Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 5 / 2

For an integer tuple $\ell =({\ell }_1,\dots ,{\ell }_m)$, the monomial alternant isNegative exponents require nonzero coordinates; all exponents in the character expansion below are nonnegative. The power-sum symmetric polynomial is $s_k(x)={\sum }_{i=1}^m{x}_i^k$. Put $\delta =(m-1,m-2,\dots ,0)$, so $A_{\delta }(x)={\prod }_{i<j}(x_i-x_j)$ is the Vandermonde determinant.

Let a conjugacy class of $S_n$ have ${\alpha }_k$ cycles of length $k$, with $\sum k{\alpha }_k=n$, and put $p_{\alpha }(x)={\prod }_k{s}_k(x{)}^{{\alpha }_k}$. The product $p_{\alpha }{A}_{\delta }$ is an alternating polynomial homogeneous of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. In an alternating polynomial, a monomial with two equal exponents has zero coefficient, since interchanging those variables fixes the monomial and reverses its sign. Grouping the remaining monomials by their permutation orbits gives a unique expansion in alternants $A_{\ell }$ with ${\ell }_1>\cdots >{\ell }_m\ge 0$.

Such tuples of the indicated total degree are exactly ${\ell }_i={\lambda }_i+m-i$ for partitions of an integer $\lambda$ of $n$ with at most $m$ parts. Define the class functionThe displayed monomial occurs with coefficient $1$ in $A_{\lambda +\delta }$ and in no other ordered alternant, soThis proves the expansion and explicitly defines its coefficients. They depend only on the cycle counts and hence are class functions. Identifying these coefficients with Specht module characters is the Frobenius alternant character formula, which the remaining parts allow us to assume.

Take $m=2$ and $\lambda =(2,1)$, so $\lambda +\delta =(3,1)$. A transposition in $S_3$ has one singleton cycle and one two-cycle. The Frobenius alternant character formula therefore givesIn the coordinate permutation representation on ${\mathbb{C}}^3$, a transposition fixes one basis vector, so its character is $1$. This representation is the direct sum of the invariant line of constant vectors and the standard representation of the symmetric group, whose vectors have coordinate sum zero. Subtracting the trivial character gives ${\chi }_V((12))=1-1=0$, as required.

The Young permutation module $M^{\lambda }$ has a basis of tabloids, equivalently ordered row sets of sizes ${\lambda }_1,\dots ,{\lambda }_r$. A tabloid is fixed by $\sigma$ precisely when every cycle of $\sigma$ lies entirely in one row. Assigning a length-$q$ cycle to row $j$ contributes $x_j^q$. Distinct cycles can be assigned independently, so the character isPadding to $n$ variables with zero row sizes gives exactly the same coefficient. Explicitly, the character of a Young permutation module isHere $a_{qj}$ counts the length-$q$ cycles assigned to row $j$. Rows remain distinguished even when their sizes are equal, so there is no further division by permutations of equal rows.