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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.

Travis Willse
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user95
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1 Answers1

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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 $\mathcal{B}'$?

Travis Willse
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  • What do you mean the standard action? Simply $A.x=A\cdot x$? And what is the meaning of pseudo-orthonormal? – user95 May 31 '17 at 17:03
  • 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. – Travis Willse May 31 '17 at 17:39
  • 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. – Travis Willse May 31 '17 at 17:40
  • I've rewritten my answer to be a little more direct. – Travis Willse May 31 '17 at 17:46
  • 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. – user95 May 31 '17 at 17:57
  • @user95 The matrix should be the identity matrix. Note, for example, that (for $q > 0 $) the induced bilinear form $g$ satisfies $g(i e_{p + 1}, i e_{p + 1}) = i^2 g(e_{p + 1}, e_{p + 1}) = (-1) (-1) = 1$. As for establishing the isomorphism, remember that, in terms of the concrete definitions of your Lie algebras, we're looking for a Lie algebra isomorphism that takes a (general, complex) matrix $M$ satisfying $M^T J_{p,q} + J_{p, q} M^T = 0$ to a complex matrix satisfying $M^T + M = 0$. – Travis Willse May 31 '17 at 18:53
  • I thought the induced bilinear form is $b_{\mathbb{C}}(1\otimes a_1+i\otimes b_1,1\otimes a_2+i\otimes b_2)=b(a_1,a_2)+ib(b_1,b_2)$. But then we have for $i\in {p+1,..,p+q}$ that $b_{\mathbb{C}}(i\otimes e_i,i\otimes e_i)=ib(e_i,e_i)=i(-1)=-i$. Can you tell me where is the mistake? Or did you mean something else by the induced bilinearform? Further I still don't see how I can use that to create the isomorphism.. – user95 May 31 '17 at 20:35
  • The fact $i$ occurs twice in $b_{\Bbb C}(i \otimes e_i, i \otimes e_i)$ means that $b_{\Bbb C}(i \otimes e_i, i \otimes e_i) = i \cdot i \cdot b_{\Bbb C}(e_i, e_i) = - b_{\Bbb C}(e_i, e_i)$ – Travis Willse May 31 '17 at 21:25
  • okey, I fixed the problem. Can you give me any other hint how I can use that for constructing the isomorphism? – user95 Jun 01 '17 at 05:26
  • @TravisWillse I tried the map $M \mapsto J_{p,q}M$. This gives me a bijection, but it does not seem to be a lie algebra homomorphism? Is there a different map? – user100101212 Apr 06 '20 at 17:13
  • @user100101212 Yes, that is not the correct map; what is the change-of-basis matrix $T_{\mathcal B' \leftarrow \mathcal B}$? – Travis Willse Apr 07 '20 at 00:20