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

Proposition 1 (Aronhold method)There is a natural -isomorphism between homogeneous polynomial functions of degree on and symmetric, multilinear maps (where there are factors).

*Proof:* It is clear that, given a multilinear, symmetric map , we can get a homogeneous polynomial of degree on via 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 are the same thing as symmetric *-linear* maps ; these are naturally identified with maps . So what this proposition amounts to saying is that we have a natural isomorphism

But this is eminently reasonable, since there is a functorial isomorphism 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 -action on , one can check that the polarization and diagonal imbeddings are -equivariant.

**2. Schur-Weyl duality**

Let be a vector space. Now we take acting on a tensor power ; this is the th tensor power of the tautological representation on . However, we have on not only the natural action of , but also the action of , given by permuting the factors. These in fact commute with each other, since acts by operators of the form and acts by permuting the factors.

Now the representations of these two groups and on are both semisimple. For , it is because the group is finite, and we can invoke Maschke’s theorem. For , 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 2Let be the algebra generated by , and let be the subalgebra generated by . Then are the centralizers of each other in the endomorphism algebra.

We are primarily interested in the invariant theory of , and as a result of this, will obtain a complete description of those operators in that commute with : they are precisely the algebra spanned by . So we get a description of the invariant theory of .

*Proof:* By the “double centralizer theorem,” we need only prove that anything that commutes with all of the image of is in the span of the image of . So, let’s identify ; then we can describe a -action on that comes by conjugation.

In other words, we are putting the usual representation structure on the hom-space. However, we also have a representation of on given by interchanging the factors. A little algebra shows that these are the same action. Namely, conjugating for each by corresponds to permuting the factors.

So we need to describe the -invariants in : these are the *symmetric* tensors. In general, for any vector space , the symmetric tensors in (i.e. those invariant under the natural -action) are spanned by the tensors of the form . We’ll prove this below.

Granting this, the result follows: indeed, then the -invariants in are spanned by elements of the form for ; we can clearly take invertible (since the invertible guys are Zariski dense). In other words, we find that the -invariants are those spanned by the image of .

So we need to prove the auxiliary claim about symmetric tensors in : that they are spanned by the most obvious tensors of the form . Here we can use the Aronhold method to argue that if an element of (where can be identified with the subspace of symmetric tensors) vanished on elements of the form , then the associated polynomial function on plain old —the element in —would be trivial. This observation shows that the must be Zariski dense in, and hence span, the whole space.

**3. Invariants of **

As before, let be a finite-dimensional vector space. We have an action of on the tensor power , and a result can make into an -representation. Similarly, we can make into a -representation.

Schur-Weyl duality describes the invariants of on as those endomorphisms spanned by . So, we have our first computation of the invariants of on a space. Now we want to re-interpret it slightly: namely, we use the isomorphism of representations

Recall that if is given the dual action, then the functorial isomorphism of vector spaces is also a functorial isomorphism of representations. As a result, we can describe the -invariants of . Since this object represents maps , we can think of the invariants as invariant multilinear functions in vector variables and “covector” (dual) variables.

Namely, we just have to figure out what element of corresponds to . Then the span of these elements will be the -invariants. But one can check that is identified with the multilinear map

Here the notation is that the are vector variables, while the are covector variables. The inner product is just evaluation of a covector at a vector. This is because, for instance, the identity of is identified with the evaluation map . We have proved:

Proposition 3The space of -invariant multilinear forms of vector and covector variables is spanned by the maps above.

Note that the space of -invariant multilinear forms of vector and covector variables is trivial for . Indeed, if we had such a form, we can act by for a scalar on all variables; then the first variables are multiplied by and the last are multiplied by . Since our form is multilinear, the net action is , but it is also by invariance. This implies the form is zero.

**4. Invariant polynomial functions**

Now we wish to consider a more general question. Fix as before, take , and consider the action on . There is an induced action of on the algebra of polynomial functions on this space, i.e. polynomial functions of vector variables and covector variables.

We’d like to determine the algebra of invariants. A simple example is the following. If , then we can define a map : given vector variables, and covector variables, evaluates the th covector on the th vector. This is clearly a polynomial map on , and by definition it’s invariant. We are going to show that these generate the ring of invariants.

To do this, as before it suffices to determine what the *homogeneous* polynomials which are -invariant are. So let be a homogeneous polynomial of degree , invariant under . The first thing we can do is to polarize . As a result, we get a {symmetric} multilinear form

This is a bit confusing, so we’ll think of it as a -invariant multilinear form in a whole bunch of vector variables and a whole bunch of covector variables; we can get the original object by the diagonal embedding. Now the point is that we already know what the invariant *multilinear* maps are. Namely, these are spanned by products of inner products between vectors and covectors. So is a sum of such things. It follows that the original object —which can be obtained from via a diagonal embedding—is itself a sum of products of inner products between vectors and covectors. This means:

Theorem 4The ring of invariants of the action of on the ring of polynomial functions on is generated by evaluation maps .

In particular, it is finitely generated. Since any representation of is a direct summand of some , it follows that for any finite-dimensional representation of the general linear group, the ring of invariants of polynomial functions on is finitely generated. In fact, the ring of invariants on a polynomial ring by a reductive group is always finitely generated; this is the beginning of geometric invariant theory.

September 3, 2011 at 11:08 am

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