Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / B / 6

Let ${\pi }^{\lambda }$ be the permutation character of $M^{\lambda }$. Young's rule gives ${\pi }^{\lambda }={\chi }^{\lambda }+{\sum }_{\mu >\lambda }{K}_{\mu \lambda }{\chi }^{\mu }$, with integer Kostka multiplicities. This dominance-triangular system has diagonal entries one, so recursive subtraction expresses every irreducible character as an integer linear combination of permutation characters. The latter count fixed tabloids, proving that all ${\chi }^{\lambda }(g)$ are integers.

For a $k$-cycle, the Murnaghan–Nakayama rule reduces towhere height is the number of occupied rows minus one, and $f^{\nu }$ is the degree of the remaining Specht character at the identity. In particular a full $n$-cycle has character zero unless $\lambda =(n-r,1^r)$ is a hook, in which case the value is $(-1{)}^r$.

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.

Irreducible character orthogonality gives $⟨\chi ,\chi ⟩=1$. Under the proposed vanishing condition, only the identity contributes, soBut $\chi (1)$ is an integer dimension and $720$ is not a square. Thus no such partition exists.

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is