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GL-representations, symmetric polynomials, and geometry


One of the many applications of symmetric polynomials is to representation theory, and in this post I want to begin sketching out how.

Symmetric polynomials and the ring \Lambda are involved in the representation theory of the symmetric group S_n, and the general linear group GL_n, in related ways. The precise relationship between the representation theory of these two groups is spelled out in the Schur-Weyl Duality theorem, as well as in explicit constructions of representations of both groups.

I’m mainly interested in the Schur functors, which are representations of GL_n, so I’ll be focusing on those.

GL representations and symmetric polynomials

Let’s first see how GL_n representations are related to symmetric functions. Suppose

\displaystyle{\rho : GL_n \to GL_m}

is a continuous representation, and let \chi be the trace. (I’ll only be interested in continuous, and soon enough, smooth representations.)

If g \in GL_n is a diagonal matrix, then certainly \chi(g) is a symmetric function of the entries of g, since these can be rearranged by conjugation. More generally, if g is diagonalizable, then by conjugation-invariance again, \chi(g) is a symmetric function of the eigenvalues of g. But diagonalizable matrices are dense in GL_n, so the same holds for all elements g. (This is why continuity is important.)

Next, let’s restrict to algebraic representations. Then the function \chi(g) must be a (symmetric) polynomial in the eigenvalues! In fact, we’ll see later that it’s enough to restrict to smooth representations — all of them are rational, and most are polynomial.

In any case, the ring operations on symmetric polynomials are obviously compatible with direct sums and tensor products of representations, so we have a map

\displaystyle{R(GL_n) \to \Lambda_n,}

where R(GL_n) is the representation ring and \Lambda_n = k[x_1, \ldots, x_n]^{S_n}. Note that the latter is a polynomial ring on e_1, \ldots, e_n with no relations (it’s isomorphic to \Lambda / (e_{n+1}, e_{n+2}, \ldots).)

A few important geometric facts about smooth representations of GL_n:

Fact. Every smooth representation of GL_n is semisimple, that is, splits as a direct sum of irreducible representations.

Fact. A smooth representation of GL_n is determined up to isomorphism by its character.

The former follows by relating GL_n to the compact subgroup of unitary matrices U_n, where semisimplicity is automatic. The latter follows from the theory of Lie algebras, rephrasing characters and smooth representations in terms of weights of Lie algebra representations. I’ll come back to these later.

Also, this tells us that the map R(GL_n) \to \Lambda_n is injective.

Some Examples

Let’s see some examples of GL representations. At this point, I’ll write things in terms of GL(V) for some vector space V, rather than GL_n: this has the advantage of illustrating how all of these constructions will be functorial in V.

First, there’s the tautological representation V itself. The trace is just “the” trace,

\displaystyle{\chi(g) = x_1 + \cdots + x_n,}

which we have variously referred to as the polynomial m_1, e_1, h_1, s_1 and p_1.

Next, we have the exterior powers \bigwedge^k V. Recall that if v_1, \ldots, v_n is a basis for V, then a basis for the exterior power is given by

\displaystyle{v_{i_1} \wedge \cdots \wedge v_{i_k},}

running over all choices of size-k subsets I = \{i_1, \ldots, i_k\}. If g is diagonal with entries x_1, \ldots, x_n, then each of the above basis elements is already an eigenvector, with eigenvalue given by the monomial x^I. Hence, the trace is

\displaystyle{\chi(g) = \sum_{|I|=k} x^I = e_k,}

the elementary symmetric function! In particular, this shows that the ring map R(GL_n) \to \Lambda_n is surjective, hence an isomorphism. That is a bit of a surprise.

The other commonly-used GL representations are the symmetric powers S^k(V). Here, a basis is given by all symmetric products

\displaystyle{v_{i_1} \cdots v_{i_k},}

running over all choices of k indices with repetition. By similar reasoning to the above, the trace is

\displaystyle{\chi(g) = \sum \text{monomials of total degree } k = h_k,}

the homogeneous symmetric polynomial.

