Character of GL(V) There is a natural action of $S_n\times GL(V)$ on $V^{\otimes n},$ and we know $V^{\otimes n}=\oplus_\lambda W_\lambda\otimes L_\lambda$ where $W_\lambda$ is the Specht module of $S_n$ and $L_\lambda$ is irreducible reps of $GL(V)$. My problem is:
Suppose $V$ is a vector space over $\Bbb C$, dim$V=N, g\in GL(V),$ and $x_1,...x_N$ are eigenvalues of $g$ on $V$. We are trying to find character of $g$ on $L_\lambda$. The book try to calculate $Tr_{V^{\otimes n}}(g^{\otimes n}s), s\in S_n$ and $s$ in the conjugacy class of $(i_1,...i_n)$ where $i_l$ is the number of length $l$ cycles in $s$. The book says we easily get that $Tr_{V^{\otimes n}}(g^{\otimes n}s)=\prod_m Tr(g^m)^{i_m}$. But I don't see it why?
 A: For question 1, $S_n$ acts on $V^{\otimes n}$ by permuting terms in pure tensors (ie $\sigma . (v_1 \otimes \dots \otimes v_n) = v_{\sigma 1} \otimes \dots \otimes v_{\sigma n})$, and $GL_n$ acts diagonally on pure tensors (ie $\phi . (v_1 \otimes \dots \otimes v_n) = \phi(v_1) \otimes \dots \otimes \phi(v_n)$.
A: If $g$ acts diagonalisable, there is an eigenbasis $v_1, .., v_N$ of $V$. Therefore $\{v_{i_1} \otimes ... \otimes v_{i_n} \mid 1\leq i_1,...,i_n\leq N\}$  is a basis of $V^{\otimes n}$ which is easily seen to be an eigenbasis of $g^{\otimes n}$. 
If $\sigma\in Sym(n)$ is any permutation, we therefore have
$$(\sigma g^{\otimes n})(v_{i_1}\otimes\cdots\otimes v_{i_n}) = \lambda_{i_1}\cdots \lambda_{i_n} \sigma(v_{i_1}\otimes...\otimes v_{i_n})=\lambda_{i_1}\cdots \lambda_{i_n} v_{i_{\sigma(1)}}\otimes\cdots\otimes v_{i_{\sigma(n)}}$$
Note that the basisvector on the right hand side is the same as the vector on the LHS iff $i_1=i_{\sigma(1)} \wedge ... \wedge i_n=i_{\sigma(n)}$, i.e. if $k\mapsto i_k$ is constant on the orbits of $\sigma$ (which are precisely its cycles). If that is the case, we can write the coefficient on the right hand side as $\prod_{c\,cycle} \lambda_{i(c)}^{|c|}$.
Hence we can compute the trace:
$$\begin{align*}
tr(\sigma g^{\otimes n}) &= \sum_{1\leq i_1,...,i_n\leq N} \begin{cases} \lambda_{i_1}\cdots \lambda_{i_n} & i_1=i_{\sigma(1)}, ..., i_n=i_{\sigma(n)} \\ 0 &\text{otherwise}\end{cases} \\
&= \sum_{i: \{cycles\}\to\{1,...,N\}} \prod_{c} \lambda_{i(c)}^{|c|} \\
&= \prod_{c\,cycle} \sum_{i=1}^N \lambda_i^{|c|} \\
&= \prod_{c\,cycle} tr(g^{|c|})
\end{align*}$$
which is precisely the formula that you want.
And finally: Since this formula is continuous in $g$ and diagonalisable matrices are dense in $GL_N(\mathbb{C})$, we find that the same formula holds for all $g\in GL(V)$, not just the diagonalisable ones.
