Koszul properties of the moment map of some classical representations

Abstract

This work concerns the moment map $\mu$ associated with the standard representation of a classical Lie algebra. For applications to deformation quantization it is desirable that $S/(\mu )$, the coordinate algebra of the zero fibre of $\mu$, be Koszul. The main result is that this algebra is not Koszul for the standard representation of $\mathfrak {sl}_{n}$, and of $\mathfrak {sp}_{n}$. This is deduced from a computation of the Betti numbers of $S/(\mu )$ as an S-module, which are of interest also from the point of view of commutative algebra.

Appendices

Appendix A. Quadratic exterior forms and maximal rank

The main task of this section is to prove Proposition A.2, which was used in the proof of Theorem 5.1. Throughout k will be a field. For elements $\varvec{x}=x_1,\dots , x_u$ in a commutative ring, we write $s_i(\varvec{x})$ for the i-th elementary symmetric polynomials in $\varvec{x}$ with the convention that $s_0(\varvec{x})=1$ and $s_i(\varvec{x})=0$ for $i<0$ or $i>u$.

Lemma A.1

In $k[x_1,\dots , x_u, y_1,\dots , y_v]/J$, where $J=(y_1^2,\dots , y_v^2,x_1^2,\dots , x_u^2)$, for every $d\ge 0$ there is an equality

$$\begin{aligned} (s_1(\varvec{x})+s_1(\varvec{y}))\sum _{k=0}^d (-1)^k k! (d-k)! s_{k}(\varvec{x}) s_{d-k}(\varvec{y})=(d+1)! (s_{d+1}(\varvec{y})+(-1)^d s_{d+1}(\varvec{x})). \end{aligned}$$

Proof

It is not hard to verify that there is an equality

$$\begin{aligned} s_1(\varvec{x})s_{k}(\varvec{x})=(k+1)s_{k+1}(\varvec{x})\,; \end{aligned}$$

by symmetry, this implies also that $s_1(\varvec{y})s_{d-k}(\varvec{y})=(d-k+1)s_{d-k+1}(\varvec{y})$. Using these one gets equalities

$$\begin{aligned}&(s_1(\varvec{x})+s_1(\varvec{y}))\sum _{k=0}^d (-1)^k k! (d-k)! s_{k}(\varvec{x})s_{d-k}(\varvec{y}) \\&\quad = \sum _{k=0}^d (-1)^k k! (d+1-k)! s_{k}(\varvec{x})s_{d-k+1}(\varvec{y}) \\&\qquad +\,\sum _{k=0}^d (-1)^k (k+1)! (d-k)! s_{k+1}(\varvec{x}) s_{d-k}(\varvec{y})\\&\quad =(d+1)! (s_{d+1}(\varvec{y})+(-1)^d s_{d+1}(\varvec{x})) \end{aligned}$$

$\square$

Let V be a k-vector space of dimension 2n, and $\Lambda$ the exterior algebra on V. Let $e_1,\dots , e_n, f_1,\dots , f_n$ be a basis of V and set

$$\begin{aligned} w:=\sum _{i=1}^n e_i f_i\,; \end{aligned}$$

this is an element in $\Lambda _{2}$. The result below is well known when ${\text {char}}k=0$ and can be deduced as a special case of the Hard Lefschetz Theorem; see [9, page 122]. However, we have been unable to find an argument that covers also the case of positive characteristics in the literature.

Proposition A.2

If ${\text {char}}k=0$ or ${\text {char}}k>(n+1)/2$, then the multiplication map $w:\Lambda _i\rightarrow \Lambda _{i+2}$ has maximal rank for every i; in other words, $w:\Lambda _i\rightarrow \Lambda _{i+2}$ is injective for $i\le n-1$ and surjective for $i\ge n-1$.

Proof

To begin with, there is an isomorphism ${\text {Hom}}_{k}(\Lambda ,k)\cong \Lambda (2n)$ of $\Lambda$-modules and from this it follows that it suffices to verify that multiplication by w is injective in degrees $\le n-1$. Moreover, it suffices to verify injectivity for $i=n-1$.

Indeed, assume $w:\Lambda _{i}\rightarrow \Lambda _{i+2}$ is injective for some $i\le n-1$. For any nonzero element $\lambda$ of degree $i-1$, there exists an element $\nu$ in $\Lambda _{1}$ for which $\lambda \nu \ne 0$; this is because the socle of $\Lambda$ is $\Lambda _{2n}$. Then $w\lambda \nu \ne 0$ implies $w\lambda \ne 0$. An iteration yields the desired result.

To reiterate: It suffices to verify that $w:\Lambda _{n-1}\rightarrow \Lambda _{n+1}$ is injective. Since the ranks of the source and target coincide, this is equivalent to verifying that the map is surjective. We prove this by an induction on n, the base case $n=1$ being obvious.

