Four flavors of Schur-Weyl duality

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Four flavors of Schur-Weyl duality

November 13, 2012 by Qiaochu Yuan

If $V$ is a finite-dimensional complex vector space, then the symmetric group $S_n$ naturally acts on the tensor power $V^{\otimes n}$ by permuting the factors. This action of $S_n$ commutes with the action of $\text{GL}(V)$ , so all permutations $\sigma : V^{\otimes n} \to V^{\otimes n}$ are morphisms of $\text{GL}(V)$ -representations. This defines a morphism $\mathbb{C}[S_n] \to \text{End}_{\text{GL}(V)}(V^{\otimes n})$ , 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 $V^{\otimes n}$ admits a canonical decomposition

$\displaystyle V^{\otimes n} = \bigoplus_{\lambda} V_{\lambda} \otimes S_{\lambda}$

where $\lambda$ runs over partitions, $V_{\lambda}$ are some irreducible representations of $\text{GL}(V)$ , and $S_{\lambda}$ are the Specht modules, which describe all irreducible representations of $S_n$ . This gives a fundamental relationship between the representation theories of the general linear and symmetric groups; in particular, the assignment $V \mapsto V_{\lambda}$ 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 $\mathfrak{gl}(V), \text{GL}(V)$ , and (in the special case that $V$ is a complex inner product space) $\mathfrak{u}(V), \text{U}(V)$ .

Throughout this post, $V$ denotes a finite-dimensional vector space over a field $k$ of characteristic $0$ .

Schur-Weyl duality for $\mathfrak{gl}(V)$

Before proving Schur-Weyl duality for the Lie group $\text{GL}(V)$ , we first prove it for the Lie algebra $\mathfrak{gl}(V)$ . First, recall that if $V, W$ are representations of a Lie algebra $\mathfrak{g}$ , then their tensor product $V \otimes W$ is also a representation of $\mathfrak{g}$ where $X \in \mathfrak{g}$ acts by the Leibniz rule:

$\displaystyle X(v \otimes w) = Xv \otimes w + v \otimes Xw$ .

This ensures that the exponential $e^{tX}$ acts the way an element of a group acts on a tensor product:

$\displaystyle e^{tX}(v \otimes w) = (e^{tX} v) \otimes (e^{tX} w)$ .

Second, we will need some preparatory lemmas.

Lemma 1: Suppose that for all $t \in k$ , $V$ contains the element

$\displaystyle f(t) = \sum_{i=0}^d v_i t^i$

where $v_i$ are vectors in some vector space containing $V$ . Then $V$ contains $v_0, v_1, ... v_n$ .

Proof. We proceed by induction on the degree $d$ . The result is clear for $d = 0$ . In general, setting $t = 0$ we conclude that $V$ contains $v_0$ . It follows that $V$ contains

$\displaystyle \frac{f(t) - f(0)}{t} = \sum_{i=1}^d v_i t^{i-1}$

for all $t \neq 0$ . But over an infinite field, any value of a polynomial of degree $d$ is a linear combination of values that the polynomial attains at any $d + 1$ distinct points by Lagrange interpolation. It follows that $V$ contains $\frac{f(t) - f(0)}{t}$ for all $t$ , which is a polynomial of degree $d - 1$ , and the conclusion follows by induction. $\Box$

Lemma 2: The symmetric power $S^n(V)$ is spanned by elements of the form $v^n, v \in V$ .

Proof. Let $W$ be the subspace of $S^n(V)$ spanned by elements of the form $v^n$ and let $v_1, ... v_d$ be a basis of $V$ . Then $W$ contains the elements

$\displaystyle \left( \sum_{i=1}^d t_i v_i \right)^n = \sum_{\sum_i m_i = n} {n \choose m_1, ... m_d} \prod_{i=1}^d t_i^{m_i} v_i^{m_i}$

for all choices of scalars $t_i \in k$ . Applying Lemma 1 for each variable $t_i$ , it follows that $W$ contains $\prod_i v_i^{m_i}$ for all choices of $m_i \in \mathbb{Z}_{\ge 0}$ such that $\sum_i m_i = n$ , and these form a basis of $S^n(V)$ . (We need the hypothesis that $k$ has characteristic zero or else some of the coefficients above may vanish.) $\Box$

Lemma 3: Let $A$ be a finite-dimensional algebra over a field $k$ of characteristic $0$ . Then the invariant subalgebra $(A^{\otimes n})^{S_n}$ is generated as an algebra by the elements

$\displaystyle \Delta_n(a) = a \otimes \text{id} \otimes ... + \text{id} \otimes a \otimes ... + ... + \text{id} \otimes \text{id} \otimes ... \otimes a$ .

