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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.

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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.

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