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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?

Lotte
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  • What is $\omega$? Certainly not a vector, otherwise $\omega = \pm\omega$ makes no sense. So what are these "subspaces"? Are you saying: For every nonzero simple $\omega \in \mathrm{Alt}^kV$ such that $\omega = \omega$, there is a unique-up-to-sign $\omega' \in \mathrm{Alt}^kV$ such that $*\omega' = -\omega'$ and $V = [\omega]\oplus[\omega']$, with $$[\omega] = {v \in V ;:; v\wedge\omega = 0}$$ the subspace represented by a simple multivector $\omega$? Then the answer is trivial: $\dim[\omega] = \dim[\omega'] = k$. – Nicholas Todoroff Mar 22 '24 at 18:59

1 Answers1

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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} $.

Andy Zeng
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