The purpose of this post is to collect some standard results on duality and symmetric and exterior powers for representations of general and special linear groups. As a warm-up, here is a small true/false quiz with some, I think, rather unintuitive answers. Let be the algebraic closure of and let be the natural representation of .
- is an irreducible representation of .
- The restriction of to is self-dual.
- is self-dual.
- The restriction of to is self-dual.
- The restriction of to is self-dual.
- The restriction of to is self-dual, reducible and semisimple.
Answers, all of which should be clear by the end of this post: 1. False, 2. False, 3. False (but true if you interpreted duality as contragredient duality for algebraic groups), 4. True, 5. False, 6. True.
For deeper results and more of the general theory from coalgebras and algebraic groups, see Green, Polynomial representations of , Springer Lecture Notes in Mathematics 830. These notes of mine might also be of interest.
Let be an infinite field. Fix . Let for be commuting indeterminates and let be the polynomial ring they generate. Given and a matrix , let denote evaluated at the matrix coefficients for . A representation of the general linear group is said to be polynomial of degree is there exist polynomials for such that for all matrices .
The natural left -module is a polynomial representation of of dimension and degree , in which for . The symmetric power (defined as a quotient of ) is a polynomial representation of degree . For example if then is the polynomial representation of degree in which
As a check on notational conventions, observe that the first column of the matrix on the right-hand side records the image
of . Hence , and . Note that since is infinite, the degree of the matrix coefficients on the right-hand side is unambiguously two. This fails over finite fields: in fact, by polynomial interpolation, any function is a polynomial, of degree at most . Another difference between the infinite algebraic group and the finite group arises when we define duality.
Duality and symmetric powers
Mimicking the definition from finite groups, we would define the dual of a representation to be the vector space dual with the action
Let be a basis for , with dual basis . The calculation
shows that , and so (carefully remembering the convention for matrices acting on column vectors), we have . The inverse and transpose combine to make this a well-defined representation, but except when is a direct sum of trivial representations, each of degree , it is not polynomial.
Example 1 In the basis , , of , we have
where denotes the determinant. The top-left coefficient is , and this is not a polynomial in . We shall see later that when this representation is restricted to the special linear group , duality is much better behaved.
Because conventional duality takes us out of the category of polynomial representations, a different definition of duality is used. We define the contragredient dual of a polynomial representation of to be the vector space dual with the action
for and . A very similar calculation to the above shows that . Note that typically, the two transposes do not cancel. For instance,
is the matrix for but with and doubled and and halved.
To make this construction more explicit, we shall prove that , where
is the symmetric square, but now constructed as a submodule of . (The formal definition is below.) Indeed, with respect to the indicated basis of , the action is
exactly as for . For instance, the first column is now given by the image of , namely
If then , via the isomorphism defined by
(Correspondingly, changing the basis of to double the third column and halves the third row of a matrix , giving the matrix .) But if then the representations are non-isomorphic. Indeed, from the matrices above, it’s clear that has as a subrepresentation, and not hard to see that in fact this is its unique non-trivial proper subrepresentation. The quotient is the one-dimensional determinant representation. (Some calculation is required: the warning example for in Example 11 below shows that uniqueness fails if is replaced with .) Similarly, has as its unique non-trivial proper subrepresentation. Thus, as expected from duality, the head of is isomorphic to the socle of , both being the determinant representation.
More generally, has as a subrepresentation, where ; since acts transitively on the set , is irreducible. Whenever it is a proper subrepresentation of . For example, if and then is spanned by and . (This indicates the general proof.)
Returning to contragredient duality, it is a good exercise (performed in Proposition 5 below) to show that, generalizing our running example, . For this it is important to be clear on the definition of . We define it using the right action of the symmetric group by place permutation on . This is specified on basis elements by
(The inverse is correct for a right action of the symmetric group where permutations are composed by multiplying from left to right: it ensures that the basis element in position on the left-hand side appears in position on the right-hand side.)
Definition 2. Let . We call the elements of multi-indices. Given we define
where the sum is over a set of coset representatives for the stabiliser of in the place permutation action. Then
. For instance if and then
Suitable coset representatives for the stabiliser of in are , and .
Example 3. As an example of the multi-index notation, note that if and then
Summing over coset representatives in Definition 2 ensures that each summand in is a different rearrangement of and so, no matter what the characteristic, is the subspace of of fixed points for the place permutation action of . By this observation and the following lemma, whose proof is left as an exercise, is a subrepresentation of .
