Dimensions of Orthogonal Group Representations

Question

I'm aware that the irreducible representations of the orthogonal group $O(n;\mathbb{C})$ are labeled by partitions $\lambda$ such that the sum of the first two columns of $\lambda$ is at most $n$. Is there a way to determine the dimension of the $\lambda$ representation? Perhaps by counting some kind of tableau, similar to how we do for the symmetric group $S_n$?

Answer 1

Yes there is a tableau way but it doesn't involve counting tableaux. The dimensionality is best computed using the method exposed by Ron King in

King RC. The dimensions of irreducible tensor representations of the orthogonal and symplectic groups. Canadian Journal of Mathematics. 1971 Feb;23(1):176-88.

The paper is very clear so let me just provide you with an elementary example for $O(n)$. Suppose you have a valid partition with $p$ parts - say $(3,1)$ with $p=2$ parts.

One first construct the tableau for $(3,1)$, placing at the end of the first row $n-1$, at the end of the second row $n-3$ etc. In this case, this first step yields $$ \begin{array}{cccccc} *&*&n-1\\ n-3&& \end{array} $$ and then fill the * by adding one for every step in towards the left: $$ \begin{array}{cccccc} n+1&n&n-1\\ n-3&& \end{array}\, . \tag{1} $$

For your partition of $p$ parts, construct now a triangular diagram with $p$ boxes in the first row, $p-1$ in the second row etc, and fill the entries antidiagonally with the entries of your partition. Here, our partition has two parts: $3$ and $1$; so we construct a triangular tableau with two boxes on the first row, and fill up as indicated to obtain $$ \begin{array}{cccccc} 3&1\\ 1&\\ \end{array} \tag{2} $$ Add the entries in the partitions (1) and (2): $$ \begin{array}{cccccc} n+1+3&n+1&n-1\\ n-3+1 \end{array} $$ and then proceed as usual, including hook lengths in the denominator, to obtain (if I didn't mess up too much) $$ (n-1)\frac{(n+1)}{2}\frac{(n+4)}{4}(n-2) \, . $$ There is an additional twist if the triangular tableau has some rows greater than your original partition, and King shows how to handle this case (using what are now known as the King numbers).