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I am learning about symplectic reflection algebras, following the paper by Etingof and Ginzburg. Given a symplectic $\mathbb{C}$-vector space $V$ with symplectic (i.e. non-degenerate alternating bilinear) form $\omega:V\times V\rightarrow \mathbb{C}$, then $s$ is a symplectic reflection if firstly $s\in Sp(V)$ (i.e. $\omega(s(x),s(y))=\omega(x,y)\ \forall x,y\in V$), and secondly $rank(1-s)=2$.

This is likely very obvious, but in the paper it is stated that the direct sum decomposition $V=im(1-s)\oplus ker(1-s)$ is $\omega$-orthogonal. Which I take to mean for each $x\in im(1-s)$ $\omega(x,y)=0\ \forall y\in ker(1-s)$. How would one go about proving this?

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$x$ can be written as $u-s(u)$ and $s(y)=y$ then $$\omega(x,y) = \omega(u-s(u),y) = \omega(u,y)-\omega(s(u),y) = \omega(u,y)-\omega(s(u),s(y)) = \omega(u,y)-\omega(u,y) =0$$

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  • $\begingroup$ it was indeed very obvious! $\endgroup$ – Ted Jh Aug 6 '19 at 9:13

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