# Orthogonal complex structures for indefinite inner products

Suppose that $$(V,b)$$ is a real even-dimensional ($$n=2k$$) vector space with a nondenegenerate symmetric bilinear form $$b$$.

Question : Is there (always) a linear map $$J:V\to V$$ such that

1. $$J^2 = -I$$
2. $$J\in O(V,b)$$

I know that in the case of a positive-definite form $$b$$, such a (orthogonal complex) structure always exists, but I was not able to see how to extend that (or prove that we cannot extend that) to the non-positive-definite case, or find references for that question.

The existence of a complex structure depends on the signature $$(p,q)$$ of $$b$$. If $$p$$ and $$q$$ are even, then there does exist a complex structure. Within $$O(p,q)$$ you can find a copy of $$O(p) \times O(q)$$, which reduces to the case you know.
On the other hand, if $$p$$ is odd (so also $$q$$ is odd) then there won't be a complex structure.
For example, let $$(V,b)$$ be $$\mathbb{R}^2$$ with $$b(v,w) = v^T \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} w .$$ If $$J \colon V \to V$$ satisfies $$\det (J)= \pm 1$$ and $$J^t \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} J = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ then either $$J = \begin{pmatrix} a & b \\ b & a \end{pmatrix} \text{ and } \det (J)=1, \text{ or } \begin{pmatrix} a & b \\ -b & -a \end{pmatrix} \text{ and } \det(J)=-1 .$$ You can check that if you further impose $$J^2=-1$$ there will be no solutions.