1}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟐C𝟎𝟏𝟏C𝟏𝟐𝟑+f𝟏𝟑C𝟏𝟏𝟎C𝟎𝟑𝟑+f𝟐𝟎C𝟐𝟏𝟑C𝟑𝟎𝟑+f𝟑𝟏C𝟑𝟏𝟐C𝟐𝟏𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟐+af𝟏𝟑f𝟐𝟎+af𝟑𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}-f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}
f𝟐𝟎\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =f𝒌𝒓C𝒌𝟐𝒑C𝒑𝒓𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=f𝟎𝟐C𝟎𝟐𝟐C𝟐𝟐𝟎+f𝟏𝟑C𝟏𝟐𝟑C𝟑𝟑𝟎+f𝟐𝟎C𝟐𝟐𝟎C𝟎𝟎𝟎+f𝟑𝟏C𝟑𝟐𝟏C𝟏𝟏𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=bf𝟎𝟐abf𝟏𝟑+bf𝟐𝟎+abf𝟑𝟏\displaystyle=bf^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}+abf^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}
f𝟐𝟏\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =f𝒌𝒓C𝒌𝟐𝒑C𝒑𝒓𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟑C𝟎𝟐𝟐C𝟐𝟑𝟏+f𝟏𝟐C𝟏𝟐𝟑C𝟑𝟐𝟏+f𝟐𝟏C𝟐𝟐𝟎C𝟎𝟏𝟏+f𝟑𝟎C𝟑𝟐𝟏C𝟏𝟎𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=bf𝟎𝟑+bf𝟏𝟐+bf𝟐𝟏+bf𝟑𝟎\displaystyle=-bf^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}
f𝟐𝟐\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =f𝒌𝒓C𝒌𝟐𝒑C𝒑𝒓𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟎C𝟎𝟐𝟐C𝟐𝟎𝟐+f𝟏𝟏C𝟏𝟐𝟑C𝟑𝟏𝟐+f𝟐𝟐C𝟐𝟐𝟎C𝟎𝟐𝟐+f𝟑𝟑C𝟑𝟐𝟏C𝟏𝟑𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟎af𝟏𝟏+bf𝟐𝟐+abf𝟑𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}+abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}
f𝟐𝟑\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =f𝒌𝒓C𝒌𝟐𝒑C𝒑𝒓𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k2}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟏C𝟎𝟐𝟐C𝟐𝟏𝟑+f𝟏𝟎C𝟏𝟐𝟑C𝟑𝟎𝟑+f𝟐𝟑C𝟐𝟐𝟎C𝟎𝟑𝟑+f𝟑𝟐C𝟑𝟐𝟏C𝟏𝟐𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟏+f𝟏𝟎+bf𝟐𝟑+bf𝟑𝟐\displaystyle=-f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}+bf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}
f𝟑𝟎\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}} =f𝒌𝒓C𝒌𝟑𝒑C𝒑𝒓𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=f𝟎𝟑C𝟎𝟑𝟑C𝟑𝟑𝟎+f𝟏𝟐C𝟏𝟑𝟐C𝟐𝟐𝟎+f𝟐𝟏C𝟐𝟑𝟏C𝟏𝟏𝟎+f𝟑𝟎C𝟑𝟑𝟎C𝟎𝟎𝟎\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}
=abf𝟎𝟑+abf𝟏𝟐abf𝟐𝟏abf𝟑𝟎\displaystyle=-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}+abf^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}
f𝟑𝟏\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}} =f𝒌𝒓C𝒌𝟑𝒑C𝒑𝒓𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=f𝟎𝟐C𝟎𝟑𝟑C𝟑𝟐𝟏+f𝟏𝟑C𝟏𝟑𝟐C𝟐𝟑𝟏+f𝟐𝟎C𝟐𝟑𝟏C𝟏𝟎𝟏+f𝟑𝟏C𝟑𝟑𝟎C𝟎𝟏𝟏\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}
=bf𝟎𝟐abf𝟏𝟑bf𝟐𝟎abf𝟑𝟏\displaystyle=bf^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}-bf^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}
f𝟑𝟐\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}} =f𝒌𝒓C𝒌𝟑𝒑C𝒑𝒓𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=f𝟎𝟏C𝟎𝟑𝟑C𝟑𝟏𝟐+f𝟏𝟎C𝟏𝟑𝟐C𝟐𝟎𝟐+f𝟐𝟑C𝟐𝟑𝟏C𝟏𝟑𝟐+f𝟑𝟐C𝟑𝟑𝟎C𝟎𝟐𝟐\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}
=af𝟎𝟏+af𝟏𝟎abf𝟐𝟑abf𝟑𝟐\displaystyle=-af^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}+af^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}
f𝟑𝟑\displaystyle f_{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}} =f𝒌𝒓C𝒌𝟑𝒑C𝒑𝒓𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}kr}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}k3}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}p}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}pr}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟎C𝟎𝟑𝟑C𝟑𝟎𝟑+f𝟏𝟏C𝟏𝟑𝟐C𝟐𝟏𝟑+f𝟐𝟐C𝟐𝟑𝟏C𝟏𝟐𝟑+f𝟑𝟑C𝟑𝟑𝟎C𝟎𝟑𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+f^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\ C_{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}
=f𝟎𝟎af𝟏𝟏bf𝟐𝟐abf𝟑𝟑\displaystyle=f^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}-af^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}-bf^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}-abf^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}

