How to use tensor products and evaluation maps in representation theory

Quick description

Evaluation maps provide a tool for understanding how one representation (of a group, say, or of a ring) can appear in another. The basic idea is as follows: suppose that $U$ and $W$ are representations of a group $G$ (over a field $k$ , say). How could we tell if $U$ appears as a subrepresentation of $W$ ? Well, if it does, there will certainly be a $k$ -linear $G$ -equivariant homomorphism $U \rightarrow W$ (namely, the inclusion of $U$ as a subrepresentation of $W$ ). So, it makes sense to look at the $k$ -vector space $\mathrm{Hom}_{k[G]}(U,W)$ of all $k$ -linear $G$ -equivariant maps from $U$ to $W$ . Any such homomorphism may be evaluated on elements of $U$ to yield elements of $W$ , and so we get a map

$U \times \mathrm{Hom}_{k[G]}(U,W) \rightarrow W;$

precisely, this map is defined by $(u,\phi) \mapsto \phi(u).$ This map is evidently $k$ -bilinear, and so tautologically gives rise to a $k$ -linear map on the tensor product

$U\otimes_k \mathrm{Hom}_{k[G]}(U,W) \rightarrow W,$

defined by $u\otimes\phi \mapsto \phi(u)$ . We call this map the evaluation map.

Note that the source and target of this map both have $G$ -actions: on the source, there is the action induced by the action on $U$ , while on the target, there is the action on $W$ . Furthermore, one checks that the evaluation map is $G$ -equivariant. Simply compute that $g(u\otimes\phi) = (g u) \otimes \phi \mapsto \phi(g u) = g \phi(u)$ . It provides a tautological measurement of the relationship between the representations $U$ and $W$ . For example, it is non-zero if and only if some non-zero quotient of $U$ embeds as a subrepresentation of $W$ . By imposing various conditions on $U$ and $W$ , one can draw various conclusions of differing degrees of strength.

One important point is that the formation of $\mathrm{Hom}_{k[G]}(U,W)$ is functorial (equivalently, natural) in $U$ and $W$ , and so if $U$ or $W$ have any extra structure, this structure is inherited by $\mathrm{Hom}_{k[G]}(U,W).$ We will see applications of this in the examples below.

Prerequisites

Basic graduate level algebra, including tensor products and representation theory.

Example 1

Before we study more general examples of evaluation maps, it will be helpful to study the structure of the source of the evaluation map.

Suppose to begin with that $U$ is a finite-dimensional irreducible $G$ -representation over an algebraically closed field $k$ . If $V$ is a $k$ -vector space, then the tensor product $U\otimes_k V$ is equipped with a $G$ -action (via the $G$ -action on the first factor), and so is naturally a $G$ -representaiton over $k$ .

What does it look like? Well, if we choose a basis ${v_i}_{i \in I}$ of $V$ , then we obtain an isomorphism $V \cong k^{\oplus I},$ and hence an isomorphism of $G$ -representations $U\otimes_k V \cong U^{\oplus I}$ . So $U\otimes_k V$ is just a direct sum of copies of $U$ . However, the basis $v_i$ of $I$ is not canonical. On the other hand, the $k$ -vector space $V$ is canonically determined by the representation $U\otimes_k V$ , as the following result shows.

Theorem 1 There is a natural isomorphism $V \cong \mathrm{Hom}_{k[G]}(U,U\otimes_k V),$ and the evaluation map

$U\otimes_k \mathrm{Hom}_{k[G]}(U,U\otimes_k V) \rightarrow U\otimes_k V$

is an isomorphism.

Proof of Theorem 1. Tensoring with $V$ induces a canonical map

$\mathrm{Hom}_{k[G]}(U,U)\otimes_k V \rightarrow \mathrm{Hom}_{k[G]}(U, U\otimes_k V),$

which we claim is an isomorphism. To check this, we can compute this map after choosing a basis $(v_i)_{i \in I}$ of $V$ . It then becomes identified with the natural map

$\mathrm{Hom}_{k[G]}(U,U)^{\oplus I} \rightarrow \mathrm{Hom}_{k[G]}(U,U^{\oplus I}),$

which is immediately seen to be an isomorphism.