Proof. By the fundamental theorem of symmetric functions, the elementary symmetric function $e_n = x_1 x_2 ... x_n$ is some polynomial in the power symmetric functions $p_k = x_1^k + ... + x_n^k$ as $k$ ranges from $1$ to $n$ . We conclude that

$a \otimes a \otimes ... \otimes a$

is a polynomial in the elements

$\displaystyle \Delta_n(a), \Delta_n(a^2), ... \Delta_n(a^n)$ .

Over a field of characteristic $0$ we can freely identify $(A^{\otimes n})^{S_n}$ with $S^n(A)$ , and then the conclusion follows by Lemma 2. $\Box$

We are now ready to prove the theorem.

Theorem (Schur-Weyl duality): The natural map $k[S_n] \to \text{End}_{\mathfrak{gl}(V)}(V^{\otimes n})$ is surjective.

Proof. Elements $X \in \mathfrak{gl}(V)$ act on $V^{\otimes n}$ by the Leibniz rule as described above:

$\displaystyle X(v_1 \otimes ... \otimes v_n) = \sum_{i=1}^n v_1 \otimes ... \otimes Xv_i \otimes ... \otimes v_n$ .

Let $T$ be the subalgebra of $\text{End}(V^{\otimes n})$ spanned by operators like the above (equivalently, the image of the universal enveloping algebra $U(\mathfrak{gl}(V))$ in $\text{End}(V^{\otimes n})$ . We want to show that the commutant of $T$ is the image of $k[S_n]$ in $\text{End}(V^{\otimes n})$ . By Maschke’s theorem, $V^{\otimes n}$ is a finite-dimensional semisimple $k[S_n]$ -module, so by the double commutant theorem it suffices to show instead that the commutant of the image of $k[S_n]$ is $T$ .

The commutant of the image of $k[S_n]$ in $\text{End}(V^{\otimes n})$ is precisely the invariant subalgebra

$\displaystyle \text{End}(V^{\otimes n})^{S_n} \cong \left( \text{End}(V)^{\otimes n} \right)^{S_n}$

with respect to the natural action of $S_n$ on $\text{End}(V^{\otimes n}) \cong \text{End}(V)^{\otimes n}$ . By Lemma 3, $(\text{End}(V)^{\otimes n})^{S_n}$ is generated by elements of the form

$\displaystyle v_1 \otimes ... \otimes v_n \mapsto \sum_{i=1}^n v_1 \otimes ... \otimes Xv_i \otimes ... \otimes v_n$

and the conclusion follows. $\Box$

Corollary (existence of Schur functors): Let $\lambda$ be a partition describing an irreducible representation $S_{\lambda}$ of the symmetric group $S_n$ . Then there is a functor $V \mapsto V_{\lambda} \cong \text{Hom}_{S_n}(S_{\lambda}, V^{\otimes n})$ such that $V_{\lambda}$ is either an irreducible representation of $\mathfrak{gl}(V)$ or zero.

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

Example. If $S_{\lambda}$ is the trivial representation of $S_n$ , then $V_{\lambda}$ is the space of symmetric tensors $\text{Sym}^n(V)$ . In characteristic $0$ this functor is naturally isomorphic to the symmetric power functor $S^n(V)$ (which is constructed as a quotient rather than a subspace of $V^{\otimes n}$ ), but in positive characteristic the two are not isomorphic as $\text{GL}(V)$ -representations.

Example. If $S_{\lambda}$ is the sign representation of $S_n$ , then $V_{\lambda}$ is the space of antisymmetric tensors $\text{Ant}^n(V)$ . In characteristic $0$ this functor is naturally isomorphic to the exterior power functor $\Lambda^n(V)$ (which is constructed as a quotient rather than a subspace of $V^{\otimes n}$ ), 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 $\text{GL}(V)$

Theorem: The subalgebra of $\text{End}(V^{\otimes n})$ spanned by elements of $\mathfrak{gl}(V)$ is precisely the subalgebra of $\text{End}(V^{\otimes n})$ spanned by elements of $\text{GL}(V)$ .

Proof. Since the subalgebra spanned by $\text{GL}(V)$ is contained in the commutant of the image of $k[S_n]$ , by Schur-Weyl duality for $\mathfrak{gl}(V)$ it is contained in the subalgebra spanned by $\mathfrak{gl}(V)$ . To show the reverse inclusion, observe that if $X \in \mathfrak{gl}(V)$ then $(t + X)^{\otimes n}$ is in the subalgebra spanned by $\text{GL}(V)$ for all but finitely many $t$ . But by Lagrange interpolation it must in fact lie in this subalgebra for all $t$ , in particular for $t = 0$ . The result then follows from Lemma 2. $\Box$

Corollary (Schur-Weyl duality): The natural map $k[S_n] \to \text{End}_{\text{GL}(V)}(V^{\otimes n})$ is surjective.

Corollary: $V_{\lambda}$ is either an irreducible representation of $\text{GL}(V)$ or zero.

