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Enumeration of Anti-Invariant Subspaces and Touchard’s
Formula for the Entries of the qq-Hermite Catalan Matrix

Amritanshu Prasad The Institute of Mathematical Sciences, Chennai, India. Homi Bhabha National Institute, Mumbai, India. amri@imsc.res.in  and  Samrith Ram Indraprastha Institute of Information Technology Delhi, New Delhi, India. samrith@gmail.com
Abstract.

We express the number of anti-invariant subspaces for a linear operator on a finite vector space in terms of the number of its invariant subspaces. When the operator is diagonalizable with distinct eigenvalues, our formula gives a finite-field interpretation for the entries of the qq-Hermite Catalan matrix. We also obtain an interesting new proof of Touchard’s formula for these entries.

Key words and phrases:
Touchard-Riordan formula, anti-invariant subspaces, invariant subspaces, splitting subspaces, finite fields, qq-Hermite orthogonal polynomials, chord diagrams.

1. Introduction

Let qq be a prime power, and let 𝐅q\mathbf{F}_{q} denote a finite field of order qq. For nonnegative integers nn and kk, let [nk]q{n\brack k}_{q} denote the qq-binomial coefficient, which is the number of kk-dimensional subspaces of 𝐅qn\mathbf{F}_{q}^{n}. Recall that for a linear operator TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}), a subspace W𝐅qnW\subset\mathbf{F}_{q}^{n} is said to be TT-invariant if T(W)WT(W)\subset W.

Definition 1.

For a linear operator TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}), a subspace W𝐅qnW\subset\mathbf{F}_{q}^{n} is said to be TT-anti-invariant if

dim(W+TW)=2dimW.\dim(W+TW)=2\dim W.
Main Theorem.

For any TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}), the number of \ell-dimensional TT-anti-invariant subspaces of 𝐅qn\mathbf{F}_{q}^{n} is given by

(1) ωnT=q(2)j=0(1)j(XjTXj1T)[njn2]qq(j+12),\omega^{T}_{n\ell}=q^{\binom{\ell}{2}}\sum_{j=0}^{\ell}(-1)^{j}(X^{T}_{j}-X^{T}_{j-1}){n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}},

where XjTX^{T}_{j} is the number of jj-dimensional TT-invariant subspaces of 𝐅qn\mathbf{F}_{q}^{n}.

The computation of XjTX^{T}_{j} from the similarity class of TT as a polynomial in qq is easy [13, Section 2] and has been implemented in SageMath [10]. The formula in Eq. (1) can be recast in the following form, which will be used in the proof.

(2) ωnT=q(2)j=0(1)j([njn2]qq(j+12)+[nj1n2]qq(j2))XjT.\omega^{T}_{n\ell}=q^{\ell\choose 2}\sum_{j=0}^{\ell}(-1)^{j}\left({n-\ell-j\brack n-2\ell}_{q}q^{\ell-j+1\choose 2}+{n-\ell-j-1\brack n-2\ell}_{q}q^{\ell-j\choose 2}\right)X^{T}_{j}.

Anti-invariant subspaces were introduced by Barría and Halmos [3]. A matrix TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}) is said to be \ell-transitive if each ×\ell\times\ell matrix over 𝐅q\mathbf{F}_{q} appears as the upper left submatrix of some matrix similar to TT. Using the fact that TT is \ell-transitive if and only if it admits an \ell-dimensional anti-invariant subspace, they characterized \ell-transitive matrices. Sourour [14] determined the maximal dimension of an anti-invariant subspace. Knüppel and Nielsen [8] defined a subspace WW to be kk-fold anti-invariant if

dim(W+TW+TkW)=(k+1)dimW,\dim(W+TW\dotsb+T^{k}W)=(k+1)\dim W,

and determined the maximal dimension of a kk-fold anti-invariant subspace for a linear operator TT. They also showed [8, Thm. 2.1] that if d1d2dnd_{1}\leq d_{2}\leq\cdots\leq d_{n} denote the degrees of the invariant factors in the Smith normal form of xIAxI-A where AA is the matrix TT with respect to some basis, then a kk-fold anti-invariant subspace of dimension mm exists if and only if dnm+1++dn(k+1)md_{n-m+1}+\cdots+d_{n}\geq(k+1)m.

Enumerative versions of such problems can be traced back to a paper of Bender, Coley, Robbins and Rumsey [4]. In the context of pseudorandom number generation, Niederreiter [9] made the following definition: given TMmd(𝐅q)T\in M_{md}(\mathbf{F}_{q}), an mm-dimensional subspace W𝐅qmdW\subset\mathbf{F}_{q}^{md} is said to be a TT-splitting subspace of degree dd if

W+TW++Td1W=𝐅qmd.W+TW+\dotsb+T^{d-1}W=\mathbf{F}_{q}^{md}.

Niederreiter asked for the number of mm-dimensional TT-splitting subspaces of degree dd when the characteristic polynomial of TT is irreducible, unaware that the question had already been answered in [4]. The number of TT-splitting subspaces in this case is given by

qm(m1)(d1)qmd1qm1.q^{m(m-1)(d-1)}\frac{q^{md}-1}{q^{m}-1}.

Chen and Tseng [5] reproved this result by developing recurrence relations involving a larger class of combinatorial problems. Their recurrence relations are independent of the matrix TT, and reduce the enumeration of splitting subspaces to the enumeration of flags of TT-invariant subspaces. However, the recurrences are very difficult to solve in general.

Aggarwal and Ram [1] used the recurrences of Chen and Tseng to show that when TT is regular nilpotent, the number of TT-splitting subspaces of degree dd is given by qm2(d1)q^{m^{2}(d-1)}. For regular nilpotent TT, XjT=1X_{j}^{T}=1 for 0jn0\leq j\leq n, so (1) becomes

(3) ωnT=q2[nn2]q.\omega^{T}_{n\ell}=q^{\ell^{2}}{n-\ell\brack n-2\ell}_{q}.

Setting =m\ell=m and n=2mn=2m, we recover the formula of [1] in the case d=2d=2. Later, it came to light that these results in the regular nilpotent case follow from the results in [4].

Viennot’s combinatorial theory of orthogonal polynomials [16] places the moments of an orthogonal polynomial sequence in the first column of an infinite array known as the Catalan matrix. There is a connection between our main theorem and the Catalan matrix associated to the qq-Hermite orthogonal polynomial sequence as defined by Ismail, Stanton and Viennot [7]. This connection emerges from our work [11, 12] where a more general class of enumerative problems is considered.

Definition 2.

For a linear endomorphism TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}) and a partition μ=(μ1,,μk)\mu=(\mu_{1},\dotsc,\mu_{k}) of nn, a subspace W𝐅qnW\subset\mathbf{F}_{q}^{n} is said to have TT-profile μ\mu if

dim(W+TW++Ti1W)=μ1++μi for each 1ik.\dim(W+TW+\dotsb+T^{i-1}W)=\mu_{1}+\dotsb+\mu_{i}\text{ for each }1\leq i\leq k.

The number of subspaces with TT-profile μ\mu is denoted σμT\sigma^{T}_{\mu}.