We group these relationships into systems of linear equations (8.1), (8.2), (8.3), (8.4).

(8.5) is solution of system of linear equations (8.1).

(8.6) is solution of system of linear equations (8.2).

(8.7) is solution of system of linear equations (8.3).

(8.8) is solution of system of linear equations (8.4). ∎

Theorem 8.2.

For any values of parameters a0a\neq 0, b0b\neq 0, there exists one to one map between coordinates of linear function of algebra E(R,a,b)E(R,a,b) and its standard components.

Proof.

The statement of theorem is corollary of theorem 8.1. ∎

9. Regular Function

Although there is no analogue of the Cauchy-Riemann equations in quaternion algebra, in different papers mathematicians explore different sets of functions that have properties similar to properties of functions of complex variable. In [4, 5], there is definition of regular function that satisfies to equation

(9.1) fx𝟎+ifx𝟏+jfx𝟐+kfx𝟑=0\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}+i\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}+j\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+k\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}=0
Theorem 9.1.

Differential equation (9.1) is equivalent to system of differential equations

(9.2) {f𝟎x𝟎f𝟏x𝟏f𝟐x𝟐f𝟑x𝟑=0f𝟎x𝟏+f𝟏x𝟎f𝟐x𝟑+f𝟑x𝟐=0f𝟎x𝟐+f𝟏x𝟑+f𝟐x𝟎f𝟑x𝟏=0f𝟎x𝟑+f𝟐x𝟏f𝟏x𝟐+f𝟑x𝟎=0\left\{\begin{matrix}\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}=0\end{matrix}\right.
Proof.

Let us substitute

(9.3) fx𝒊=f𝟎x𝒊+f𝟏x𝒊i+f𝟐x𝒊j+f𝟑x𝒊k\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}=\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}i+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}j+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}k

into equation (9.1). We will get

(9.4) fx𝟎+ifx𝟏+jfx𝟐+kfx𝟑=\displaystyle\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}+i\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}+j\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+k\frac{\partial f}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}= f𝟎x𝟎f𝟏x𝟏f𝟐x𝟐f𝟑x𝟑\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}
+\displaystyle+ i(f𝟎x𝟏+f𝟏x𝟎f𝟐x𝟑+f𝟑x𝟐)\displaystyle i(\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}})
+\displaystyle+ j(f𝟎x𝟐+f𝟏x𝟑+f𝟐x𝟎f𝟑x𝟏)\displaystyle j(\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}})
+\displaystyle+ k(f𝟎x𝟑+f𝟐x𝟏f𝟏x𝟐+f𝟑x𝟎)=0\displaystyle k(\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}})=0

The statement of theorem follows from equation (9.4). ∎

In the paper [1], corollary 3.1.2, p. 1000, Deavours proves that the only regular quaternion functions with bounded norm is a constant.

