# Invariants of symmetric forms with respect to the symplectic group

Take a 6-dimensional vector space $$V$$ (for simplicity, over $$\mathbb{C}$$) and play the following game (for example, by employing the online Lie program): consider the 21-dimensional space $$S^2V^*$$ of symmetric two-forms on $$V$$ and decompose the space $$S^k(S^2V)$$ of degree-$$k$$ homogeneous polynomials on $$S^2V^*$$ into irreducible $$\mathsf{SL}_6$$-modules and, simultaneously, into irreducible $$\mathsf{Sp}_6$$-module, with $$k=1,2,3,4,5,6$$. The number of one-dimensional constituents you'll obtain is the following:

• For $$\mathsf{SL}_6$$ there is a unique one-dimensional constituent $$\langle d\rangle$$, that appears when $$k=6$$;
• For $$\mathsf{Sp}_6$$ the first one-dimensional constituent $$\langle p\rangle$$ pops up with $$k=2$$, then a second one $$\langle q\rangle$$ with $$k=4$$, accompained by $$\langle p^2\rangle$$, and, finally, for $$k=6$$, there are three one-dimensional constituents: $$\langle p^3\rangle$$, $$\langle p q\rangle$$ and $$\langle d\rangle$$.

Now it is well known that $$d$$ is the determinant.

QUESTION: what about the $$\mathsf{Sp}_6$$-invariants $$p$$ and $$q$$ of a symmetric two-form $$\alpha$$ on $$V$$? Can we read them off from the characteristic polynomial of a suitable endomorphism of $$V$$ related to $$\alpha$$? Does anybody know where precisely in the literature this is discussed? (Should be classical.)

In particular, I'm interested in the normal forms of elements $$\alpha\in S^2V^*$$ with rispect to the symplectic group: in the case of the linear group, the normal form of $$\alpha$$ is simply a diagonal matrix with as many 1's on the diagonal as the rank of $$\alpha$$ - but if the group is smaller I expect a more involved outcome.

• Isn't this isn't just the standard result from Lie theory? It's well-known that, in this case, $S^2(V)\simeq S^2(V^*)$ is isomorphic as an $\mathsf{Sp}(V)$-module to the adjoint representation on $\mathfrak{sp}(V)$ itself. Since this is a simple Lie algebra, the ring $R$ of polynomial invariants $p(x)$ for $x\in\mathfrak{sp}(V)$ is generated by the coefficients of the characteristic polynomial of $\mathrm{ad}(x)$. In particular, the generic element is conjugate to an element of a maximal torus, and this immediately gives that $R$ is freely generated by $n$ elements of degrees $2,4,\ldots, 2n$. – Robert Bryant Aug 4 '20 at 14:56
• @RobertBryant you're right, as usual: I forgot that $S^2V^*$ is nothing but the Lie algebra of $\mathsf{Sp}(V)$; this answers the question about the ring of polynomial invariants. Nevertheless, what interests me more is a list of normal forms: from Collingwood and McGovern's book "Nilpotent Orbits in Semisimple Lie Algebras" I've learnead that there are 8 nilpotent orbits and a three-parametric family of semi-simple ones; however, nowhere in the literature I've found them recast in terms of quadratic forms on $V$, neither (which is the true problem) any hint about the "mixed" orbits (those ... – Giovanni Moreno Sep 4 '20 at 10:47
• ... that are neither semi-simple neither nilpotent). Do you know of some reference where I can see an explicit list of such normal forms or some paper/book explaining how to classify these "mixed" orbits? – Giovanni Moreno Sep 4 '20 at 10:49

Nice question!

More generally, let $$V=\mathbb{C}^{2n}$$. Consider the $$2n\times 2n$$ matrix $$\varepsilon=\begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$$ and the symplectic group $${\mathsf{S}\mathsf{p}}_{2n}$$ which preserves the fundamental alternating bilinear form with matrix $$\varepsilon$$. An element $$F$$ of the symmetric power $$S^p(V^{\vee})$$ can be seen as a homogeneous polynomial $$F(x)$$ of degree $$p$$ in the variable $$x=(x_1,\ldots,x_{2n})$$. It also corresponds to a unique symmetric array $$(F_{i_1,\ldots,i_p})_{(i_1,\ldots,i_p)\in [2n]^p}$$ where $$[2n]$$ denotes the set of allowed index values $$\{1,2,\ldots,2n\}$$. Symmetric means the entries stay the same if one permutes the $$p$$ indices. The correspondence is so that the identity $$F(x)= F_{i_1,\ldots,i_p} x_{i_1}\cdots x_{i_p}$$ holds. Note that I used Einstein's convention where indices $$i_1,\ldots,i_p$$ are to be summed independently over the set $$[2n]$$. I will keep using this convention below.

Now for integers $$q,r,\ell$$ with $$0\le \ell\le\min(q,r)$$, one can define a "symplectic transvectant" which is a $${\mathsf{S}\mathsf{p}}_{2n}$$-equivariant map $$S^q(V^{\vee})\times S^r(V^{\vee})\rightarrow S^{q+r-2\ell}(V^{\vee})$$. To a pair of forms $$F$$, $$G$$, we associate the new form $$H(x)= F_{i_1,\ldots,i_q} G_{j_1,\ldots,j_r} \varepsilon_{i_1,j_1}\cdots \varepsilon_{i_{\ell},j_{\ell}}\ x_{i_{\ell+1}}\cdots x_{i_q}\ x_{j_{\ell+1}}\cdots x_{j_r}$$ I will write $$(F,G)_{\ell}$$ for this new form $$H$$.