When n=m+n=m+\ell, a subspace WW has profile (m,)(m,\ell) if WW is mm-dimensional, and W+TW=𝐅qnW+TW=\mathbf{F}_{q}^{n}. Let TT^{*} denote the transpose of TT. For a subspace W𝐅qnW\subseteq\mathbf{F}_{q}^{n}, let W0W^{0} denote its annihilator in the linear dual of 𝐅qn\mathbf{F}_{q}^{n}. When WW has profile (m,)(m,\ell), dim(W0)=l\dim(W^{0})=l. Also W+TW=𝐅qnW+TW=\mathbf{F}_{q}^{n} if and only if W0(TW)0=(W+TW)0={0}W^{0}\cap(TW)^{0}=(W+TW)^{0}=\{0\}. By using the identity

(4) dim(US1U)=2dimUdim(U+SU),\dim(U\cap S^{-1}U)=2\dim U-\dim(U+SU),

for SMn(𝐅q)S\in M_{n}(\mathbf{F}_{q}) and each subspace U𝐅qnU\subset\mathbf{F}_{q}^{n}, it follows that

dim(W0(TW)0)=dim(W0(T)1W0)=2dimW0dim(W0+TW0).\dim(W^{0}\cap(TW)^{0})=\dim(W^{0}\cap(T^{*})^{-1}W^{0})=2\dim W^{0}-\dim(W^{0}+T^{*}W^{0}).

Therefore, WW has profile (m,)(m,\ell) if and only if W0W^{0} is an \ell-dimensional TT^{*}-anti-invariant subspace. Since TT is similar to TT^{*}, we have

(5) ωnT=ωnT=σ(m,)T.\omega_{n\ell}^{T}=\omega_{n\ell}^{T^{*}}=\sigma_{(m,\ell)}^{T}.

Let [n][n] denote the set consisting of the first nn positive integers. Denote by Π(μ)\Pi(\mu^{\prime}) the set of partitions of [n][n] whose block sizes are the parts of the integer partition conjugate to μ\mu. Suppose TT is a diagonalizable matrix with distinct eigenvalues in 𝐅q\mathbf{F}_{q}. One of the main results of [12] is a combinatorial formula for σμT\sigma_{\mu}^{T} in terms of a statistic vv on set partitions known as the interlacing number:

(6) σμT=(q1)j2μjqj2(μj2)𝒜Π(μ)qv(𝒜).\sigma^{T}_{\mu}=(q-1)^{\sum_{j\geq 2}\mu_{j}}q^{\sum_{j\geq 2}\binom{\mu_{j}}{2}}\sum_{\mathcal{A}\in\Pi(\mu^{\prime})}q^{v(\mathcal{A})}.

From (5), we obtain

(7) ωnT=(q1)q(2)an,n2,\omega_{n\ell}^{T}=(q-1)^{\ell}q^{\binom{\ell}{2}}a_{n,n-2\ell},

where an,n2=𝒜qv(𝒜)a_{n,n-2\ell}=\sum_{\mathcal{A}}q^{v(\mathcal{A})}, a sum is over partitions of [n][n] with \ell blocks of size 22 and n2n-2\ell singleton blocks. In Section 5, we show that an,n2a_{n,n-2\ell} coincides with polynomials defined recursively by Touchard [15] in the context of the stamp-folding problem. Touchard [15, Eq. (28)] showed that

(8) (q1)an,n2=j=0(1)j[(nj)(nj1)][njn2]qq(j+12).(q-1)^{\ell}a_{n,n-2\ell}=\sum_{j=0}^{\ell}(-1)^{j}\left[\binom{n}{j}-\binom{n}{j-1}\right]{n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}}.

When TT is a diagonalizable matrix with distinct eigenvalues in 𝐅q\mathbf{F}_{q}, XjT=(nj)X^{T}_{j}=\binom{n}{j}. Substituting this into the formula (1) of our main theorem gives

(9) ωnT=q(2)j=0(1)j[(nj)(nj1)][njn2]qq(j+12).\omega_{n\ell}^{T}=q^{\binom{\ell}{2}}\sum_{j=0}^{\ell}(-1)^{j}\left[\binom{n}{j}-\binom{n}{j-1}\right]{n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}}.

Comparing the expressions in (7) and (9) gives a new, linear-algebraic proof of Touchard’s formula (8).

A road map

The proof of the main theorem consists of three parts which take up the next three sections of this article.

In the first part (Section 2) we establish the existence of a formula of the form (12) for the number ωnT\omega_{n\ell}^{T} of anti-invariant subspaces of a given dimension as a linear combination of the numbers XjTX_{j}^{T} of invariant subspaces, whose coefficients are independent of TT. It remains to show that the coefficients of these linear combinations are as in Eq. (2).

In the second part (Section 3) we consider a family of matrices for which the number of anti-invariant subspaces is known. Each matrix in this family gives rise to an equation in the coefficients of (12). Theorem 6 establishes that this system of equations has a unique solution. Thus in order to show that the coefficients are exactly the ones given in (2), it suffices to show that the identity (1) holds for each matrix in the family. This can be be expressed as the family (17) of identities.

In the third part (Section 4) we prove the identities (17) by reducing them to Heine’s transformations for qq-hypergeometric functions.

The proof strategy in this article closely follows ideas in [13]; the main distinction is that the more general identity (17) needed here requires a very different approach.

The final section (Section 5) of this article is devoted to the case where TT is a diagonal matrix with distinct diagonal entries, and the connection to the Catalan matrix of qq-Hermite orthogonal polynomials.

2. Existence of a Universal Formula

In this section we prove the existence of a universal formula for the number of anti-invariant subspaces of a given dimension for an arbitrary operator TT. The main step is Lemma 3, which is a special case of the recurrence of Chen and Tseng [5, Lemma 2.7]. We begin by introducing some notation.

Given TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}) and sets AA and BB of subspaces of 𝐅qn\mathbf{F}_{q}^{n}, define

β(A,B)\displaystyle\beta(A,B) :={WAWT1WB},\displaystyle:=\{W\in A\mid W\cap T^{-1}W\in B\},
γ(A,B)\displaystyle\gamma(A,B) :={(W1,W2)W1A,W2B, and W1T1W1W2}.\displaystyle:=\{(W_{1},W_{2})\mid W_{1}\in A,W_{2}\in B,\text{ and }W_{1}\cap T^{-1}W_{1}\supset W_{2}\}.

For integers a,ba,b, we also write β(a,b)\beta(a,b) for the set of aa-dimensional subspaces WW of 𝐅qn\mathbf{F}_{q}^{n} such that WT1WW\cap T^{-1}W has dimension bb. The quantity γ(a,b)\gamma(a,b) is defined analogously. For example, using identity (4), β(a,a)\beta(a,a) denotes the set of aa-dimensional TT-invariant subspaces, whereas β(a,0)\beta(a,0) denotes the set of aa-dimensional TT-anti-invariant subspaces. To explicitly specify the linear operator TT, we also write βT(A,B)\beta^{T}(A,B) and γT(A,B)\gamma^{T}(A,B).

Lemma 3.

For each TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}) and 0an0\leq a\leq n,

|β(a,b)|\displaystyle|\beta(a,b)| =XbT[nbab]qXaT[ab]q\displaystyle=X_{b}^{T}{n-b\brack a-b}_{q}-X_{a}^{T}{a\brack b}_{q}
+j=0b1|β(b,j)|[n2b+ja2b+j]qk=b+1a1|β(a,k)|[kb]q.\displaystyle+\sum_{j=0}^{b-1}|\beta(b,j)|{n-2b+j\brack a-2b+j}_{q}-\sum_{k=b+1}^{a-1}|\beta(a,k)|{k\brack b}_{q}.
Proof.

