Cartan subalgebra and roots of sl_2

Killing-form computation: ${\mathrm{ad}}_h=\mathrm{diag}(0,2,-2)$ on the basis $(h,e,f)$, ${\mathrm{ad}}_e(h)=-2e$, ${\mathrm{ad}}_e(e)=0$, ${\mathrm{ad}}_e(f)=h$, and ${\mathrm{ad}}_f(h)=2f$, ${\mathrm{ad}}_f(e)=-h$, ${\mathrm{ad}}_f(f)=0$. Hence $B(h,h)=8$, $B(e,f)=B(f,e)=4$, and all other basis pairings vanish. The Killing matrix $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$ has determinant $-128\mathrm{\ne }0$, so [L1] proves that ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ is semisimple.

The subspace $\mathfrak{h}=\mathbb{C}h$ is a Cartan subalgebra: it is one-dimensional, hence abelian and nilpotent, and $N_{\mathfrak{g}}(\mathfrak{h})=\{ah+be+cf:[ah+be+cf,h]\in \mathbb{C}h\}$ equals $\mathbb{C}h$, because $[h,h]=0$, $[e,h]=-2e$ and $[f,h]=2f$, so a normalizing element has $b=c=0$.

The eigenspaces of ${\mathrm{ad}}_h$ are $\mathbb{C}h$ with eigenvalue $0$, $\mathbb{C}e$ with eigenvalue $2$ and $\mathbb{C}f$ with eigenvalue $-2$. Defining $\alpha \in {\mathfrak{h}}^{\ast }$ by $\alpha (h)=2$ and using Root and root space, the nonzero eigenspaces are ${\mathfrak{g}}_{\alpha }=\mathbb{C}e$ and ${\mathfrak{g}}_{-\alpha }=\mathbb{C}f$, so $\mathrm{\Phi }=\{\pm \alpha \}$ and the root-space decomposition is the displayed one.

The dual vector $H_{\alpha }$ satisfies $B(H_{\alpha },h)=\alpha (h)=2$; writing $H_{\alpha }=th$ gives $8t=2$, so $t=\frac{1}{4}$ and $H_{\alpha }=\frac{1}{4}h$; then $\alpha (H_{\alpha })=\frac{1}{4}\cdot 2=\frac{1}{2}=B(H_{\alpha },H_{\alpha })$, so the coroot is $h_{\alpha }=2H_{\alpha }\mathrm{/}\alpha (H_{\alpha })=2\cdot \frac{1}{4}h\mathrm{/}\frac{1}{2}=h$.