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This question was asked on MSE some time ago, here, but got no attention.

The Schur functions are characters of irreps of the unitary group, $s_\lambda(U)=Tr(R_\lambda(U))$. They are symmetric functions of the eigenvalues of $U$ and can be written in terms of power sum symmetric functions, $s_\lambda(U)=\sum_\mu c_{\lambda\mu}p_\mu(U)$. The coefficients are the characters of the permutation group.

My question is how this translates for the orthogonal group. If the character of an irreducible representation, $Tr(R_\lambda(O))$, is written in terms of power sums, $\sum_\mu d_{\lambda\mu}p_\mu(O)$, what is known about the coefficients?

Same question for writing Schur functions in terms of orthogonal characters, and vice-versa.

References would be appreciated.

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2 Answers 2

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I might be totally wrong, but the analogues of Schur functions in the orthogonal case, the so called orthogonal characters are not polynomials in just the $x_i$, but polynomials in $x_i^{\pm 1}$. You can perhaps treat the negative alphabet separately, and expand in say $p_{\lambda}(x_1,x_2,\dotsc,x_n)p_\mu(x_1^{-1},x_2^{-1},\dotsc,x_n^{-1})$.

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  • $\begingroup$ Wait, the formula you give for $o_\lambda$ in your page as a determinant of a difference of $h$-functions is symmetric in all its arguments. I don't understand. $\endgroup$
    – Marcel
    Commented Jun 14, 2019 at 17:14
  • $\begingroup$ @Marcel : ah, i should add a warning - on the top of the page, it is stated what the alphabet is, (which includes the negative powers!). $\endgroup$
    – Per Alexandersson
    Commented Jun 14, 2019 at 17:42
  • $\begingroup$ Exactly, so the function $o_\lambda(x)$ has $2n$ arguments and has full $S_{2n}$ symmetry with respect to them, yes? So why did you suggest in your answer to treat the positive and negative alphabets separately? $\endgroup$
    – Marcel
    Commented Jun 14, 2019 at 17:46
  • $\begingroup$ @Marcel: Ah you are right of course! Hm, so there is a nice relationship using $\omega$ between orthogonal and symplectic Schur, so if there is a Murnaghan-Nakayama rule for one of these, then there is one for the other. $\endgroup$
    – Per Alexandersson
    Commented Jun 14, 2019 at 19:05
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I don't know about writing ${\rm Tr}(R_\lambda(O))$ in terms of power sums, but the reverse procedure can be carried out using the so-called "characters of the Brauer algebra" (they are not really characters).

This theory is developed by Arun Ram in two papers:

  • "Characters of Brauer's centralizer algebra", Pacific J. Math. 169, p.173, 1995
  • "A ‘Second Orthogonality Relation’ for Characters of Brauer Algebras", Europ . J . Combinatorics 18, p. 685, 1997

In those works he gives some combinatorial way to compute such characters and also provides a formula for them in terms of the characters of the permutation group.

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