Since the collection of all aa-dimensional subspaces of 𝐅qn\mathbf{F}_{q}^{n} is the disjoint union 0kaβ(a,k)\coprod_{0\leq k\leq a}\beta(a,k), we have

γ(a,b)=0kaγ(β(a,k),b).\displaystyle\gamma(a,b)=\coprod_{0\leq k\leq a}\gamma(\beta(a,k),b).

To count pairs of subspaces (W1,W2)γ(β(a,k),b)(W_{1},W_{2})\in\gamma(\beta(a,k),b), first choose W1β(a,k)W_{1}\in\beta(a,k) and then choose W2W_{2} to be an arbitrary bb-dimensional subspace of W1T1W1.W_{1}\cap T^{-1}W_{1}. It follows that

|γ(a,b)|\displaystyle|\gamma(a,b)| =k=0a|γ(β(a,k),b)|=k=ba|β(a,k)|[kb]q\displaystyle=\sum_{k=0}^{a}|\gamma(\beta(a,k),b)|=\sum_{k=b}^{a}|\beta(a,k)|{k\brack b}_{q}
(10) =|β(a,b)|+k=b+1a|β(a,k)|[kb]q.\displaystyle=|\beta(a,b)|+\sum_{k=b+1}^{a}|\beta(a,k)|{k\brack b}_{q}.

Similarly, the set of all bb-dimensional subspaces of 𝐅qn\mathbf{F}_{q}^{n} equals the disjoint union 0jbβ(b,j)\coprod_{0\leq j\leq b}\beta(b,j). Therefore

γ(a,b)=0jbγ(a,β(b,j)).\displaystyle\gamma(a,b)=\coprod_{0\leq j\leq b}\gamma(a,\beta(b,j)).

To count pairs (W1,W2)γ(a,β(b,j)),(W_{1},W_{2})\in\gamma(a,\beta(b,j)), first choose W2β(b,j)W_{2}\in\beta(b,j) and note that dim(W2+TW2)=2bj\dim(W_{2}+TW_{2})=2b-j by (4).

Given W2W_{2}, a pair (W1,W2)(W_{1},W_{2}) belongs to γ(a,β(b,j))\gamma(a,\beta(b,j)) if and only if W1W_{1} is an aa-dimensional subspace that contains W2+TW2W_{2}+TW_{2}. Therefore, the number of choices for W1W_{1} is [n(2bj)a(2bj)]q{n-(2b-j)\brack a-(2b-j)}_{q}. Consequently,

|γ(a,b)|\displaystyle|\gamma(a,b)| =j=0b|γ(a,β(b,j))|\displaystyle=\sum_{j=0}^{b}|\gamma(a,\beta(b,j))|
(11) =j=0b|β(b,j)|[n(2bj)a(2bj)]q.\displaystyle=\sum_{j=0}^{b}|\beta(b,j)|{n-(2b-j)\brack a-(2b-j)}_{q}.

The lemma now follows from Eqs. (10) and (11), and the fact that |β(a,a)|=XaT|\beta(a,a)|=X_{a}^{T}. ∎

Proposition 4.

Given integers nn, aa, bb, there exist polynomials pj(t)𝐙[t]p_{j}(t)\in\mathbf{Z}[t] (0ja)(0\leq j\leq a), such that, for every prime power qq and every TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}),

|βT(a,b)|=j=0apj(q)XjT.\displaystyle|\beta^{T}(a,b)|=\sum_{j=0}^{a}p_{j}(q)X_{j}^{T}.
Proof.

Lemma 3 expands |βT(a,b)||\beta^{T}(a,b)| in terms of XaTX_{a}^{T}, XbTX_{b}^{T}, and |βT(a,b)||\beta^{T}(a^{\prime},b^{\prime})| where either a<aa^{\prime}<a, or a=aa^{\prime}=a and ab<aba^{\prime}-b^{\prime}<a-b. The coefficients are polynomials in qq that are independent of TT. Thus repeated application of Lemma 3 results in an expression of the stated form in finitely many steps. ∎

The following corollary shows the existence of a universal formula for the number of anti-invariant subspaces of a given dimension.

Corollary 5.

For all integers n20n\geq 2\ell\geq 0, there exist polynomials pj(t)𝐙[t]p_{j}(t)\in\mathbf{Z}[t] (0j)(0\leq j\leq\ell) such that, for every prime power qq and every TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}),

(12) ωnT=j=0pj(q)XjT.\omega^{T}_{n\ell}=\sum_{j=0}^{\ell}p_{j}(q)X_{j}^{T}.
Proof.

Set a=a=\ell and b=0b=0 in Proposition 4. ∎

3. Determination of Coefficients in the Universal Formula

We set up a system of linear equations which completely determine the polynomials pj(t)p_{j}(t) (0j)(0\leq j\leq\ell) in Corollary 5 by constructing, for each prime power qq, a sequence of matrices T0(q),,T(q)T_{0}(q),\dotsc,T_{\ell}(q) such that the following conditions are satisfied.

  1. (1)

    For each 0i,j0\leq i,j\leq\ell, there exists a polynomial Xij(t)X_{ij}(t) such that XjTi(q)=Xij(q)X^{T_{i}(q)}_{j}=X_{ij}(q) for all prime powers qq.

  2. (2)

    The determinant of the matrix X(t):=(Xij(t))0i,jX(t):=(X_{ij}(t))_{0\leq i,j\leq\ell} is a non-zero element of 𝐙[t]\mathbf{Z}[t].

  3. (3)

    The identity (1) holds for Ti(q)T_{i}(q) for i=0,,i=0,\dotsc,\ell and all prime powers qq.

In effect, we have the following result.

Theorem 6.

For each prime power qq, the system of linear equations

(13) ωnTi(q)=j=0pj(q)Xij(q),0i,\omega^{T_{i}(q)}_{n\ell}=\sum_{j=0}^{\ell}p_{j}(q)X_{ij}(q),\quad 0\leq i\leq\ell,

in the variables p0(q),,p(q)p_{0}(q),\dotsc,p_{\ell}(q) has a solution given by

(14) pj(q)=(1)jq(2)([njn2]qq(j+12)+[nj1n2]qq(j2)).\displaystyle p_{j}(q)=(-1)^{j}q^{\ell\choose 2}\left({n-\ell-j\brack n-2\ell}_{q}q^{\ell-j+1\choose 2}+{n-\ell-j-1\brack n-2\ell}_{q}q^{\ell-j\choose 2}\right).

This solution is unique for sufficiently large prime powers qq and hence uniquely determines the polynomials pj(t)p_{j}(t) for 0jl0\leq j\leq l.

We now proceed with the construction of the matrices Ti(q)(0j)T_{i}(q)(0\leq j\leq\ell) above. For each qq, let T0(q)T_{0}(q) be any matrix in Mn(𝐅q)M_{n}(\mathbf{F}_{q}) with irreducible characteristic polynomial. For i=1,,i=1,\dotsc,\ell, take Ti(q)T_{i}(q) to be the n×nn\times n matrix with block decomposition

Ti(q)=(𝟎𝟎𝟎Ti(q)),T_{i}(q)=\begin{pmatrix}{\bf 0}&{\bf 0}\\ {\bf 0}&T^{\prime}_{i}(q)\end{pmatrix},

where Ti(q)T^{\prime}_{i}(q) is a nonsingular (i)×(i)(\ell-i)\times(\ell-i) matrix with irreducible characteristic polynomial. We have

XjTi(q)={δ0j+δnji=0,[n+ij]q+[n+ij+i]q1i<,[nj]qi=.\displaystyle X_{j}^{T_{i}(q)}=\begin{cases}\delta_{0j}+\delta_{nj}&i=0,\\ {n-\ell+i\brack j}_{q}+{n-\ell+i\brack j-\ell+i}_{q}&1\leq i<\ell,\\ {n\brack j}_{q}&i=\ell.\end{cases}

Therefore we can take

(15) Xij(t)={δ0j+δnji=0,[n+ij]t+[n+ij+i]t1i<,[nj]ti=.\displaystyle X_{ij}(t)=\begin{cases}\delta_{0j}+\delta_{nj}&i=0,\\ {n-\ell+i\brack j}_{t}+{n-\ell+i\brack j-\ell+i}_{t}&1\leq i<\ell,\\ {n\brack j}_{t}&i=\ell.\end{cases}
Lemma 7.

