4.2 The structure of finite dimensional complex semi-simple Lie subalgebra

TLDR

Suppose $h\subseteq g$ is a Cartan subalgebra, then weights of $h\stackrel{\mathrm{ad}}{\curvearrowright }g$ are same up to a rational scalar. Moreover, if $g$ is complex finite dimensional semi-simple Lie algebra, then $h$ and $\Phi$ are roughly dual. They have lots of properties that force $\Phi$ to be a reduced root system in the future.

Weights are same up to rational scalar

Let $g$ be a finite dimensional Lie algebra. After choosing a Cartan subalgebra $h$, we obtain a decomposition $g=h\oplus ({\oplus }_{\alpha \in \Phi }{L}_{\alpha })$ where $\Phi$ is the set of non-zero weights w.r.t. the action $h\stackrel{\mathrm{ad}}{\curvearrowright }g$ and $L_{\alpha }$ is the generalized weight space with weight $\alpha$. We call this decomposition the Cartan decomposition.

Remark. $[L_{\alpha },L_{\beta }]\subseteq L_{\alpha +\beta }$ is a information not come from the action $h\stackrel{\mathrm{ad}}{\curvearrowright }g$.

Observation 0. There are $p,q$ s.t.

$$0\leftarrow L_{-p\alpha +\beta }\leftarrow \cdots \leftarrow L_{\beta }\to L_{\alpha +\beta }\to \cdots \to L_{q\alpha +\beta }\to 0$$

is a sequence of finite length, and the action of $L_{\alpha },L_{-\alpha },h$ on $V={\oplus }_{i=-p}^q{L}_{i\alpha +\beta }$ should be compatible.

For $X\in L_{\alpha }$ and $Y\in L_{-\alpha }$, $\mathrm{ad}([X,Y])=[\mathrm{ad}X,\mathrm{ad}Y]\in \mathrm{End}(V)$ is trace $0$. Since

$$0=\mathrm{tr}(\mathrm{ad}[X,Y]{|}_V)=\sum _{i=-p}^q\dim L_{i\alpha +\beta }(i\alpha ([X,Y])+\beta ([X,Y])),$$

there is

$$\beta ([X,Y])=-\frac{{\sum }_{i=-p}^{q}i\dim L_{i\alpha +\beta }}{{\sum }_{i=-p}^q\dim L_{i\alpha +\beta }}\alpha ([X,Y]).$$

Remark. When $g$ is semi-simple, such a type of rigidity would force $g$ falling into only a few patterns.

Decomposition of the Killing form

Now suppose $g$ is semi-simple, then the Killing form $⟨\cdot ,\cdot ⟩$ is non-degenerated.

Observation 1. For $\alpha ,\beta \in \Phi$, if $\alpha +\beta \ne 0$, then $L_{\alpha }\perp L_{\beta }$.

\begin{proof} For $X\in L_{\alpha },Y\in L_{\beta }$, $⟨X,Y⟩=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$. Since $\mathrm{ad}X\mathrm{ad}Y{L}_{\gamma }\in L_{\alpha +\beta +\gamma }$, if $\alpha +\beta \ne 0$ then $L_{\gamma }\ne L_{\alpha +\beta +\gamma }$. After we choose a basis for each $L_{\gamma }$, the map $\mathrm{ad}X\mathrm{ad}Y$ is expressed by matrices of the shape $\left[\begin{array}{cc}0& \ast \\ \ast & 0\end{array}\right]$. Thus $\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)=0$. \end{proof}

By Observation 1, the Killing form can be decomposed

$$⟨\cdot ,\cdot {⟩}_{g\times g}=⟨\cdot ,\cdot ⟩{|}_{h\times h}\oplus ({\oplus }_{\Phi \backslash \{0\}}⟨\cdot ,\cdot ⟩{|}_{(L_{\alpha }\oplus L_{-\alpha })\times (L_{-\alpha }\oplus L_{\alpha })}).$$

So $⟨\cdot ,\cdot ⟩{|}_{h\times h}$ is non-degenerated and $⟨\cdot ,\cdot ⟩{|}_{L_{\alpha }\times L_{-\alpha }}$ is a perfect pairing.