Now suppose $$p$$ is even. Then for any $$m\ge \frac{p}{2}$$, one has a linear endomorphism $$\begin{array}{cccc} \mathcal{L}_{n}^{F}: & S^{m}(V^{\vee}) & \longrightarrow & S^{m}(V^{\vee}) \\ \ & G & \longmapsto & (F,G)_{\frac{p}{2}} \end{array}$$ which depends on the choice of $$F$$. Let $$\mathscr{H}_{m,s}(F)$$ denote the coefficient of $$\lambda^s$$ in essentially the characteristic polynomial $${\rm det}(Id-\lambda \mathcal{L}_{n}^{F})$$. Alternatively, let $$\mathscr{P}_{m,s}(F)$$ denote the trace of the $$s$$-th power of $$\mathcal{L}_{n}^{F}$$. It is not hard to see that $$\mathscr{H}_{m,s}(F)$$ and $$\mathscr{P}_{m,s}(F)$$ are $${\mathsf{S}\mathsf{p}}_{2n}$$-invariants of $$F$$. They give you one-dimensional submodules in $$S^{s}(S^{p}(V))$$.

The above is a trivial generalization to the symplectic context of a construction in the invariant theory of binary forms (the $${\mathsf{S}\mathsf{p}}_{2}={\mathsf{S}\mathsf{L}}_{2}$$ case) due to Hilbert in his Königsberg Habilitationsschrift. I studied these concrete invariants in my recent article

"An algebraic independence result related to a conjecture of Dixmier on binary form invariants" in Res. Math. Sci. 2019. The preprint version is here. The main result I proved in that article is that for $$n=1$$, and for $$p=2k$$ with $$k$$ even, the invariants $$\mathscr{P}_{k,2},\mathscr{P}_{k,3},\ldots,\mathscr{P}_{k,k+1}$$ are algebraically independent. Note that this trivially shows the same holds true for any $$n\ge 1$$, by specializing to a generic form $$F$$ which only depends on the variables $$x_1,x_{n+1}$$.

Note that one can also represent the invariants graphically, as in the picture

which is taken from the above article. In the left picture, the lines with arrows correspond to $$\varepsilon$$'s, and the boxes correspond to symmetrizations.

Now take $$n=3$$, $$p=2$$, $$m=\frac{p}{2}=1$$, which gives $$\mathscr{P}_{1,s}(F)={\rm tr}((\varepsilon F)^s)$$, where $$F$$ is viewed as a $$6\times 6$$ symmetric matrix. These are the invariants you see in the Lie program calculations. Clearly, they vanish unless $$s\ge 2$$ is even.

For $$p=2$$, general $$n$$. The first fundamental theorem (FFT) of invariant theory for $${\mathsf{S}\mathsf{p}}_{2n}$$ easily implies that the particular invariants $$\mathscr{P}_{1,s}$$, $$s\ge 1$$ generate the ring of invariants. Because of the relations between power sum symmetric functions, and the remark about parity, one has for this ring the list of generators $$\mathscr{P}_{1,2},\mathscr{P}_{1,4},\mathscr{P}_{1,6},\ldots,\mathscr{P}_{1,2n}.$$ They are algebraically independent. Indeed, take $$F$$ to be the quadratic form with matrix $$\begin{pmatrix} 0 & D \\ D & 0 \end{pmatrix}$$ where $$D$$ is the diagonal matrix with entries $$y_1,\ldots,y_n$$. Then the above invariants specialize to the power sums in the variables $$y_1^2,\ldots,y_n^2$$. So this gives a complete description of the ring of invariants.

For a quick sketch of a proof of the FFT for $${\mathsf{S}\mathsf{p}}_{2n}$$ see:

Invariants for the exceptional complex simple Lie algebra $F_4$

It proceeds by reduction to the FFT for $${\mathsf{S}\mathsf{L}}$$ and/or $${\mathsf{G}\mathsf{L}}$$ which are proved in

How to constructively/combinatorially prove Schur-Weyl duality?

and

How to constructively/combinatorially prove Schur-Weyl duality?

• That's one hell of an answer! Two minor questions: when you wrote "for any $m>0$" you actually meant $m\geq\tfrac{p}{2}$, right? and in the formula of the characteristic polynomial $\lambda$ should be in front of $Id$, or not? Frankly speaking, I was expecting something more digestible than Hilbert's habilitation thesis - but it's good to know where it all began: now I'll try to dig out what I need. If you have some remark about the normal form issue, I'll be happy to see it! – Giovanni Moreno Feb 12 '20 at 17:48
• 1) yes $m\ge p/2$. other wise one can define the transvectant to just be identically zero and this becomes a dissertation about zero. 2) normally $\lambda$ is by $Id$ and that's why I said "essentially", but this would just change the labeling. I want the subscript $s$ to correspond to the degree of the invariant in $F$. BTW, I just realized I had this labeling wrong in my paper for the H's, but not the P's. – Abdelmalek Abdesselam Feb 12 '20 at 18:08