If TMn(𝐅q)T\in M_{n}(\mathbf{F}_{q}) has irreducible characteristic polynomial, then

(16) σμT=qn1qμ11i2qμi2μi[μi1μi]q.\sigma^{T}_{\mu}=\frac{q^{n}-1}{q^{\mu_{1}}-1}\prod_{i\geq 2}q^{\mu_{i}^{2}-\mu_{i}}{\mu_{i-1}\brack\mu_{i}}_{q}.
Proof.

Follows from [12, Prop. 4.6] and [5, Thm. 3.3]. ∎

Lemma 8.

The equation (13) holds for i=0i=0.

Proof.

Take μ=(n,)\mu=(n-\ell,\ell) in (16). The left hand side of (13) is given by

ωnT0(q)=σμT0(q)=qn1qn1q2[n]q.\omega^{T_{0}(q)}_{n\ell}=\sigma^{T_{0}(q)}_{\mu}=\frac{q^{n}-1}{q^{n-\ell}-1}q^{\ell^{2}-\ell}{n-\ell\brack\ell}_{q}.

On the other hand, the right hand side of (13) becomes

q(2)(q(+12)[n]q+q(2)[n11]q)\displaystyle q^{\ell\choose 2}\left(q^{\ell+1\choose 2}{n-\ell\brack\ell}_{q}+q^{\ell\choose 2}{n-\ell-1\brack\ell-1}_{q}\right)
=q2[n]q(q+[]q[n]q)\displaystyle=q^{\ell^{2}-\ell}{n-\ell\brack\ell}_{q}\left(q^{\ell}+\frac{[\ell]_{q}}{[n-\ell]_{q}}\right)
=q2[n]q[n]q[n]q=qn1qn1q2[nn2]q,\displaystyle=q^{\ell^{2}-\ell}{n-\ell\brack\ell}_{q}\frac{[n]_{q}}{[n-\ell]_{q}}=\frac{q^{n}-1}{q^{n-\ell}-1}q^{\ell^{2}-\ell}{n-\ell\brack n-2\ell}_{q},

establishing (13) for i=0i=0. ∎

It remains to show that the values of pj(q)p_{j}(q) given by (14) are solutions to (13) for 1i1\leq i\leq\ell. We begin by showing that the left hand side of (13) vanishes in these cases.

Lemma 9.

For 1i1\leq i\leq\ell, we have ωnTi(q)=0\omega^{T_{i}(q)}_{n\ell}=0.

Proof.

The 𝐅q[t]\mathbf{F}_{q}[t]-module Mi=𝐅qnM_{i}=\mathbf{F}_{q}^{n}, where tt acts by Ti(q)T_{i}(q) is of the form

Mi={𝐅qn+i𝐅q[t]/fi(t)i=1,,1,𝐅qni=.M_{i}=\begin{cases}\mathbf{F}_{q}^{n-\ell+i}\oplus\mathbf{F}_{q}[t]/f_{i}(t)&i=1,\dotsc,\ell-1,\\ \mathbf{F}_{q}^{n}&i=\ell.\end{cases}

Here fi(t)f_{i}(t) denotes the characteristic polynomial of Ti(q)T^{\prime}_{i}(q) for i=1,,1i=1,\dotsc,\ell-1. Since none of the modules MiM_{i} (1i)(1\leq i\leq\ell) can be generated by n+1n-\ell+1 or fewer generators, Ti(q)T_{i}(q) does not admit a subspace with profile (n,)(n-\ell,\ell). In other words, ωnTi(q)=σ(n,)Ti(q)=0\omega^{T_{i}(q)}_{n\ell}=\sigma^{T_{i}(q)}_{(n-\ell,\ell)}=0 for i=1,,i=1,\dotsc,\ell. ∎

In view of Lemma 9, in order to establish (13) for Ti(q)T_{i}(q) (1i)(1\leq i\leq\ell), it suffices to prove the identity

j=0(1)j(Xij(q)Xi,j1(q))[njn2]qq(j+12)=0.\sum_{j=0}^{\ell}(-1)^{j}(X_{ij}(q)-X_{i,j-1}(q)){n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}}=0.

Let

Y(n,,i,j)=[n+ij]q+[n+ij+i]q[n+ij1]q[n+ij1+i]q.Y(n,\ell,i,j)={n-\ell+i\brack j}_{q}+{n-\ell+i\brack j-\ell+i}_{q}-{n-\ell+i\brack j-1}_{q}-{n-\ell+i\brack j-1-\ell+i}_{q}.

Then by Eq. (15),

Y(n,,i,j)={Xij(q)Xi,j1(q)for 1i<2(Xij(q)Xi,j1(q))for i=.Y(n,\ell,i,j)=\begin{cases}X_{ij}(q)-X_{i,j-1}(q)&\text{for }1\leq i<\ell\\ 2(X_{ij}(q)-X_{i,j-1}(q))&\text{for }i=\ell.\end{cases}

Therefore, it suffices to show that, for 1i1\leq i\leq\ell,

(17) j=0(1)jY(n,,i,j)[njn2]qq(j+12)=0.\sum_{j=0}^{\ell}(-1)^{j}Y(n,\ell,i,j){n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}}=0.

The proof uses techniques from the theory of qq-hypergeometric series, and appears in Section 4.

The non-singularity of X(t)=(Xij(t))0i,jX(t)=(X_{ij}(t))_{0\leq i,j\leq\ell} is proved by using inequalities satisfied by the degrees of its entries. We recall [13, Lemma 4.4].

Lemma 10.

Let (aij)n×n(a_{ij})_{n\times n} be a real matrix such that whenever i<ki<k and j<kj<k,

aikaij<akkakj.a_{ik}-a_{ij}<a_{kk}-a_{kj}.

Then the sum S(σ)=1inaiσ(i)S(\sigma)=\sum_{1\leq i\leq n}a_{i\sigma(i)} attains its maximum value precisely when σ\sigma is the identity permutation.

Proposition 11.

For all n2l0n\geq 2l\geq 0, the determinant of the matrix X(t)X(t) is non-zero. Therefore, for sufficiently large prime powers qq, detX(q)0\det X(q)\neq 0.

Proof.

By Eq. (15), the first row of X(t)X(t) is the unit vector (1,0,,0)(1,0,\dotsc,0). Therefore it suffices to show that the determinant of the submatrix X(t)=(Xij(t))1i,jX^{\prime}(t)=(X_{ij}(t))_{1\leq i,j\leq\ell} is non-zero. Let aij=degXij(t)a_{ij}=\deg X_{ij}(t). Since deg[nk]t=k(nk)\deg{n\brack k}_{t}=k(n-k), it follows from (15) that, for 1i,j1\leq i,j\leq\ell,

aij=max{j(n+ij),(j+i)(nj)}=j(n+ij),a_{ij}=\max\{j(n-\ell+i-j),(j-\ell+i)(n-j)\}=j(n-\ell+i-j),

since

j(n+ij)(j+i)(nj)=(i)(n2j)0.\displaystyle j(n-\ell+i-j)-(j-\ell+i)(n-j)=(\ell-i)(n-2j)\geq 0.