Theorem 9.2.

Components of the Gâteaux derivative of regular quaternion function satisfy to equations

(9.5) 𝟎𝟎yx+𝟏𝟏yx+𝟐𝟐yx+𝟑𝟑yx=0𝟎𝟏yx+𝟏𝟎yx+𝟐𝟑yx𝟑𝟐yx=0𝟎𝟐yx𝟏𝟑yx+𝟐𝟎yx+𝟑𝟏yx=0𝟎𝟑yx𝟏𝟐yx+𝟐𝟏yx+𝟑𝟎yx=0\begin{array}[]{r}\displaystyle\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}=0\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}=0\end{array}
Proof.

Let us substitute equations [2]-(LABEL:0812.4763-English-eq:_quaternion_over_real_field,_derivative,_1,_0), [2]-(LABEL:0812.4763-English-eq:_quaternion_over_real_field,_derivative,_1,_1), [2]-(LABEL:0812.4763-English-eq:_quaternion_over_real_field,_derivative,_1,_2), [2]-(LABEL:0812.4763-English-eq:_quaternion_over_real_field,_derivative,_1,_3) into equation (9.2). We will get

f𝟎x𝟎f𝟏x𝟏f𝟐x𝟐f𝟑x𝟑\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}
=\displaystyle= 𝟎𝟎yx𝟏𝟏yx𝟐𝟐yx𝟑𝟑yx(𝟎𝟎yx𝟏𝟏yx+𝟐𝟐yx+𝟑𝟑yx)\displaystyle\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}-\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}\right)
\displaystyle- (𝟎𝟎yx+𝟏𝟏yx𝟐𝟐yx+𝟑𝟑yx)(𝟎𝟎yx+𝟏𝟏yx+𝟐𝟐yx𝟑𝟑yx)\displaystyle\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}\right)-\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}\right)
=\displaystyle= 2𝟎𝟎yx2𝟏𝟏yx2𝟐𝟐yx2𝟑𝟑yx\displaystyle-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}00}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}
=\displaystyle= 0\displaystyle 0
f𝟎x𝟏+f𝟏x𝟎f𝟐x𝟑+f𝟑x𝟐\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}
=\displaystyle= 𝟎𝟏yx𝟏𝟎yx+𝟐𝟑yx𝟑𝟐yx+(𝟎𝟏yx+𝟏𝟎yx+𝟐𝟑yx𝟑𝟐yx)\displaystyle-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}+\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}\right)
\displaystyle- (𝟎𝟏yx𝟏𝟎yx𝟐𝟑yx𝟑𝟐yx)+(𝟎𝟏yx+𝟏𝟎yx𝟐𝟑yx𝟑𝟐yx)\displaystyle\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}\right)+\left(-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}\right)
=\displaystyle= 2𝟎𝟏yx+2𝟏𝟎yx+2𝟐𝟑yx2𝟑𝟐yx\displaystyle-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}01}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}
=\displaystyle= 0\displaystyle 0
f𝟎x𝟐+f𝟏x𝟑+f𝟐x𝟎f𝟑x𝟏\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}
=\displaystyle= 𝟎𝟐yx𝟏𝟑yx𝟐𝟎yx+𝟑𝟏yx+(𝟎𝟐yx𝟏𝟑yx+𝟐𝟎yx𝟑𝟏yx)\displaystyle-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}+\left(-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}\right)
+\displaystyle+ (𝟎𝟐yx𝟏𝟑yx+𝟐𝟎yx+𝟑𝟏yx)(𝟎𝟐yx𝟏𝟑yx𝟐𝟎yx𝟑𝟏yx)\displaystyle\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}\right)-\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}\right)
=\displaystyle= 2𝟎𝟐yx2𝟏𝟑yx+2𝟐𝟎yx+2𝟑𝟏yx\displaystyle-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}02}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}
=\displaystyle= 0\displaystyle 0
f𝟎x𝟑+f𝟐x𝟏f𝟏x𝟐+f𝟑x𝟎\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}-\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}+\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}
=\displaystyle= 𝟎𝟑yx+𝟏𝟐yx𝟐𝟏yx𝟑𝟎yx(𝟎𝟑yx𝟏𝟐yx𝟐𝟏yx+𝟑𝟎yx)\displaystyle-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}-\left(-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}\right)
\displaystyle- (𝟎𝟑yx𝟏𝟐yx𝟐𝟏yx𝟑𝟎yx)(𝟎𝟑yx+𝟏𝟐yx𝟐𝟏yx+𝟑𝟎yx)\displaystyle\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}\right)-\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}\right)
=\displaystyle= 2𝟎𝟑yx2𝟏𝟐yx+2𝟐𝟏yx+2𝟑𝟎yx\displaystyle-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}03}}}y}{\partial x}-2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}+2\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}
=\displaystyle= 0\displaystyle 0

