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I'm quite new to Lie Algebras, and so there's a lot of easy stuff that I'm probably missing. Anyway following Kac notes I'm asked to compute the center of $\mathfrak{gl}(n,\Bbb K)$, and I've done it in the usual way (defining a bases for two matrices and working it out with brute force).

Anyway I was wondering: since the kernel of the adjoint representation is the center of the group, i.e. $ker(ad)=Z(\mathfrak{gl}(n,\Bbb K))$, is there any more elegant way to arrive to the result, i.e. $Z(\mathfrak{gl}(n,\Bbb K))= \lambda I$, starting from some proprieties of the adojoint representation?

Dac0
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  • I don't think there is going to be any very nice proof along these lines that does not essentially replicate your arguments anyway. The reason is that we obviously need to know something about the specific Lie algebra we have to know the center. – Tobias Kildetoft Feb 05 '16 at 13:35
  • you men you need to know the field K? – Dac0 Feb 05 '16 at 13:45
  • No, I mean we need to use something about the specific Lie algebra (I mean, we cannot make any sort of general conclusion about the size of the center of a Lie algebra without further assumptions anyway). – Tobias Kildetoft Feb 05 '16 at 13:46
  • well the Lie algebra in question is $\mathfrak{gl}(n,\Bbb K)$ with the canonical brackets... should I specify something I'm missing? – Dac0 Feb 05 '16 at 14:10

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I also think that the adjoint representation will not give a "more elegant proof". The computation of the kernel amounts again to the problem to show that $[A,B]=AB-BA=0$ for all $B$ implies that $A=\lambda I$. I do not agree that the proof here necessarily needs to be "brute force". There are several proofs given, among them such which try to avoid matrix computations, see these answers. Also, the usual proof is not so bad.

Community
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Dietrich Burde
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    I believe that Dac0 understands that the center is the kernel of $ad$. I think the question is whether one can use the adjoint representation to find the center without resorting to computations. – David Hill Feb 05 '16 at 16:12
  • I also think that Dac0 understands that the center is $ker(ad)$. The question is perhaps more about the proof in general concerning the center of $\mathfrak{gl}_n(K)$. Here I think $ker(ad)$ is just a reformulation, and does not help. – Dietrich Burde Feb 05 '16 at 21:28