If i<ki<k and j<kj<k, then

aikaij\displaystyle a_{ik}-a_{ij} =k(n+ik)j(n+ij)\displaystyle=k(n-\ell+i-k)-j(n-\ell+i-j)
=(kj)(n+ikj)\displaystyle=(k-j)(n-\ell+i-k-j)
<(kj)(nj)\displaystyle<(k-j)(n-\ell-j)
=akkakj.\displaystyle=a_{kk}-a_{kj}.

Now Lemma 10 implies that detX(t)\det X^{\prime}(t) has degree i=1aii>0\sum_{i=1}^{\ell}a_{ii}>0 and is thus non-zero. ∎

This completes all steps in the proof of Theorem 6 except for the identity (17).

4. Reduction to Heine’s Transformations

In this section, we prove the identity (17) encountered in the proof of Theorem 6 by using a Heine transformation for qq-hypergeometric series. Accordingly, define

Y1(n,,i,j):=[n+ij]q,\displaystyle Y_{1}(n,\ell,i,j):={n-\ell+i\brack j}_{q},\quad Y2(n,,i,j):=[n+ij+i]q,\displaystyle Y_{2}(n,\ell,i,j):={n-\ell+i\brack j-\ell+i}_{q},
Y3(n,,i,j):=[n+ij1]q,\displaystyle Y_{3}(n,\ell,i,j):={n-\ell+i\brack j-1}_{q},\quad Y4(n,,i,j):=[n+ij1+i]q.\displaystyle Y_{4}(n,\ell,i,j):={n-\ell+i\brack j-1-\ell+i}_{q}.

Let Sr(n,,i)S_{r}(n,\ell,i) denote the sum obtained in (17) by replacing Y(n,,i,j)Y(n,\ell,i,j) by Yr(n,,i,j)Y_{r}(n,\ell,i,j). We need to show that S1+S2S3S4=0S_{1}+S_{2}-S_{3}-S_{4}=0. We will show that S1=S4S_{1}=S_{4} while S2=S3S_{2}=S_{3} by expressing the sums SiS_{i} as qq-hypergeometric series.

Define, as usual, the qq-Pochhammer symbols

(a;q)=k=0(1aqk) and (a;q)n=(a;q)(aqn;q)=k=0n1(1aqk).(a;q)_{\infty}=\prod_{k=0}^{\infty}(1-aq^{k})\text{ and }(a;q)_{n}=\frac{(a;q)_{\infty}}{(aq^{n};q)_{\infty}}=\prod_{k=0}^{n-1}(1-aq^{k}).

For convenience, we will also use the notation

(a,b;q)n=(a;q)n(b;q)n.(a,b;q)_{n}=(a;q)_{n}(b;q)_{n}.

Heine (see Gasper and Rahman [6]) defined the qq-hypergeometric series

ϕ12(a,b;c;q,z)=n0(a,b;q)n(q,c;q)nzn.{}_{2}\phi_{1}(a,b;c;q,z)=\sum_{n\geq 0}\frac{(a,b;q)_{n}}{(q,c;q)_{n}}z^{n}.
Lemma 12.

Let m=nm=n-\ell. For all 1i1\leq i\leq\ell, we have

(S1S1) S1(n,,i)\displaystyle S_{1}(n,\ell,i) =(1)(qm+i+1;q)(q;q)ϕ12(q,qm+1;qm+i+1;q,q+1),\displaystyle=(-1)^{\ell}\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\;{}_{2}\phi_{1}(q^{-\ell},q^{m-\ell+1};q^{m+i-\ell+1};q,q^{\ell+1}),
(S2S2) S2(n,,i)\displaystyle S_{2}(n,\ell,i) =(1)(qm+1;q)i(q;q)iϕ12(qi,qm+1;qm+1;q,qi+1),\displaystyle=(-1)^{\ell}\frac{(q^{m+1};q)_{i}}{(q;q)_{i}}\;{}_{2}\phi_{1}(q^{-i},q^{m-\ell+1};q^{m+1};q,q^{i+1}),
(S3S3) S3(n,,i)\displaystyle S_{3}(n,\ell,i) =(1)(qm+i+2;q)1(q;q)1ϕ12(q1,qm+1;qm+i+2;q,q),\displaystyle=(-1)^{\ell}\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\;{}_{2}\phi_{1}(q^{1-\ell},q^{m-\ell+1};q^{m+i-\ell+2};q,q^{\ell}),
(S4S4) S4(n,,i)\displaystyle S_{4}(n,\ell,i) =(1)(qm+2;q)i1(q;q)i1ϕ12(q1i,qm+1;qm+2;q,qi).\displaystyle=(-1)^{\ell}\frac{(q^{m+2};q)_{i-1}}{(q;q)_{i-1}}\;{}_{2}\phi_{1}(q^{1-i},q^{m-\ell+1};q^{m+2};q,q^{i}).

We will see that Heine’s transformation formula [6, Eq. (III.2)] transforms S1(n,,i)S_{1}(n,\ell,i) into S4(n,,i)S_{4}(n,\ell,i) and S2(n,,i)S_{2}(n,\ell,i) into S3(n,,i)S_{3}(n,\ell,i); therefore (17) will follow from the lemma.

Proof of the lemma.

We will use the following identities (equation numbers refer to Gasper and Rahman [6, Appendix I])

(I.7) (a;q)n=(q1n/a;q)n(a)nq(n2),\displaystyle(a;q)_{n}=(q^{1-n}/a;q)_{n}(-a)^{n}q^{\binom{n}{2}},
(I.10) (a;q)nk=(a;q)n(q1n/a;q)k(qa)kq(k2)nk,\displaystyle(a;q)_{n-k}=\frac{(a;q)_{n}}{(q^{1-n}/a;q)_{k}}\left(-\frac{q}{a}\right)^{k}q^{\binom{k}{2}-nk},
(I.42) [αk]q=(qα;q)k(q;q)k(qα)kq(k2).\displaystyle{\alpha\brack k}_{q}=\frac{(q^{-\alpha};q)_{k}}{(q;q)_{k}}(-q^{\alpha})^{k}q^{-\binom{k}{2}}.

Replacing jj by j\ell-j in the sum

S1(n,,i)=j=0(1)j[m+ij]q[mjm]qq(j+12)S_{1}(n,\ell,i)=\sum_{j=0}^{\ell}(-1)^{j}{m+i\brack j}_{q}{m-j\brack m-\ell}_{q}q^{\binom{\ell-j+1}{2}}

gives

j=0(1)j[m+ij]q[m+jj]qq(j+12).\sum_{j=0}^{\ell}(-1)^{\ell-j}{m+i\brack\ell-j}_{q}{m-\ell+j\brack j}_{q}q^{\binom{j+1}{2}}.

The identity (I.42) allows us to write

[m+ij]q[m+jj]q=(1)(qmi;q)j(q;q)j(qmj;q)j(q;q)jq(m+i)(2)ij.{m+i\brack\ell-j}_{q}{m-\ell+j\brack j}_{q}=(-1)^{\ell}\frac{(q^{-m-i};q)_{\ell-j}}{(q;q)_{\ell-j}}\frac{(q^{\ell-m-j};q)_{j}}{(q;q)_{j}}q^{\ell(m+i)-\binom{\ell}{2}-ij}.

