Dimension of "self-dual" and "anti-self-dual"

Question

Let $V$ a $2k$-dimensional nondegenerate quadratic space and assume that the index is such that $**=\text{Id}: \text{Alt}^kV \rightarrow \text{Alt}^kV$ (Hodge star operator). Then $V$ is the direct sum of the subspaces "self-dual" ($*\omega=\omega$) and "anti-self-dual" ($*\omega=-\omega$). How can I calculate the dimensions of these subspaces?

Answer 1

Note that,$** = (-1)^{k^2}Id$ on a $2k$ dimensional space, which does not always gives you self-dual and anti-self-dual decompostion. Now assume that $k^2$ is even. $\omega = \frac{1}{2}(\omega - \star\omega) + \frac{1}{2}(\omega + \star\omega)$, which gives you an orthogonal decomposition of $\Lambda^k V$, hence $\dim V^+ = \dim V^- = \frac{1}{2} \dim \Lambda^k V = \frac{1}{2} {\dim V \choose k} = \frac{1}{2} {2k \choose k} $.