A new SU(2/1) supergroup with determinant 1 explains many mysteries of the weak interactions.

The standard model of the strong, weak and electromagnetic interactions is very successful. The fundamental forces are fully specified by postulating the $SU(3)SU(2)U(1)$ gauge symmetry of the Yang-Mills Lagrangian. Yet, the properties of the elementary Fermions remain mysterious. Why do the weak interactions break parity? Why right neutrinos? Why several generations? Why fractional charges? Why do quarks and leptons succeed to cancel the quantum field theory (QFT) anomalies? Unfortunately, neither supersymmetry nor string theory directly addresses any of these fundamental questions. As a new classification principle completing the standard model, we propose to consider representations of a new compact form of the $SU(2/1)$ supergroup .

Two pioneers, Ne’eman N1 and Fairlie F1 , explored in 1979 the simple Lie-Kac superalgebra $su(2/1)$ Kac1 ; Kac2 because its even subalgebra $su(2)\oplus u(1)$ coincides with the electroweak gauge algebra. It was found that the Higgs fields have the quantum numbers of the odd generators N1 ; F1 , and that the helicities and weak hypercharges of the leptons N1 ; F1 and the quarks DJ ; NTM1 , graded by chirality, fit its smallest irreducible representations. Furthermore the existence of the 3 generations of leptons and quarks becomes natural COQ0 ; HS98 , because superalgebras admit indecomposable representations Marcu80 .

But all these desirable properties do no explain the cancellation of the Adler-Bell-Jackiw anomalies Adler ; BJ , without which gauge invariance is lost and the theory is not renormalizable. The quarks must, in a subtle way called the Bouchiat, Iliopoulos and Meyer (BIM) mechanism BIM , compensate in the quantum loops the deleterious effect of the leptons. This balance cannot be derived from the Lie algebra mathematics because their finite dimensional representations are fully reducible. Our main objective is to show that a composite set of quartet representations of $sl(2/1,\mathbb{C})$ with rational Dynkin numbers admits a real form which exponentiates to a new form of the $SU(2/1)$ supergroup with determinant 1 if and only if the trace of the $U(1)$ weak hypercharge $Y$ vanishes, $Tr(Y)=0$, and that the corresponding quantum field theory (QFT) is anomaly free.

This article is written from the point of view of particle physics but contains several new results in mathematics. Fixing some confusion in the physics literature, we carefully distinguish the complex superalgebra $sl(2/1,\mathbb{C})$, its real form $su(2/1,\mathbb{R})$, and the supergroup $SU(2/1)$. In section 2, all finite dimensional Kac modules Kac2 of $sl(2/1,\mathbb{C})$ are constructed. In section 3, the cubic super-Casimir tensor of an $sl(2/1,\mathbb{C})$ Kac module is shown to be proportional to $Tr(Y)$. In section 4, a new generalization of the Berezin super-Hermitian conjugation and new representations of $su(2/1,\mathbb{R})$ are defined. In section 5 new nested indecomposable representations of $su(2/1,\mathbb{R})$ are constructed. In section 6, a new form of the $SU(2/1)$ supergroup is constructed by exponentiation and shown to be compact and to have unit Berezinian and unit determinant if $Tr(Y)=0$. In section 7, it is shown that the even sector of the super-Casimir tensor provides an algebraic meaning to the quantum field theory anomalies Adler ; BJ . Finally, in section 8, it is shown that the BIM composite module, 3 quarks per lepton, satisfies the selection rule $Tr(Y)=0$. We therefore conclude that the vanishing of the trace of the weak hypercharge $Y$ implies at the same time a supergroup with determinant 1 and the consistency of the quantum field theory.

The simple Lie algebra $sl(2,\mathbb{C})$ admits 3 Chevalley generators $(f,h,e)$ with commutators

A finite representation of dimension $(a+1)$ is provided by the iterated action of $f$ on a highest weight state $|a>$, where $a$ is the non-negative integer Dynkin number of the representation, modulo the vector space generated by $\{f^{a+i}|a>,i>0\}$.

