1
$\begingroup$

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 :)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.