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I'm just spending some of my spare time trying to learn about Lie algebras... took 2 semesters of linear algebra and abstract algebra back in college but since i've had kids i haven't had much time to learn more math which is sad because I love it!!

Anyway... I'm trying to compute the killing form of $sl(2,\mathbb{C})$. So, this killing form is a symmetric bilinear form on a 3-dimensional vector space, so I'm getting to get a $3 \times 3$ symmetric matrix, right??? Can somebody show me how to check that the killing form is non-degenerate? I feel like I'm good at following arguements by other people but can never but the basics together myself!!!

Yeah, then I was just wondering, is the killing form of $gl(2,\mathbb{C})$ non-degenerate? The exercise I'm trying to figure out says that it is for $sl(2,\mathbb{C})$, but i'm not sure about $gl$!!!

Thanks, i really appreciate you guys here!! I never knew there were so many genius's in the world hahahaha!

FireFenix777
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    It's necessary and sufficient to check that the determinant of that symmetric matrix is nonzero. The Killing form of $\mathfrak{gl}_2(\mathbb{C})$ has to be degenerate because nondegeneracy would imply that $\mathfrak{gl}_2(\mathbb{C})$ is semisimple, which it isn't. – Qiaochu Yuan Oct 16 '20 at 19:52
  • Check the following easy fact: the center of every Lie algebra is contained in the kernel of its Killing form. – YCor Oct 21 '20 at 14:36

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The Killing form of $\mathfrak{sl}_2(K)$ with respect to the standard basis $(e,f,h)$ is given by the symmetric matrix $$ \kappa=\begin{pmatrix} 0 & 4 & 0 \cr 4 & 0 & 0 \cr 0 & 0 & 8\end{pmatrix} $$ Its determinant is nonzero over a field $K$ of characteristic different from $2$, hence it is non-degenerate in this case.

For $\mathfrak{gl}_2(K)$ the symmetric matrix for the Killing form has a zero row and zero column (the center), with respect to the basis $(e,f,h,id)$, so its determinant is zero.

References:

Determinant of Killing form of sl_n

Dietrich Burde
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