# Root spaces are 1-dimensional

## Recap

If $\mathfrak{g}$ is the complexified Lie algebra of a compact semisimple group $K$ then $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in R}\mathfrak{g}_{\alpha}$ where:

• $\mathfrak{h}$ is the complexified Lie algebra of a maximal torus $T\subset K$ ,

• $\mathfrak{g}_{\alpha}=\{X\in\mathfrak{g}\ :\ \mathrm{ad}_{H}X=\alpha(H)X\ % \forall H\in\mathfrak{h}\}$ is the root space corresponding to the root $\alpha\in\mathfrak{h}^{*}_{\mathbf{Z}}\subset\mathfrak{h}^{*}_{\mathbf{R}}$ .

The figure below shows the root diagram of $\mathfrak{sl}(3,\mathbf{C})$ . We've been gradually discovering that various nice features of this Lie algebra and its root diagram carry over to root diagrams in this more general context. These nice features include:

• If $\alpha$ is a root then so is $-\alpha$ .

• If $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$ then $X$ , $Y$ and $H_{\alpha}=[X,Y]$ span a subalgebra $S_{\alpha}\cong\mathfrak{sl}(2,\mathbf{C})$ .

• Partway through the proof of the previous point, we saw that $[X,Y]=K(X,Y)\alpha^{\sharp}\in\mathfrak{h}$ , where $\alpha^{\sharp}$ is Killing-dual to $\alpha$ . This depends on $X$ and $Y$ only through their Killing pairing $K(X,Y)$ . We also saw that if we pick $X$ and $Y$ suitably (so that $K(X,Y)=2/K(\alpha^{\sharp},\alpha^{\sharp})$ ) then $H_{\alpha}=[X,Y]=\frac{2\alpha^{\sharp}}{K(\alpha^{\sharp},\alpha^{\sharp})}$ satisfies the commutation relations $[H_{\alpha},X]=2X,\quad[H_{\alpha},Y]=-2Y,\quad[X,Y]=H_{\alpha}.$

• In this video, we will show that:

Proposition:

All the root spaces $\mathfrak{g}_{\alpha}$ are 1-dimensional.

## Proof

Fix a root $\alpha$ and consider the line through $\alpha$ . We will sum some of the root spaces along this line:

• over $0$ , we take the span of $H_{\alpha}$ .

• we take the root spaces $\mathfrak{g}_{\alpha}$ , $\mathfrak{g}_{2\alpha}$ , etc and $\mathfrak{g}_{-\alpha}$ , $\mathfrak{g}_{-2\alpha}$ etc, that is root spaces corresponding to integer multiples of $\alpha$ .

The result is a subspace $V=\mathbf{C}\cdot H_{\alpha}\oplus\bigoplus_{k\in\mathbf{Z}\setminus\{0\}}% \mathfrak{g}_{k\alpha}$

Lemma:

$V$ is preserved by the action of the subalgebra $S_{\alpha}$ . Moreover, it is irreducible as a representation of $S_{\alpha}$ .

Remark:

Irreducibility is the key thing, because this implies that the root spaces $\mathfrak{g}_{k\alpha}$ are all 1-dimensional.

Proof:

We need to show that $\mathrm{ad}_{X}$ , $\mathrm{ad}_{Y}$ and $\mathrm{ad}_{H_{\alpha}}$ preserve $V$ . Let's just do $\mathrm{ad}_{X}$ (the others are similar/easier).

• We have $\mathrm{ad}_{X}H_{\alpha}=[X,H_{\alpha}]=-[H_{\alpha},X]=-2X\in\mathfrak{g}_{% \alpha}\subset V$ , so $\mathrm{ad}_{X}$ sends $\mathbf{C}\cdot H_{\alpha}$ to something in $V$ .

• We have $\mathrm{ad}_{X}\colon\mathfrak{g}_{k\alpha}\to\mathfrak{g}_{(k+1)\alpha}$ because $X\in\mathfrak{g}_{\alpha}$ .

• If $k+1\neq 0$ we have $\mathfrak{g}_{(k+1)\alpha}\subset V$ and we're done.

• If $k+1=0$ then $\mathfrak{g}_{(k+1)\alpha}=\mathfrak{g}_{0}$ , which is not contained in $V$ . In fact, we have $V\cap\mathfrak{g}_{0}=\mathbf{C}\cdot H_{\alpha}$ , so we need to show that $\mathrm{ad}_{X}Y$ is a multiple of $H_{\alpha}$ for any $Y\in\mathfrak{g}_{-\alpha}$ . But we saw that if $X\in\mathfrak{g}_{\alpha}$ and $Y\in\mathfrak{g}_{-\alpha}$ then $[X,Y]=K(X,Y)\alpha^{\sharp}$ , and $\alpha^{\sharp}$ is just a multiple of $H_{\alpha}$ , so $\mathrm{ad}_{X}(\mathfrak{g}_{-\alpha})=\mathbf{C}\cdot H_{\alpha}\subset V$ and we're done.

How do we see that this representation is irreducible? Let's understand the weight space decomposition of $V$ under the action of $S_{\alpha}$ . I claim that $\mathfrak{g}_{k\alpha}$ is a weight space with weight $2k$ for the action of $H_{\alpha}$ .

This is because if $Z\in\mathfrak{g}_{k\alpha}$ , we have $\mathrm{ad}_{H_{\alpha}}Z=k\alpha(H_{\alpha})Z$ and, since $H_{\alpha}=2\alpha^{\sharp}/K(\alpha^{\sharp},\alpha^{\sharp})$ , we have $\alpha(H_{\alpha})=2\alpha(\alpha^{\sharp})/K(\alpha^{\sharp},\alpha^{\sharp})$ , but $\alpha(\alpha^{\sharp})=K(\alpha^{\sharp},\alpha^{\sharp})$ , so $\mathrm{ad}_{H_{\alpha}}Z=2kZ$ and $Z$ has weight $2k$ under the action of $H_{\alpha}$ as required.

So our weight diagram looks like this: $\cdots\oplus\mathfrak{g}_{-2\alpha}\oplus\mathfrak{g}_{-\alpha}\oplus\mathbf{C% }\cdot H_{\alpha}\oplus\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{2\alpha}\oplus\cdots$ in weights $\cdots\quad-4,\quad-2,\quad 0,\quad 2,\quad 4,\quad\cdots$ respectively. In particular, the weight space with weight zero is $\mathbf{C}\cdot H_{\alpha}$ , which is 1-dimensional. If we decompose $V$ into irreducibles then these irreducibles all have even highest weight (there are only even weights) and in particular they will all have 1-dimensional weight space in weight zero. But there is only room for one such irreducible piece, so $V$ must itself be irreducible.