Theorem 9.3.

The Gâteaux differential of regular function over quaternion algebra has form

(9.6) (𝟏𝟏yx+𝟐𝟐yx+𝟑𝟑yx)dx+𝟏𝟏yxidxi+𝟐𝟐yxjdxj+𝟑𝟑yxkdxk+(𝟏𝟎yx+𝟐𝟑yx𝟑𝟐yx)dxi+𝟏𝟎yxidx+𝟐𝟑yxjdxk𝟑𝟐yxkdxi+(𝟏𝟑yx+𝟐𝟎yx+𝟑𝟏yx)dxj𝟏𝟑yxidxk+𝟐𝟎yxjdx+𝟑𝟏yxkdxi+(𝟏𝟐yx+𝟐𝟏yx+𝟑𝟎yx)dxk𝟏𝟐yxidxj+𝟐𝟏yxjdxi+𝟑𝟎yxkdx\begin{array}[]{r}\displaystyle-\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}\right)dx+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}11}}}y}{\partial x}idxi+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}22}}}y}{\partial x}jdxj+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}33}}}y}{\partial x}kdxk\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}+\left(\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}\right)dxi+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}10}}}y}{\partial x}idx+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}23}}}y}{\partial x}jdxk-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}32}}}y}{\partial x}kdxi\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}+\left(-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}\right)dxj-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}13}}}y}{\partial x}idxk+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}20}}}y}{\partial x}jdx+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}31}}}y}{\partial x}kdxi\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}+\left(-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}\right)dxk-\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}12}}}y}{\partial x}idxj+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}21}}}y}{\partial x}jdxi+\frac{\partial^{\boldsymbol{{\color[rgb]{1,.3,.6}30}}}y}{\partial x}kdx\end{array}
Proof.

The statement of theorem is corollary of theorem [2]-LABEL:0812.4763-English-theorem:_Gateaux_differential,_standard_form,_division_ring. ∎

Equation (9.1) is equivalent to equation

(9.7) (1ijk)(f𝟎x𝟎f𝟎x𝟏f𝟎x𝟐f𝟎x𝟑f𝟏x𝟎f𝟏x𝟏f𝟏x𝟐f𝟏x𝟑f𝟐x𝟎f𝟐x𝟏f𝟐x𝟐f𝟐x𝟑f𝟑x𝟎f𝟑x𝟏f𝟑x𝟐f𝟑x𝟑)(1ijk)=0\begin{pmatrix}1&i&j&k\end{pmatrix}{}_{*}{}^{*}\begin{pmatrix}\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}\\ \displaystyle\vphantom{\overset{\rightarrow}{\frac{1}{1}}^{\frac{1}{1}}}\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}&\displaystyle\frac{\partial f^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}\end{pmatrix}{}_{*}{}^{*}\begin{pmatrix}1\\ i\\ j\\ k\end{pmatrix}=0

10. Instead of an Epilogue

The complex field and quaternion algebra have both common properties and differences. These differences make it harder to identify in the quaternion algebra patterns similar to those we have observed in the complex field. Therefore, it is very important to understand these differences.

