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Let $\mathfrak{so}(2n,\mathbb{C})$ be all the matrices in the form of $g=\begin{pmatrix}A & B\\ C & -A^T\end{pmatrix}$, where $B$ and $C$ are skew symmetric and $A,B,C$ are $n\times n$ matrices. Show that $\mathfrak{so}(2n,\mathbb{C})$ is a simple Lie algebra.

By some simple calculation I can deduce that if $r$ is a non-zero ideal of $\mathfrak{so}(2n,\mathbb{C})$ containing an element in the form $\begin{pmatrix}A & B\\ C & -A^T\end{pmatrix}$ then there must be an element in $r$ such that $B\neq 0,C\neq 0$. Also the elements $\begin{pmatrix}A & 0\\ 0 & -A^T\end{pmatrix}$ and $\begin{pmatrix}0 & \pm B\\ \pm C & 0\end{pmatrix}$ are in $r$. But how can I go ahead?

Li Li
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2 Answers2

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It is possible to show directly that $\mathfrak{so}(2n,\mathbb{C})$ is a simple Lie algebra for $n\ge 3$. One starts with a non-zero ideal $I$ and applying successively adjoint operators one shows that all basis vectors have to lie in $I$, so that $I=\mathfrak{so}(2n,\mathbb{C})$. Hence the algebra is simple. This is proved in detail in the book "naive Lie theory" by John Stillway, see the answer here.

However, there are much more elegant ways to show this, e.g., by using the Killing form $$\kappa (x,y)=tr(ad(x)ad(y)).$$

It is easy to see that the Killing form of $\mathfrak{so}(2n,\mathbb{C})$ is a non-zero multiple of the trace form and is non-degenerate. Then it follows that $\mathfrak{so}(2n,\mathbb{C})$ is semisimple. Since the root system for $n>2$ is indecomposable, it is also simple.

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Dietrich Burde
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  • And to connect to key-words, the implication that non-degenerate Killing form implies (semi-) simplicity is "Cartan's criterion". – paul garrett Mar 18 '17 at 12:49
  • But the non-degeneration of the Killing form only implies the semi-simplicity. How can I deduce the simplicity? – Li Li Mar 18 '17 at 16:19
  • Fixed. And no need for more effort, the direct proof (John Stillwell) is also fine. – Dietrich Burde Mar 18 '17 at 20:38
  • But how to identify my space with the $\mathfrak{so}(2n,\mathbb{C})$ in John Stillway's book which is the space of all $2n\times 2n$ skew symmetric matrices?The linear isomorphism $a\mapsto Ba$ does not preserve the Lie bracket, where $B$ is the matrix $\begin{pmatrix}0 & I\ I & 0\end{pmatrix}$. – Li Li Mar 19 '17 at 03:03
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    This is explained in books on Lie algebras, e.g., in the book of K. Erdmann and M. Wildon, page $33$. – Dietrich Burde Mar 19 '17 at 20:11
  • How can I get this Killing form, it is hard to compute for general n. – Goldmund Mar 19 '23 at 13:30
  • @Goldmund No, it is not hard at all to compute - see for example wikipedia. – Dietrich Burde Mar 19 '23 at 13:34
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Thanks to the hints by Dietrich Burde, I can give a sketch of proof. First, for $gl_S(n)=\{x\in gl(n)|x^TS+xS=0\}$, we can show that $gl_S(n)\cong gl_T(n)$ if there exists a complex invertible matrix $P$ such that $P^TSP=T$(the isomorphism given by $A\mapsto P^{-1}AP$). Then we observe that there exists such a $P$ satisfying $P^T\begin{pmatrix}0 & I\\ I & 0\end{pmatrix}P=I_{2n}$. Hence we turn to show that the space of $2n\times2n$ complex skew-symmetric matrices is a simple Lie algebra. Then using the method(just some calculation) given by chapter 6.5 of John Stillwell's book Naive Lie Theory, we can prove it. Of course, as Qiaochu mentioned, $n$ has to be more than or equal $3$.

Li Li
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