# Peter-Weyl vs. Schur-Weyl theorem

Let $V$ be a finite dimensional complex vector space.

According to the Peter-Weyl theorem there is a decomposition $\mathcal O(\mathrm{GL}(V)) \cong \bigoplus_\lambda V_\lambda \otimes V_\lambda^\ast$ of the algebraic coordinate ring of $\mathrm{GL}(V)$ into a direct sum indexed by partitions, where $V_\lambda$ denotes the representation of highest weight $\lambda$.

According to Schur-Weyl duality there is a decomposition $T(V) \cong \bigoplus_{\lambda} V_\lambda \otimes \sigma_\lambda$ of the tensor algebra on $V$, where $\sigma_\lambda$ now denotes the Specht module associated to a partition $\lambda$.

The two statements look very similar. Is there a direct relation between the commutative ring $\mathcal O(\mathrm{GL}(V))$ and the associative algebra $T(V)$? E.g. a map between them that behaves nicely w.r.t. the decompositions?

• What is the Specht module associated to lambda? – Claudio Gorodski Apr 18 '17 at 17:29
• Is it too obvious to note that each of $\mathcal{O}(\mathrm{GL}(V))$ and $V^{\otimes n}$ carry distinct commuting group actions? Namely the first one has induced actions from both the left and right group multiplication of $\mathrm{GL}(V)$ on itself, while the $V^{\otimes n}$ summand of $T(V) = \bigoplus_{n=0 }^\infty V^{\otimes n}$ is acted on by $\mathrm{GL}(V)$ and $S_n$, where the former the former is the obvious action on the tensor product and the latter permutes the tensor factors. – Igor Khavkine Apr 18 '17 at 19:25
• @ClaudioGorodski There is a bijection between irreducible representations of $S_n$ in characteristic zero and partitions $\lambda \vdash n$. The representation corresponding to a partition $\lambda$ is called a Specht module. – Dan Petersen Apr 19 '17 at 5:33

Yes. In combinatorics this is known as Robinson-Schensted-Knuth vs. just Robinson-Schensted. (Properly speaking the latter is about a yet smaller duality, $\mathbb C[S_n] = \bigoplus_{\lambda\vdash n} \sigma_\lambda \otimes \sigma_\lambda^*$.)

First, shrink the Peter-Weyl result from $\mathcal O(GL(n))$ (you overuse $V$, I feel) to the slightly smaller $\mathcal O(M_n)$. Then the RHS shrinks to $\oplus_\lambda V_\lambda \otimes V_\lambda^*$, where $\lambda$ now runs over partitions $(\lambda_1 \geq \ldots \geq \lambda_n \geq 0)$ instead of all dominant weights $(\lambda_1 \geq \ldots \geq \lambda_n)$.

Then generalize to other matrix spaces, not just square matrices, obtaining $\mathcal O(M_{a\times b}) \cong \bigoplus_\lambda V^a_\lambda \otimes (V^b_\lambda)^*$, the sum now over partitions of height $\leq \min(a,b)$.

(The combinatorial statement, RSK, is a bijective proof of two different character formulae for this representation. The obvious weight basis is given by monomials in the matrix entries, equivalently listed as $M_{a\times b}(\mathbb N)$. On the RHS we have pairs of same-shape SSYT. Under the bijection the row and column sums of the matrix in $M_{a\times b}(\mathbb N)$ go to the contents, i.e. entry multiplicities, of the two SSYT.)

Now, consider functions on $M_{a\times b}$ of weight $(1,1,\ldots,1)$ under the $T^a \leq GL(a)$ action. Since that's $S_a$-invariant and $S_a$ normalizes $T^a$, this weight space will have a $S_a \times GL(b)$ action.

The LHS will be made of functions that are multilinear in the rows, i.e. $(\mathbb C^b)^{\otimes a}$. The representation $V^a_\lambda$ has a $(1,1,\ldots,1)$ weight space iff $\lambda$ is a partition of $a$, and in that case, the $S_a$ action on it is the Specht irrep $\sigma_\lambda$ of $S_a$. Which is to say, the RHS has become $\oplus_{\lambda \vdash a} \sigma_\lambda \otimes (V_\lambda)^*$ like you wanted. QED.

(Now we're insisting that the row sums are all $1$. On the RHS, one of the SSYT is an SYT. If you go further and ask that the column sums be all $1$ also, then the LHS becomes just permutation matrices, the RHS pairs of same-shape SYT, and the correspondence is just Robinson-Schensted no Knuth.)

As I recently learned from Martin Kassabov, you can run this in reverse: take two copies of the Schur-Weyl isomorphism, reverse one, and tensor them together over $\mathbb C[S_n]$ to get the Peter-Weyl (for matrices) result. So it's a matter of taste deciding which one is the more fundamental.