Suppose $n\ge 2$. Then $\Lambda _{n+1}$ has a k-basis consisting of monomials

$$\begin{aligned} \mu =e_{i_1} \dots e_{i_a} f_{j_1} \dots f_{j_b} \end{aligned}$$

where $1\le a,b\le n$ with $a+b=n+1$, and $i_1<\dots <i_a$ and $j_1<\dots j_b$,

If for some h we have $i_h\not \in \{j_1,\dots , j_b\}$, then $e_{i_h}w=e_{i_h}w_1$ with

$$\begin{aligned} w_1=\sum _{v=1, v\ne i_h}^n e_v f_v \end{aligned}$$

and, by induction on n, there exists $\lambda \in \Lambda _{n-2}(U)$ such that

$$\begin{aligned} \lambda w_1=e_{i_1} \dots \hat{e}_{i_h} \dots e_{i_a} f_{j_1} \dots f_{j_b}. \end{aligned}$$

Here U is the k-subspace of V generated by $\{e_1,\dots , e_n, f_1,\dots , f_n\} \setminus \{e_{i_h}, f_{i_h}\}$. Then

$$\begin{aligned} \nu e_{i_h} w= \nu e_{i_h} w_1=\pm e_{i_h} \nu w_1=\pm \mu . \end{aligned}$$

A similar argument settles the case when there exists h with $j_h\not \in \{i_1,\dots , i_a\}$.

It remains to consider the case $a=b$ and $\{i_1,\dots , i_a\}=\{j_1,\dots ,j_b\}$. Set $m=a=b$ so that $n=2m-1$. We can assume (renaming the indices) that $i_h=j_h=h$ for $h=1,\dots , m$, that is to say, $\mu =\prod _{i=1}^m e_i f_i$.

We apply Lemma A.1 as follows: we set $v=m$, $u=m-1$ and $d=m-1$ and

$$\begin{aligned} \begin{array}{llll} &{}x_i\rightarrow e_{m+i} f_{m+i}&{} \quad &{}\text { for }i=1,\dots , m-1\\ &{}y_i\rightarrow e_i f_i&{} \quad &{}\text { for } i=1,\dots , m. \end{array} \end{aligned}$$

Such a specialization makes sense because the elements $x_{i}y_i$ have square zero and commute among themselves. Since $s_1(s')+s_1(t')=w$, $s_{m}(t')=\prod _{i=1}^m s_i f_i$ and $s_{m}(s')=0$ we obtain:

$$\begin{aligned} w\sum _{k=0}^{m-1} (-1)^k k! (m-1-k)! s_{k}(s') s_{m-1-k}(t')=m! \prod _{i=1}^m s_i f_i. \end{aligned}$$

Since we are assuming that ${\text {char}}k=0$ or ${\text {char}}k>(n+1)/2$ we have that m! is invertible and hence we may conclude that $w\nu =\prod _{i=1}^m s_i f_i$ with

$$\begin{aligned} \nu =\frac{1}{(m!)} \sum _{k=0}^{m-1} (-1)^k k! (m-1-k)! s_{k}(s') s_{m-1-k}(t'). \end{aligned}$$

This completes the proof. $\square$

Appendix B. Betti tables and numerology

In this section we collect some observation, and questions, concerning the algebras studied in this work. To begin with, consider the following Betti tables for the representations $\mathfrak {sl}_n:k^n$, computed in Theorem 5.1, for the first few values of n.

The framed numbers in the table are the Catalan numbers; see [17, A000108]. This is true for each n, and what is more, the Catalan triangle introduced by Shapiro [20], appears as well.

Indeed, the sequence $(C_n)_{n\ge 0}$ of Catalan numbers, which has numerous combinatorial interpretations [14, 23], can be defined as follows:

$$\begin{aligned} C_n:={\frac{1}{n+1}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) =\left( {\begin{array}{c}2n\\ n\end{array}}\right) - \left( {\begin{array}{c}2n\\ n+1\end{array}}\right) . \end{aligned}$$

The Segner’s recursion formula $C_{n+1}=\sum _{i,j:i+j=n} C_i C_j$ can be rewritten in terms of the generating function $C(x)=\sum _{n\ge 0} C_n x^n$ as follows: $x\,C^2(x)=C(x)-C_0$.

The entries B(N, r) of the Catalan triangle, with $N,r \ge 1$, are defined by the higher moments $\gamma ^r(x)=:\sum _{N\ge 1}B(N,r)\,x^N$ of the generating function $\gamma (x)=C(x)-C_0=C(x)-1$. In other words,

$$\begin{aligned} B(N,r)= {\mathop {\mathop {\sum }\limits _{i_1,i_2,\dots , i_r\ge 1}}\limits _{\sum _j i_j=N}}\, C_{i_1}C_{i_2}\cdots C_{i_r}. \end{aligned}$$

It has been shown in [20] that $B(N,r)= \frac{r}{N} \left( {\begin{array}{c}2N\\ N-r\end{array}}\right)$, which also makes sense for $r<0$.

