0
$\begingroup$

$X$ is a linear space over $\mathbb C$,$q$ is a nondegenerate $(X^q=0)$ Hermitian form on $X$. $V$ is a subspace of $X$,$V^q= \{x \in X|q(x,y)=0, \forall y \in V \}$,$dim X/(V+V^q) $ is finite, $(V \cap V^q)^q=V+V^q$,$V={v^q}^q$ ,$m^-(q)$ is the dimension of the maximal negative definite subspace of $q$.

Please prove :$m^-(q)=m^-(q|_V)+m^-(q|_{V^q}))+dim (V \cap V^q)$

I have got no idea about the proof, so I choose to get some help here. Thanks for your instruction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.