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Let $G$ be a group and let

$\displaystyle V = \bigoplus_{n \ge 0} V_n$

be a graded representation of $G$, i.e. a functor from $G$ to the category of graded vector spaces with each piece finite-dimensional. Thus $G$ acts on each graded piece $V_i$ individually, each of which is an ordinary finite-dimensional representation. We want to define a character associated to a graded representation, but if a character is to have any hope of uniquely describing a representation it must contain information about the character on every finite-dimensional piece simultaneously. The natural definition here is the graded trace

$\displaystyle \chi_V(g) = \sum_{n \ge 0} \chi_{V_n}(g) t^n$.

In particular, the graded trace of the identity is the graded dimension or Hilbert series of $V$.

Classically a case of particular interest is when $V_n = \text{Sym}^n(W^{*})$ for some fixed representation $W$, since $V = \text{Sym}(W^{*})$ is the symmetric algebra (in particular, commutative ring) of polynomial functions on $W$ invariant under $G$. In the nicest cases (for example when $G$ is finite), $V$ is finitely generated, hence Noetherian, and $\text{Spec } V$ is a variety which describes the quotient $W/G$.

In a previous post we discussed instead the case where $V_n = (W^{*})^{\otimes n}$ for some fixed representation $W$, hence $V$ is the tensor algebra of functions on $W$. I thought it might be interesting to discuss some generalities about these graded representations, so that’s what we’ll be doing today.

Molien’s theorem and pals

Let $T$ denote a linear transformation on a finite-dimensional complex vector space $V$ of dimension $n$, and let $\lambda_1, ... \lambda_n$ denote its eigenvalues. $T$ acts diagonally on the tensor powers of $V$, hence on the symmetric powers of $V$.

Lemma: The trace of $T$ acting on $\text{Sym}^k(V)$ is given by $h_k(\lambda_1, ... \lambda_n)$ where $h_k$ is the complete homogeneous symmetric polynomial of degree $k$.

Proof. First suppose that $T$ has a full set of eigenvectors $\mathbf{v}_1, ... \mathbf{v}_n$. Then $\text{Sym}^k(V)$ is spanned by the monomials of degree $k$ in these eigenvectors. Any such monomial $\prod \mathbf{v}_i^{e_i}$ is an eigenvector for the action of $T$ on $V$ with eigenvalue $\prod \lambda_i^{e_i}$, and this gives a full set of eigenvectors for the action of $T$ on $\text{Sym}^k(T)$. The sum of these eigenvalues is $h_k(\lambda_1, ... \lambda_n)$ as desired. Since the set of $T$ with a full set of eigenvectors is dense in $\mathcal{M}_n(\mathbb{C})$, the general result follows by continuity.

As a corollary, the graded trace of $T$ acting on $\text{Sym}(V)$ is $\frac{1}{\det(\mathbf{I} - \mathbf{T} t)}$. Compare with the following result: since the trace is multiplicative under tensor product, the trace of $T$ acting on $V^{\otimes k}$ is $(\lambda_1 + ... + \lambda_n)^k$, hence the graded trace of $T$ acting on $T(V)$ is $\frac{1}{1 - (\text{tr } \mathbf{T}) t}$.

Now let $G$ be a compact group. The category of graded unitary representations of $G$ with each graded piece finite-dimensional has an internal Hom $\textbf{Hom}(V, W)$ given by the space of grade-preserving linear transformations from $V$ to $W$, where the graded piece of degree $n$ consists of the direct sum of the linear transformations from $V_i$ to $W_{n-i}, i = 0, 1, ... n$. As in the case of ordinary representations, $\textbf{Hom}(V, W)$ is canonically isomorphic to $V^{*} \otimes W$, which has graded character $\overline{\chi_V}(g) \chi_W(g)$. Again as in the case of ordinary representations, the external Hom $\text{Hom}(V, W)$, given by the space of grade-preserving linear transformations from $V$ to $W$ which respect the action of $G$, is precisely the direct sum of the copies of the trivial representation in $V^{*} \otimes W$. By inspecting each graded piece, it follows that

$\displaystyle \dim \text{Hom}(V, W) = \int_G \overline{\chi_V}(g) \chi_W(g) \, d \mu$

which exactly mimics the statement in the ordinary representation case. In particular, the graded dimension of the trivial part of a graded representation $W$ is

$\displaystyle \int_G \chi_W(g) \, d \mu$

which corresponds to the case where $V$ is the trivial representation in degree zero. As a corollary, we obtain the following results.

Theorem (Molien): The graded dimension of the trivial part of $\text{Sym}(V)$ is

$\displaystyle \int_G \frac{1}{\det(\mathbf{I} - gt)} \, d \mu$.

In particular, when $G$ is finite this is a rational function. Molien’s theorem is, as I understand it, used to help describe explicitly the structure of the ring of polynomial invariants; we’ll give some examples below. Similarly, for the tensor algebra we have the following.

