I am trying to gain a better understanding of Schur-Weyl duality specifically applied to symmetric functions. My motivating example is trying to understand the Frobenius character of the multilinear component of the free Lie algebra (Theorem 8.1 in Reutenauer's book on the subject), but my general confusion is more than just that example. I feel comfortable with representation theory of finite groups and less so with representation theory of Lie groups.

In several sources, the statement of Schur-Weyl duality is framed as the $\operatorname{GL}(V)\times \mathfrak{S}_k$ isomorphism of the tensor algebra $V^{\otimes k}$ with the direct sum of tensor products $\mathbb{S}^\lambda(V) \otimes M^\lambda$ of Specht modules with Schur functors indexed by partitions of $k$ into at most $\dim(V)$-many parts. As I understand it, one can then determine the multiplicity of either a Schur functor (by looking at the dimension of the corresponding Specht module) or the multiplicity of a Specht module (by looking at the dimension of the corresponding Schur functor).

I also seem to understand that, as explained in the second appendix of Stanley's Enumerative Combinatorics Vol. 2, the character value of $A \in \operatorname{GL}(V)$ for an irreducible rational $\operatorname{GL}(V)$ representation $\mathbb{S}^\lambda$ is given by evaluating the Schur polynomial on the eigenvalues of $A$.

Reutenauer's proof of theorem 8.1 (that two particular symmetric functions are equal) is as follows:

This is the Schur-Weyl duality between the representations of the symmetric group and the general linear group (Weyl 1946, Theorem 7.6.F; Macdonald 1979, A7 in Chapter 1).

My first confusion is that the references I find refer to Schur-Weyl duality as a statement about the tensor algebra, not any other algebra. The universal enveloping algebra of the free Lie algebra can be constructed as a quotient of the tensor algebra, so I'm willing to suspend some disbelief that the statement is still applicable here.

Even with that, I am having trouble connecting the dots here. Reutenauer sets up a $\operatorname{End}(V) \times \mathfrak{S}_k$-representation, and then looks at the action of a copy of $\mathfrak{S}_k$ living inside of $\operatorname{End}(V)$ and considers the Frobenius characteristic of the $\mathfrak{S}_k$-representation found by restricting the action of $\operatorname{End}(V)$. He then compares this to the multivariate generating function of dimensions of homogeneous polynomials of multidegree $\alpha$ which are stable under the $\mathfrak{S}_k \subseteq \operatorname{End}(V)$ action and asserts they are the same via the proof quoted above.

The character value of the restriction of a representation is just the character value of the original representation. The eigenvalues of any $A \in \mathfrak{S} \leq \operatorname{End}(V)$ should be all $1$. Then I think the character value of $A$ should be the number of semistandard Young tableaux of shape $\lambda$ with entries at most $\dim(V)$ (i.e. $s_\lambda(1,1,\ldots,1)$). But I am struggling to see why this proves Reutenauer's claim.

Any help or general information about the relationship between Schur-Weyl duality and symmetric functions you could provide would be greatly appreciated.

Edit: I made some mistakes in my question including:

  • The universal enveloping algebra is the tensor algebra, not a (nontrivial) quotient.
  • Eigenvalues of permutation matrices are roots of unity, not 1.
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    $\begingroup$ The universal enveloping algebra of the free Lie algebra is the tensor algebra. I haven't looked at the rest of this but maybe that's already helpful. In any case, so that answering this question does not require having a copy of Reutenauer, it might be a good idea for you to include more of the text, e.g. the statement of the Theorem and a relevant section of the proof. $\endgroup$ Jan 1 at 23:38
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    $\begingroup$ Also, permutation matrices do not have all eigenvalues $1$. But again maybe I am misunderstanding some of your assertions because I can't look at the text you're trying to understand. $\endgroup$ Jan 2 at 2:29
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    $\begingroup$ The problem is that whatever Reutenauer uses doesn't seem to be standard Schur-Weyl duality. I got confused by his symmetric group too -- his symmetric group $S_n$ acts diagonally through $\operatorname{GL}\left(V\right)$, not by permuting the tensor factors! The latter action wouldn't preserve the free Lie subalgebra (viewed as a subspace of the tensor algebra), so this is not surprising in hindsight, but it makes me wonder if Schur-Weyl duality is applicable at all. $\endgroup$ Jan 2 at 10:56
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    $\begingroup$ @darijgrinberg: At the beginning of Chapter 8, before the statement of Theorem 8.1, Reutenauer in fact considers both of these actions of the symmetric group on the tensor product space (he calls them the "place permutation action" and the "variable permutation action"). $\endgroup$ Jan 2 at 18:01
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    $\begingroup$ @TrevorK: Indeed, although your inquiries seem more related to a different open problem about symmetric functions, namely "Thrall's problem" (see www-users.cse.umn.edu/~reiner/Talks/ThrallsProblem.pdf) $\endgroup$ Jan 5 at 19:57

2 Answers 2


This answer is a response the the prompt

Any help or general information about the relationship between Schur-Weyl duality and symmetric functions you could provide would be greatly appreciated.

