sl(2,C) subalgebras, 2

Finding sl(2,C) subalgebras

Statement of theorem

This time, we will prove:

Theorem:

0.00 If $X\in\mathfrak{g}_{\alpha}$ , $Y\in\mathfrak{g}_{-\alpha}$ are nonzero then $X$ , $Y$ and $H=[X,Y]$ will span a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbf{C})$ .

Lemma 1

We start by proving a lemma which identifies $[X,Y]$ .

Lemma:

If $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$ then $[X,Y]=K(X,Y)\alpha^{\sharp}$ .

Remark:

Here $\alpha\in\mathfrak{h}^{*}$ and $\alpha^{\sharp}\in\mathfrak{h}$ is dual to $\alpha$ under the Killing form. Recall that:

$\mathfrak{h}$ is the complexification of the Lie algebra of the maximal torus.
$\alpha^{\sharp}$ is the unique vector such that $K(\alpha^{\sharp},v)=\alpha(v)$ for all $v\in\mathfrak{h}$ . Uniqueness follows from nondegeneracy of $K$ .

Proof of Lemma 1

1.15 For a start, let's show that $[X,Y]\in\mathfrak{h}$ . This is because $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$ , so $[X,Y]\in\mathfrak{g}_{\alpha-\alpha}=\mathfrak{g}_{0}$ and we proved last time that $\mathfrak{g}_{0}=\mathfrak{h}$ .

2.20 To show that $[X,Y]=K(X,Y)\alpha^{\sharp}$ , we therefore need to prove that $K([X,Y],Z)=K(X,Y)\alpha(Z)$ for all $Z\in\mathfrak{h}$ . We have $K([X,Y],Z)=\mathrm{Tr}(\mathrm{ad}_{[X,Y]}\mathrm{ad}_{Z}).$

3.10 Since $\mathrm{ad}$ is a Lie algebra representation, $\mathrm{ad}_{[X,Y]}=\mathrm{ad}_{X}\mathrm{ad}_{Y}-\mathrm{ad}_{Y}\mathrm{ad}_% {X}$ . Therefore $K([X,Y],Z)=\mathrm{Tr}(\mathrm{ad}_{X}\mathrm{ad}_{Y}\mathrm{ad}_{Z}-\mathrm{% ad}_{Y}\mathrm{ad}_{X}\mathrm{ad}_{Z})$

3.45 Since $\mathrm{Tr}(PQR)=\mathrm{Tr}(RPQ)=\mathrm{Tr}(QRP)$ for any three matrices $P, Q, R$ , we can cyclically permute $\mathrm{ad}_{Y}\mathrm{ad}_{X}\mathrm{ad}_{Z}$ to get $\mathrm{ad}_{X}\mathrm{ad}_{Z}\mathrm{ad}_{Y}$ , so overall: $K([X,Y],Z)=\mathrm{Tr}(\mathrm{ad}_{X}\mathrm{ad}_{Y}\mathrm{ad}_{Z}-\mathrm{% ad}_{X}\mathrm{ad}_{Z}\mathrm{ad}_{Y})=\mathrm{Tr}(\mathrm{ad}_{X}\mathrm{ad}_% {[Y,Z]})$

4.30 which is equal to $K(X,[Y,Z])$ . Since $[Y,Z]=-[Z,Y]=-\mathrm{ad}_{Z}Y=\alpha(Z)Y$ because $Y\in\mathfrak{g}_{-\alpha}$ and $Z\in\mathfrak{h}$ . This proves $K([X,Y],Z)=\alpha(Z)K(X,Y)$ as required.

Remark:

5.55 Interestingly, the only way that $[X,Y]$ depends on $X$ and $Y$ through the scalar factor $K(X,Y)$ . This will become important later.

Proof of theorem

Pick $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$ and let $H:=[X,Y]=K(X,Y)\alpha^{\sharp}$ .

6.40 In fact, by rescaling our choice of $Y$ (and hence linearly rescaling $K(X,Y)$ ), we can assume that $K(X,Y)=\frac{2}{K(\alpha^{\sharp},\alpha^{\sharp})}$ . Note that this makes sense because $K(\alpha^{\sharp},\alpha^{\sharp})=K^{*}(\alpha,\alpha)\neq 0$ because $\alpha\in\mathfrak{h}_{\mathbf{R}}^{*}$ and $K^{*}$ is positive definite on $\mathfrak{h}_{\mathbf{R}}^{*}$ (because our group is compact and semisimple).

8.10 To check that the subalgebra spanned by $X$ , $Y$ and $\alpha^{\sharp}$ is isomorphic to $\mathfrak{sl}(2,\mathbf{C})$ , we just need to check that the standard commutation relations hold: $[H,X]=2X,\quad[H,Y]=-2Y,\quad[X,Y]=H.$

8.40 $[X,Y]=H$ holds by definition.

$X\in\mathfrak{g}_{\alpha}$ so $[H,X]=\alpha(H)X$ and $\alpha(H)=K(X,Y)\alpha(\alpha^{\sharp})=\frac{2}{K(\alpha^{\sharp},\alpha^{% \sharp})}\alpha(\alpha^{\sharp})$ but $\alpha(\alpha^{\sharp})=K(\alpha^{\sharp},\alpha^{\sharp})$ because $\alpha^{\sharp}$ was defined by the equation $K(\alpha^{\sharp},v)=\alpha(v),\ \forall v\in\mathfrak{h}.$

10.05 Therefore $[H,X]=2X$ . The proof of $[H,Y]=-2Y$ is similar.

Remark:

Here, we used the fact that $K$ is positive definite on $\mathfrak{h}_{\mathbf{R}}$ , for which we appealed to the fact that our Lie algebra is the Lie algebra of a compact group. We don't need to do this: it is possible to show that $K(\alpha^{\sharp},\alpha^{\sharp})\neq 0$ assuming only semisimplicity, but you need to work harder.