Invariant theory for unitary groups $\mathcal{U}(n)$

Question

I'm trying to understand the invariant theory of the unitary groups $\mathcal{U}(n)$ on tensor powers of their standard representations $V^{\otimes p} \otimes (V^*)^{\otimes q}$. Let $\mathcal{U}(n)$ have the standard representation on a complex vector space $V$ of dimension $n$ and inner product; it acts on $V$ by multiplication by $U \in \mathcal{U}(n)$ and on $V^*$ by multiplication by $U^\dagger$. We call a function on a vector space a polynomial function if it is a member of the ring of functions generated by the dual space. A polynomial map is one $$ f:V^{\otimes p} \otimes (V^*)^{\otimes q} \to V^{\otimes r} \otimes (V^*)^{\otimes s} $$ where the components of the codomain are polynomial functions. A concomitant of $\mathcal{U}(n)$ is a polynomial map such that $$ f(U \cdot w_1,\ldots, U \cdot w_p, U \cdot v_1,\ldots, U \cdot v_q) = (\underbrace{U \otimes \cdots \otimes U}_{r+s}) \cdot f(w_1,\ldots, w_p, v_1,\ldots, v_q) $$ meaning the action commutes with $f$. If $r=s=0$ then $\mathcal{U}(n)$ acts trivially on $\mathbb{C}$ and $f$ is a $\mathcal{U}(n)$-invariant polynomial function.

Question: What are the $\mathcal{U}(n)$ concomitants for a given $p,q,r,s,n$?

It seems like the answer should be well known. I think there is an answer in Procesi's The invariant theory of n × n matrices, in particular they focus on invariant of matrix tuples ($p=q$ generic matrices, $r=s=0$) where the polynomial functions are generated by functions of the form $$ X_1,\ldots,X_p \mapsto \text{Tr}( X_{i_1}^{\varepsilon_1} X_{i_2}^{\varepsilon_2} \cdots X_{i_k}^{\varepsilon_k}) $$ for $\varepsilon_i \in \{1,\dagger\}$. This is very elegant and independent of $n$, but it is also a little confusing. For instance, $\text{Tr}(X^\dagger) = \sum_{i=1}^n \overline{[X]}_{ii}$ is not strictly speaking an element of $\mathbb{C}[ (x_{ij})_{i,j=1}^n ]$ because it involves conjugation, is it implied that the base field is somehow $\mathbb{R}$?

My suspicion is that the $p=q$, $r=s$ "even" concomitants can be described by Jones' planar algebras, but I'm surprised I cannot find this mentioned explicitly in the papers I'm reading on them. But there are also many other cases, for instance the map $(A,b) \mapsto Ab$ for a matrix $A \in V\otimes V^*$ and vector $b \in V$ is a simple concomitant for $p=2, q=1, r=1, s=0$ that occurs in the "odd" case. Is there a diagrammatic way to understand concomitants for all $p,q,r,s$ independent of $n$?

Answer 1

A $\mathcal U(n)$-equivariant map of (finite-dimensional) representations $X \to Y$ is the same as a $\mathcal U(n)$-invariant of $X^* \otimes Y$, or equivalently a $\mathcal U(n)$-invariant function $X \otimes Y^* \to \mathbb C$. Note that

$$ (V^{\otimes p} \otimes (V^*)^{\otimes q})^* \otimes V^{\otimes r} \otimes (V^*)^{\otimes s} \cong V^{\otimes (r + q)} \otimes (V^*)^{\otimes (s+p)}.$$

Weyl's fundamental theorem of invariant theory says that invariant linear functions on $V^{\otimes k} \otimes (V^*)^{\otimes \ell}$ exist if and only if $k=\ell$, and those that exist are generated by the canonical pairing $V^* \otimes V \to \mathbb C$. You can check that taking the trace of a product of matrices is a product of these canonical pairings.

For your example of $(A,b) \mapsto Ab$, you can check that under the isomorphism of $V \otimes V^*$ with matrices that the map $(V \otimes V^*) \otimes V \to V$ is exactly the canonical pairing in the second and third arguments multiplied by the identity $V \to V$.

Note that for $x \in \mathcal U(n)$, $x^\dagger = x^{-1}$, so the entries of $x^\dagger$ are polynomials in the entries of $x$ and $1/\det(x)$.