If you have more questions (e.g. about the free Lie algebra), feel free to ask.

For convenience I will let $n = \dim(V)$.

Step 1: A bitrace formula.

As discussed, we have an action of $GL(V) \times S_k$ on $V^{\otimes k}$. We will compute the trace of an element $(M, g) \in GL(V) \otimes S_k$ on $V^{\otimes k}$. Conceptually we will think about $GL(V)$ and $S_k$ as separate (rather than combined into $GL(V) \otimes S_k$) which is why I use the term "bitrace" (it is the synthesis of two traces).

We compute the trace directly. The matrix $M$ has a Jordan decomposition $M = S + N$ ($S$ semisimple, $N$ nilpotent), and the action of $(M,g)$ on $V^{\otimes k}$ is the sum of the actions of $(S,g)$ and $(N,g)$; since $N$ is nilpotent, $(N,g)$ is not actually an element of $GL(V)\times S_k$, nevertheless the trace is well defined and equal to zero because it is nilpotent. I say all this only to justify restricting to diagonalisable matrices.

Now, let $v_1, \ldots, v_n$ be an eigenbasis of $M$, so that $Mv_i = x_i v_i$ for some complex numbers $x_i$ (i.e. $M = diag(x_1,\ldots,x_n)$).

This induces a basis $V_I = v_{i_1} \otimes \cdots \otimes v_{i_k}$ of $V^{\otimes k}$ indexed by words $I = (i_1, \ldots, i_k)$ (where $1 \leq i_j \leq n$). Conveniently, the action of $(M,g)$ on $V_I$ is easy to compute: $$ (M,g) \cdot (v_{i_1} \otimes \cdots \otimes v_{i_k}) = g \cdot (Mv_{i_1} \otimes \cdots \otimes Mv_{i_k}) = g \cdot (x_{i_1}v_{i_1} \otimes \cdots \otimes x_{i_k}v_{i_k}) \\ = x_{i_1}x_{i_2}\cdots x_{i_k} (v_{g^{-1}(1)} \otimes \cdots \otimes v_{g^{-1}(n)}) $$ Side note: whether you apply $g^{-1}$ or $g$ to the indices depends on whether you view the symmetric group as having a left or right action, it's not really important.

This value of $(M,g) \cdot v_I$ is a scalar multiple of another basis element, which we might write $v_{g(I)}$ by using the induced action of $g \in S_k$ on tuples of length $k$. To compute the trace of $(M,g)$ we need to sum the "diagonal entries", i.e. the scalars corresponding to those $I$ with $v_I = v_{g(I)}$. This computation becomes $$ \sum_{g(I) = I} x_{i_1} \cdots x_{i_k}. $$ Now, suppose for example that $(2,5,6)$ is a cycle of $g \in S_k$. Then the condition $g(I) = I$ implies that the equality of indices $i_2 = i_5 = i_6$, but the actual value could be anything in $1, \ldots, n$. This condition also implies that $x_{i_2} = x_{i_5} = x_{i_6}$. This same reasoning extends to cycles of all sizes. The only nonzero "diagonal terms" to be summed are those where all indices acted on by a cycle of $g$ are the same. The actual index associated to each cycle is arbitrary, we we need to sum over those. If the cycles of $g$ have sizes $\mu_1, \ldots, \mu_l$, the trace becomes $$ \sum_{j_1=1}^{n} \cdots \sum_{j_l=1}^{n} x_{j_1}^{\mu_1} \cdots x_{j_l}^{\mu_l} = (\sum_{j_1=1}^{n} x_{j_1}^{\mu_1}) \cdots (\sum_{j_l=1}^{n} x_{j_l}^{\mu_l}) = p_{\mu_1}(x) \cdots p_{\mu_l}(x) = p_\mu(x) $$ where I am now using the standard notation for power-sum symmetric functions.

