# Center of Lie subalgebra of $\mathfrak{gl}_n$

My Lie algebra theory is quite rusty, and I have problems in proving the following or giving a counterexample.

Let $$L$$ be a non-abelian Lie subalgebra of $$\mathfrak{gl}_n$$ such that the bilinear form given by $$b(x,y) = tr(xy)$$ is nondegenerate. Then any matrix in the center of $$L$$ is diagonal.

Certainly the result is false if you omit the "non-abelian", but I've not been able to prove it nor to find a counterexample for non-abelian subalgebras.

• Why should any matrix be diagonal in the center? If we change the basis, they are no longer all diagonal. Do you mean "diagonalizable"? – Dietrich Burde Mar 22 at 13:27
• You're right. The problem says "diagonal", but maybe they meant "diagonalizable". Or maybe it consists only of multiples of the identity like the center of $gl_n$. I don't really know. – Charlie Mar 22 at 14:11

The result as stated is false. To see it, we can construct a counterexample as follows. Consider a $$2\times 2$$ matrix $$A$$ such that $$tr(A^2)\neq 0$$, for instance: $$A = \left( \begin{array}{cc} 1& 1 \\ 0 & 1 \end{array} \right)$$. Let $$\mathfrak{a}$$ the abelian Lie algebra generated by $$A$$, and realize the Lie algebra $$\mathfrak{sl}_2 \oplus \mathfrak{a}$$ as a subalgebra of $$\mathfrak{gl}_4$$ by considering diagonal block matrices of the form: $$\left( \begin{array}{c|c} B & 0 \\ \hline 0 & \lambda A \end{array} \right)$$ where $$B \in \mathfrak{sl}_2$$ and $$\lambda \in \mathbb{R}$$. It can be seen that the trace form is non-degenerate in this algebra (as it is non-degenerate in each factor: in $$\mathfrak{sl}_2$$ because it is a multiple of the Killing form and the algebra is semisimple, and in $$\mathfrak{a}$$ by construction) and it contains the non-diagonalizable matrix $$\left( \begin{array}{c|c} 0& 0 \\ \hline 0 & A \end{array} \right)$$ in its center.
• Neat. I though about your question a bit and thought I'd reached some partial results (centre of $\mathfrak{g}$ cannot contain non-zero nilpotent elements; $\mathfrak{g}$ is reductive (?); result is true for solvable Lie algebras via Lie's theorem). Seeing a counterexample puts my mind at ease, also it does not contradict those partial positive results. – Torsten Schoeneberg Mar 25 at 4:39
• @Torsten Schoeneberg: how do you show $\mathfrak{g}$ is reductive? – Charlie Mar 26 at 6:47
• Thanks for asking, and good that I put a question mark there, because on second thought my proof fell flat. The idea was that with the other results we can show that the intersection of the centre of $\mathfrak{g}$ with $[\mathfrak r, \mathfrak r]$ (where $\mathfrak{r}$ is the radical of $\mathfrak{g}$) is trivial. But then I must have mixed up the centre of $\mathfrak{g}$ with the centre of $\mathfrak{r}$ and somehow concluded that the radical is the centre, for which I see no reason now. On the other hand, I have no counterexample either. Maybe there's an easy upgrade of yours? – Torsten Schoeneberg Mar 27 at 3:10
• Bourbaki's "Groupes et Algèbres de Lie", chapter 1, §6, prop. 5 states that $\mathfrak{g}$ is reductive if and only if it has a finite-dimensional representation such that the associated bilinear form is non-degenerate. Here, $\mathfrak{g}$ has a natural representation on $k^n$ (as a subalgebra of $\mathfrak{gl}_n$), and the associated bilinear form is exactly $b$, so $\mathfrak{g}$ is reductive. – Robin Carlier Mar 28 at 15:59