The Catalan triangle can be recovered when taking differences of columns in the Pascal triangle:

$$\begin{aligned} \left( {\begin{array}{c}2n\\ i\end{array}}\right) - \left( {\begin{array}{c}2n\\ i+2\end{array}}\right)&= \left( 1- \frac{(2n-i)(2n-i-1)}{(i+1)(i+2)}\right) \left( {\begin{array}{c}2n\\ i\end{array}}\right) \\&=2\, \frac{(i+1-n)(2n+1)}{(i+1)(i+2)}\, \left( {\begin{array}{c}2n\\ i\end{array}}\right) \\&= \frac{i+1-n}{n+1}\, \frac{(2n+1)(2n+2)}{(i+1)(i+2)}\, \left( {\begin{array}{c}2n\\ i\end{array}}\right) \\&=\frac{i+1-n}{n+1}\, \left( {\begin{array}{c}2n+2\\ i+2\end{array}}\right) \\&=B(n+1,i+1-n). \end{aligned}$$

This proves the occurrence of the Catalan triangle in the Betti table of $\mathfrak {sl}_n:k^n$; compare Theorem 5.1. The many combinatorial interpretations of the Catalan triangle [14, 20] raise the question: Is there a minimal free resolution of the moment map of $\mathfrak {sl}_n:k^n$ that underlies the occurrence of the Catalan triangles?

The Betti tables of the moment map for the representations $\mathfrak {sp}_n:k^{2n}$, computed in Theorem 6.3, are also worth pointing to. For $n=1$ the Betti table is the same as that of $\mathfrak {sl}_2:k^{2}$. For $n=2,3$ one gets the following.

The noteworthy feature here is that the Betti table has only two strands, and the jump from one to the next occurs after the second step.

Poincaré series. Finally we discuss the Poincaré series of k over the coordinate algebra of the zero fibre. Let R be a standard graded algebra, as in Sect. 3. For simplicity, we focus on the ${\mathbb N}$-graded case. We are interested in the Betti numbers of k as an R-module. A basic question is when the corresponding Poincaré series ${\text {P}}^{R}_{k}(u)$ is rational; it is not so for a general graded ring R; see [2, sect. 4.3]. What about for algebras of the form $S/(\mu )$, for a moment map $\mu ?$

The Poincaré series for a complete intersection is rational [2, Theorem 9.2.1], so this takes care of 0-modular representations. Thus, as for the Koszul property, the question is moot only for small representations. Rationality also holds when R is Koszul, for then there is an equality ${\text {P}}^{R}_{k}(s,u)= {\text {Hilb}}_{R}(-su)^{-1}$. This settles the case of the standard representations $\mathfrak {gl}_{n}:k^{n}$ and $\mathfrak {so}_{n}:k^{n}$.

We do not know the answer for $\mathfrak {sl}_{n}:k^{n}$ and $\mathfrak {sp}_{n}:k^{n}$. For $\mathfrak {sl}_2:k^2$ Gröbner basis calculations (using Macaulay2) suggest that the Betti table of the minimal free resolution of k is upper triangular, with a new strand appearing after every 3 steps in the homological degree. The following formula for the Poincaré series best fits the available numerical data

$$\begin{aligned} {\text {P}}^{S/(\mu )}_{k}(s,u) ={(1+us)^2\over (1-us)^3(1+us)-2u^3 s^4}. \end{aligned}$$

For $\mathfrak {sl}_3:k^3$ we also encounter a triangular shape but we have been unable to guess what the Poincaré series might be.

Following the arXiv posting of an earlier version of this manuscript, Jan-Erik Roos wrote to us (email, dated 17 June 2017) that he could answer some of the questions posed in the preceding paragraphs. Pointing to his paper [19], he writes “I have in principle solved “all” cases of the Homological Behaviour of families of quadratic forms in four variables...” Roos notes that for $\mathfrak {sl}_2:k^2$ the ring $S/(\mu )$ is isomorphic to the one in [19, Case 7, pp. 427], and that the corresponding Poincaré series is precisely the one we proposed above. Furthermore, he proposes the following formula for the Poincaré series for the case of $\mathfrak {sl}_{3}:k^{3}$

$$\begin{aligned} {\text {P}}^{S/(\mu )}_{k}(s,u) ={(1+us)^3\over (1-us)(1-2us-4u^2s^2-2u^3s^3+ u^4s^4)-2u^4 s^5}. \end{aligned}$$

and suggests a method to tackle $\mathfrak {sl}_n:k^n$, for arbitrary n; we hope to develop these ideas in due course.