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The irreducible characters of the orthogonal group $O(2N)$ are given by $$ o_\lambda(x_1,x_1^{-1},...x_N,x_N^{-1})=\frac{\det(x_j^{N+\lambda_i-i}+x_j^{-(N+\lambda_i-i)})}{\det(x_j^{N-i}+x_j^{-(N-i)})}$$

I was playing with them as basis for the space of homogeneous symmetric polynomials. I wanted to write the function $$p_3=\sum_{i=1}^N (x_i^3+x_i^{-3})$$ as a linear combination of $o_\lambda$'s.

I started with $N=2$ and in that case I found that $$ p_3=o_{(3)}-2o_{(2,1)}+o_{(1)}.$$

However, I then tried with $N=4$ and in that case I found that $$ p_3=o_{(3)}-o_{(2,1)}+o_{(1,1,1)}.$$

I was expecting the coefficients to be the same, i.e. to be independent of $N$. (Independence of $N$ indeed holds when using characters of the unitary group, in which case the coefficients are the characters of the permutation group.)

Have I made some mistake or are the coefficients indeed dependent on $N$?

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The coefficients do depend on $N$. A way to get around this and deal with "universal characters" was found by Koike and Terada (Young-diagrammatic methods for the representation theory of the classical groups of type $B_n$, $C_n$, $D_n$).

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