Theorem: The graded dimension of the trivial part of $T(V)$ is

$\displaystyle \int_G \frac{1}{1 - \chi_V(g) t}\, d \mu$.

Examples

Let $G = SO(2), V = \mathbb{C}^2$ acting by rotation matrices. Then the graded dimension of the trivial part of $\text{Sym}(V^{*})$, hence the Hilbert series of $\mathbb{C}[x, y]^G$, is

$\displaystyle \int_0^1 \frac{1}{1 - 2t \cos 2 \pi x + t^2} \, dx.$

Using the complex-analytic technique described, for example, here, this integral evaluates to $\frac{1}{1 - t^2}$; in other words, $\mathbb{C}[x, y]^G$ is generated by $x^2 + y^2$.

The graded dimension of the trivial part of $T(V^{*})$, on the other hand, is

$\displaystyle \int_0^1 \frac{1}{1 - 2t \cos 2 \pi x} \, dx$

which, again using the complex-analytic method, evaluates to $\frac{1}{ \sqrt{1 - 4t^2} }$; this agrees with the combinatorial answer we found in the previous post by counting walks on the infinite cycle graph.

Let $G = \mathbb{Z}/n\mathbb{Z}, V = \mathbb{C}^2$ acting by rotation matrices. Then the Hilbert series of $\mathbb{C}[x, y]^G$ is

$\displaystyle \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{1 - 2t \cos \frac{2 \pi k}{n} + t^2}$.

It’s not hard to see that $x^2 + y^2, (x + yi)^n, (x - yi)^n$ are always polynomial invariants, so we expect the denominator of the above to have a $1 - t^2$ and a $1 - t^n$ in it. In fact, it’s not hard to show that

$\displaystyle \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{1 - 2t \cos \frac{2 \pi k}{n} + t^2} = \frac{1 + t^n}{(1 - t^2)(1 - t^n)}$

by comparing the residue of both sides at an $n^{th}$ root of unity and at $-1$. This will let us describe the structure of $\mathbb{C}[x, y]^{\mathbb{Z}/n\mathbb{Z}}$ as follows. The three invariants $x^2 + y^2, (x + yi)^n, (x - yi)^n$ satisfy the relation

$(x^2 + y^2)^n = (x + iy)^n (x - iy)^n$

so the ring they generate is isomorphic to $\mathbb{C}[u, v, w]/(u^n - vw)$. By always replacing $vw$ with $u^n$, every monomial in this ring is either a monomial in $u, v$ or a monomial in $u, w$, and since $u$ has degree $2$ and $v, w$ have degree $n$ the Hilbert series of this ring agrees with the Hilbert series of $\mathbb{C}[x, y]^G$; in other words, the relation we’ve identified is the only relation and

$\mathbb{C}[x, y]^{\mathbb{Z}/n\mathbb{Z}} \simeq \mathbb{C}[u, v, w]/(u^n - vw)$.

This exhibits the quotient of $\mathbb{C}^2$ by the action of $\mathbb{Z}/n\mathbb{Z}$ as a variety $\text{MaxSpec } \mathbb{C}[u, v, w]/(u^n - vw)$ embedded in $\mathbb{C}^3$.

Similarly, the graded dimension of the trivial part of $T(V^{*})$ is equal to

$\displaystyle \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{1 - 2t \cos \frac{2 \pi k}{n} }$

and it has the same interpretation as before in terms of counting walks on the cycle graph with $n$ vertices.

Isomorphisms

Since the character of a finite-dimensional unitary representation of a compact group $G$ uniquely determines it, it follows that the graded character of a graded representation also uniquely determines it. This implies that it should be possible to interpret identities between certain generating functions in terms of isomorphisms between certain graded representations.

So let $G = SU(2), V = \mathbb{C}^2$ with the defining representation yet again. Since we know the symmetric powers of $V^{*}$ are irreducible, they contain no copy of the trivial representation, so $\mathbb{C}[x, y]^G$ consists only of the constant functions. We know, however, that the graded dimension of the trivial part $W$ of $T(V^{*})$ is

$\displaystyle\int_{0}^{1} \frac{2 \sin^2 \pi x}{1 - 2t \cos \pi x} \, dx = \frac{1 - \sqrt{1 - 4t^2}}{2t^2}$

which is the generating function for the Catalan numbers. This generating function satisfies the Catalan relation $C = 1 + t^2 C^2$, which implies an isomorphism

$\displaystyle W \simeq I \oplus (T^{\otimes 2} \otimes W^{\otimes 2})$

of graded vector spaces, where $I$ denotes the trivial representation in degree zero and $T$ denotes the trivial representation in degree one. I am very curious as to whether this isomorphism can be written down explicitly (or, better, canonically); if so, it should be possible to construct a basis for $W$ which is in bijection with binary trees. Presumably one can wrestle such a basis out of the description of the graded parts of $W$ given in the last post, but it would be interesting if there were a general construction which specialized to this.