Now since $k$ is algebraically closed and $U$ if finite-dimensional, Schur's lemma shows that $\mathrm{Hom}_{k[G]}(U,U) = k$ . Since $k\otimes_k V = V$ , we obtain the required isomorphism $V \cong \mathrm{Hom}_{k[G]}(U,U\otimes_k V).$ After identifying $V$ and $\mathrm{Hom}_{k[G]}(U,U\otimes_k V)$ via this isomorphism, the evaluation map becomes the identity map $U\otimes_k V = U\otimes_k V$ , and thus the evaluation map is seen to be an isomorphism, as claimed.□

We have seen that unlike the indexing set $I$ , the vector space $V$ is canonically determined by $U\otimes_k V$ . Still, it is reasonable to ask what advantage comes from describing our representation in the form $U\otimes_k V$ , rather than just explicitly as the direct sum $U^{\oplus I}$ ? Well, here is one advantage: suppose that $W$ is any subrepresentation of $U\otimes_k V$ . Then one can show that $W$ is also isomorphic to a direct sum of copies of $U$ , say $W \cong U^{\oplus J}$ . But the subset $J$ need not be in any natural way a subset of $I$ . (Just as we may not in general be able to choose the basis of a subspace of a given vector space to be a subset of some particular fixed basis of the entire space.) On the other hand, we have the following result.

Theorem 2 Any subrepresentation of $U \otimes_k V$ is of the form $U\otimes_k S$ , where $S$ is a subspace of $V$ .

Proof of Theorem 2. Let $W$ be a subrepresentation. Since (as we remarked above) $W$ is isomorphic to a direct sum of copies of $U$ , we may write $W \cong U^{\oplus J} \cong U \otimes_k k^{\oplus J}$ for some set $J$ . Thus Theorem 1 shows that the evaluation map $U\otimes_k \mathrm{Hom}_{k[G]}(U,W) \rightarrow W$ is an isomorphism. On the other hand, since $W \subset U\otimes_k V,$ we have a corresponding inclusion $\mathrm{Hom}_{k[G]}(U,W) \subset \mathrm{Hom}_{k[G]}(U,U\otimes_k V).$ Hence, if we write $= \mathrm{Hom}_{k[G]}(U,W),$ and use Theorem 1 to identify $\mathrm{Hom}_{k[G]}(U,U\otimes_k V)$ ^◊ with $U$ , we see that we may regard $S$ as a subspace of $U$ , and that the evaluation map identifies $U\otimes_k S$ with $W$ , as claimed.□

Example 2

Suppose that $V$ and $W$ are representation of the group $G$ over an algebraically closed field $k$ , with $V$ irreducible. Then $\mathrm{Hom}_{k[G]}(V,W)$ is called the multiplicity space of $V$ in $W$ . It measures how many independent copies of $V$ appears as subrepresentations of $W$ . More precisely, we have the following theorem.

Theorem 3 The evaluation map

$V \otimes_k \mathrm{Hom}_{k[G]}(V,W) \rightarrow W$

is injective, and its image is the sum of all the irreducible subrepresentations of $W$ that are isomorphic to $V$ , or equivalently, the maximal subrepresentation of $W$ that is isomorphic to a direct sum of copies of $V$ .

Proof of Theorem 3. If $T$ is any irreducible subrepresentation of $W$ isomorphic to $U$ , then there is an isomorphism $U \cong T\subset W.$ Thus the image of $U\otimes \imath$ under the evaluation map is equal to $T$ , and thus the image of the evaluation map contains the sum of all subobjects isomorphic to a direct sum of copies of $V$ . Thus it contains their sum. Denote this sum by $X$ ; it is a subrepresentation of $W$ contained in the image of the evaluation map.

On the other hand, $U\otimes_k \mathrm{Hom}_{k[G]}(U,V)$ is isomorphic to a direct sum of copies of $U$ , and so certainly its image under the evaluation map is a sum of copies of subrepresentations isomorphic to $U$ . Hence its image is contained in $X$ . Consequently, we see that the image of the evaluation map is precisely equal to $X$ . We also see that the inclusion $\mathrm{Hom}_{k[G]}(U,X) \subset \mathrm{Hom}_{k[G]}(U,V)$ is in fact an equality. Thus to complete the proof of the theorem, it suffices to show that $X$ is isomorphic to a direct sum of copies of $U$ , since Theorem 1 will then imply that the evaluation map

$U\otimes_k \mathrm{Hom}_{k[G]}(U,X) \rightarrow X$

is an isomorphism.

