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.

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    $\begingroup$ 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 $2,4,\ldots, 2n$. $\endgroup$ – Robert Bryant Aug 4 '20 at 14:56
  • $\begingroup$ @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 ... $\endgroup$ – Giovanni Moreno Sep 4 '20 at 10:47
  • $\begingroup$ ... 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? $\endgroup$ – 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

enter image description here

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?


How to constructively/combinatorially prove Schur-Weyl duality?

  • $\begingroup$ 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! $\endgroup$ – Giovanni Moreno Feb 12 '20 at 17:48
  • $\begingroup$ 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. $\endgroup$ – Abdelmalek Abdesselam Feb 12 '20 at 18:08

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