To construct the Lie superalgebra $sl(2/1,\mathbb{C})$, add a Cartan generator $k$ and a pair of anticommuting odd generators $(u,v)$ satisfying

A Kac module Kac2 is provided by the iterated action of $(f,v)$ on a highest weight state $\Lambda_{0}=|a,b>$ where $a$ is a non negative integer and $b$ is complex:

modulo the $a$-negative even submodules. Since $v^{2}=0$, the complex number $b$, called the odd Dynkin weight, is not quantized: if $a\in\mathbb{N}$, $\forall b\in\mathbb{C}$, the representation has finite dimension $4(a+1)$. Using (1-4), if $a>0$, the four weights

are highest weight of the even subalgebra: $e\Lambda_{i}=0,\;\forall i=0,1,2,3$. If $a=0$, $\Lambda_{2}$ is removed. The generator $Y=2k-h$ is called the hypercharge. It commutes with $(f,h,e)$ and generates the center $\mathbb{C}$ of the even subalgebra $sl(2)\oplus\mathbb{C}$. On the 4 even submodules, $Y$ has eigenvalues $(y=2b-a,y-1,y-1,y-2)$. The adjoint representation $a=b=1$ and all other Kac modules satisfying $y=2b-a=1$ are real in the sense that they are invariant if we flip the signs of $(h,k)$, exchange $(e,f)$ and map $(u,v)$ to $(iv,iu)$.

There are 2 special cases, called atypical Kac2 . Since $u\Lambda_{1}=b\Lambda_{0}$, if $b=0$ then $\Lambda_{1}$ is an $sl(2/1)$ highest weight and the Kac module is indecomposable rather than irreducible. This case is called atypical 1. Similarly, since $(eu-ue)\Lambda_{3}=(b-a-1)\Lambda_{1}$, if $b=a+1,a>0$ then $\Lambda_{2}$ (or $\Lambda_{3}$ if $b=1,a=0$) is an $sl(2/1)$ highest weight and this case is called atypical 2. Yet, these states are not quotiented out of the Kac module and the total dimension remains $4(a+1)$.

This formal study defines the matrices $\forall\;a,b$, and will be used in section 3. To better understand the situation and for the application to particle physics, it is helpful to visualize the 8 generators $\lambda$ in the case $a=0$, $y=2b$. This fundamental quartet was first constructed by Nahm, Scheunert and Rittenberg in 1977 SNR77 . The 4 even matrices are

$(f,e,h)$ generate $sl(2,\mathbb{C})$ and commute with $Y$. $u,v,w=[e,u],x=-[f,v]$ are odd:

They form by commutation with $(f,h,e,Y)$ a complex doublet of $sl(2)\oplus\mathbb{C}$ and close by anticommutation on the even matrices. The indecomposable structure if $b=0$ or $b=1$ is very visible. In addition, the superidentity $\chi=diag(-1,1,1,-1)$ defines the supertrace and the grading. $\chi$ commutes with the even generators $\lambda_{a}$ and anticommutes with the odd generators $\lambda_{i}$

All these matrices have vanishing supertrace, they also all have vanishing trace except the hypercharge for which $Tr(Y)=4(y-1)$. This crucial detail will play a fundamental role in our analysis.

The cubic super-Casimir tensor $C$ is defined as the supertrace of the correctly symmetrized products of 3 matrices. There are 2 non trivial sectors, even-even-even and even-odd-odd, or rather 5 sectors if the hypercharge $Y$ is distinguished from the $sl(2,\mathbb{C})$ generators $(\lambda_{a}=f,h,e)$ and from the odd generators $(\lambda_{i}=u,v,x,y)$.

To evaluate these supertraces over a Kac module with highest weight $(a\in\mathbb{N},y\in\mathbb{C})$, first compute the traces over its 4 submodules $\{\Lambda_{i},i=1.2.3.4\}$ (5) with highest weights $(a^{\prime},y^{\prime})=\{(a,y),(a+1,y-1),(a-1,y-1),(a,y-2)\}$ and super-dimensions $\{-(a+1),a,a+2,-(a+1)\}$. Then add up. By inspection $Tr(Y)=4(a+1)(y-1)$.