• Thanks, this is really helpful! So what you're saying is that $(\mathbb C^b)^{\otimes a} = \bigoplus_{\lambda \vdash a} \sigma_\lambda \otimes (\mathbb C^b)_\lambda$ sits inside of $\mathcal O(M_{a \times b}) = \bigoplus_{\lambda} (\mathbb C^a)_\lambda \otimes (\mathbb C^b)^\ast_\lambda$ in a nice and conceptual way that respects the decompositions. Is there a statement that also relates the multiplication in the tensor algebra and the multiplication in the coordinate ring? – Dan Petersen Apr 19 '17 at 5:28
• I added a different answer to the question - if you have any input it would be appreciated! – Dan Petersen Apr 19 '17 at 9:00
• For the tensor algebra, you need to combine different $a$. Then, you're asking for a statement on $\bigoplus_a \mathcal O(M_{a\times b})$, as in why should that thing have another, noncommutative, multiplication. What seems more reasonable is that it have a comultiplication, i.e. $\coprod_a M_{a\times b}$ has a monoid operation "concatenate vertically". – Allen Knutson Apr 20 '17 at 10:58

Allen's nice answer led me in a slightly different direction. Let me try to give another answer to the question.

Let's start from the Cauchy identities, $$\prod_{i,j \geq 0} (1-x_i y_j)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y)$$ which is an equality between bisymmetric functions in infinitely many variables $\{x_i\}$ and $\{y_i\}$.

Now recall that there is a correspondence between symmetric functions of degree $n$, representations of $S_n$, and polynomial functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$ of degree $n$. Passing to the completion of the ring of symmetric functions wrt degree gives instead a correspondence between (completed) symmetric functions, sequences of representations of $S_n$ (i.e. "tensorial species") and analytic functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$.

Using this we can give three different interpretations of the Cauchy identities:

(1) Consider both the $x$- and $y$-variables as corresponding to representations of the symmetric groups. The Cauchy identities become $$\bigoplus_{n \geq 0} \mathbb C[S_n] = \bigoplus_{\lambda} \sigma_\lambda \otimes \sigma_\lambda,$$ i.e. the Peter-Weyl theorem for $S_n$.

(2) Consider the $x$-variables as corresponding to an analytic functor and the $y$-variables as corresponding to a sequence of representations. Then the left hand side becomes the analytic functor $V \mapsto T(V)$ and the right hand side becomes $V \mapsto \bigoplus_\lambda V_\lambda \otimes \sigma_\lambda$.

(3) Consider both $x$- and $y$-variables as corresponding to analytic functors. The left hand side becomes the analytic functor $(V,W) \mapsto \mathcal O(V\otimes W)$ and the right hand side becomes $(V,W) \mapsto \bigoplus_\lambda V_\lambda \otimes W_\lambda$.

Specializing to $W = V^\ast$ in (3) gives the coordinate ring of the matrix space as in Allen's answer.

The three interpretations of the Cauchy identities can be seen as equalities between sequences of representations of $S_n \times S_n$, sequences of polynomial functors of degree $n$ into $S_n$-representations, and analytic functors of two variables, respectively. But in all cases there is also an obvious multiplication on the left hand side: given by the inclusion $\mathbb C[S_n] \otimes \mathbb C[S_m] \to \mathbb C[S_{n+m}]$, the multiplication in the tensor algebra, and the multiplication in the coordinate ring, respectively. This is because in all three cases we have a commutative algebra object in the respective symmetric monoidal category, and the structure of commutative algebra object gets transferred along the different equivalences of categories. For example, a commutative algebra object in the category of tensorial species is what's usually called a twisted commutative algebra, so we are saying that the tensor algebra $T(V)$ is a twisted commutative algebra (even though the multiplication in $T(V)$ is certainly not commutative), and so on. So the multiplication in $T(V)$ is in a precise sense "the same" as the multiplication in the coordinate ring of $V \otimes W$!

PS - I certainly hope all the above is correct. But I am confused about the fact that what appears is $\sigma_\lambda \otimes \sigma_\lambda$ in case (1), rather than $\sigma_\lambda \otimes \sigma_\lambda^\ast$ which would be more expected. (Of course $\sigma_\lambda \cong \sigma_\lambda^\ast$, but I would still like the dual to be there!)

• Actually I guess the dual should be there because in all three cases we have commuting left and right actions. Then to be consistent one should also write the Schur-Weyl duality as $V^{\otimes n} = \bigoplus_{\lambda \vdash n} V_\lambda \otimes \sigma_\lambda^\ast$. – Dan Petersen Apr 19 '17 at 11:18