A little earlier, we studied invariant theory for the general linear group ${GL(V)}$ for a finite-dimensional vector space ${V}$ over ${\mathbb{C}}$. We considered the canonical representation on ${V^{\otimes p} \otimes V^{* \otimes q}}$ and studied “invariant polynomials” on this space: that is, polynomials ${P: V^{\otimes p} \otimes V^{* \otimes q} \rightarrow \mathbb{C}}$ constant on orbits. We showed that these formed a finitely generated ${\mathbb{C}}$-algebra, and indeed gave a set of generators: these were given by pairing a factor of ${V}$ with a factor of ${V^*}$ with respect to the evaluation pairing. This is not, of course, a linear map, but it is a well-defined polynomial map of ${p}$ vector and ${q}$ covector variables.

1. Introduction

Now we want to consider a more general question. Let ${G}$ be an (affine) algebraic group over ${\mathbb{C}}$, acting on the finite-dimensional vector space ${V}$. We’d like to ask what the invariant polynomials on ${V}$ are, or in other words what is ${(\mathrm{Sym} V^*)^G}$. It was a Hilbert problem to show that this “ring of invariants” is finitely generated. The general answer turns out to be no, but we will show that it is the case when ${G}$ is reductive.

What is a reductive group? For our purposes, a reductive group over ${\mathbb{C}}$ is an algebraic group ${G}$ such that the category ${\mathrm{Rep}(G)}$ of (algebraic) finite-dimensional representations is semisimple. In other words, the analog of Maschke’s theorem is true for ${G}$. The “classical groups” (the general linear, special linear, orthogonal, and symplectic groups) are all reductive. There is a geometric definition (which works in characteristic ${p}$ too), but we will just take this semisimplicity as the definition.

The semisimplicity is quite a surprising phenomenon, because the method of proof of Maschke’s theorem—the averaging process—fails for reductive groups, which are never compact in the complex topology (as then they would not be affine varieties). However, it turns out that a reductive group ${G}$ over ${\mathbb{C}}$ contains a maximal compact Lie subgroup ${K}$ (which is not algebraic, e.g. the unitary group in ${GL_n}$), and the category of algebraic representations of ${G}$ is equivalent (in the natural way) to the category of continuous representations of ${K}$. Since the category of continuous representations is always semisimple (by the same averaging idea as in Maschke’s theorem, with a Haar measure on ${K}$), ${\mathrm{Rep}(K)}$ is clearly semisimple. But this is ${\mathrm{Rep}(G)}$.

Anyway, here’s what we wish to prove:

Theorem 1 Let ${G}$ be a reductive group over ${\mathbb{C}}$ acting on the finite-dimensional vector space ${V}$. Then the algebra of invariant polynomials on ${V}$ is finitely generated. (more…)

With the semester about to start, I have been trying to catch up on more classical material. In this post, I’d like to discuss a foundational result on the ring of invariants of the general linear group acting on polynomial rings: that is, a description of generators for the ring of invariants.

1. The Aronhold method

Let ${G}$ be a group acting on a finite-dimensional vector space ${V}$ over an algebraically closed field ${k}$ of characteristic zero. We are interested in studying the invariants of the ring of polynomial functions on ${V}$. That is, we consider the algebra ${\mathrm{Sym} V^*}$, which has a natural ${G}$-action, and the subalgebra ${(\mathrm{Sym} V^*)^G}$. Clearly, we can reduce to considering homogeneous polynomials, because the action of ${G}$ on polynomials preserves degree.

Proposition 1 (Aronhold method) There is a natural ${G}$-isomorphism between homogeneous polynomial functions of degree ${m}$ on ${V}$ and symmetric, multilinear maps ${V \times \dots \times V \rightarrow k}$ (where there are ${m}$ factors).

Proof: It is clear that, given a multilinear, symmetric map ${g: V \times \dots \times V \rightarrow k}$, we can get a homogeneous polynomial of degree ${m}$ on ${V}$ via ${v \mapsto g(v, v, \dots, v)}$ by the diagonal imbedding. The inverse operation is called polarization. I don’t much feel like writing out, so here’s a hand-wavy argument.

Or we can think of it more functorially. Symmetric, multilinear maps ${V \times \dots \times V \rightarrow k}$ are the same thing as symmetric ${k}$-linear maps ${V^{\otimes m} \rightarrow k}$; these are naturally identified with maps ${\mathrm{Sym}^m V \rightarrow k}$. So what this proposition amounts to saying is that we have a natural isomorphism $\displaystyle \mathrm{Sym}^m V^* \simeq (\mathrm{Sym}^m V)^*.$

But this is eminently reasonable, since there is a functorial isomorphism ${(V^{\otimes m})^* \simeq (V^{*})^{\otimes m}}$ functorially, and replacing with the symmetric algebra can be interpreted either as taking invariants or coinvariants for the symmetric group action. Now, if we are given the ${G}$-action on ${V}$, one can check that the polarization and diagonal imbeddings are ${G}$-equivariant. $\Box$

2. Schur-Weyl duality

Let ${V}$ be a vector space. Now we take ${G = GL(V)}$ acting on a tensor power ${V^{\otimes m}}$; this is the ${m}$th tensor power of the tautological representation on ${V}$. However, we have on ${V^{\otimes m}}$ not only the natural action of ${GL(V)}$, but also the action of ${S_m}$, given by permuting the factors. These in fact commute with each other, since ${GL(V)}$ acts by operators of the form ${ A\otimes A \otimes \dots \otimes A}$ and ${S_m}$ acts by permuting the factors.

Now the representations of these two groups ${GL(V)}$ and ${S_m}$ on ${V^{\otimes m}}$ are both semisimple. For ${S_m}$, it is because the group is finite, and we can invoke Maschke’s theorem. For ${GL(V)}$, it is because the group is reductive, although we won’t need this fact. In fact, the two representations are complementary to each other in some sense.

Proposition 2 Let ${A \subset \mathrm{End}(V^{\otimes m})}$ be the algebra generated by ${GL(V)}$, and let ${B \subset \mathrm{End}(V^{\otimes m})}$ be the subalgebra generated by ${S_m}$. Then ${A, B}$ are the centralizers of each other in the endomorphism algebra. (more…)

There are quite a few more tools to tell whether a ring is Noetherian. In this post, I’ll discuss another basic tool: integrality. I’ll discuss the application to invariant theory for finite groups.

Subrings

In general, it is not true that a subring of a Noetherian ring is Noetherian. For instance, let ${A := k[X_1, X_2, \dots]}$ be the polynomial ring in infinitely many variables over a field ${k}$. Then ${A}$ is not Noetherian because of the ascending chain $\displaystyle (X_0) \subset (X_0, X_1) \subset (X_0, X_1, X_2) \subset \dots.$

However, the quotient field of ${A}$ is Noetherian. This applies to any non-Noetherian integral domain.

There are special cases where we can conclude a subring of a Noetherian ring is Noetherian.