One of the research directions is finding an analog to the Cauchy-Riemann equation in quaternion algebra. In this paper, I reviewed some studies in this area.

According to the theorem [2]-LABEL:0812.4763-English-theorem:_complex_field_over_real_field, linear mapping has matrix

(a𝟎a𝟏a𝟏a𝟎)\begin{pmatrix}a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{pmatrix}

This mapping corresponds to multiplication by the number a=a𝟎+a𝟏ia=a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i. The statement follows from equations

(a𝟎+a𝟏i)(x𝟎+x𝟏i)=a𝟎x𝟎a𝟏x𝟏+(a𝟎x𝟏+a𝟏x𝟎)i(a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i)(x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}i)=a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+(a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}+a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})i
(a𝟎a𝟏a𝟏a𝟎)(x𝟎x𝟏)=(a𝟎x𝟎a𝟏x𝟏a𝟏x𝟎+a𝟎x𝟏)\begin{pmatrix}a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{pmatrix}\begin{pmatrix}x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\\ x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\end{pmatrix}=\begin{pmatrix}a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}-a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ a_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}+a_{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x_{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\end{pmatrix}

I decided to consider a similar class of functions of quaternions. The linear mapping of quaternion algebra

xaxx\rightarrow ax

has matrix

(10.1) (a𝟎a𝟏a𝟐a𝟑a𝟏a𝟎a𝟑a𝟐a𝟐a𝟑a𝟎a𝟏a𝟑a𝟐a𝟏a𝟎)\left(\begin{array}[]{rr@{}rr@{}rr@{}r}a^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ a^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\\ a^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ a^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&-&a^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&&a^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{array}\right)

It is interesting to consider the class of quaternion functions which has derivative of similar structure. However I think that the structure of the matrix (10.1) is too restrictive for derivative of functions of quaternions and I little relaxed the requirement. I assumed that the derivative satisfies to following equations

(10.2) y𝟎x𝟎=y𝟏x𝟏=y𝟐x𝟐=y𝟑x𝟑\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}=\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}=\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}}=\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}}
(10.3) y𝒊x𝒋=y𝒋x𝒊𝒊𝒋\begin{matrix}\displaystyle\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}}=-\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}j}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}i}}}}&\boldsymbol{{\color[rgb]{1,.3,.6}i}}\neq\boldsymbol{{\color[rgb]{1,.3,.6}j}}\end{matrix}

It is easy to see that derivative of function like

y=axy=xay=ax\ \ \ y=xa

satisfies to equations (10.2), (10.3).

Consider the function

y=x2y=x^{2}

Direct calculation gives

(10.4) y𝟎=(x𝟎)2(x𝟏)2(x𝟐)2(x𝟑)2y𝟏=2x𝟎x𝟏y𝟐=2x𝟎x𝟐y𝟑=2x𝟎x𝟑\begin{array}[]{r@{}l}y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&=(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\\ y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&=2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}\\ y^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&=2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}\\ y^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&=2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\end{array}

The derivative of the mapping (10.4) has matrix

(10.5) (2x𝟎2x𝟏2x𝟐2x𝟑2x𝟏2x𝟎002x𝟐02x𝟎02x𝟑002x𝟎)\begin{pmatrix}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ 2x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&0&0\\ 2x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}&0&2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&0\\ 2x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}&0&0&2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}\end{pmatrix}

Therefore, the matrix (10.5) satisfies to equations (10.2), (10.3).

Theorem 10.1.

Derivative of conjugation does not satisfy equations (10.2), (10.3).

Proof.