If we let ${T_j}_{j \in J}$ denote the collection of all subrepresentations $T$ of $V$ that are isomorphic to $U$ (thus $J$ is an index set labelling all such $T$ ), then there is a canonical surjection

$X.$

Now $\bigoplus_{j \in J} T_j \cong U^{\oplus J} \cong U\otimes_k k^{\oplus J},$ and thus Theorem 2 shows that the kernel of the above surjection has the form $U\otimes_k S$ for some subspace $S \subset k^{\oplus J}$ . Thus $X$ is isomorphic to

$(U\otimes_k k^{oplus J})/U\otimes _S \cong U\otimes_k(k^{oplus J}/S),$

and so is indeed isomorphic to a direct sum of copies of $U$ . As we already remarked, this completes the proof.□

Example 3

Suppose that $G$ and $H$ are two groups.

Theorem 4 If $U$ (respectively $V$ ) is a finite-dimensional irreducible representation of $G$ (respectively of $H$ ) over an algebraically closed field $k$ , then the tensor product $U\otimes_k V$ , with the natural $G\times H$ -action induced by the $G$ -action on $U$ and the $H$ -action on $V$ , is an irreducible $G\times H$ -representation.

Proof of Theorem 4. Suppose that $W$ is a $G\times H$ -subrepresentation of $U\otimes_k V$ . Thinking of $W$ simply as a $G$ -subrepresentation for the moment, Theorem 2 shows that $W$ has the form $U\otimes_k S$ for some $S \subset V$ . Moreover, since $W$ is in fact a $G\times H$ -subrepresentation of $U\otimes_k V$ , we see that $S$ must be an $H$ -invariant subspace of $V$ , i.e. an $H$ -subrepresentation of $V$ . Since $V$ is irreducible by assumption, we see that either $S = 0$ or $S = V$ . Thus either $W = 0$ or $W = U\otimes_k V$ , and so $U\otimes_k V$ is an irreducible $G\times H$ -representation, as claimed.□

We now prove the converse of Theorem 4.

Theorem 5 Suppose that $W$ is an irreducible finite-dimensional representation of a product of groups $G \times H$ over an algebraically closed field $k$ . Then $W$ is isomorphic to the tensor product over $k$ of an irreducible representation of $G$ and an irreducible representation of $H$ .

Proof of Theorem 5. To begin with, think of $W$ just as a $G$ -representation. It may no longer be irreducible, but we may find an irreducible $G$ -subrepresentation, say $U$ . (This is where we use finite-dimensionality.) As above, we construct the evaluation map

$U\otimes_k \mathrm{Hom}_{k[G]}(U,W)\rightarrow W.$

Now since $W$ is a $G\times H$ -representation, it has an action of $H$ that commutes with the action of $G$ . (To say that $W$ is endowed with a $G\times H$ -action is just to say that it is endowed with a $G$ -action and an $H$ -action that commute with one another.) Thus this $H$ -action on $W$ induces an $H$ -action on $\mathrm{Hom}_{k[G]}(U,W)$ , i.e. $\mathrm{Hom}_{k[G]}(U,W)$ is naturally an $H$ -representation. The source of the evaluation map is then a $G\times H$ -representation (via the $G$ -action on the first factor and the $H$ -action on the second factor).

Since $\mathrm{Hom}_{k[G]}(U,W)$ is a subspace of the space $\mathrm{Hom}_k(U,W)$ of all (not necessarily $G$ -equivariant) $k$ -linear maps from $U$ to $W$ , it is finite dimensional, and so contains an irreducible $H$ -representation, say $V$ . The tensor product $U\otimes_k V$ is a $G\times H$ -subrepresentation of $U\otimes_k \mathrm{Hom}_{k[G]}(U,W),$ and so the evaluation map restricts to give a $G\times H$ -equivariant map

$U\otimes_k V \rightarrow W.$

We claim that this map is non-zero: if $\phi \in V$ is non-zero, then it is a non-zero map from $U$ to $W$ (by definition), and thus we may find an element $u \in U$ such that $\phi(u) \neq 0$ . The evaluation map then sends $u \otimes \phi$ to $\phi(u) \neq 0,$ and so we see that indeed we have a non-zero map.