Applying (I.10) to (qmi;q)j(q^{-m-i};q)_{\ell-j} and (q;q)j(q;q)_{\ell-j} gives

(1)(qmi;q)(q;q)(q;q)j(qm+i+1;q)j(qmj;q)j(q;q)jq(m+i)+(m+1)j(2).(-1)^{\ell}\frac{(q^{-m-i};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{-\ell};q)_{j}}{(q^{m+i-\ell+1};q)_{j}}\frac{(q^{\ell-m-j};q)_{j}}{(q;q)_{j}}q^{\ell(m+i)+(m+1)j-\binom{\ell}{2}}.

Applying (I.7) to (qmi;q)(q^{-m-i};q)_{\ell} and (qmj;q)j(q^{\ell-m-j};q)_{j} gives

(18) [m+ij]q[m+jj]q=(1)j(qm+i+1;q)(q;q)(q,qm+1;q)j(q,qm+i+1;q)jqj(j2).{m+i\brack\ell-j}_{q}{m-\ell+j\brack j}_{q}=(-1)^{j}\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{-\ell},q^{m-\ell+1};q)_{j}}{(q,q^{m+i-\ell+1};q)_{j}}q^{\ell j-\binom{j}{2}}.

Thus

S1(n,,i)=(1)(qm+i+1;q)(q;q)j=0(q,qm+1;q)j(q,qm+i+1;q)jq(+1)j.S_{1}(n,\ell,i)=(-1)^{\ell}\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\sum_{j=0}^{\ell}\frac{(q^{-\ell},q^{m-\ell+1};q)_{j}}{(q,q^{m+i-\ell+1};q)_{j}}q^{(\ell+1)j}.

Since (q;q)j=0(q^{-\ell};q)_{j}=0 for j>j>\ell, the sum can be extended to infinity, giving (S1S1).

In order to prove (S3S3), observe that

[m+ij1]q[m+jj]q=[m+ij]q[m+jj]q1qj1qm+i+j+1.\displaystyle{m+i\brack\ell-j-1}_{q}{m-\ell+j\brack j}_{q}={m+i\brack\ell-j}_{q}{m-\ell+j\brack j}_{q}\frac{1-q^{\ell-j}}{1-q^{m+i-\ell+j+1}}.

Apply (18) and rewrite the right hand side as

(1)j(qm+i+1;q)(q;q)(q,qm+1;q)j(q,qm+i+1;q)j1qj1qm+i+j+1qj(j2).(-1)^{j}\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{-\ell},q^{m-\ell+1};q)_{j}}{(q,q^{m+i-\ell+1};q)_{j}}\frac{1-q^{\ell-j}}{1-q^{m+i-\ell+j+1}}q^{\ell j-\binom{j}{2}}.

Making the substitutions

(qm+i+1;q)(q;q)\displaystyle\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}} =(qm+i+2;q)1(q;q)11qm+i+11q,\displaystyle=\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\frac{1-q^{m+i-\ell+1}}{1-q^{\ell}},
(q;q)j(qm+i+1;q)j\displaystyle\frac{(q^{-\ell};q)_{j}}{(q^{m+i-\ell+1};q)_{j}} =1q1qj(q1;q)j(qm+i+2;q)j1qm+i+j+11qm+i+1,\displaystyle=\frac{1-q^{-\ell}}{1-q^{j-\ell}}\frac{(q^{1-\ell};q)_{j}}{(q^{m+i-\ell+2};q)_{j}}\frac{1-q^{m+i-\ell+j+1}}{1-q^{m+i-\ell+1}},

and cancelling out common factors gives

(19) [m+ij1]q[m+jj]q=(1)j(qm+i+2;q)1(q;q)1(q1,qm+1;q)j(q,qm+i+2;q)jq(1)j(j2).{m+i\brack\ell-j-1}_{q}{m-\ell+j\brack j}_{q}=(-1)^{j}\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\frac{(q^{1-\ell},q^{m-\ell+1};q)_{j}}{(q,q^{m+i-\ell+2};q)_{j}}q^{(\ell-1)j-\binom{j}{2}}.

Evaluating S3(n,,i)S_{3}(n,\ell,i) after replacing jj by j\ell-j in the sum and using (19) gives (S3S3).

To prove (S4S4), we proceed as in the proof of (S1S1).

S4(n,,i)\displaystyle S_{4}(n,\ell,i) =j=0[m+ij1+i]q[mjm]qq(j+12)\displaystyle=\sum_{j=0}^{\ell}{m+i\brack j-1-\ell+i}_{q}{m-j\brack m-\ell}_{q}q^{\binom{\ell-j+1}{2}}
=j=0(1)j[m+ii1j]q[m+jj]qq(j+12).\displaystyle=\sum_{j=0}^{\ell}(-1)^{\ell-j}{m+i\brack i-1-j}_{q}{m-\ell+j\brack j}_{q}q^{\binom{j+1}{2}}.

The identity (I.42) allows us to write

[m+ii1j]q[m+jj]q=(1)i1(qmi;q)i1j(q;q)i1j(qmj;q)j(q;q)jq(m+1)(i1)+(i2)(+1)j.{m+i\brack i-1-j}_{q}{m-\ell+j\brack j}_{q}\\ =(-1)^{i-1}\frac{(q^{-m-i};q)_{i-1-j}}{(q;q)_{i-1-j}}\frac{(q^{\ell-m-j};q)_{j}}{(q;q)_{j}}q^{(m+1)(i-1)+\binom{i}{2}-(\ell+1)j}.

Applying (I.10) to (qmi;q)i1j(q^{-m-i};q)_{i-1-j} and (q;q)i1j(q;q)_{i-1-j} gives

(1)i1(qmi;q)i1(q;q)i1(q1i;q)j(qm+2;q)j(qmj;q)j(q;q)jq(m+1)(i1)+(i2)+(m+i)j.(-1)^{i-1}\frac{(q^{-m-i};q)_{i-1}}{(q;q)_{i-1}}\frac{(q^{1-i};q)_{j}}{(q^{m+2};q)_{j}}\frac{(q^{\ell-m-j};q)_{j}}{(q;q)_{j}}q^{(m+1)(i-1)+\binom{i}{2}+(m+i-\ell)j}.

Finally, applying (I.7) to the terms (qmi;q)i1(q^{-m-i};q)_{i-1} and (qmj;q)j(q^{\ell-m-j};q)_{j} gives

(20) [m+ii1j]q[m+jj]q=(1)j(qm+2;q)i1(q;q)i1(q1i,qm+1;q)j(q,qm+2;q)jq(j+12)+ij{m+i\brack i-1-j}_{q}{m-\ell+j\brack j}_{q}=(-1)^{j}\frac{(q^{m+2};q)_{i-1}}{(q;q)_{i-1}}\frac{(q^{1-i},q^{m-\ell+1};q)_{j}}{(q,q^{m+2};q)_{j}}q^{-\binom{j+1}{2}+ij}

Evaluating S4(n,,i)S_{4}(n,\ell,i) using (20) gives (S4S4).

In order to prove (S2S2), observe that

[m+iij]q[m+jj]q=[m+iij1]q[m+jj]q1qm+j+11qij.{m+i\brack i-j}_{q}{m-\ell+j\brack j}_{q}={m+i\brack i-j-1}_{q}{m-\ell+j\brack j}_{q}\frac{1-q^{m+j+1}}{1-q^{i-j}}.

