# Show $\left<h\right>_{\mathbb{C}}$ is a Cartan subalgebra of $\mathfrak{sl}\left( 2,\mathbb{C}\right)$.

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 $$h$$, $$\left_{\mathbb{C}}$$ 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) $$\left_{\mathbb{C}}$$ abelian

(ii) Every non-trivial element of $$\left_{\mathbb{C}}$$ is semisimple.

(iii) $$\left_{\mathbb{C}}$$ 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 $$C_L( \left_{\mathbb{C}})= \left_{\mathbb{C}}$$ we get the maximal property as there is no larger abelian subalgebra which contains $$\left_{\mathbb{C}}$$.

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

## 1 Answer

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.