The preceding example shows that $U\otimes_k V$ is irreducible subrepresentation of $G\times H$ . Thus we have a non-zero map between irreducible subrepresentations of $G\times H$ , which must thus be an isomorphism. Since our map is a non-zero $G\times H$ -equivariant map, its image is a non-zero $G\times H$ -equivariant subrepresentation of $W$ . Since $W$ was assumed to be irreducible over $G\times H$ , it has no non-zero $G\times H$ -subrepresentations other than itself, and so our map must in fact be surjective. Since it is non-zero, its kernel is a proper $G\times H$ -submodule of $U\otimes_k V$ . As this is again an irreducible $G\times H$ -representation, the only proper subrepresentation that it contains is the zero representation. Thus our map has vanishing kernel, and so is also injective. This completes the proof.□

Remark

The proof of Theorem 5 brings out an important point, namely that if $W$ is a representation of $G\times H$ , and $U$ is a $G$ -representation, then the space $\mathrm{Hom}_{k[G]}(U,W)$ is naturally an $H$ -representation. This construction is frequently used (by making a careful choice of $W$ ) to create interesting correspondences between representations of a group $G$ and of a group $H$ .

We illustrate this remark further with some more examples.

Example 4

This article is incomplete. This example could be fleshed out, or perhaps linked to another page which does the fleshing out.

Let $= \mathrm{GL}_n(\C)$ , the matrix group of invertible $n\times n$ -matrices over $\C$ , and $H$ to be the symmetric group $\mathrm{Sym}_k,$ for some $k \geq 1$ . The group $G$ has a natural representation on $\C^n$ , namely the usual action of $n\times n$ -matrices on length $n$ column vectors. If we take $W = (k^n)^{\otimes k}$ , then permuting the factors in the tensor product gives a representation of $= \mathrm{Sym}_k$ on $W$ , commuting with the $G$ -action. Thus $W$ is a $G \times H$ -representation, and the map $U \mapsto \mathrm{Hom}_{\C[G]}(U,W)$ gives a map from (isomorphism classes of) irreducible $G$ -representations to (isomorphism classes of) $H$ -representations. It turns out that the multiplicity space $\mathrm{Hom}_{\C[G]}(U,W)$ is irreducible as an $H$ -representation (if it is non-zero), and this construction gives a bijection bewteen those irreducible representations of $G$ which embed into $W$ , and certain irreducible representations of $H$ . (If $n \geq k,$ then in fact every irreducible representation of $H$ arises in this way.) This is known as Schur-Weyl duality.

Example 5

Here are two more very briefly sketched examples, both at a much more advanced level than the rest of this article:

In the context of the Langlands program, one can take $G$ to be the finite adélic points of some reductive linear algebraic group over $\Q$ , $H$ to be an appropriately chosen Galois group, and $W$ to be a representation of $G\times H$ constructed out of the cohomology of a Shimura variety attached to $G$ . One will take $U$ to be (the finite part of) an irreducible automorphic representation of the group $G$ . The passage from $U$ to $\mathrm{Hom}_{k[G]}(U,W)$ will then give a construction of Galois representations associated to automorphic representations.

In the general theory of theta functions, one takes $G$ and $H$ to be the members of a dual reductive pair of groups, and $W$ is a certain Weil representation. The resulting passage from $G$ -representations to $H$ -representations is then referred to as the theta correspondence.

Comments

"The proof of Theorem 5

Sun, 26/04/2009 - 13:58 — Anonymous (not verified)

"The proof of Theorem 5 brings out an important point, namely that if W is a representation of G\times H, and U is an irreducible G-representation, then"

irreducible* isn't neccessary for this, is it?

You are correct, but then one

Mon, 27/04/2009 - 02:26 — emerton

You are correct, but then one probably shouldn't use the term "multiplicity space" to describe $\mathrm{Hom}_{k[G]}(U,W)$ . I will make an edit that reflects your comment.

Does Theorem 1 require that U

Tue, 26/05/2009 - 13:57 — Anonymous (not verified)

Does Theorem 1 require that U is simple? The proof (use of Schur's lemma) seems to assume this.

Yes. The preamble to the

Mon, 01/06/2009 - 03:18 — emerton

Yes. The preamble to the theorem states that $U$ is finite dimensional and irreducible. If you want to makes these conditions explicit in the statement of the theorem, feel free to add them.

Inline comments

The following comments were made inline in the article. You can click on 'view commented text' to see precisely where they were made.

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2 typos in Proof of Theorem 2

Wed, 09/09/2009 - 18:44 — maexe (not verified)

In Proof of Theorem 2:
... use Theorem 1 to identify \mathrm{Hom}_{k(G)}(U,U \otimes_k V) with V (not U),
we see that we may regard S as a subspace of V (not U).
Beyond this, a very nice article.