How to prove $\text{Sym}^{m}(\text{Sym}^{n}V) \cong \text{Sym}^{n}(\text{Sym}^{m}V)$?

Question

This question is about exercise 1.13 of Fulton and Harris book(page 11): $$\text{Sym}^{m}(\text{Sym}^{n}V) \cong \text{Sym}^{n}(\text{Sym}^{m}V)$$

It seems to me the isomorphism is "obvious". Since I feel like the elements of $\text{Sym}^{m}(\text{Sym}^{n}V)$ look like $(v_{11}\cdots v_{1n})\cdots (v_{m1}\cdots v_{mn})$ and somehow one can "rearrange" it to make it an element on the RHS. However, this is just an idea and I don't know how to prove it rigorously. Moreover, my "argument" seems to show that
$$\text{Sym}^{m}(\text{Sym}^{n}V) \cong \text{Sym}^{n}(\text{Sym}^{m}V) \cong \text{Sym}^{mn}(V).$$ But the last statement isn't correct(here). So can anybody show me a proof of $$\text{Sym}^{m}(\text{Sym}^{n}V) \cong \text{Sym}^{n}(\text{Sym}^{m}V)?$$

Since this problem is in the their Lecture 1, I hope the proof doesn't involve tools that are too sophisticated...But every solution will be appreciated!

Answer 1

Hint. Prove that $$\dim \text{Sym}^{m}(\text{Sym}^{n}V) = \dim \text{Sym}^{n}(\text{Sym}^{m}V)$$

Answer 2

The standard representation on $\mathcal{S}_3$ and symmetric powers

The group $\mathcal{S}_3$ is generated by the transposition $\sigma = (12)$ and the cycle $\tau = (123)$.

Let $V$ be the standard representation of $\mathcal{S}_3$. We can take $V := \{ (z_1,z_2,z_3) \in \mathbb{C}^3 : z_1 + z_2 + z_3 = 0 \}$. Let $\omega := e^{2 \pi i /3}$. Then $V$ has a basis $\{\alpha,\beta\}$ where $\alpha := (\omega,1,\omega^2)$ and $\beta = (1,\omega,\omega^2)$. The action of $\mathcal{S}_3$ on $V$ is then determined by \begin{align*} \tau \alpha = \omega \alpha, \tau \beta = \omega^2 \beta, \sigma \alpha = \beta, \sigma \beta = \alpha. \end{align*} We now consider the symmetric powers. A basis of $\mathrm{Sym}^n(V)$ is then given by $\{ \alpha^j \beta^{n-j} : 0 \leq j \leq n \}$. In particular, a basis for $\mathrm{Sym}^m(\mathrm{Sym}^n(V))$ is then given by $\{ \alpha^{\lambda_1} \beta^{n-\lambda_1} \cdot \ldots \cdot \alpha^{\lambda_m} \beta^{n-\lambda_m} : n \geq \lambda_1 \geq \ldots \geq \lambda_m \geq 0 \}$.

Operations on integer partitions

$\lambda =(\lambda_1 \geq\ldots \geq \lambda_m \geq 0)$ is an integer partition into $m$ (possibly empty) blocks of size at most $n$, and sits inside the $m \times n$ rectangle. We define two new integer partitions, $\lambda'$ and $\lambda^*$ as follows.

Let $\lambda'$ be the dual partition of $\lambda$. This is the partition obtained by flipping $\lambda$ along its diagonal. More specifically, $\lambda'_i := \# \{ j : \lambda_j \geq i \}$. $\lambda'$ has at most $n$ blocks of size at most $m$.

Let $\lambda^*$ be the flipped complement of $\lambda$ in the $m \times n$ rectangle. That is, $\lambda^*_i := n - \lambda_{m-i+1}$.

A crucial observation here is that these operations commute: $(\lambda')^* = (\lambda^*)'$.

Let $|\lambda| := \sum_{i \geq 1} \lambda_i$ denote the number of blocks in the partition. Then $|\lambda'| = |\lambda|$ and $|\lambda^*| = mn - |\lambda|$.

The action of $\mathcal{S}_3$ on $\mathrm{Sym}^m(\mathrm{Sym}^n(V))$.

Consider now that $\tau(\alpha^j\beta^{n-j}) = \omega^{n-j} \alpha^j \beta^j$ and $\sigma(\alpha^j \beta^{n-j} ) = \alpha^{n-j} \beta^j$.

Write $\lambda$ as shorthand for the element $\alpha^{\lambda_1} \beta^{n-\lambda_1} \cdot \ldots \cdot \alpha^{\lambda_m} \beta^{n-\lambda_m} $ of $\mathrm{Sym}^m(\mathrm{Sym}^n(V))$. It follows that \begin{align*} \tau(\lambda) = \omega^{mn - |\lambda| } \lambda, \sigma(\lambda) = \lambda^*. \end{align*}

Isomorphism of representations

Consider now the vector space isomorphism $\phi:\mathrm{Sym}^m(\mathrm{Sym}^n(V)) \to \mathrm{Sym}^n(\mathrm{Sym}^m(V))$ given by $\phi(\lambda) = \lambda'$, where of course we are using the shorthand $\lambda'$ for the element $\alpha^{\lambda_1'}\beta^{m-\lambda_1'} \cdot \ldots \cdot \alpha^{\lambda_n'} \beta^{m- \lambda_m'}$ of $\mathrm{Sym}^n(\mathrm{Sym}^m(V))$. We claim that this is an isomorphism of $\mathcal{S}_3$ representations, i.e. $g\phi = \phi g$ for all $g \in \mathcal{S}_3$. To verify this, it is sufficient to check that $\phi$ commutes with the generators $\sigma$ and $\tau$ of $\mathcal{S}_3$.

We begin by noting that $\phi \sigma \lambda = \phi( \lambda^*) = (\lambda^*)'$, where as $\sigma \phi \lambda = \sigma ( \lambda') = (\lambda')^*$. Since $(\lambda')^* = (\lambda^*)'$, we have $\phi \sigma \lambda = \sigma \phi \lambda$.

Considering now $\tau$, we have $\phi \tau \lambda = \omega^{ mn - |\lambda|} \phi \lambda = \omega^{mn-|\lambda|} \lambda'$. On the other hand, $\tau \phi \lambda = \tau \lambda' = \omega^{mn - |\lambda'|} \lambda'$. Since $|\lambda'| = |\lambda|$, we have $\phi \tau \lambda = \tau \phi \lambda$.

It follows that $\phi$ is an isomorphism of representations.