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Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$. How do I proove that: $[\mathfrak{gl}(n,\mathbb{K}),\mathfrak{gl}(n,\mathbb{K})] = \mathfrak{sl}(n,\mathbb{K})$?

I know that it is easy to see that $[\mathfrak{gl}(n,\mathbb{K}),\mathfrak{gl}(n,\mathbb{K})] \subset \mathfrak{sl}(n,\mathbb{K})$, since the trace vanishes over the product Lie Algebra, however the other inclusion does not work out for me. I know that I could go ahead and compute a minmal generator out of the set $[v_i,v_j]$, where $v_i$ is a basis for $\mathfrak{gl}(n,\mathbb{K})$, but this does not seem like a beautiful proove to me. Is there an elegant way to see this?

QED
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2 Answers2

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Using $\mathfrak{gl}_n(K)\cong \mathfrak{sl}_n(K)\oplus Z$, see below, where $Z$ is the center of $\mathfrak{gl}_n(K)$, we obtain $$ [\mathfrak{gl}_n(K),\mathfrak{gl}_n(K)]=[\mathfrak{sl}_n(K),\mathfrak{sl}_n(K)]\oplus[\mathfrak{sl}_n(K),Z]\oplus [Z,Z]=\mathfrak{sl}_n(K). $$ Here the commutator of $\mathfrak{sl}_n(K)$ is $\mathfrak{sl}_n(K)$, because the Lie algebra is simple and the commutator is a nonzero ideal.

Proving that $\mathfrak{gl}(n,\mathbb{F})=\mathfrak{sl}(n, \mathbb{F}) \oplus \mathfrak{s}(n,\mathbb{F})$.

Dietrich Burde
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  • How do you show $\mathfrak{sl}_n$ is simple? – Kenta S Jul 02 '22 at 20:09
  • @KentaS There is an elementary proof, see this post. – Dietrich Burde Jul 02 '22 at 20:11
  • By the way, you could have just said $[\mathfrak{gl}_n,\mathfrak{gl}_n]\subset\mathfrak{sl}_n$ is an ideal. – Kenta S Jul 02 '22 at 20:17
  • Yes, but the fact $\mathfrak{gl}_n(K)\cong \mathfrak{sl}_n(K)\oplus Z$ is of independent interest. It implies this and many other things easily. – Dietrich Burde Jul 02 '22 at 20:34
  • I think this is a matter of philosophy, whether or not one should use certain facts in proofs. I personally believe when first learning a subject, a more direct argument is oftentimes more illuminating. – Kenta S Jul 02 '22 at 20:37
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    Yes, but the OP explicitly asked not only the direct way, but also "Is there an elegant way to see this?". After you have explained the direct way, there was only to add some "certain facts", as you say. And I still think, they are helpful for the further study. – Dietrich Burde Jul 02 '22 at 20:39
  • Indeed this answer has answered me another question, why $[\mathfrak{sl}_n(K), \mathfrak{sl}_n(K)] = \mathfrak{sl}_n(K)$ without showing that the basis is included, so thank you for that! Even though I kind of need to learn further now, since these examples in my book were given before we talked about simple lie algebras. – QED Jul 03 '22 at 10:24
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Let $e_{ij}\in\mathfrak{gl}(n,\mathbb K)$ be the matrix with a single $1$ in the $(i,j)$-component. Then, it suffices to check that $e_{ij},e_{ii}-e_{jj}\in[\mathfrak{gl}(n,\mathbb K),\mathfrak{gl}(n,\mathbb K)]$ (where $i\ne j$).

We have $e_{ii}-e_{jj}=[e_{ij},e_{ji}]$. I will leave expressing $e_{ij}$ similarly in terms of brackets to you as an exercise.

Kenta S
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    We get that $e_{ij}= [e_{ii},e_{ij}]$ for $i \neq j$, from which we can conclude the result since we now have a basis of $\mathfrak{sl}$ in our set, correct? – QED Jul 02 '22 at 16:13
  • Looks correct to me. – Kenta S Jul 02 '22 at 20:08