Are top Brauer characters bounded?

Question

Let $p_\lambda$ be power sum symmetric functions. Let $s_\lambda$ and $o_\lambda$ be irreducible characters of the unitary and orthogonal groups $U(N)$ and $O(N)$, respectively (the $s$ are the Schur functions).

Then $$p_\mu(X)=\sum_{|\lambda|=n}\chi_\lambda(\mu)s_\lambda(X)$$ and $$p_\mu(X)=\sum_{|\lambda|\leq n}b_{N,\lambda}(\mu)o_\lambda(X),$$ where $\chi_\lambda(\mu)$ are irreducible $S_n$ characters and $b_{N,\lambda}(\mu)$ are what I am calling Brauer characters (in contrast with the former, the latter depend on $N$, the dimension of the matrix $X$).

For the special case when $\mu$ is the singleton, $\mu=(n)$, it is known that $\chi_\lambda(n)$ is non-zero only if $\lambda$ is a hook, and it is $\pm 1$ depending on the size of the hook. In particular, this character is bounded, i.e. $|\chi_\lambda(n)|\leq 1$ for any $\lambda$.

My question is if the corresponding Brauer characters are also bounded, $|b_{N,\lambda}(n)|\leq 1$ for any $\lambda$ and $N$?

Answer 1

The branching rule for hook partitions is multiplicity free and pretty easy to describe:

$$s_{(k,1^j)} = \sum_{i < k/2}o_{(k-2i,1^j)} + \sum_{i < k/2}o_{(k-2i-1,1^{j-1})}$$

That is: You either remove an even number of boxes from the first row of the hook and leave the first column as is, or you remove one box from the first column and an odd number of boxes from the first row.

If you substitute this into your first formula when $\mu = (n)$ each orthogonal group character seems to appear in at most 2 terms in the sum - the hook with the same first column size and the one with column size one larger. Moreover when there are two terms they have opposite signs and cancel out.