Using the standard basis elements $$e=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, f=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},h=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ I want to show that the span of $h$, $\left<h\right>_{\mathbb{C}}$ is a Cartan subalgebra of $L=\mathfrak{sl}\left( 2,\mathbb{C}\right)$.

To do this I understand from the definition I must show $3$ things:

(i) $\left<h\right>_{\mathbb{C}}$ abelian

(ii) Every non-trivial element of $\left<h\right>_{\mathbb{C}}$ is semisimple.

(iii) $\left<h\right>_{\mathbb{C}}$ is maximal w.r.t both (i) and (ii)

Now for (i) showing the abelian property is straightforward, and for (iii) I believe that by showing that $$C_L( \left<h\right>_{\mathbb{C}})= \left<h\right>_{\mathbb{C}} $$ we get the maximal property as there is no larger abelian subalgebra which contains $\left<h\right>_{\mathbb{C}}$.

However I am struggling with showing (ii), by definition I would need to show for all $x \in \left<h\right>_{\mathbb{C}}$ we have that $x$ is semisimple. I believe from the definition this means showing that $ad(x)$ is diagonlisable for all $x \in \left<h\right>_{\mathbb{C}}$ but I am unsure how to do this, any help would be appreciated thanks :)


Note that:

  • $\operatorname{ad}(h)(x)=hx-xh=x$;
  • $\operatorname{ad}(h)(h)=hh-hh=0$;
  • $\operatorname{ad}(h)(h)=hy-yh=-y$.

So, yes, $\operatorname{ad}(h)$ is diagonalisable. And, if $\lambda\in\mathbb C$, $\operatorname{ad}(\lambda h)=\lambda\operatorname{ad}(h)$, and therefore $\operatorname{ad}(\lambda h)$ is diagonalizable too.


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