Applying (20) allows us to write the right hand side as

(1)j(qm+2;q)i1(q;q)i1(q1i,qm+1;q)j(q,qm+2;q)j1qm+j+11qijq(j+12)+ij.(-1)^{j}\frac{(q^{m+2};q)_{i-1}}{(q;q)_{i-1}}\frac{(q^{1-i},q^{m-\ell+1};q)_{j}}{(q,q^{m+2};q)_{j}}\frac{1-q^{m+j+1}}{1-q^{i-j}}q^{-\binom{j+1}{2}+ij}.

Making the substitutions

(qm+2;q)i1(q;q)i1\displaystyle\frac{(q^{m+2};q)_{i-1}}{(q;q)_{i-1}} =(qm+1;q)i(q;q)i1qi1qm+1,\displaystyle=\frac{(q^{m+1};q)_{i}}{(q;q)_{i}}\frac{1-q^{i}}{1-q^{m+1}},
(q1i;q)j(qm+2;q)j\displaystyle\frac{(q^{1-i};q)_{j}}{(q^{m+2};q)_{j}} =1qji1qi(qi;q)j(qm+1;q)j1qm+11qm+j+1\displaystyle=\frac{1-q^{j-i}}{1-q^{-i}}\frac{(q^{-i};q)_{j}}{(q^{m+1};q)_{j}}\frac{1-q^{m+1}}{1-q^{m+j+1}}

and cancelling out common factors gives

(21) [m+iij]q[m+jj]q=(1)j(qm+1;q)i(q;q)i(qi,qm+1;q)j(q,qm+1;q)jq(j+12)+(i+1)j.{m+i\brack i-j}_{q}{m-\ell+j\brack j}_{q}=(-1)^{j}\frac{(q^{m+1};q)_{i}}{(q;q)_{i}}\frac{(q^{-i},q^{m-\ell+1};q)_{j}}{(q,q^{m+1};q)_{j}}q^{-\binom{j+1}{2}+(i+1)j}.

Evaluating S2(n,,i)S_{2}(n,\ell,i) using (21) after replacing jj by j\ell-j in the sum gives (S2S2). ∎

Proof that S1=S4S_{1}=S_{4}.

Applying Heine’s transformation formula [6, Eq. (III.2)]

(22) ϕ12(a,b;c;q,z)=(c/b,bz;q)(c,z;q)ϕ12(abz/c,b;bz;q,c/b){}_{2}\phi_{1}(a,b;c;q,z)=\frac{(c/b,bz;q)_{\infty}}{(c,z;q)_{\infty}}\;{}_{2}\phi_{1}(abz/c,b;bz;q,c/b)

with (a,b,c,z)=(q,qm+1,qm+i+1,q+1)(a,b,c,z)=(q^{-\ell},q^{m-\ell+1},q^{m+i-\ell+1},q^{\ell+1}) gives

S1(n,,i)=(1)(qm+i+1;q)(q;q)(qi,qm+2;q)(qm+i+1,q+1;q)ϕ12(q1i,qm+1;qm+2;q,qi).S_{1}(n,\ell,i)=(-1)^{\ell}\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{i},q^{m+2};q)_{\infty}}{(q^{m+i-\ell+1},q^{\ell+1};q)_{\infty}}{}_{2}\phi_{1}(q^{1-i},q^{m-\ell+1};q^{m+2};q,q^{i}).

Observe that

(qm+i+1;q)(q;q)(qi,qm+2;q)(qm+i+1,q+1;q)\displaystyle\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{i},q^{m+2};q)_{\infty}}{(q^{m+i-\ell+1},q^{\ell+1};q)_{\infty}} =(qm+i+1;q)(q;q)(qi;q)i+1(qm+i+1;q)i+1\displaystyle=\frac{(q^{m+i-\ell+1};q)_{\ell}}{(q;q)_{\ell}}\frac{(q^{i};q)_{\ell-i+1}}{(q^{m+i-\ell+1};q)_{\ell-i+1}}
=(qm+2;q)i1(q;q)i1,\displaystyle=\frac{(q^{m+2};q)_{i-1}}{(q;q)_{i-1}},

which gives S1(n,,i)=S4(n,,i)S_{1}(n,\ell,i)=S_{4}(n,\ell,i). ∎

Proof that S2=S3S_{2}=S_{3}.

Heine’s transformation formula (22) with (a,b,c,z)=(q1,qm+1,qm+i+2,q)(a,b,c,z)=(q^{1-\ell},q^{m-\ell+1},q^{m+i-\ell+2},q^{\ell}) gives

S3(n,,i)=(1)(qm+i+2;q)1(q;q)1(qi+1,qm+1;q)(qm+i+2,q;q)ϕ12(qi,qm+1;qm+1;q,qi+1).S_{3}(n,\ell,i)=(-1)^{\ell}\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\frac{(q^{i+1},q^{m+1};q)_{\infty}}{(q^{m+i-\ell+2},q^{\ell};q)_{\infty}}{}_{2}\phi_{1}(q^{-i},q^{m-\ell+1};q^{m+1};q,q^{i+1}).

Observe that

(qm+i+2;q)1(q;q)1(qi+1,qm+1;q)(qm+i+2,q;q)\displaystyle\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\frac{(q^{i+1},q^{m+1};q)_{\infty}}{(q^{m+i-\ell+2},q^{\ell};q)_{\infty}} =(qm+i+2;q)1(q;q)1(qi+1;q)i1(qm+i+2;q)i1\displaystyle=\frac{(q^{m+i-\ell+2};q)_{\ell-1}}{(q;q)_{\ell-1}}\frac{(q^{i+1};q)_{\ell-i-1}}{(q^{m+i-\ell+2};q)_{\ell-i-1}}
=(qm+1;q)i(q;q)i,\displaystyle=\frac{(q^{m+1};q)_{i}}{(q;q)_{i}},

giving S2(n,,i)=S3(n,,i)S_{2}(n,\ell,i)=S_{3}(n,\ell,i). ∎

This completes the proof of the main theorem.

5. The qq-Hermite Catalan matrix

In this section, we discuss the connection between our main theorem and Touchard’s formula for the entries of the qq-Hermite Catalan matrix.

An extended chord diagram is a visual representation of an involution σ\sigma on [n][n]. Arrange nn nodes labelled 1,,n1,\dotsc,n along the XX-axis. To their right, add a node labelled \infty. A circular arc lying above the XX-axis is used to connect the elements of each 22-cycle of σ\sigma. Each fixed point of σ\sigma is connected to the node \infty. The extended chord diagram of the involution (1,4)(2,6)(7,8)(1,4)(2,6)(7,8) on the set [8][8] is shown below:

1122334455667788\infty

A crossing is a pair of arcs [(i,j),(k,)][(i,j),(k,\ell)] such that i<k<j<i<k<j<\ell. The extended chord diagram above has four crossings, namely [(1,4),(2,6)][(1,4),(2,6)], [(1,4),(3,)][(1,4),(3,\infty)], [(2,6),(3,)][(2,6),(3,\infty)], and [(2,6),(5,)][(2,6),(5,\infty)]. Let v(σ)v(\sigma) denote the number of crossings of the extended chord diagram of an involution σ\sigma.

