Symmetric power and characters

Question

Let $V$ be a 2 dimensional vector space over $\mathbb{C}$. Then $W := Sym^{n}(Sym^{m}V)$ is a representation of $GL(V)$. For $g \in GL(V)$, I consider $\chi_{W}(g)$. Let $x$ and $y$ denote the eigenvalues of $g$. Why is $\chi_{W}(g) = \sum_{k=0}^{nm} f(k)x^{k}y^{nm - k}$ where $f(k)$ is the numbers of ways to partition $k$ into at most $n$ integers each at most $m$? I know that $\chi_{W}(g) = Tr(g)$, but how do I see the above summation from this?

Answer 1

Remember that a basis for $Sym^m V$ is the $m$-fold product of basis vectors for $V$ (here "product" means in the symmetric space, so order of the factors doesn't matter). If $x$ and $y$ are eigenvalues of $g$ acting on $V$, then we can see from this that the eigenvalues for $g$ acting on $Sym^m V$ will be of the form $x^ky^{m-k}$ for $k = 0, \dots, m$: if $v_1, \dots, v_m$ are eigenvectors of $g$ with eigenvectors $\lambda_i$, then $$ g \cdot (v_1\otimes \dots \otimes v_m) = \lambda_1v_1 \otimes \dots \otimes \lambda_m v_m = \lambda_1 \dots \lambda_m v_1\otimes \dots \otimes v_m $$ To determine the eigenvalues for $g$ acting on $Sym^n(Sym^m V)$, we now take $n$-fold products of these elements. Any such eigenvalue will be of the form $x^k y^{nm-k}$, and a particular $k$ shows up each time $k$ is expressible as a partition of at most $n$ integers each at most $m$, as you say. Then the character $\chi_W(g)$ is, exactly as you said, just the sum of these numbers.