Invariants of $V^{\otimes N}$. Let $V$ be a finite dimensional complex vector space, and $G = SL(V)$ be the group of linear transformations of $V$ with determinant $1$.
(a) Show that $V^{\otimes N}$ contains a nonzero $G$-invariant if and only if $N$ is a multiple of $\dim V$.
(b) If $N = m\dim V$ where $m$ is a positive integer, compute the dimension of the space of invariants $\left(V^{\otimes N}\right)^G$.
 A: Let $d = \dim V$. By Schur-Weyl duality, as a representation of $S_N \times GL(V)$, $V^{\otimes N}$ decomposes as$$\bigoplus_{\lambda\text{ partition of }N} V_\lambda \otimes L_\lambda,$$where $V_\lambda$ is the Specht module and $L_\lambda$ is a representation of $GL(V)$. As a representation of $GL(V)$, the above collapses to$$\bigoplus_{\lambda\text{ partition of }N} \dim(V_\lambda)L_\lambda.\tag*{(1)}$$We know $L_\lambda = 0$ if $\lambda$ has more than $d$ rows. Otherwise, the dimension of $L_\lambda$ is given by$$\dim L_\lambda = \prod_{1 \le i < j \le d} {{\lambda_i - \lambda_j + j - i}\over{j-i}}.$$This is equal to $1$ only when $\lambda_1 = \dots = \lambda_d$, i.e. $\lambda_i = N/d$ for all $i$. If $v \in V^{\otimes N}$ is a nonzero $G$-invariant, then $\text{span}(v) \in \text{dim}(V_\lambda)L_\lambda$ is the decomposition $(1)$ with $\lambda$ defined as above. This is because $GL(V)v = \mathbb{C}SL(V)v = \text{span}(v)$, so $\text{span}(v)$ is a one-dimensional subrepresentation of $V^{\otimes N}$ under $GL(V)$. hence by Schur's Lemma it must actually be contained in $\dim(V_\lambda)L_\lambda$.
In particular, if $N$ is not a multiple of $d$, then $V$ has no invariants. From now on, suppose $N = dm$. The character of $L_\lambda$ is$$\chi_{L_\lambda}(g) = {{\det\left[x_i^{\lambda_j + d - j}\right]_{i, j}}\over{\det\left[x_i^{d-j}\right]_{i, j}}} = (x_1\dots x_d)^m,$$where $x_i$ are the eigenvalues of $g$; when $g \in SL(V)$ the above equals $1$. Hence $SL(V)$ acts trivially on $L_\lambda$.
By $(1)$, the dimension of the space of invariants is $\dim V_\lambda$ where $\lambda = (\underbrace{m,\dots, m}_d)$. By the hook length formula $($see Pavel Etingof's notes on representation theory$)$,$$\begin{align*}
\dim(V_{\lambda})=\frac{n!}{\prod_{i\leq \lambda_j} h(i,j)}
=\begin{cases}
\frac{(md)!}{(m+d-1)(m+d-2)^2\dots m^d (m-1)^d\dots (d+1)^dd^d (d-1)^{d-1}\dots 1^1} & \text{if }d\leq m\\
\frac{(md)!}{(m+d-1)(m+d-2)^2\dots d^m (d-1)^m\dots (m+1)^mm^m (m-1)^{m-1}\dots 1^1} & \text{if }d> m
\end{cases}
\end{align*}.
$$In particular, there exists a nonzero invariant.
