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I have the same question as this one from a long time ago. Is there an easy way to see that the Killing form on $\mathfrak{sp}(2n)$ is $\kappa(x,y) = (4n+2) \mathrm{tr}(xy)$?

For example, the Killing form for $\mathfrak{sl}(n)$ can be found from the easier to calculate case of $\mathfrak{gl}(n)$ using the decomposition $$\mathfrak{gl}(n) = \mathfrak{sl}(n) \oplus \mathbb{C}.$$ I don't know a similar argument for $\mathfrak{sp}(2n)$ though.

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user342127
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  • I think this is worked out in detail in Fulton&Harris. You can just take a look. – Bombyx mori May 25 '16 at 04:16
  • @Bombyxmori It looks like they describe the Killing form only on diagonal matrices. Also they get the different answer $(4n+4) \mathrm{tr}(xy)$ and it doesn't look like they have a different convention for $n$. The proof is the usual "One can compute..." I'm not sure who to believe. – user342127 May 25 '16 at 04:21
  • You should believe Fulton&Harris. I went through their computations for $\mathcal{sl}(n)$ and they were correct. I went through some of their computation for other cases, and they were correct as well. The proofs may not be the cleanest, but for classical Lie groups you can always do everything by hand. – Bombyx mori May 25 '16 at 04:23
  • Also the adjoint map for low dimensional Lie algebras on itself is not that difficult to compute. I remember I used to compute $sl(n)$ simply using definition. There may be tricks and various short cuts you can take - but if I could do it, I think you can do it too. – Bombyx mori May 25 '16 at 04:26
  • @Bombyxmori I'll keep trying it but it looks real ugly. $\mathfrak{gl}(n)$ has an easy basis and $\mathfrak{sl}(n)$ is just the restriction because of the decomposition above. $\mathfrak{so}(n)$ is easy because diagonal elements are $0$. $\mathfrak{sp}(n)$ is a huge pain in the pooper to do by hand though as far as I can tell. – user342127 May 25 '16 at 04:29
  • I have not touched the material for many years, but the $sl(n)$ case is really no more difficult than $gl(n)$ once you fix a basis. I suspect the $sp(n)$ case should not be that difficult either. But I agree it is the most complicated of the four $A,B,C,D$ cases. – Bombyx mori May 25 '16 at 04:34

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For the computation of the Killing form of the classical Lie algebras, the crucial thing is to note that a bilinear form $\rho: {\mathfrak g}\times {\mathfrak g}\to {\mathbb C}$ which is non-degenerate and associative - i.e. $\rho([X,Y],Z)=\rho(X,[Y,Z])$ - is unique up to scalar. The Killing form is associative in general and non-degenerate for semisimple Lie algebras, while for the trace form you can proceed directly.

With this at hand, it suffices to find any non-zero value of the forms under consideration to compute the scalar factor between them. For example, it is convenient to pick elements from a Cartan subalgebra because for them the Killing form is easy to compute.

Does this help?

Hanno
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  • OK I understand this argument. Don't you need to know in advance that $\mathfrak{sp}(2n)$ is simple to know that its Killing form is determined up to scalar by being associative though? – user342127 May 25 '16 at 04:45
  • Yes, you need the knowledge that ${\mathfrak s}{\mathfrak p}(2n)$ is simple. If you don't know or don't want to use this, you might argue ad hoc by noting that the radical of the correctly scaled difference of the two forms is firstly an ideal and secondly contains some diagonal matrix. Now you can proceed constructively to show that from any nonzero diagonal matrix you can build everything else by linear combination and bracketing. – Hanno May 25 '16 at 05:08
  • For the latter, the root space decomposition is again the crucial point, as it was in the computation of example values of the Killing form. – Hanno May 25 '16 at 05:13
  • OK, thanks! I "know" $\mathfrak{sp}(2n)$ is simple but the only way I knew how to prove it involved showing that its Killing form is nondegenerate. I think I can work with your explanation though. – user342127 May 25 '16 at 05:23