Let Inv(n,k)\operatorname{Inv}(n,k) denote the set of involutions in SnS_{n} with kk fixed points. Define

(23) ank(q)=σInv(n,k)qv(σ).a_{nk}(q)=\sum_{\sigma\in\operatorname{Inv}(n,k)}q^{v(\sigma)}.

If nkn-k is odd then Inv(n,k)=\operatorname{Inv}(n,k)=\emptyset, so ank(q)=0a_{nk}(q)=0. If nkn-k is even, then an element of Inv(n,k)\operatorname{Inv}(n,k) has l:=(nk)/2l:=(n-k)/2 cycles of length two.

For each non-negative integer kk, let [k]q[k]_{q} denote the qq-integer 1+q++qk11+q+\dotsb+q^{k-1}.

Lemma 13.

We have

a00(q)=1,a0k=0 for k>0,\displaystyle a_{00}(q)=1,\quad a_{0k}=0\text{ for }k>0,
ank(q)=an1,k1(q)+[k+1]qan1,k+1(q)(n>0).\displaystyle a_{nk}(q)=a_{n-1,k-1}(q)+[k+1]_{q}a_{n-1,k+1}(q)\quad(n>0).
Proof.

Each involution σInv(n1,k1)\sigma\in\operatorname{Inv}(n-1,k-1) can be extended to an element of Inv(n,k)\operatorname{Inv}(n,k) by adding nn as a fixed point. Furthermore, each σInv(n1,k+1)\sigma\in\operatorname{Inv}(n-1,k+1) can be extended to an element of Inv(n,k)\operatorname{Inv}(n,k) in k+1k+1 different ways: any one of its k+1k+1 fixed points can be paired with nn. Pairing the rr-th fixed point from right to left with nn results in r1r-1 new crossings. Taken together, these k+1k+1 choices contribute (1+q++qk)qv(σ)=[k+1]qqv(σ)(1+q+\dotsb+q^{k})q^{v(\sigma)}=[k+1]_{q}q^{v(\sigma)} to ank(q)a_{nk}(q). Since every element of Inv(n,k)\operatorname{Inv}(n,k) can be obtained uniquely by one of these methods, the identity of the lemma follows. ∎

Touchard [15] studied the polynomials

Tm(q)=a2m,0(q)=σInv(2m,0)qv(σ),T_{m}(q)=a_{2m,0}(q)=\sum_{\sigma\in\operatorname{Inv}(2m,0)}q^{v(\sigma)},

which admit a simple (but subtle) formula

(24) (q1)mTm(q)=j=0m(1)j[(2mj)(2mj1)]q(mj+12),(q-1)^{m}T_{m}(q)=\sum_{j=0}^{m}(-1)^{j}\left[\binom{2m}{j}-\binom{2m}{j-1}\right]q^{\binom{m-j+1}{2}},

known as the Touchard-Riordan formula. The polynomials an,n2a_{n,n-2\ell} are precisely the entries of the Catalan matrix associated to the normalized qq-Hermite orthogonal polynomials of Ismail, Stanton and Viennot [7]. See Aigner [2, Chapter 7] for a comprehensive exposition.

The combinatorial theory of orthogonal polynomials [16] places the moments of an orthogonal polynomial sequence in the first column of a Catalan matrix: if an orthogonal polynomial sequence {Pk(x)}k0\{P_{k}(x)\}_{k\geq 0} satisfies the three-term recurrence relation

Pk+1(x)=(xbk)Pk(x)λkPk1(x) for k1,\displaystyle P_{k+1}(x)=(x-b_{k})P_{k}(x)-\lambda_{k}P_{k-1}(x)\text{ for }k\geq 1,
with P0(x)=1,P1(x)=xb0,\displaystyle\text{with }P_{0}(x)=1,\quad P_{1}(x)=x-b_{0},

for some {bk}k0\{b_{k}\}_{k\geq 0} and {λk}k1\{\lambda_{k}\}_{k\geq 1}, with λk0\lambda_{k}\neq 0, the entries of the Catalan matrix (cnk)n,k0(c_{nk})_{n,k\geq 0} are given by

c00=1,c0k=0 for k>0,\displaystyle c_{00}=1,\quad c_{0k}=0\text{ for }k>0,
cnk=cn1,k1+bkcn1,k+λk+1cn1,k+1.\displaystyle c_{nk}=c_{n-1,k-1}+b_{k}c_{n-1,k}+\lambda_{k+1}c_{n-1,k+1}.

The moments of the orthogonal polynomial sequence are

μn=cn0 for n0.\mu_{n}=c_{n0}\text{ for }n\geq 0.

Lemma 13 implies that the polynomials ank(q)a_{nk}(q) of (23) are the entries of the Catalan matrix with bk=0b_{k}=0 and λk=[k]q\lambda_{k}=[k]_{q}, which correspond to the combinatorial version of the qq-Hermite orthogonal polynomial sequence [7, Eq. (2.11)]. Thus Touchard’s polynomials Tm(q)T_{m}(q) are the even moments of the qq-Hermite orthogonal polynomial sequence (the odd moments being 0). This is well-known and plays a role in the proof of the Touchard-Riordan formula [2, Chapter 7].

Let TT be an n×nn\times n diagonalizable matrix with distinct eigenvalues in 𝐅q\mathbf{F}_{q}. By (5), the number of \ell-dimensional TT-anti-invariant subspaces is equal to the number of subspaces with TT-profile (m,)(m,\ell), where m=nm=n-\ell. The conjugate of the partition μ=(m,)\mu=(m,\ell) is the partition μ=(2,1m)\mu^{\prime}=(2^{\ell},1^{m-\ell}). Set partitions with \ell blocks of size 22 and mm-\ell blocks of size 11 can be identified with involutions on [n][n] with mm-\ell fixed points. The interlacing number [12, Defn. 3.3] of such a set partition reduces to the number of crossings of the extended chord diagram on the corresponding involution. Therefore, the formula (6) can be rephrased as follows.

Theorem 14.

For all integers n20n\geq 2\ell\geq 0,

ωnT=(q1)q(2)an,n2(q).\omega_{n\ell}^{T}=(q-1)^{\ell}q^{\binom{\ell}{2}}a_{n,n-2\ell}(q).

Since all TT-invariant subspaces are direct sums of eigenspaces, XjTX^{T}_{j} is just the binomial coefficient (nj)\binom{n}{j}. Combining Theorem 14 with our formula (1) for ωnT\omega^{T}_{n\ell} gives a new proof of Touchard’s formula [15, Eq. (28)] for an,n2(q)a_{n,n-2\ell}(q).

Theorem 15 (Touchard’s formula).

For all integers n20n\geq 2\ell\geq 0,

(q1)an,n2(q)=j=0(1)j[(nj)(nj1)][njn2]qq(j+12).(q-1)^{\ell}a_{n,n-2\ell}(q)=\sum_{j=0}^{\ell}(-1)^{j}\left[\binom{n}{j}-\binom{n}{j-1}\right]{n-\ell-j\brack n-2\ell}_{q}q^{\binom{\ell-j+1}{2}}.

Specializing to n=2n=2\ell recovers the Touchard-Riordan formula (24).

6. Acknowledgements

We are indebted to an anonymous referee for several comments and suggestions that helped improve the overall presentation of this paper. We thank Divya Aggarwal for her comments on an earlier draft of this manuscript. We thank Michael Schlosser for suggesting the method used in the proof of the main identity (17). The second author was partially supported by a MATRICS grant MTR/2017/000794 awarded by the Science and Engineering Research Board and an Indo-Russian project DST/INT/RUS/RSF/P41/2021.

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