According to the proof of the theorem [2]-LABEL:0812.4763-English-theorem:_quaternion_conjugation, derivative of conjugation does not satisfy equation (10.2). ∎

However, the set of functions whose derivative satisfies to equations (10.2), (10.3) is not sufficiently large.

Theorem 10.2.

The mapping

(10.6) y=x3y=x^{3}

does not satisfy to equation (10.2).

Proof.

We can represent the mapping (10.6) in the following form

y=xx2y=x\,x^{2}

This mapping has coordinates

(10.12) y𝟎=x𝟎((x𝟎)2(x𝟏)2(x𝟐)2(x𝟑)2)x𝟏2x𝟎x𝟏x𝟐2x𝟎x𝟐x𝟑2x𝟎x𝟑=(x𝟎)3x𝟎(x𝟏)2x𝟎(x𝟐)2x𝟎(x𝟑)22x𝟎(x𝟏)22x𝟎(x𝟐)22x𝟎(x𝟑)2=(x𝟎)33x𝟎(x𝟏)23x𝟎(x𝟐)23x𝟎(x𝟑)2\displaystyle\begin{array}[]{r@{\,}l}y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}&=x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}((x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2})\\ &-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ &=(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{3}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\\ &-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\\ &=(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{3}-3x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-3x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-3x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\end{array}
(10.18) y𝟏=x𝟏((x𝟎)2(x𝟏)2(x𝟐)2(x𝟑)2)+x𝟎2x𝟎x𝟏x𝟑2x𝟎x𝟐+x𝟐2x𝟎x𝟑=x𝟏(x𝟎)2(x𝟏)𝟑x𝟏(x𝟐)2x𝟏(x𝟑)2+2(x𝟎)2x𝟏2x𝟎x𝟐x𝟑+2x𝟎x𝟐x𝟑=3x𝟏(x𝟎)2(x𝟏)3x𝟏(x𝟐)2x𝟏(x𝟑)2\displaystyle\begin{array}[]{r@{\,}l}y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}&=x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}((x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2})\\ &+x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}+x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ &=x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\\ &+2(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}-2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}+2x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}}x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}}\\ &=3x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{3}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}\end{array}

From the equation (10.12), it follows that

(10.19) y𝟎x𝟎=3(x𝟎)23(x𝟏)23(x𝟐)23(x𝟑)2\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}}}=3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}

From the equation (10.18), it follows that

(10.20) y𝟏x𝟏=3(x𝟎)23(x𝟏)2(x𝟐)2(x𝟑)2\frac{\partial y^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}{\partial x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}}}=3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}0}}})^{2}-3(x^{\boldsymbol{{\color[rgb]{1,.3,.6}1}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}2}}})^{2}-(x^{\boldsymbol{{\color[rgb]{1,.3,.6}3}}})^{2}

The statement of the theorem follows from equations (10.19), (10.20). ∎

Without a doubt, this is only the beginning of the study, and many questions must be answered.

11. References

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    C.A. Deavours, The Quaternion Calculus, American Mathematical Monthly, 80 (1973), pp. 995 - 1008

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    Aleks Kleyn, Introduction into Calculus over Division Ring,
    eprint arXiv:0812.4763 (2010)

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    I. M. Gelfand, M. I. Graev, Representation of Quaternion Groups over Localy Compact and Functional Fields,
    Funct. Anal. Appl. 2 (1968) 19 - 33;
    Izrail Moiseevich Gelfand, Semen Grigorevich Gindikin,
    Izrail M. Gelfand: Collected Papers, volume II, 435 - 449,
    Springer, 1989

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    Fueter, R. Die Funktionentheorie der Differentialgleichungen Δu=0\Delta u=0 und ΔΔu=0\Delta\Delta u=0 mit vier reellen Variablen. Comment. Math. Helv. 7 (1935), 307-330

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    A. Sudbery, Quaternionic Analysis, Math. Proc. Camb. Phil. Soc. (1979), 85, 199 - 225

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    Sir William Rowan Hamilton, The Mathematical Papers, Vol. III, Algebra,
    Cambridge at the University Press, 1967