In each even sector $C_{abc}(a^{\prime},y^{\prime})=0$ because $su(2)$ does not admit a cubic-Casimir tensor. By inspection $C_{yab}(a^{\prime},y^{\prime})=a^{\prime}(a^{\prime}+1)(a^{\prime}+2)y^{\prime}/3$, and $C_{yyy}(a^{\prime},y^{\prime})=(a^{\prime}+1)y^{\prime 3}$. Super-adding the 4 sectors with signs $(-1,1,1,-1)$ then gives

The odd sector can be evaluated using the definitions (1-4), with patience or a computer, on enough examples TMJG22 to saturate a polynomial of degree at most 4 in $(a,y)$, giving

In conclusion, all components of the $sl(2/1,\mathbb{C})$ cubic super-Casimir tensor super-traced over a Kac module are linear in $Tr(Y)=4(a+1)(y-1)$ with no other dependence in the Dynkin numbers $(a,b)$. This result also applies to the trace metric and to several other trace tensors appearing in the calculation of the Feynman diagrams of the scalar-vector-tensor super-chiral quantum field theory TM20b ; TMJ21a

By introducing an appropriate phase co-factor $\eta=exp(i\pi/4)$ and by symmetrization, one can extract from any Kac module representation of $sl(2/1,\mathbb{C})$ with real Dynkin numbers, a set of zeta-Hermitian matrices

where $\zeta$ is a real diagonal matrix, commuting with the even subalgebra and satisfying $\zeta^{2}=Id$. The signs of the $\pm 1$ eigenvalues along the diagonal are controlled by the signs of the Kac atypicality conditions $(b)$ and $(b-a-1)$. This construction is valid for all typical or singly-atypical $sl(m/n)$ Kac modules (in preparation). As an example, we give the $\zeta$ matrix and the odd matrices of the $a=0$ Kac module. We keep (6) and modify (7):

One can then construct ‘anti‘-$\zeta$-Hermitian matrices

where $(\lambda_{a},\lambda_{i})$ are respectively symmetrized forms of $(f,h,e,k)$ and $(\lambda_{4},\lambda_{5},\lambda_{6},\lambda_{7})$ The even $\mu_{a}$ matrices represent $su(2,\mathbb{R})$. By inspection, the odd $\mu$ matrices close under super-commutation using real structure constants. Therefore, the $\mu$ matrices represent the real superalgebra $su(2/1,\mathbb{R})$. The co-factor $\eta$ was introduced in the case $a=0$ by Furutsu Furutsu88 ; Furutsu89 , but he wrongly concluded that his construction of a real form of $sl(2/1,\mathbb{C})$ was only valid for $b>1$ or $b<0$. Indeed, if $b>1$, the $\zeta$ matrix is the identity: $\zeta=Id$ and the $\lambda$ matrices are all Hermitian: $\lambda^{\#}=\lambda^{\dagger}$. If $b<0$, the $\zeta$ matrix (14) coincides, up to a sign choice, with the super-identity $\zeta=-\chi$ (8), and the $\lambda$ matrices are super-Hermitian in the sense of Berezin: $\lambda^{\#}=\chi\lambda^{\dagger}\chi$. But no one realized that when $0<b<a+1$ one could interpolate between $Id$ and $\chi$ of Berezin and define $\zeta$-Hermitian matrices: $\lambda^{\#}=\zeta\lambda^{\dagger}\zeta$.

Let $\lambda$ be as above a set of square matrices of size $4(a+1)$ representing an $sl(2/1,\mathbb{C})$ or $su(2/1,\mathbb{R})$ Kac module with highest weight $|a,b>,y=2b-a$. Let $N$ be a non negative integer. Construct a banded lower triangular block matrix $\Lambda$ of size $4N(a+1)$ such that the $n$-$th$ column coincides with the first $N-n+1$ terms of the Taylor development of $\lambda(y)$ relative to $y$. For example if $N=3$

