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 $N$?


1 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.

  • $\begingroup$ Could you point to a reference? Also, very important: does your equation hold for any $N$? Or does it assume $N>k+j$? I ask because I have seen formulas that relate $s$ functions to $o$ functions, but they assume $N$ sufficiently large. $\endgroup$
    – thedude
    Feb 9, 2021 at 17:51
  • $\begingroup$ I think this branching holds without conditions on $N$. The general branching rule for going from $GL_n$ to $O(n)$ can be found in chapter 25 of Fulton and Harris, and this is just an easy case of that. $\endgroup$
    – Nate
    Feb 9, 2021 at 17:58
  • $\begingroup$ But, if the expansion of $p$'s into $s$'s does not depend on $N$, and the expansion of $s$'s into $o$'s also does not depend on $N$, as you suggest, how do you explain that the expansion of $p$'s into $o$'s does depend on $N$? This dependence was confirmed by Richard Stanley in his answer to this question:mathoverflow.net/questions/351648/… $\endgroup$
    – thedude
    Feb 10, 2021 at 0:27

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