If is a finite-dimensional complex vector space, then the symmetric group naturally acts on the tensor power by permuting the factors. This action of commutes with the action of , so all permutations are morphisms of -representations. This defines a morphism , and a natural question to ask is whether this map is surjective.

Part of Schur-Weyl duality asserts that the answer is yes. The double commutant theorem plays an important role in the proof and also highlights an important corollary, namely that admits a canonical decomposition

where runs over partitions, are some irreducible representations of , and are the Specht modules, which describe all irreducible representations of . This gives a fundamental relationship between the representation theories of the general linear and symmetric groups; in particular, the assignment can be upgraded to a functor called a Schur functor, generalizing the construction of the exterior and symmetric products.

The proof below is more or less from Etingof’s notes on representation theory (Section 4.18). We will prove four versions of Schur-Weyl duality involving , and (in the special case that is a complex inner product space) .

Throughout this post, denotes a finite-dimensional vector space over a field of characteristic .

**Schur-Weyl duality for **

Before proving Schur-Weyl duality for the Lie group , we first prove it for the Lie algebra . First, recall that if are representations of a Lie algebra , then their tensor product is also a representation of where acts by the Leibniz rule:

.

This ensures that the exponential acts the way an element of a group acts on a tensor product:

.

Second, we will need some preparatory lemmas.

**Lemma 1:** Suppose that for all , contains the element

where are vectors in some vector space containing . Then contains .

*Proof.* We proceed by induction on the degree . The result is clear for . In general, setting we conclude that contains . It follows that contains

for all . But over an infinite field, any value of a polynomial of degree is a linear combination of values that the polynomial attains at any distinct points by Lagrange interpolation. It follows that contains for all , which is a polynomial of degree , and the conclusion follows by induction.

**Lemma 2:** The symmetric power is spanned by elements of the form .

*Proof.* Let be the subspace of spanned by elements of the form and let be a basis of . Then contains the elements

for all choices of scalars . Applying Lemma 1 for each variable , it follows that contains for all choices of such that , and these form a basis of . (We need the hypothesis that has characteristic zero or else some of the coefficients above may vanish.)

**Lemma 3:** Let be a finite-dimensional algebra over a field of characteristic . Then the invariant subalgebra is generated as an algebra by the elements

.

*Proof.* By the fundamental theorem of symmetric functions, the elementary symmetric function is some polynomial in the power symmetric functions as ranges from to . We conclude that

is a polynomial in the elements

.

Over a field of characteristic we can freely identify with , and then the conclusion follows by Lemma 2.

We are now ready to prove the theorem.

**Theorem (Schur-Weyl duality):** The natural map is surjective.

*Proof.* Elements act on by the Leibniz rule as described above:

.

Let be the subalgebra of spanned by operators like the above (equivalently, the image of the universal enveloping algebra in . We want to show that the commutant of is the image of in . By Maschke’s theorem, is a finite-dimensional semisimple -module, so by the double commutant theorem it suffices to show instead that the commutant of the image of is .

The commutant of the image of in is precisely the invariant subalgebra

with respect to the natural action of on . By Lemma 3, is generated by elements of the form

and the conclusion follows.

**Corollary (existence of Schur functors):** Let be a partition describing an irreducible representation of the symmetric group . Then there is a functor such that is either an irreducible representation of or zero.

To make sense of this result it is not necessary to understand in detail the classification of the irreducible representations of , but it is helpful to know that they can all be realized over , and consequently the description of the Schur functors above does not depend on the base field .

*Example.* If is the trivial representation of , then is the space of symmetric tensors . In characteristic this functor is naturally isomorphic to the symmetric power functor (which is constructed as a quotient rather than a subspace of ), but in positive characteristic the two are not isomorphic as -representations.

*Example.* If is the sign representation of , then is the space of antisymmetric tensors . In characteristic this functor is naturally isomorphic to the exterior power functor (which is constructed as a quotient rather than a subspace of ), but in positive characteristic the definitions of symmetric and antisymmetric tensor coincide, although there is still a usable salvage of the exterior power functor.

**Schur-Weyl duality for **

**Theorem:** The subalgebra of spanned by elements of is precisely the subalgebra of spanned by elements of .

*Proof.* Since the subalgebra spanned by is contained in the commutant of the image of , by Schur-Weyl duality for it is contained in the subalgebra spanned by . To show the reverse inclusion, observe that if then is in the subalgebra spanned by for all but finitely many . But by Lagrange interpolation it must in fact lie in this subalgebra for all , in particular for . The result then follows from Lemma 2.

**Corollary (Schur-Weyl duality):** The natural map is surjective.

**Corollary:** is either an irreducible representation of or zero.

**Schur-Weyl duality for **

We now take and equip with an inner product. This allows us to replace with the unitary Lie algebra

of skew-adjoint operators because we have a natural isomorphism . More explicitly, are precisely the self-adjoint operators, and any element is uniquely a linear combination of a skew-adjoint and a self-adjoint operator, namely

.

In other words, is a real form of . It is not the only real form; if , then is of course also a real form.

In any case, it follows that the span of the image of in the space of endomorphisms of any representation functorially constructed from is the same as the span of the image of .

**Corollary (Schur-Weyl duality):** The natural map is surjective.

**Corollary:** is either an irreducible representation of or zero.

**Schur-Weyl duality for **

As for the general linear group, we can now lift the previous result to the unitary group .

**Theorem:** The subalgebra of spanned by elements of is precisely the subalgebra of spanned by elements of .

*Proof.* Given , we observed previously that the exponential of the action of on is the action of the exponential of on . Since the exponential map is surjective, this establishes one inclusion. To establish the other, given we can differentiate the action of to obtain the action of .

