What is the derived algebra of $\mathfrak{sl}(n,\mathbb C)$?

Question

I believe that the derived algebra of gl(n,C) =[gln(C),gln(C)]=[sln(C),Sln(C)]=Sln(C) but what is the derived algebra of Sln(C)? Is it just [sln(c),Sln(C)].

Answer 1

Since $\mathfrak{sl}(n,\mathbb C)$ is a simple Lie algebra and its derived algebra is an ideal which is not $\{0\}$, $\bigl[\mathfrak{sl}(n,\mathbb C),\mathfrak{sl}(n,\mathbb C)\bigr]=\mathfrak{sl}(n,\mathbb C)$.

Answer 2

Another response not using the fact of simlicity of $\mathfrak{sl}(n,\mathbb{C})$. I will prove it for $\mathfrak{sl}(2,\mathbb{C})$ and the same thing for $\mathfrak{sl}(n,\mathbb{C})$ .

If we considere $(e,f,h)$ is the basis of $\mathfrak{sl}(2,\mathbb{C})$ as $[e,f]=h,[h,e]=2e,[h,f]=-2f$.

In one hand: $\forall x,y\in \mathfrak{sl}(2,\mathbb{C}), [x,y]\in \mathfrak{sl}(2,\mathbb{C})$ by the definition of the brakets, i.e. $[\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})]\subseteq \mathfrak{sl}(2,\mathbb{C})$.

In the other hand, $e=[h,\frac{1}{2}e]\in [\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})], f=[h,\frac{-1}{2}f]\in [\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})]$ and $h=[e,f]\in [\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})]$.

Recall that $(h,e,f)$ is the basis, hence $\mathfrak{sl}(2,\mathbb{C})\subseteq[\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})]$.

As conculsion: $\mathfrak{sl}(2,\mathbb{C}) = [\mathfrak{sl}(2,\mathbb{C}),\mathfrak{sl}(2,\mathbb{C})]$.