I'm trying to make sense of this statement that appears on wiki:
"The group O(p,q) is defined for vector spaces over the reals. For complex spaces, all groups O(p,q; C) are isomorphic to the usual orthogonal group O(p + q; C), since the transform $z_j \mapsto iz_j$ changes the signature of a form."
I don't quite get it. Can anybody give me a proof? And also, does it work for SO(p,q) as well?
Thanks
Let $$Q_{p,q}(z,w) = \sum_{i = 1}^{p} z_i w_i - \sum_{i=p+1}^{p+q} z_i w_i.$$ This is the standard symmetric bilinear form of signature $(p,q)$ on $\Bbb C^{p+q}$ (note that the definition of indefinite orthogonal groups uses bilinear forms, not sesquilinear). The complex indefinite orthogonal group is then defined by $$\mathrm{O}(p,q; \Bbb C) = \{A \in \mathrm{GL}_{p+q}(\Bbb C) : Q_{p,q}(Az, Aw) = Q_{p,q}(z,w)\}.$$ If we take $\{e_i\}$ to be the standard basis of $\Bbb C^{p+q}$ and define $$\varphi(e_i) = \begin{cases} e_i, & 1 \leq i \leq p, \\ ie_i, & p+1 \leq i \leq q \end{cases}$$ then we have that $$Q_{p,q}(\varphi(z), \varphi(w)) = \sum_{i = 1}^p z_i w_i + \sum_{i = p+1}^{p+q} z_i w_i = \sum_{i = 1}^{p+q} z_i w_i = Q_{p+q,0}(z,w).$$ Hence $\varphi$ is an isomorphism between the symmetric bilinear forms $Q_{p,q}$ and $Q_{p+q,0}$. Then we get an induced isomorphism $$\mathrm{O}(p,q; \Bbb C) \xrightarrow{~\cong~} \mathrm{O}(p+q; \Bbb C),$$ $$A \mapsto A \circ \varphi.$$
This of course works for $\mathrm{SO}(p,q; \Bbb C)$ as well since we just need to take the determinant $1$ elements for that.