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I want to prove in several different ways that $\mathfrak{sl}_4(\mathbb{C})$ is semisimple, where $$\mathfrak{sl}_4(\mathbb{C}) = \{ x \in \mathfrak{g}_4(\mathbb{C})\, | \, Tr(x)=0\}.$$

I know there exist a lot of criterions of semisemplicity. In particular, for $L=\mathfrak{sl}_4(\mathbb{C})$, I can use:

  1. Brute force
  2. $Rad(L) = Z(L) = 0$
  3. Killing form is non degenerate
  4. $L$ is simple
  5. $L$ doesn't have abelian ideal

4)There exists a basis of $L$ of the form $\{ e_{ij}$ with $i \neq j$ and $h_i=e_{ii}-e_{i+1,i+1} \}$. Let $I$ be a non-empty ideal of $L$, I want to show $I=L$. Indeed, let $0 \neq x \in I$ then $[e_{ij},x] \in I$ such that $x = \sum_{i=1}^4 \sum_{j \neq i} \alpha_{ij}[e_{mn},e_{ij}]+\sum_{i=1}^3 \beta_i[e_{mn},h_i]$. Thus, applying ad $e_{mn}$ twice, I get $-2\alpha_{nm}e_{mn} \in I$, and $-2 \alpha_{mn}e_{nm} \in I$ by the same calculation. However, applying ad $h_m$ twice I obtain $4(\alpha_{m,m+1}e_{m,m+1}+\alpha_{m+1,m}e_{m+1,m}) \in I$. Since one of those $\alpha_{ij}, \beta_i$ must be non-zero, then I can say $I=L$, right?

What about the other methods? In particular 3), I can't figure out the solution.

Davide Motta
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    I also think that it is quite easy to show directly that $\mathfrak{sl}_n(\Bbb C)$ is simple, see this question. For the Killing form, there are several duplicates here on this site, e.g. this one. – Dietrich Burde Jun 04 '19 at 15:02
  • It's maybe worthwhile to note that none of these proofs should use any property of $\Bbb C$ besides that it's a field of characteristic $0$. -- I wonder what exactly is meant by method 1, "brute force" though. You still got to compute something? – Torsten Schoeneberg Jun 04 '19 at 17:11
  • With “brute force” i mean “try to compute $Rad(L)$ directly – Davide Motta Jun 04 '19 at 19:43

1 Answers1

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The Killing form can be explicitly calculated. Let $A,B\in \mathfrak{sl}_n(\mathbb{C})$, Killing form is $$\kappa(A,B) = 2n\text{tr}(AB)$$ it is straightforward to show this is non-degenerate. Briefly, if $A$ is such that $\text{tr}(AB) = 0$ for all $B\in \mathfrak{sl}_n(\mathbb{C})$, then testing with $B=e_{ij}, i\neq j$ shows $a_{ji}=0$, testing with $B=e_{ii}-e_{jj}$ shows $a_{ii}=a_{jj}$, this forces all entries of $A$ to be $0$.

To see why the Killing form is this one, let $V=\mathbb{C}^n$, view elements of $V$ as column vectors. $\text{ad}_A\text{ad}_B$ is a operator sending $\text{End}(V)$ to $\text{End}(V)$, sending $X\mapsto ABX-AXB-BXA+XBA$, we need to compute its trace.

It suffices to find trace of the operator: $T:X\mapsto CXD$. View $\text{End}(V)$ as $V\otimes V$, where $v_1\otimes v_2$ corresponds to $v_1 v_2^T$. Then under this identification, $T$ is $C\otimes D^T$ (check this!). Therefore $\text{tr}(T) = \text{tr}(C)\text{tr}(D)$.

For our original $X\mapsto ABX-AXB-BXA+XBA$, it is the sum of four operators. The first one take $C=AB, D=I$, the second one take $C=A,D=B$ etc. The trace of $\text{ad}_A\text{ad}_B$ is then $$\text{tr}(AB)\text{tr}(I) - \text{tr}(A)\text{tr}(B)- \text{tr}(B)\text{tr}(A)+\text{tr}(I)\text{tr}(BA) = 2n \text{tr}(AB)-2 \text{tr}(A)\text{tr}(B)$$

Restricting on $\mathfrak{sl}_n(\mathbb{C})\subset \text{End}(V)$ gives the result.

pisco
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