**Corollary (Schur-Weyl duality):** The natural map is surjective.

**Corollary:** is either an irreducible representation of or zero.

on September 15, 2019 at 12:45 pm |AritraIn Lemma 2, what is the assumption on dimension of $V$? Is it same as the $n$ of the symmetric power taken?

on September 22, 2019 at 7:16 pm |Qiaochu YuanThere’s no assumption.

on September 26, 2019 at 3:24 pm |justanothermathstudentIt’s confusing because you have taken an n-element basis of V

on September 26, 2019 at 9:37 pm |Qiaochu YuanAh, sorry, that’s a typo. Fixed. Thanks!

on January 18, 2013 at 6:08 pm |artinHow do we show the equivalence of V_\lambda and Hom_{S_n}(S^\lambda, V^{\otimes n})

on January 19, 2013 at 7:36 pm |Qiaochu YuanThis is either by definition or it follows from the discussion in the previous post on the double commutant theorem.

on January 13, 2013 at 6:52 am |timHey, i read Etingof’s notes and now your blog. It really helps me out!

I have a question regarding the map K[S_n] –> End(V^n). I think its an embedding. But Etingof states (Schur Weyl duality) that some of the representations L_lambda may be (0). On the other hand the double centralizer theorem says that this can not happen. Do you see my mistake?

Thank you

on January 13, 2013 at 3:01 pm |Qiaochu YuanIt is not an embedding if is small compared to because but . Even when it is an embedding, this doesn’t imply that every irreducible representation of occurs as a direct summand of .

on January 13, 2013 at 7:30 pm |djimNotice that the double centralizer theorem states that the bijection between irreducible representations happens between what Etingof calls A,and B ( subalgebras of End E ). Thus, in the case of Schur-Weyl duality, A becomes the image of k[Sn] in End( E ), and B = End_A ( E ) is shown to be the image of GL in End( E ) under the natural action. The point being that the double centralizer theorem tells you that there is a bijection between reps of A and B , in this case, means that there is a bijection between irreps of GL(V) appearing with non zero multiplicity in E and irreps of Sn appearing with non zero multiplicity in E. This is because irreps of the image in End ( E ) are equivalent to irreps of the group algebra that appear with non zero multiplicity in E

on January 13, 2013 at 7:34 pm |djimIn fact, you can show that as an Sn representation, the specht module S^lambda appears with nonzero multiplicity in V^tensor(n) only if lambda is a partition of n with less than or equal to dim V parts

on January 12, 2013 at 9:08 pm |ScavHi, maybe this is a silly question, but I am just learning about algebras:

Given that the group algebra of Sn is a semi-simple algebra, how do we know that the image of the group algebra in End E is also semi-simple? ( So we can use the double centralizer theorem )

Thank you

on January 12, 2013 at 9:13 pm |Qiaochu YuanAny quotient of a semisimple ring is semisimple.

on January 1, 2013 at 8:24 am |harrypHi,

i really like the way lemma 2 works. I already read your reply to djim, but still have a question about lemma 1.

First of all the elements v1,…,vd should come from a space W containing V, right? My big problem in understanding is, that i dont see how the Lagrange interpolation works in this special case. Why is the constructed polynomial equal to

f(t)-f(0)/ t for all t in K? (not only the distinct points). Dont you need kind “Identity theorem” for elements of V[X]?

Thank you

on January 1, 2013 at 1:23 pm |Qiaochu YuanAh. Yes, as currently written Lemma 1 is vacuous. I’ll fix that.

Yes, I do need an identity theorem, and I’m proving it. The point is that by Lagrange interpolation, I only need to show that two polynomials of degree agree at points to show that they agree identically (the usual argument by factorization has to be applied to each component of these polynomials separately because they are really vector-valued polynomial functions, but the Lagrange interpolation argument applies without a need to break into components).

on January 1, 2013 at 3:06 pm |harrypthank you!

on December 23, 2012 at 6:51 am |djimI see it now, thank you.

on December 16, 2012 at 9:55 pm |djimHello, thank you for the post. I am reading it alongside the section in Etingof.

I would like to know if there is a good “intrisinic” reason why it is a good idea to study Schur-Weyl duality for the lie algebra gl(V), and then deducing it for GL(V) afterwards? ( other than the fact that this way makes a nice proof )

Thank you

on December 16, 2012 at 10:03 pm |Qiaochu YuanI don’t have much to say beyond what’s in the proof itself.

on December 22, 2012 at 3:54 pm |djimThanks. Sorry, but i have one other question. In the proof of the Schur-Weyl Duality for GL(V), I am not sure how Lagrange interpolation implies that the polynomial lies in the subalgebra for all t. Could you elaborate?

Thank you

on December 22, 2012 at 3:59 pm |Qiaochu YuanIt’s essentially the same argument as in the other use of Lagrange interpolation. You have a subspace of a vector space and a polynomial which takes values in that subspace for all but finitely many values of the input . But by Lagrange interpolation the values a polynomial takes at arbitrary points are linear combinations of the values it takes at any given set of points.

on November 14, 2012 at 6:05 am |Allen KnutsonSome more things that come in this package:

1) For any GL(n)-rep V_lambda, the (1,1,1,…,1) weight space carries an action of S_n. Unfortunately this space is usually zero. It’s nonzero exactly if |lambda|=n, and then it’s the Specht module.

2) Instead of looking at just V^@n, where V is k-dimensional, look at the space of functions on kxn matrices. This carries a GL(k)xGL(n) action, and a canonical decomposition just like the one you mentioned at the beginning, now a sum over all partitions with at most min(k,n) rows. In the case k=n, you can take the larger space of functions on invertible nxn matrices, and this decomposition into V @ V^* is the Peter-Weyl theorem.

1+2) Inside each V^*, pick out the Specht module as in (1). This recovers your decomposition as a tiny piece.