Definition

A pairing $B(\cdot ,\cdot ):U\times V\to \mathbb{C}$ is called perfect if it induces isomorphisms

$$\begin{array}{rl}U& \stackrel{\sim }{\to }{V}^{\ast },u\mapsto B(u,\cdot ),\\ V& \stackrel{\sim }{\to }{U}^{\ast },v\mapsto B(\cdot ,v).\end{array}$$

Dual relation between $h$ and $\Phi$

Observation 2. We have the following conclusions:

• $h$ is abelian.
• $\Phi$ span $h^{\ast }$.

\begin{proof} i) For $X,Y\in h$, $⟨X,Y⟩=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)={\sum }_{\alpha \in \Phi }\dim L_{\alpha }\alpha (X)\alpha (Y)$. Since all $\alpha \in \Phi$ are representations of dimension $1$, they all vanish on $[h,h]$. By Observation 1, we have $[h,h]\perp h$. This would force $[h,h]=0$ because $⟨\cdot ,\cdot ⟩{|}_{h\times h}$ is non-degenerate. So $h$ is abelian.

ii) If $h^{\ast }\ne \mathrm{span}\{\alpha :\alpha \in \Phi \}$, then there exists non-zero $x\in h$ such that all $\alpha \in \Phi$ vanish on $x$. Whence $x\perp h$, contradiction. \end{proof}

Observation 3. Define $\phi :h\to h^{\ast },x\mapsto ⟨x,\cdot ⟩$. By Observation 2, for any $\alpha \in \Phi$, there is $h_{\alpha }^{\prime }\in h$ s.t. $\phi (h_{\alpha }^{\prime })=\alpha$. Then:

• $h_{\alpha }^{\prime }\in [L_{\alpha },L_{-\alpha }]$;
• $⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩\ne 0$;
• $\dim L_{\alpha }=\dim L_{-\alpha }=1$.

\begin{proof}i) Let $e_{\alpha }\in L_{\alpha }$ be a non-zero vector of weight $\alpha$. For $Y\in L_{-\alpha }$, assume $Z\in h$, then we have

$$\begin{array}{rl}⟨[e_{\alpha },Y],Z⟩& =⟨Y,[Z,e_{\alpha }]⟩=⟨Y,\alpha (Z)e_{\alpha }⟩\\ & =\alpha (Z)⟨Y,e_{\alpha }⟩=⟨Z,h_{\alpha }^{\prime }⟩⟨Y,e_{\alpha }⟩\\ & =⟨Z,⟨Y,e_{\alpha }⟩h_{\alpha }^{\prime }⟩.\end{array}$$

Thus $[e_{\alpha },Y]=⟨Y,e_{\alpha }⟩h_{\alpha }^{\prime }$. Because $⟨\cdot ,\cdot ⟩{|}_{L_{\alpha }\times L_{-\alpha }}$ is perfect, there is a $Y_0\in L_{-\alpha }$ s.t. $⟨Y_0,e_{\alpha }⟩\ne 0$. Then

$$h_{\alpha }^{\prime }=\frac{[e_{\alpha },Y_0]}{⟨Y_0,e_{\alpha }⟩}\in [L_{\alpha },L_{-\alpha }].$$

Remark. Note that $[Z,e_{\alpha }]=\alpha (Z)e_{\alpha }$ because $[Z,e_{\alpha }]=\mathrm{ad}Z(e_{\alpha })$.

ii) Otherwise, if $⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩=\alpha (h_{\alpha }^{\prime })=0$, then by Observation 0, $\beta (h_{\alpha }^{\prime })=(\text{some rational number})\ast \alpha (h_{\alpha }^{\prime })=0$ for all $\beta \in \Phi$. Since $h^{\ast }$ is spanned by $\Phi$, $h_{\alpha }^{\prime }=0$, contradiction.

iii) Consider

$$V:=\mathbb{C}{e}_{\alpha }\oplus [e_{\alpha },L_{-\alpha }]\oplus L_{-\alpha }\oplus \cdots \oplus L_{-p\alpha },$$

where $L_{-p\alpha }\ne 0$, $L_{-(p+1)\alpha }=0$. Note that $V$ is invariant under the action of $\mathbb{C}{e}_{\alpha },[e_{\alpha },L_{-\alpha }],L_{-\alpha }$. Thus for $Y\in L_{-\alpha }$,

$$0=\mathrm{tr}(\mathrm{ad}[e_{\alpha },Y]{|}_V)=\alpha ([e_{\alpha },Y])+\sum _i\dim L_{-i\alpha }(-i\alpha ([e_{\alpha },Y])).$$

