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I'm working on a homework assignment in which part of the question statement says that each of the classical Lie algebras can be described as the set of all matrices $X \in gl(n,\mathbb{C})$ satisfying

$$XM + MX^T = 0$$

for some particular choice of $M$. Specifically, it says that if $M$ is the $n\times n$ unit matrix $\mathbb{I}_n$, then $X \in A_{(n-1)} = su(n)$. But that doesn't seem to make sense. For $M = \mathbb{I}_n$, if my math is correct, the equation above indicates that $X$ should be limited to $n\times n$ antisymmetric matrices, which is not the same set as the $su(n)$ Lie algebra. So it seems that there is a mistake in the question statement. Is that correct?

As a secondary question, it'd be nice to get confirmation for the given definitions of the other classical algebras:

  • Setting $M = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & \mathbb{I}_n \\ 0 & \mathbb{I}_n & 0\end{pmatrix}$ implies $X \in B_n = so(2n+1)$
  • Setting $M = \begin{pmatrix}0 & -\mathbb{I}_n \\ \mathbb{I}_n & 0\end{pmatrix}$ implies $X \in C_n = sp(n)$
  • Setting $M = \begin{pmatrix}\mathbb{I}_n & 0 \\ 0 & \mathbb{I}_n\end{pmatrix}$ implies $X \in D_n = so(2n)$

I didn't tag this [homework] because the question I'm asking is not the question I'm supposed to be solving in the assignment. I'm just looking for clarification on the instructions. (Also, I'm asking here because the instructor isn't available to contact right now)

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  • $\begingroup$ The Wikipedia page on classical Lie groups explains that the Lie groups of classical type are those preserving certain bilinear or sesquibilinear nondegenerate forms. The corresponding result for Lie algebras can be deduced by differentiation. $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 10, 2011 at 3:33
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    $\begingroup$ That transpose should be a complex-conjugate-transpose. It is then true that the compact forms of the classical Lie algebras are all of this form. $\endgroup$
    – David E Speyer
    Feb 10, 2011 at 16:56
  • $\begingroup$ @David: thanks, I had a feeling that would be the case. If you'd like to make that an answer I'll be happy to mark it as accepted. $\endgroup$
    – David Z
    Feb 10, 2011 at 18:31
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    $\begingroup$ @DavidSpeyer Maybe you want to follow the OP's suggestion of converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$
    – Julian Kuelshammer
    Jun 15, 2013 at 7:09
  • $\begingroup$ @JulianKuelshammer I'm no longer sure what I was thinking when I wrote this comment; either it is false or it is true for a non-obvious reason I don't remember. (The OP seemed to think it was true, though.) $\endgroup$
    – David E Speyer
    Jun 22, 2013 at 1:38

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