# Show that $\mathfrak{o}(2l+1)$ coincides with its derived algebra

Let $$\mathfrak{o}(2l+1)$$ the (odd) orthogonal Lie algebra defined as the Lie algebra of matrices $$x$$ such that $$sx+x's=0$$ ($$x'$$ is the transpose), where

$$s=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_l\\ 0 & I_l & 0 \end{pmatrix}.$$

I want to show that $$\mathfrak{o}(2l+1)\subseteq [\mathfrak{o}(2l+1),\mathfrak{o}(2l+1)]$$.

From the definition it is not hard to find a basis of $$\mathfrak{o}(2l+1)$$ as vector space, so one strategy is to compute the brackets of the elements of the basis in order to find that all of them can be expressed in terms of brackets of the basis. This is not conceptually hard but it's certainly tedious, so my question is the following:

Is there a more conceptual way to show $$\mathfrak{o}(2l+1)\subseteq [\mathfrak{o}(2l+1),\mathfrak{o}(2l+1)]$$ using the implicit definition of the algebra?

## 1 Answer

It is sufficient to notice that this algebra is semisimple, which may be verified by checking that its Killing form is nondegenerate. Of course this requires some computation also. Another route is to find a Cartan subalgebra and the corresponding root system. This is done in many textbooks in Lie theory. I would be very happy to learn about a way to get the desired result with no computations at all, but I'm affraid this may be impossible.

Another way to show that the algebra is semisimple would be to construct any nondegenerate invariant bilinear form - and it is obvious that the Hilbert-Schmidt form does the job. Then you know that it is reductive, so it remains to show that the center is trivial.

• Why is it sufficient to notice that this algebra is semisimple? Semisimple means that it has $0$ radical or equivalently no proper abelian ideals. I'm afraid I can't see the connection – Javi Jun 12 at 14:56
• Dear Javi, the following conditions are equivalent for a finite-dimensional Lie algebra $\mathfrak g$: (1) $\mathfrak g$ is a direct sum of some simple Lie algebras, (2) Killing form of $\mathfrak g$ is nondegenerate, (3) $\mathfrak g$ has no solvable ideals, (4) $\mathfrak g$ has no abelian ideals, (5) $\mathfrak g$ is reductive with trivial center. Among implications between these five conditions, proofs of some are simple exercises. On the other hand proofs of some of the other implications require considerable amount of Lie theory (which can be found in textbooks). – Blazej Jun 12 at 19:37
• It is a fact that if $\mathfrak g$ is a semisimple Lie algebra, then $[\mathfrak g, \mathfrak g] = \mathfrak g$ (it is a nice exercise to find a counterexample for the converse statement). I agree that this fact may not be immediately obvious if you start with the definition (3) or (4) of semisimplicity, but can you see it if you start from the definition (1)? – Blazej Jun 12 at 19:41