Lemma 4. The right action of on commutes with the left action of .
Proposition 5. .
Proof. Fix and . By the remark about rearrangements above, , where, as a standing convention, the sum is over all distinct rearrangements of . By Example 3 we have
The right-hand side is known to be in . Looking at the coefficient of where we see that the coefficient of in the right hand side is
We now turn to . Given define
Observe that the for are a basis of . Let denote the corresponding element of the dual basis of . We have
Since , the coefficient of in the image of is
Noting the transpose in , this agrees with the coefficient of in the image of . Therefore the map defines a isomorphism of representations .
The proposition is an interesting example of a `self-generalizing’ result. To be entirely honest, some human intervention is required, but it is essentially a matter of book-keeping.
Let be a representation of . Then
Proof. Let . By Proposition 5, taking , there is a matrix such that
for all matrices . Hence, setting , we have
for all matrices . In the representation the matrix representing is, by definition, , while in the representation , the matrix representing is, again by definition, . By the previous displayed equation, these representing matrices are conjugate by the fixed matrix . .
Perhaps surprisingly, copying the construction of for exterior powers does not give a new representation. Let . We first find the matrices for .
Lemma 7. Let . Let be the matrix representing in its action on , with respect to the basis of all where . Then
Proof. We have
Given , let
Observe that if the multi-index has two equal entries. (This holds even if .) Let be the subspace of with basis all for such that . It follows from Lemma 4 that is a subrepresentation of . Let . By Example 3,
The right-hand side is known to be in . The coefficient of where in the right-hand side is
Comparing with Lemma 7, we see that the matrices are the same. Therefore .
Similarly one can show that . For a more conceptual proof of this, one could argue that since is generated by the highest weight vector , of weight , it is irreducible; over any field the irreducible polynomial representations of are self-dual.
Special linear groups
We saw above that the contragredient duality used for representations of the infinite algebraic group does not, in general, agree with duality as defined for representations of finite groups. There is an interesting exception to this which occurs when and we restrict to the action of . Write for an isomorphism that holds after this restriction. Here it is in our running example.
Example 8. Let . We saw in Example 1 that
If then . Changing the basis of by swapping the first and second basis vectors and negating the third we get an isomorphic representation in which
This agrees with the matrix representing in its action on . Therefore .
More generally we have the following proposition.
Proposition 9. Let be a representation of . Then .
Proof. Let . By the following calculation
for any matrix , the matrices and are conjugate by the fixed matrix . Hence so are and , and since transposition preserves conjugacy, so are and . The proposition follows. .
If has characteristic zero then it is known that
Example 11. We shall prove that when , this isomorphism holds for arbitrary , provided that one side is replaced with its contragredient dual. By Lemma 7, given a matrix acting on a basis , the matrix representing in its action on with respect to the indicated basis is where
Applying this result when acts on with respect to the basis , and setting and , we find that
Now use that for matrices in and change basis by negating the third basis element to get the conjugate matrix on the left below
The matrix on the right is its conjugate by a permutation of the basis. We saw above in the example of contragredient duality that
The matrices agree and so . This isomorphism can be recast in various ways, for example, or .
Let . By Proposition 9 any isomorphism of -representations restricts to an isomorphism of -representations, provided we systematically replace contragredient duality with conventional duality . For instance, Example 10 implies that
Here is a warning example showing that some features of the infinite case are not preserved by restriction.
Example 11. Let . We have
and, in the usual basis of ,
Observe that the sums of the rows of both -matrices are constantly . Therefore in the alternative basis we have
Hence is reducible and semisimple, as claimed in the answer to the final quiz question.
To end here is a problem combining ordinary and contragredient duality, to which I do not immediately have a complete solution.
Problem 12. Let be a finite subgroup of and let be a representation of . Let be the representation defined by . What is the relationship between , and ?
For instance, if is a subfield of the real numbers then there is an orthogonal form on invariant under . (This is part of the standard proof of Maschke’s Theorem.) Hence in this case we may assume that is a subgroup of the orthogonal group, and so for all . It follows that
for all , and so . The example , where is the natural representation of and , has . It is also possible to have . For example, this is clearly the case if is a finite subgroup of not contained in and is the natural -dimensional representation of .