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 $N$?



Using irreducible characters of the orthogonal group as basis for homogeneous symmetric polynomials
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 $B\_n$, $C\_n$, $D\_n$](https://www.sciencedirect.com/science/article/pii/0021869387900998)).