Conclusion: if the eigenvalues of $M$ are $x_i$ and $g$ has cycle type $\mu$, then the bitrace of $(M,g)$ acting on $V^{\otimes k}$ is $p_\mu(x)$.

Step 2: Frobenius characteristic and Cauchy identity.

The Frobenius characteristic, $ch$, is an isomorphism between the Grothendieck group of class functions on $S_k$ and symmetric functions of degree $k$ (here we work over $\mathbb{C}$). It is convenient to define $ch(f)$ for all functions on $S_k$ (not just class functions) by saying that if $g^*$ is the indicator function of $g \in S_k$, then $ch(g^*) = \frac{1}{k!} p_\mu(y)$, where $\mu$ is the cycle type of $g$, and I write $y$ for the symmetric function variables in order to distinguish them from the discussion in the previous step. So for example, if $C_{\mu}^*$ is the indicator function of the conjugacy class $C_\mu$ of elements of cycle type $\mu$, then $ch(C_{\mu}^*) = \frac{|C_\mu|}{k!} p_\mu(y) = \frac{1}{z_\mu}p_{\mu}(y)$, where $z_\mu$ has its usual meaning.

Now if we fix $M \in GL(V)$, then the bitrace of $(M,g)$ may be viewed as a (class) function on $S_k$, and so we may apply the Frobenius characteristic. If we write $cyc(g)$ for the cycle type of $g$, the result of this calculation is $$ ch(tr(M,g)) = \sum_{g \in S_k} p_{cyc(g)}(x) \frac{1}{k!} p_{cyc(g)}(y) = \sum_{\mu \vdash k} \frac{1}{z_\mu} p_\mu (x) p_\mu(y). $$ Now, the famous Cauchy identity implies that we have $$ ch(tr(M,g)) = \sum_{\mu \vdash k} \frac{1}{z_\mu} p_\mu (x) p_\mu(y) = \sum_{\lambda \vdash k} s_\lambda(x) s_\lambda(y), $$ where $s_\lambda$ is the Schur function indexed by $\lambda$.

This may be viewed as a symmetric-function-theoretic formulation of Schur-Weyl duality for the following reasons. Suppose we know that the Frobenius characteristic of the Specht module $S^\lambda$ is the Schur function $s_\lambda(y)$, and the character of the Schur functor $\mathbb{S}^\lambda(V)$ is the Schur function $s_\lambda(x)$ (the meaning of which is discussed in the comments to the original post). Then we have found that the bitrace (i.e. $GL(V) \times S_k$-character) of $V^{\otimes k}$ agrees with that of $\bigoplus_{\lambda \vdash k} \mathbb{S}^\lambda(V) \otimes S^\lambda$. By semisimplicity, these must be isomorphic.

For example, we can recover the multiplicity of the Specht module $S^\lambda$ in $V^{\otimes l}$ by computing $\dim(\mathbb{S}^\lambda(V))$, which is nothing but the trace of the identity of $GL(V)$. But the identity element of $GL(V)$ (viewed as a matrix) has $1$ as an eigenvalue repeated $n$ times, so the dimension is the evaluation $s_\lambda(1,\ldots,1)$ (where there are $n$ $1$-s), as you mention in your post.

  • $\begingroup$ This very directly points out the relationship between Schur-Weyl duality and symmetric functions. It also is nice that if you are dealing with symmetric polynomials (rather than functions) for the $s_\lambda(x)$, then it automatically enforces the limit on the length of the indexing partition. $\endgroup$
    – Trevor K
    Jan 6 at 18:52

In your context, you want to think of the Schur-Weyl duality as a way to construct representations of $GL(V)$ out of representations of symmetric groups.

To give a precise answer along these lines that works well for your original motivation: let us denote by $Lie(n)$ the multilinear component in the free Lie algebra with $n$ generators. It has indeed an $S_n$-action coming from embedding $S_n$ into $GL_n$; you should think of this space as the space of abstract Lie operations that have $n$ arguments and are multilinear in them - forget about any free algebras for now. The point now (very well understood in the context of PI-algebras and of operads) is that out of the collection $\{Lie(n)\}_{n\ge1}$ you can construct free Lie algebras by the formula $$ Lie(V)=\bigoplus_{n\ge1}Lie(n)\otimes_{S_n}V^{\otimes n}, $$ where $S_n$ acts on $V^{\otimes n}$ by permutations of tensor factors.

It takes time to get used to the idea, but it has many powerful implications so it's totally worth it.


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