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Consider the algebra $L=\mathfrak{so}(6,\mathbb{C})=\{x\in\mathfrak{gl}(6,\mathbb{C}): x^tJ+Jx=0\}$, where $J_3=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}$ and $J=\begin{bmatrix} 0_3 & J_3 \\ J_3 & O_3 \\ \end{bmatrix}$. Show that $L$ is semisimple.

What I did was find the basis of the algebra, $\{e_{ij} - e_{kl}: i + k = 7 = j + l\}$, where $e_{ij}$ is the matrix with 1 in the entry $ij$ and $0$ in the others. Then I tried to calculate the associated matrix to the killing form to show that it has determinant non zero. However, I think that this is a really long and tedious way of show that L is semisimple.

So, how else can I prove that L is semisimple?

SSG19
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  • The trace form $(X,Y)={\rm trace}(XY)$ is non-degenerate. Hence the Killing form is also non-degenerate, so that $L$ is semisimple. – Dietrich Burde Sep 14 '21 at 13:44
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    In that post semi simplicity is shown using the killing form that is exactly what you say about the trace form. My question is if there is another way to prove semi simplicity without use the trace form. – SSG19 Sep 14 '21 at 13:53
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    Yes, the direct way by John Stillway, in my answer at the duplicate. Assume that $I$ is any nonzero ideal. Then do some caclulations, and then $I=\mathfrak{so}(2n)$. Hence the Lie algebra is simple. – Dietrich Burde Sep 14 '21 at 13:55
  • And it is not at all "tedious" to see that the trace form is non-degenerate. And this is enough. It is very easy. We only use the result on the Killing form in an abstract way. It is easy to see that the trace form and the Killing form vary only by a nonzero scalar. After all, why ignoring such a convenient tool as the trace form? – Dietrich Burde Sep 14 '21 at 13:57
  • I say that is tedious because the matrix of the killing form is of size 6x6. So calculate its determinant without a computer requires some calculations. – SSG19 Sep 14 '21 at 14:01
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    No, not at all. You only need the trace form. The Killing form then just is a fixed multiple of it. No need to compute the Killing form directly. In fact $B(x,y)=(2n-2){\rm trace}(XY)$ for the Killing form. It is obviously non-degenerate, so we are done without calculation, and $\mathfrak{so}(2n)$ is semisimple. – Dietrich Burde Sep 14 '21 at 14:05
  • Sorry, I notice that I really don't know the definition of trace form. In your notation, who are X and Y? – SSG19 Sep 14 '21 at 14:17
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    $X,Y\in \mathfrak{so}(2n)$. – Dietrich Burde Sep 14 '21 at 14:25

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