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Definition: $\mathfrak{so}(2n)$ is the subspace of matrices $A \in \mathfrak{gl}(2n)$ such that $AJ + JA^{t} = 0$, with $J \in \mathfrak{gl}(2n)$ and $$J= \begin{pmatrix} 0 & 0 & ... & 0 & 1 \\ 0 & 0 & ... & 1 & 0\\ \vdots & \vdots & \ddots & \vdots& \vdots \\ 0 & 1 & ... & 0 & 0 \\ 1 & 0 & ... & 0 & 0 \\ \end{pmatrix}$$

In other words, $J$ has the secondary diagonal with all entries equal to 1 and all the other entries are zero.

Questions

  1. Prove that $\mathfrak{so}(2n)$ has a subalgebra isomorphic to $\mathfrak{gl}(n)$.

  2. Prove that $\mathfrak{so}(2n)$ doesn't have any subalgebra isomorphic to $\mathfrak{gl}(n+1)$.

I have very few ideas on how to prove this: I tried to use the fact that, for example, for $n=2$, a matrix $A \in \mathfrak{so}(4)$ is of the following form $A=\begin{pmatrix} t_1 & a & z & 0 \\ b & t_2 & 0 & -z \\ u & 0 & -t_2 & -a \\ 0 & -u & -b & -t_1 \\ \end{pmatrix}$. That may suggest that the top left $2\times 2$ block is in $\mathfrak{gl}(2)$, but I'm not sure this is the right way to tackle the problem. Any hints?

C.F.G
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cip
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  • Per the discussion in the comments on your previous question, your definition of $\mathfrak{so}(2n)$ is confusing since it doesn't correspond to the usual definition. It would be helpful if you said what your name for $\mathfrak{so}(2n)$ is so that we can figure out what it's usually called. – Ben Grossmann Feb 07 '20 at 14:55
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    @Omnomnomnom: OP clarified in a comment there that they work over an algebraically closed field; once this is clarified here, the notation and definition might still be uncommon, but it makes sense and is equivalent to whatever usual one people have. -- nereide: I think your argument for 1) easily generalises to all $n$ and then suffices to prove that. For 2), that looks harder except for small $n$ where there's an easy dimension argument. I feel like what we're getting at here is that the root system $D_n$ contains $A_{n-1}$ but not $A_n$, but things get subtle on the algebra level. – Torsten Schoeneberg Feb 07 '20 at 16:25
  • @TorstenSchoeneberg in the usual definition, $A \in \mathfrak{so}$ if and only if $A$ is skew-symmetric. With this definition, we have $A \in \mathfrak{so}$ if and only if $JA$ is skew-symmetric. I'm not sure if these algebras are isomorphic. – Ben Grossmann Feb 07 '20 at 17:25
  • @TorstenSchoeneberg Would it be sufficient to prove this via inclusion of the Dynkin diagrams, as you suggested? I know that inclusion of Dynkin diagrams corresponds to that of root systems: does this imply that the corresponding Lie algebras are included one into the other? – cip Feb 07 '20 at 18:25
  • Maybe this is true for semisimple Lie algebras, so it is not applicable to the case of $\mathfrak{gl}(n)$ – cip Feb 07 '20 at 19:04
  • @Omnomnomnom: These algebras are isomorphic over algebraically closed fields of characteristic $0$. I appreciate your skepticism though, maybe you want to support my point in the comments to answers to https://math.stackexchange.com/q/323801/96384, where I basically make the same point as you, if the ground field is not algebraically closed. – Torsten Schoeneberg Feb 08 '20 at 05:10
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    @nereide: It is true that an inclusion of Dynkin diagrams implies an inclusion of the corresponding semisimple Lie algebras, although already that is not as trivial as it sounds, and the reverse implication is far from true; also, as you notice yourself, we are not actually dealing with semisimple Lie algebras here. So your proof idea for 1 is much better here than going via Dynkin diagrams, and I guess there must be a similar idea for 2 which just works with the matrices, although I fail to see it. – Torsten Schoeneberg Feb 08 '20 at 05:13
  • @T Thanks for the link, I'll take a look. – Ben Grossmann Feb 08 '20 at 09:24

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