Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 15F (Representation Theory)

The symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ identifying tensors which differ by permutation of the factors. The exterior power ${\Lambda }^{n}V$ is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a basis $e_1,\dots ,e_d$, bases consist respectively of monomials $e_1^{a_1}\cdots e_d^{a_d}$ with $\sum a_i=n$, and wedge products $e_{i_1}\wedge \cdots \wedge e_{i_n}$ with $i_1<\cdots <i_n$. Thuswith the second dimension zero for $n>d$.

A finite-order group element is diagonalizable over $\mathbb{C}$, because its minimal polynomial divides $X^m-1$, which has no repeated roots. Choose an eigenbasis with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. The induced eigenvalues, including multiplicities, areThe zeroth powers are the one-dimensional trivial representation, with the empty product equal to $1$.

The character of a representation is its trace. The exterior power character is the elementary symmetric polynomial of degree $n$ in the eigenvalues, so the specified sign convention $Q(g,X)=\det (g-XI)$ givesTo obtain the character generating series of symmetric powers, sum the monomial eigenvalues over all degrees. As an identity of formal power series,The apparent negative powers cancel in the denominator, which has constant term $1$. Equivalently, this series is the reciprocal of ${\sum }_{j=0}^d(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g)t^j$. Comparing coefficients gives ${\sum }_{j=0}^{\min (d,n)}(-1{)}^j{\chi }_{{\Lambda }^{j}V}(g){\chi }_{S^{n-j}V}(g)=0$ for $n>0$, an explicit recurrence computing the symmetric-power characters from $Q$.