# How to write $SO(2n)$ characters in terms of rotation angles?

Say one is working in a representation of $SO(2n)$ such that it has the highest weights $(h_1,...,h_n)$. And let $\{H_i\}_{i=1}^{n}$ be a basis in the Cartan of $so(2n) = Lie(SO(2n))$. Now one says that $\{t_i\}_{i=1}^{n}$ are numbers such that the given a $g\in SO(2n)$ it is conjugate to $e^{\sum_{i=1}^n t_i H_i}$

Then apparently the character of this $g$ in this representation is given as,

$$\chi (h_i,t_i) = \frac{\det (\sinh [ t_i(h_j + n-j)]) + \det (\cosh [ t_i(h_j + n-j)]) }{\det (\sinh [t_i(n-j)]) }$$

But when actually trying to put this into use I am faced with two confusions,

• Given a $g$ how do I determine "a" set of $t_i$? Is there a standard choice of $H_i$ if I am given $g$ in its "defining" representation of a rotation matrix in $\mathbb{R}^{2n}$?

I would like to know of a general method of getting "a" required set of $t_i$ from a given (defining) matrix representation of $g$.

For concreteness (and my immediate necessity!) you can consider for $g$ the rotation matrix in $\mathbb{R}^{2n}$ which rotates by an angle $\alpha$ in the $1$-$2$ plane and keeps everything else fixed. Then the $g$ given is a $2n \times 2n$ matrix such that the first $2\times 2$ block on the diagonal is $\{\{\cos \alpha, \sin \alpha\},\{-\sin \alpha, \cos \alpha \}\}$ and all other entries are $0$.

• Is there some symmetry here which ensures that the WCF will give the same answer for whatever is the choice of basis in the Cartan of the Lie algebra? (..these different choice of basis will clearly change the values of $t_i$..)

• If someone can give a reference to this particular formula for the character then that would be great.

In this MO discussion about this ARupanski had suggested the following way to get a valid set of $t_i$,

Take a $SO(2n)$ matrix say $g$ and then all $2n$ eigenvalues come in pairs $\{a_i, 1/a_i\}$. Now consider the n-vector $v = \{a_i\}$

• Is there a way in which one can think of this $v$ to be sitting in the span of the fundamental weights of $SO(2n)$?

I think the claim is that $t_i$ is determined by writing $v = \sum_{i=1} ^n t_i w_i$ where $w_i$ are the fundamental weights.

• Can someone kindly help understand as to why this is supposed to give a required set of $t_i$?

• Does this work beyond $SO(2n)$ groups?

• – Zev Chonoles Sep 2 '13 at 23:19
• Look up Weyl character formula from Humphreys' (or other books) on representations of Lie algebras. You can just plug in the eigenvalues of a matrix to the variables of a formal character. Don't know how the hyperbolic functions appear here. – Jyrki Lahtonen Sep 7 '13 at 5:37
• @ZevChonoles I am not happy with the MO answer. Its quite confusing and quite a bit of non-standard terminology. I have kind of rewritten the question here. – user6818 Sep 18 '13 at 4:40

## 1 Answer

The $t_1$ is the rotation angle in the 12 plane, $t_2$ is the rotation in the 34 plane etc. If you have an integer entry weight $(h_1,h_2,\ldots)$ this means that acting on vector such as $({\bf e}_1 +i{\bf e}_2)^{h_1} \otimes ({\bf e}_3+i{\bf e}_4)^{h_2}\cdots$ (powers are tensor products) you get a factor of $e^{h_1t_1+h_2 t_2+\cdots}$ on the diagonal of the representation matrix. The character formula sums all these diagonal entries over all elements in the basis. This also works for the spin represntations that have half integer weights.

If you change you choice of Cartan algebra, you are just changing what you mean by the 12 plane the 34 plane etc.