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I want to show that the complexification of $$\mathfrak{so}(p,q)=\{M \in Mat(p+q,\mathbb{R}) | M^TJ_{p,q}+J_{p,q}M=0\}$$ is isomorphic to $$\mathfrak{so}(p+q,\mathbb{C})=\{M \in Mat(p+q, \mathbb{C}) | M^T+M=0\} .$$ Here, $J_{p,q}$ is the diagonal $n\times n$-matrix with p times 1 and q times -1 on the diagonal. But I don't know how I should start the proof.

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  • $\begingroup$ Use the conjugation map. $\endgroup$ – Aolong Li Oct 13 '17 at 2:08
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Hint Pick a basis ${\mathcal B} = (e_1, \ldots, e_p, e_{p + 1}, \ldots, e_{p + q})$ of $\Bbb R^{p + q}$ so that the bilinear form preserved by $\mathfrak{so}(p, q)$ has matrix representation $J_{p, q}$ with respect to $\mathcal B$.

Now, consider the basis $${\mathcal B}' := (e_1 \otimes 1, \ldots, e_p \otimes 1, e_{p + 1} \otimes i, \ldots, e_q \otimes i)$$ of $\Bbb R^{p + q} \otimes \Bbb C \cong \Bbb C^{p + q}$. What is the matrix representation of the induced bilinear form with respect to $\mathcal{B}'$?

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  • $\begingroup$ What do you mean the standard action? Simply $A.x=A\cdot x$? And what is the meaning of pseudo-orthonormal? $\endgroup$ – user95 May 31 '17 at 17:03
  • $\begingroup$ More concretely I mean the bilinear form $g$ such that that $g(Ax, Ay) = g(x, y)$---this is uniquely determined by the Lie algebra up to an overall scale. $\endgroup$ – Travis Willse May 31 '17 at 17:39
  • $\begingroup$ By pseudo-orthonormal I just mean that $g(e_a, e_a) = 1$ for $1 \leq a \leq p$ and $g(e_a, e_a) = -1$ for $p < a \leq p + q$. In the case that $q = 0$ this is just an orthonormal basis. $\endgroup$ – Travis Willse May 31 '17 at 17:40
  • $\begingroup$ I've rewritten my answer to be a little more direct. $\endgroup$ – Travis Willse May 31 '17 at 17:46
  • $\begingroup$ I calculated that the representation matrix of the induced bilinear form is simply the diagonal matrix with 1 in the first p rows and i in the other rows. But I still don't see how this will help me to prove the isomorphy. $\endgroup$ – user95 May 31 '17 at 17:57

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