By construction, the matrices $\Lambda$ have the same commutation rules as the matrices $\lambda$ because the commutators of the Taylor expansion coincide with the Leibniz development of the derivatives of the commutators $[a,b]^{{}^{\prime\prime}}=[a^{{}^{\prime\prime}},b]+2[a^{\prime},b^{\prime}]+[a,b^{{}^{\prime\prime}}]$. In a Lie algebra, the Dynkin numbers are integral, such derivatives cannot be computed, and all finite dimensional representations are fully reducible. But in $sl(2/1)$, $y$ is not quantized and the matrices $\Lambda$ generate recursively nested indecomposable representations where each generations is coupled via mouse trap coefficients $(\lambda^{\prime},\lambda^{\prime\prime},\dots)$ to the upper generations. Notice that the even matrices $(f,h,e)$ representing $sl(2)$ are block diagonal, because their elements (2) are independent of $y$. The hypercharge only contributes to the main and second diagonal since $Y$ is linear in $y$ (4). However, in the $su(2/1,\mathbb{R})$ $\zeta$-Hermitian case (14), the $n$-$th$ $y$-derivatives of the odd matrices $(\mu_{i},\;i=4,5,6,7)$ do not vanish and fill the whole lower triangle. Such mixing is specific of superalgebras Todorov23 .

We call these representations matryoshkas because they are recursively nested like Russian dolls. The $N=3$ representation of $sl(2/1,\mathbb{C})$ was first constructed by Marcu Marcu80 . Germoni proved that these are the only indecomposable representations of $sl(2/1)$ corresponding to semi-direct sums of Kac modules Germoni98 . With Jarvis, Germoni and Gorelik, we proved recently TMJG23 that such representations exist for any $N$ and any Kac module of a type 1 superalgebra, the series $sl(m/n,\mathbb{C})$ and $osp(2/2n,\mathbb{C})$, but did not study $su(2/1,\mathbb{R})$.

Consider a quartet representation $\mu$ of $su(2/1,\mathbb{R})$ (15). The exponential $exp(x^{a}\mu_{a})$ of the anti-Hermitian even matrices $\mu_{a},\;a=1,2,3$ is periodic in $x^{a}$. To ensure the existence of a single neutral element, $x$ must be quotiented modulo the period. If all $su(2)$ Dynkin number are even, the period is $\pi$ giving the compact Lie group $SO(3)$, otherwise the period is $2\pi$ giving its double cover $SU(2)$. If and only if $y=p/q$ is rational, $exp(ixY)$ with $Y$ eigenvalues $(y,y-1,y-2)$ (6) is periodic with period $2q\pi$. Modulo the period, $iY$ exponentiates to the compact $U(1)$ group. Henceif $y\in\mathbb{Q}$ the exponential map of the even generators gives the $SU(2)U(1)$ compact group.

The exponential of the odd generators $\mu_{i}=\eta\lambda_{i},\;i=4,5,6,7$ (14) using 4 distinct anticommuting Grassmann parameters $\theta^{i}$ Nieto93 is discrete and finite.

Hence the whole $SU(2/1)$ supergroup is compact. Because of the presence of Grassmann parameters, the Berezinian is multiplicative $Ber(MN)=Ber(M)Ber(N)$, but the determinant is not $det(MN)\neq det(M)det(N)$. However, in the quartet case (14,15) a direct calculation respecting the order of the columns shows that:

Since the determinant is linear in $Tr(Y)$, the determinant of the exponential map of a direct sum of quartets, i.e. of a set of block diagonal matrices, acts additively mimicking the even exponential map $det(e^{iY})=e^{iTr(Y)}$. Therefore if $Y$ traced over a composite Kac module vanishes $Tr(Y)=0$, the exponential map has determinant 1

We propose to call the group $\{g\}$ the ’super-special’ $SU(2/1)$ supergroup.

The key of this construction is the linearity of $e^{X}$ in $Tr(Y)$. Our proof holds for the fundamental Kac quartet $(a=0,\;b\in\mathbb{Q})$, and for their semi-direct indecomposable sums (16) as lower triangular elements do not contribute to determinants. In the shited adjoint case $(a=1)$ TMJG22 , $det(g)$ in optimal column order is also linear in $Tr(Y)$. An open problem is to extend the proof to all Kac modules $(a\in\mathbb{N},\;b\in\mathbb{Q})$.

The vanishing of the term in $\theta^{2}$ in (18) matches the vanishing of the anomalous trace term of the tensor propagator of the $su(2/1)$ super-chiral model (eq.(3.7) in TMJ21a ). The vanishing of the term in $\theta^{4}$ in (18) is related to the vanishing of the anomalous trace terms of the Feynman diagrams with 4 external legs (eq.(2.12) in TM20b ).