Schur-Weyl duality for $\mathfrak{u}(V)$

We now take $k = \mathbb{C}$ and equip $V$ with an inner product. This allows us to replace $\mathfrak{gl}(V)$ with the unitary Lie algebra

$\displaystyle \mathfrak{u}(V) = \{ X \in \mathfrak{gl}(V) : X^{\dagger} = -X \}$

of skew-adjoint operators because we have a natural isomorphism $\mathfrak{u}(V) \otimes \mathbb{C} \cong \mathfrak{gl}(V)$ . More explicitly, $iX, X \in \mathfrak{u}(V)$ are precisely the self-adjoint operators, and any element is uniquely a linear combination of a skew-adjoint and a self-adjoint operator, namely

$\displaystyle X = \frac{X + X^{\dagger}}{2} + \frac{X - X^{\dagger}}{2}$ .

In other words, $\mathfrak{u}(V)$ is a real form of $\mathfrak{gl}(V)$ . It is not the only real form; if $\dim V = n$ , then $\mathfrak{gl}_n(\mathbb{R})$ is of course also a real form.

In any case, it follows that the span of the image of $\mathfrak{gl}(V)$ in the space of endomorphisms of any representation functorially constructed from $V$ is the same as the span of the image of $\mathfrak{u}(V)$ .

Corollary (Schur-Weyl duality): The natural map $\mathbb{C}[S_n] \to \text{End}_{\mathfrak{u}(V)}(V^{\otimes n})$ is surjective.

Corollary: $V_{\lambda}$ is either an irreducible representation of $\mathfrak{u}(V)$ or zero.

Schur-Weyl duality for $\text{U}(V)$

As for the general linear group, we can now lift the previous result to the unitary group $\text{U}(V)$ .

Theorem: The subalgebra of $\text{End}(V^{\otimes n})$ spanned by elements of $\mathfrak{u}(V)$ is precisely the subalgebra of $\text{End}(V^{\otimes n})$ spanned by elements of $\text{U}(V)$ .

Proof. Given $X \in \mathfrak{u}(V)$ , we observed previously that the exponential of the action of $X$ on $V^{\otimes n}$ is the action of the exponential of $X$ on $V^{\otimes n}$ . Since the exponential map $\mathfrak{u}(V) \to \text{U}(V)$ is surjective, this establishes one inclusion. To establish the other, given $X \in \mathfrak{u}(V)$ we can differentiate the action of $e^{tX}$ to obtain the action of $X$ . $\Box$

Corollary (Schur-Weyl duality): The natural map $\mathbb{C}[S_n] \to \text{End}_{\text{U}(V)}(V^{\otimes n})$ is surjective.

Corollary: $V_{\lambda}$ is either an irreducible representation of $\text{U}(V)$ or zero.

Posted in math.RT | Tagged MaBloWriMo, symmetric functions | 21 Comments

21 Responses

on September 15, 2019 at 12:45 pm | Reply Aritra

In 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 | Reply Qiaochu Yuan
  
  There’s no assumption.
  - on September 26, 2019 at 3:24 pm | Reply justanothermathstudent
    
    It’s confusing because you have taken an n-element basis of V
    - on September 26, 2019 at 9:37 pm | Reply Qiaochu Yuan
      
      Ah, sorry, that’s a typo. Fixed. Thanks!
on January 18, 2013 at 6:08 pm | Reply artin

How 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 | Reply Qiaochu Yuan
  
  This 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 | Reply tim

Hey, 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 | Reply Qiaochu Yuan
  
  It is not an embedding if $\dim V$ is small compared to $n$ because $\dim k[S_n] = n!$ but $\dim \text{End}(V^{\otimes n}) = (\dim V)^{2n}$ . Even when it is an embedding, this doesn’t imply that every irreducible representation of $S_n$ occurs as a direct summand of $V^{\otimes n}$ .
- on January 13, 2013 at 7:30 pm | Reply djim
  
  Notice 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 | Reply djim
    
    In 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 | Reply Scav

Hi, 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 | Reply Qiaochu Yuan
  
  Any quotient of a semisimple ring is semisimple.
on January 1, 2013 at 8:24 am | Reply harryp

Hi,

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 | Reply Qiaochu Yuan
  
  Ah. 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 $d$ agree at $d + 1$ 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 | Reply harryp
    
    thank you!
on December 23, 2012 at 6:51 am | Reply djim

I see it now, thank you.
on December 16, 2012 at 9:55 pm | Reply djim

Hello, 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 | Reply Qiaochu Yuan
  
  I don’t have much to say beyond what’s in the proof itself.
  - on December 22, 2012 at 3:54 pm | Reply djim
    
    Thanks. 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 | Reply Qiaochu Yuan
    
    It’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 $t$ . 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 $d + 1$ points.
on November 14, 2012 at 6:05 am | Reply Allen Knutson

Some 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.

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