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It seems that there are two ways to define the Lie algebra $\mathrm{so}_{2n}(\mathbb{C})$. The first one is $\mathrm{so}_{2n}(\mathbb{C})_{(1)}:=\{M \in \mathrm{gl}_{2n}(\mathbb{C}) \ | \ M + M^t = 0\}$, the second one is $\mathrm{so}_{2n}(\mathbb{C})_{(2)}:=\{M \in \mathrm{gl}_{2n}(\mathbb{C}) \ | \ BM + M^tB = 0\}$ where $B = \begin{bmatrix} 0 & I_n \\ I_n & 0\end{bmatrix}$.

In A. Kirillov's notes on page 128, it is stated that these two Lie agebras provide two alternative descriptions of $D_n$, and that they are isomorphic via $M \mapsto MB$. However, this is not true, the commutator is not preserved: $[M_1,M_2]\cdot B \neq [M_1\cdot B,M_2\cdot B]$ for skew-symmetric $M_1$ and $M_2$. What's the problem here?

Grabovsky
  • 151
  • I agree that this isn't an isomorphism of Lie algebras although it is an isomorphism as vector spaces. They are isomorphic as Lie algebras but not like this. Indeed I doubt you can base the isomorphism on a real valued matrix as the corresponding real Lie algebras defined this way are not isomorphic – Callum Nov 20 '23 at 15:12

1 Answers1

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This question was answered in by Torsten Schoeneberg in here, I'll just give explicit formulas. So, $B$ has $n$ eigenvalues $-1$ and $n$ eigenvalues $+1$. Set $V:=\frac{1}{\sqrt{2}}\begin{bmatrix}I & I\\ -I & I \end{bmatrix}$, $V^{-1} = V^t = \frac{1}{\sqrt{2}}\begin{bmatrix} I & -I \\ I & I\end{bmatrix}$.

Then $V^t B V = \begin{bmatrix} -I & 0 \\ 0 & I\end{bmatrix}$. Now set $J:=\begin{bmatrix}iI & 0 \\ 0 & I \end{bmatrix}$.

Then $J^2 = \begin{bmatrix} -I & 0 \\ 0 & I\end{bmatrix}$.

Finally, set $P:= VJ$. Then $P^t B P = I$ and $\bar{P}^t P = P\bar{P}^t = I$. The desired isomorphism is given by, for example, $M \mapsto \bar{P}^t M P$.

Grabovsky
  • 151