Consider a set of massless Fermions belonging to a semi-direct sum of Kac modules of the superalgebra $su(2/1,\mathbb{R})$ graded by chirality. That is assume that the superidentity grading operator $\chi$ of $su(2/1)$ (8) has eigenvalue $+1$ on the left Fermion states and $-1$ on the right Fermion states. Add in $su(2)\oplus u(1)$ Yang-Mills vectors coupled to the Fermions using the even matrices $\mu_{a}$ (15) of $su(2/1,\mathbb{R})$: $\not{A}=\not{A}^{a}\mu_{a}$. Because such Fermions are graded by chirality, the supertrace in the sense of the superalgebra $STr(...)=Tr(\chi\;...)$ is equivalent to the $\gamma^{5}$ trace $\Gamma Tr(...)=Tr(\gamma^{5}\;...)$ which occurs when computing the Adler-Bell-Jackiw Adler ; BJ triangle anomaly $d_{abc}=\Gamma Tr(\mu_{a}\{\mu_{b},\mu_{c}\})$. This presentation provides an algebraic meaning to the anomaly tensor $d_{abc}$ which now coincides with the even sector of the cubic super-Casimir tensor of the real simple superalgebra $su(2/1,\mathbb{R})$. As shown above, if $Tr(Y)=0$, the super-Casimir tensor vanishes, hence the $su(2)^{2}Y$ and $Y^{3}$ tensors both vanish (10) and the theory is anomaly free.

If the $SU(3)$ color group gauging the strong interactions is added to this construction, there is a further potential anomaly $su(3)^{2}Y$. But since $su(3)$ commutes with $su(2/1)$, the $su(3)^{2}Y$ anomaly factorizes and vanishes because $STr(Y)=0$ on any $su(2/1)$ representation.

The electron and neutrino N1 ; F1 graded by chirality $(\nu_{R};\nu_{L},e_{L};e_{R})$ have the quantum numbers of the fundamental Kac module $(a=0,y=0)$ of $su(2/1,\mathbb{R})$ (6) with electric charge $Q=(Y-h)/2$. The top singlet $(0,0)$ represents the right neutrino $\nu_{R}$. The quarks DJ ; NTM1 graded by chirality $(u_{R};u_{L},d_{L};d_{R})$ have the quantum numbers of the irreducible typical $(a=0,y=4/3)$ $su(2/1,\mathbb{R})$ representation (15). In both cases, $Tr(Y)=4(y-1)$, respectively $-4$ for the leptons and $+4/3$ for the quarks. Therefore, as there exist 3 quark colors in each lepton family, $Tr(Y)=0$, the theory is anomaly free (section 7), and the hydrogen atom $(e^{-1},u^{2/3},u^{2/3},d^{-1/3})$ has electric charge $Q=0$.

We have thus discovered a new implication of the celebrated BIM mechanism BIM , which allowed to predict the existence of the charm and the top quarks as soon as the strange and bottom quarks were recognized: the BIM mechanism also allows exponentiation to a new form of the $SU(2/1)$ supergroup with determinant one (section 6).

Furthermore, the 3 generations of leptons, and separately the 3 generations of quarks, can each be grouped into a single 12 dimensional nested indecomposable representation TMJG23 . Since the $\lambda^{\prime},\lambda^{\prime\prime},\dots$ blocks of the $\Lambda$ matrices (16) are below the main diagonal, the 3-generations model remains anomaly free. Cherry on the cake, the existence of three generations of charged leptons implies the existence of three generations of right neutrinos $\nu_{R}$, because without them the remaining atypical irreducible fundamental lepton triplet $(\nu_{L},e_{L}/e_{R})$ considered in N1 ; F1 is quantized ($y=0$), the $\lambda^{\prime}$ derivatives cannot be computed, and the 3 generations indecomposable representation $\Lambda$ cannot be constructed. This impossibility can also be proved by cohomology JTM22 .

Considered as a classification paradigm, the $SU(2/1)$ supergroup with determinant $1$ constructed in section 6 answers all the questions stated in the introduction.

