Invariants of $\mathrm{GL}_n$ representations $\DeclareMathOperator\GL{GL}$Let $V=\mathbb C^n$ be the natural representation of $\GL_n(\mathbb C)$ and let $W=\operatorname{Sym}^2(V)$ be the symmetric square representation. Let $W^k$ denote the direct sum of $k$-copies of $W$. A version of the first fundamental theorem gives a generating set for $\mathbb C[V^k \oplus (V^*)^k]^{\GL_n(\mathbb C)}$. Is a generating set for the ring of invariants $\mathbb C[W^k \oplus (W^*)^k]^{\GL_n(\mathbb C)}$ known?
 A: When $k=1$ the ring of invariants of degree $d$ separately in each of $W = \mathrm{Sym}^2 V$ and $W^\star = \mathrm{Sym^2} V^\star$ has dimension equal to the number of partitions of $d$ with at most $n$ parts. The full ring of invariants $\mathbb{C}[\mathrm{Sym}^2 V \oplus \mathrm{Sym}^2 V^\star]^{\mathrm{GL}_n(\mathbb{C})}$ has one generator in each of degrees $1, 2, \ldots, n$. While this may sounds like the Fundamental Theorem for $V$, one should note that the invariant ring when $k=1$ for $\mathbb{C}[V \oplus V^\star]^{\mathrm{GL}_n(\mathbb{C})}$ has a single generator in degree $1$, so the comparison is a bit misleading.
Proof. It is well known that
$$\mathrm{Sym}^d \mathrm{Sym}^2 V \cong \bigoplus_{\lambda \in \mathrm{Par}_{\le n}(d)} \nabla^{2\lambda} V$$
where $\nabla^\mu$ is the Schur functor for the partition $\mu$ and $2\lambda$ is obtained from $\lambda$ by doubling each part. To make this explicit, we take a basis $e_1, \ldots, e_n$ for $V$ and define for each $\ell \in \mathbb{N}$ a vector $W(\ell) \in \mathrm{Sym}^\ell \mathrm{Sym}^2 V$ by
$$W(\ell) = \sum_{\sigma \in S_\ell} \mathrm{sgn}(\sigma) (e_1 e_{1\sigma}) \ldots (e_\ell e_{\ell \sigma}).$$
One can show that $W(\ell)$ is a highest-weight vector generating $\nabla^{(2^{\ell})}V$: see Example 1.10 in Plethysms of symmetric functions and highest weight representations by Melanie de Boeck, Rowena Paget and me. More generally, $\nabla^{2\lambda}V$ is generated by the product of highest-weight vectors $W(\lambda_1')\ldots W(\lambda_a')$ where $\lambda_1 = a$. Now chasing through the isomorphisms in
$$\begin{aligned} \mathrm{Sym}^{2d}&(\mathrm{Sym}^2 V \oplus \mathrm{Sym}^2 V^\star)^{\mathrm{GL}_n(\mathbb{C})} \\&\cong \mathrm{Hom}_{\mathrm{GL}_n(\mathbb{C})}(\mathrm{Sym}^d (\mathrm{Sym}^2 V) \otimes \mathrm{Sym}^d (\mathrm{Sym}^2 V)^\star, \mathbb{C}) \\ 
& \cong \mathrm{Hom}_{\mathrm{GL}_n(\mathbb{C})} (\mathrm{Sym}^d \mathrm{Sym}^2 V, \mathrm{Sym}^d \mathrm{Sym}^2 V) \\
& \cong \mathrm{Hom}_{\mathrm{GL}_n(\mathbb{C})}(\bigoplus_{\lambda \in \mathrm{Par}_{\le n}(d)} \nabla^{2\lambda}V, \bigoplus_{\lambda \in \mathrm{Par}_{\le n}(d)} \nabla^{2\lambda}V) \\
& \cong \bigoplus_{\lambda \in \mathrm{Par}_{\le n}(d)} \mathrm{Hom}_{\mathrm{GL}_n(\mathbb{C})} (\nabla^{2\lambda}V, \nabla^{2\lambda}V ) 
\end{aligned}
$$
we get the dimension result. (Note in the first line we use that the degrees of an invariant with respect to $V$ and $V^\star$ must agree.) Moreover, this shows that there is an invariant in degree $d$ for each $d \le n$ defined (up to a duality) by
$$W(d) \otimes W(d)^\star \mapsto 1.$$
Since $\nabla^{(2^d)} V$ appears first in degree $d$, this invariant is not in the subalgebra of the invariant ring generated by invariants of lower degree. This gives algebraically independent generators in each degree $1,2, \ldots, n$. Finally by dimension counting, again using the isomorphism above, one sees these generate the full invariant ring. $\Box$
For general $k$ one can in principle determine the dimensions using symmetric functions calculations. Maybe the highest weight vectors $W(\ell)$ can still be used to define some of the invariants, but I'm not sure this will give a complete set.
A: In classical terms, this is about joint invariants of $k$ quadratic forms $Q^{(1)}(x),\ldots,Q^{(k)}(x)$ in point coordinates $x=(x_1,\ldots,x_n)$ as well as $k$ quadratic forms $R^{(1)}(u),\ldots,R^{(k)}(u)$ in dual coordinates $u=(u_1,\ldots,u_n)$. By the FFT for $GL_n$, these invariants are linear combinations of graphs which tell you how to build invariant polynomials by contracting indices (along edges). Here the latter are made of two types of vertices of degree two (two incident edges), ones for the $Q$'s and ones for the $R$'s. Moreover all edges are $QR$, i.e., no $QQ$ or $RR$ edges. So basically these are disjoint collections of cycles of even length made by alternating $Q$'s and $R$'s.
For a proof of the relevant FFT see my two answers to
How to constructively/combinatorially prove Schur-Weyl duality?
Another description of these invariants involves introducing the $k^2$ matrices
$A^{(a,b)}$ with matrix elements
$$
A^{(a,b)}_{ij}=\sum_{\ell=1}^{n} Q_{i\ell}^{(a)} R_{\ell j}^{(b)}\ .
$$
Then the basic invariants are traces of arbitrary products or words made of $A$ matrices.
In the $k=1$ case, there is only one matrix $A$ and the generators are simply the traces of the powers $A,A^2,\ldots,A^n$ and they are algebraically independent because one can take $R$ to be the identity, and $Q$ some generic diagonal matrix and the result follows from the algebraic independence of the elementary symmetric functions. So one recovers the result in Mark's answer.
For general $k$ and $n$, finding a minimal set of (algebra) generators and a description of their relations, sounds to me like a hopeless task. Note that one may get some information via the relation to invariants under conjugation of $k^2$ matrices. This subject has been extensively investigated in work of Procesi, Rasmyslov, Formanek, etc.