The Schur Functors

Here’s an example of a representation that doesn’t fall into either of the above categories. Let

\displaystyle{W = \text{Span}\big\{v_1 \otimes (v_2 \wedge v_3)  : v_1 \in \text{Span}(v_2,v_3) \big\} \subset V \otimes \bigwedge\ ^2 V.}

In other words, thinking of a pure wedge v_2 \wedge v_3 as a 2-plane, W is spanned by “flag tensors” of lines contained in 2-planes.

This is a valid (nonzero) subrepresentation of V \otimes \bigwedge^2 V, since the definition is GL(V)-invariant. It’s also clearly functorial in V, so it is a “natural” representation. That said, none of the following questions have obvious answers:

  • Is W proper, or is it the entirety of V \otimes \bigwedge^2 V?
  • how can we find a basis for W?
  • what is the trace of W?

The first answer has a simple enough answer:

Proposition. The representation W is proper.

Proof. Suppose v_1 \otimes v_2 \wedge v_3 is a generator, and v_1 = a v_2 + b v_3, and without loss of generality suppose a \ne 0. Then, by various linearity properties,

\begin{array}{rl}  v_1 \otimes v_2 \wedge v_3 & = v_1 \otimes (\tfrac{1}{a}v_1 - \tfrac{b}{a} v_3) \wedge v_3 \\  &= \tfrac{1}{a}v_1 \otimes v_1 \wedge v_3 \\  &= v_2 \otimes v_1 \wedge v_3 \ +\ v_3 \otimes v_1 \wedge \tfrac{b}{a} v_3 \\  &= v_2 \otimes v_1 \wedge v_3 \ +\ v_3 \otimes v_1 \wedge (\tfrac{1}{a} v_1 - v_2) \\  &= v_2 \otimes v_1 \wedge v_3 \ +\ v_3 \otimes v_1 \wedge v_2.  \end{array}

Of course, the same thing holds if instead b \ne 0, and therefore holds on all of W. But, of course, the relation

v_1 \otimes v_2 \wedge v_3 = v_2 \otimes v_1 \wedge v_3 \ +\ v_3 \otimes v_1 \wedge v_2

does not hold on all of V \otimes \bigwedge^2 V, where a basis is given by all choices of v_i \otimes v_j \wedge v_k.

So this is something different. It’s called the (2,1) Schur functor, or \mathbb{S}^{2,1}(V) for short, and the relation above is called an exchange relation, since it only involved permuting the vectors v_i.

Similar constructions for arbitrary “flag tensors” will give rise to a range of other new representations, such as

\mathbb{S}^{5,4,1}(V) \subset V \otimes \bigwedge\ ^4(V) \otimes \bigwedge\ ^5(V).

Of course, these are indexed by partitions, but so far only partitions with all distinct parts have shown up. Repeated parts will be slightly more subtle: if \lambda = (3,3,1), for example, we’ll take the second symmetric power of the 3 part:

\mathbb{S}^{3,3}(V) \subset V \otimes S^2(\bigwedge\ ^3(V)).

These are all, it turns out, new.

A Geometric Perspective

My choice of terminology, “flag tensors”, was deliberate. What’s really going on here, invoking some algebraic geometry, is that the flag manifold Fl(V) has a canonical Plücker embedding

\displaystyle{Fl(V) \to \prod_{k=1}^n Gr(k,V) \to \prod_{k=1}^n \mathbb{P}(\bigwedge^k (V)).}

This embedding gives us a multihomogeneous coordinate ring, which is multigraded (from each of the factors in the product). Moreover, the multigraded components of this ring really are “flag tensors” in the form I described above. In other words:

Theorem. The Schur functors are the multigraded components of the coordinate ring of the (complete) flag variety in its (product) Plücker embedding.

That’s enough for now. Next post, I’ll talk further about the actual construction of the Schur functors.

(Note: I’m being slightly non-canonical at the end here, since I described the Schur functors above as subrepresentations, but the multigraded components of the coordinate ring are quotient representations, since the ring is a quotient of the multigraded coordinate ring of the ambient projective space. To fix this, I should have used the dual flag variety Fl(V^*).)



  1. […] GL-representations, symmetric polynomials, and geometry […]

  2. mathpanda says:

    In the proof of “Proposition. The representation W is proper.” should the last equation contains a minus -?

  3. Oops, you are right! Thanks for catching that.

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