Hence $\alpha ([e_{\alpha },Y])={\sum }_{i=1}^{p}i\alpha ([e_{\alpha },Y])\dim L_{-i\alpha }$. When $Y$ satisfies $⟨e_{\alpha },Y⟩\ne 0$, then by i) $\alpha ([e_{\alpha },Y])\ne 0$. Then

$$1=\sum _{i}i\dim L_{-i\alpha }$$

and so $\dim L_{-\alpha }=1$ and $p=1$. \end{proof}

Remark. I think this proof is too tricky to understand… there is a more understandable proof on the textbook.

Corollary

By Observation 3 i), for any $X\in L_{\alpha }$, $Y\in L_{-\alpha }$,

$$[X,Y]=⟨X,Y⟩h_{\alpha }^{\prime }.$$

To make $\Phi$ a root system

Observation 4. Consider the sequence in Observation 0. There is

$$⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩=\beta (h_{\alpha }^{\prime })=-\frac{{\sum }_{-p}^{q}i\dim L_{i\alpha +\beta }}{{\sum }_{-p}^q\dim L_{i\alpha +\beta }}\alpha (h_{\alpha }^{\prime })=\frac{p-q}{2}⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩.$$

And so

$$2\frac{⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩}{⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩}=p-q.$$

Corollary

Suppose $\alpha \in \Phi$, then $\mathbb{C}\alpha \cap \Phi =\{\pm \alpha \}$.

\begin{proof}Suppose $c\alpha \in \Phi$ where $c\in \mathbb{C}\backslash \{0,\pm 1\}$. Then observation 4 tells us $2c\in p-q$ and so $c\in \mathbb{Z}/2$. When $c\in \mathbb{Z}$, by Observation 3. iii) we know $c=1$. When $c=(2k+1)/2,k\in \mathbb{Z}$, the chain we have is

$$L_{-\frac{p+q}{2}\alpha },\cdots ,L_{\frac{p+q}{2}\alpha }.$$

So $L_{\alpha /2}$ must show up in the chain and $\alpha /2\in \Phi$, which contradicts with $\alpha \in \Phi$. \end{proof}

Corollary

For any $\alpha ,\beta \in \Phi$, $⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩\in \mathbb{Q}$.

\begin{proof}By Observation 4, it suffices to show $⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩\in \mathbb{Q}$. Since $⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩={\sum }_{\beta \in \Phi }\beta (h_{\alpha }^{\prime }{)}^2\dim L_{\beta }$, we have

$$1/⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩=\sum _{\beta \in \Phi }(\frac{⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩}{⟨h_{\alpha }^{\prime },h_{\alpha }^{\prime }⟩}{)}^2\dim L_{\beta }\in \mathbb{Q}.$$

Now we finish the proof. \end{proof}

Thus, put $h_{\mathbb{R}}={\mathrm{span}}_{\mathbb{R}}⟨h_{\alpha }^{\prime }:\alpha \in \Phi ⟩$, then $h_{\mathbb{R}}^{\ast }{\otimes }_{\mathbb{R}}\mathbb{C}\stackrel{\sim }{\to }{h}^{\ast }$. Here $h_{\mathbb{R}}^{\ast }\stackrel{\sim }{\to }{\mathrm{span}}_{\mathbb{R}}\Phi \subseteq h^{\ast }$.

Observation 5. For $x\in h_{\mathbb{R}}$,

$$⟨x,x⟩=\sum _{\alpha \in \Phi }{\alpha }^2(x)\dim L_{\alpha }\ge 0.$$

So $⟨\cdot ,\cdot ⟩{|}_{h_{\mathbb{R}}\times h_{\mathbb{R}}}$ is positive definite and hence an inner product. So now we have a finite dimensional Euclidean space $h_{\mathbb{R}}$ with a finite set contained there satisfying nice properties.