The basic observation is that the $SU(2)U(1)$ even subgroup coincides with the electroweak gauge group of the standard model. The leptons and the quarks fit the fundamental representation (6,14) of the real $su(2/1,\mathbb{R})$ superalgebra N1 ; F1 ; DJ ; NTM1 , but since all these particles are chiral Fermions, it is natural to grade $su(2/1)$ by chirality, relating left states to right states, rather than relating Bosons to Fermions as in traditional supersymmetric models. Then, thanks to mathematics, the desired results follow.

Why is parity broken? Maximal parity breaking, the deepest mystery of the weak interactions, is implied by the grading. The fundamental $su(2/1)$ quartet $(1;2;1)$ is asymmetric (6). The weak $su(2)$ algebra only interacts with the central doublet and ignores the singlets. Therefore, grading by chirality $(1_{R};2_{L};1_{R})$ fully breaks parity.

Why fractional charges? As observed experimentally, the $u(1)$ weak hypercharge $Y$ (6) steps by full units $(y;y-1,y-1;y-2)$, so the righ t neutrino $\nu^{e}_{R}$ is neutral $(y=0)$ because the electron $e^{-}$ and the weak vector Boson $W^{-}_{\mu}$ have the same electric charge $(-1)$ if and only if $(y=0)$. Yet $Y$ is not quantized and the fractional charge $y=4/3$ of the $u_{R}$ quark is allowed (6). Our new generalization of the Berezin super-conjugation (13) then provides a real form of the quark representation (15), a required property since the photon is its own anti-particle, hence the $u(1)$ subalgebra must be real.

Why several generations, why right neutrinos? The existence of several generations of leptons and quarks, a hard problem Todorov23 , becomes natural because $su(2/1)$ admits (15,16) nested indecomposable matryoshka representations Marcu80 ; COQ0 ; HS98 ; TMJG23 . In turn, section 8, the existence of the three generations of heavier Fermions implies the existence of the three right neutrinos, because the irreducible triplet $(\nu_{L},e_{L};e_{R})$ does not admit a multi-generations indecomposable extension JTM22 .

Why do anomalies cancel out? The vanishing of the Adler-Bell-Jackiw anomalies, and this is probably the most striking result of this study, is implied because this form of the $SU(2/1)$ supergroup has determinant 1. A simple calculation (10) shows that the symmetric tensor $d_{\{abc\}}$ traced over a semi-direct sum of Kac modules vanishes, $Tr(\Lambda_{a}\;\{\Lambda_{b},\Lambda_{c}\})=0$, if and only if $Tr(Y)=0$. We then observed (19) that the superalgebra matrices (6,14) exponentiate to a matrix group with determinant 1 if and only if $Tr(Y)=0$. But since the superalgebra is graded by chirality, section 7, its supertrace coincides with the $\gamma_{5}$ trace appearing in quantum field theory when evaluating the anomalies. Therefore, exponentiation to the ’super-special’ $SU(2/1)$ supergroup is equivalent to the vanishing of the anomalies and both follow from the simple selection rule $Tr(Y)=0$.

Guided by particle physics, we computed en passant new tensor identities (10,12), defined new $\zeta$-Hermitian extensions (13) of the Berezin super-transposition, found contrary to the literature new representations (13) of the real superalgebra (14) with hypercharge $0<y<1$, constructed in a simple way (16) new nested indecomposable representations of $sl(2,1,\mathbb{R})$, and discovered the surprising existence, not reported in the recent review of Fioresi and Gavarini Fioresi23 or in Saleur and Schomerus Saleur07 , of a super-special supergroup (19) whose elements have determinant and super-determinant (Berezinian) equal to 1. This double structure might be related to the existence of two parallel transports on a supergroup, via the adjoint and via the alternative adjoint action of Arnaudon, Bauer and Frappat Arnaudon97 , and to their ghost Casimir generalized by Gorelik Gorelik2000 . We conjecture that both transports are consistent if and only if the corresponding quantum field theory is anomaly free.

Thanks to these new results, one can, for the first time, regard the electroweak interactions as structured by the $SU(2/1)$ supergroup.