4.3 Root system

TLDR

For complex finite dimensional semi-simple Lie algebra, the set of weights forms a “root system”. We define “dual pairing” w.r.t dual space. As $h=\mathrm{span}\{h_{\alpha }^{\prime }:\alpha \in \Phi \}$ and $h^{\ast }=\mathrm{span}\{\alpha :\alpha \in \Phi \}$, both $\{h_{\alpha }^{\prime }:\alpha \in \Phi \}$ and $\Phi$ are root systems. Angle and ratio of length between roots are in finite cases, which makes possible to classify complex semi-simple Lie algebras.

Motivation

Observation 0. Using conclusions of last section, we find that $\{h_{\alpha }^{\prime }:\alpha \in \Phi \}$ should in the following conditions: inner products are rational, invariance under reflections.

• Suppose $V$ is a finite dimensional Euclidean space, and $0\ne v\in V$. Define

  $$H=\{x\in V:(x,v)=0\}$$

  and $S_v$ is the reflection w.r.t. $H_v$, that is, $S_v(u)=u-2\frac{(u,v)}{(v,v)}v$.

• Recall for any $\alpha ,\beta \in \Phi$ we have the following chain:

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

  Then $\beta -(p-q)\alpha$ and $\beta$ are symmetric w.r.t. the center of the chain.

• We can choose a maximal linearly independent subset $\{{\alpha }_1,\cdots ,{\alpha }_n\}$ of $\Phi$, and then

  $$\{h_{\alpha }^{\prime }:\alpha \in \Phi \}\subseteq {\mathrm{span}}_{\mathbb{Q}}\{h_{{\alpha }_1}^{\prime },\cdots ,h_{{\alpha }_n}^{\prime }\}$$

  by $⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩\in \mathbb{Q}$. So

  $$\begin{array}{rl}{h}_{\mathbb{R}}:=& {\mathrm{span}}_{\mathbb{R}}\{h_{\alpha }^{\prime }:\alpha \in \Phi \}\\ =& {\mathrm{span}}_{\mathbb{Q}}\{h_{\alpha }^{\prime }:\alpha \in \Phi \}{\otimes }_{\mathbb{Q}}\mathbb{R}.\end{array}$$

  And $h=h_{\mathbb{R}}{\otimes }_{\mathbb{R}}\mathbb{C}$, $h^{\ast }=h_{\mathbb{R}}^{\ast }{\otimes }_{\mathbb{R}}\mathbb{C}$.

• At this moment, for $\alpha ,\beta \in \Phi$,

  $$S_{h_{\alpha }^{\prime }}(h_{\beta }^{\prime })=h_{\beta }^{\prime }-2\frac{⟨h_{\beta }^{\prime },h_{\beta }^{\prime }⟩}{⟨h_{\alpha }^{\prime },h_{\beta }^{\prime }⟩}{h}_{\alpha }^{\prime }=h_{\beta -(p-q)\alpha }^{\prime },$$

  that is, $\{h_{\alpha }^{\prime }:\alpha \in \Phi \}$ is invariant under $S_{h_{\alpha }^{\prime }}$.

root system

A root system is a finite dimensional Euclidean space $V$ together a finite subset $R\subseteq V$ s.t.

• $0\notin V$;
• $R$ spans $V$;
• for each $\alpha \in R$, $S_{\alpha }(R)=R$;
• for $\alpha ,\beta \in R$, $2\frac{(\alpha ,\beta )}{(\alpha ,\alpha )}\in \mathbb{Z}$.

If a root system $(V,R)$ satisfies a further condition:

• for $\alpha \in \Phi$, $\mathbb{Z}\alpha \cap \Phi =\{\pm \alpha \}$, then it is called a reduced root system.

Example. $\Phi$ is a reduced root system.

Dual pairing

dual pairing

Let $g$ be a finite dimensional complex semi-simple Lie algebra and $h$ be a Cartan subalgebra of $g$. Let $g=h\oplus ({\oplus }_{\alpha \in \Phi }{L}_{\alpha })$ be a Cartan decomposition. Then the Killing form induces an identification

$$\begin{array}{rl}h& \stackrel{\mathcal{L}}{\to }{h}^{\ast }\\ x& \mapsto ⟨x,\cdot ⟩\end{array}$$

and a pairing on $h^{\ast }$ which is defined as

$$(l_1^{\ast },l_2^{\ast })=⟨{\mathcal{L}}^{-1}(l_1),{\mathcal{L}}^{-1}(l_2)⟩.$$

Note $(\cdot ,\cdot )$ is an inner product on $h_{\mathbb{R}}$.

An example: $g{l}_{n,\mathbb{C}}$

Define $g{l}_{n,\mathbb{C}}=M_{n\times n}(\mathbb{C})$ and $s{l}_{n,\mathbb{C}}=\{x\in M_{n\times n}(\mathbb{C}):\mathrm{tr}(x)=0\}$. We can prove that $s{l}_{n,\mathbb{C}}$ is a simple Lie algebra with $n\ge 2$, see here or textbook. So $s{l}_{n,\mathbb{C}}$ is semi-simple.

Now let $g:=s{l}_{n,\mathbb{C}}$. Since $h=\{\mathrm{diag}(a_1,\cdots ,a_n):a_i\in \mathbb{C},\sum a_i=0\}$ is a Cartan subalgebra of $g$, then

$$g=h\oplus ({\oplus }_{i\ne j,i,j\in \{1,\cdots ,n\}}\mathbb{C}{E}_{ij}).$$

View of $h$

Suppose $H=\mathrm{diag}(a_1,\cdots ,a_n)$. Then $[H,E_{ij}]=H{E}_{ij}-E_{ij}H=(a_i-a_j)E_{ij}={\epsilon }_{ij}(H)E_{ij}$, where ${\epsilon }_{ij}={\epsilon }_i-{\epsilon }_j$ and ${\epsilon }_i:\mathrm{diag}(a_1,\cdots ,a_n)\mapsto a_i$. Thus the collection of roots is $\{{\epsilon }_{ij}:i\ne j\}$ and $h^{\ast }=\mathrm{span}\{{\epsilon }_{ij}:i\ne j\}=\{{\sum }_{i=1}^n\lambda {\epsilon }_i:{\sum }_{i=1}^n{\lambda }_i=0\}$.

Since for $H_1,H_2\in h$ with $H_1=\mathrm{diag}(a_1,\cdots ,a_n)$ and $H_2=\mathrm{diag}(b_1,\cdots ,b_n)$, there is

$$\begin{array}{rl}⟨H_1,H_2⟩& =\sum _{\alpha \in \Phi }\alpha (H_1)\alpha (H_2)\\ & =\sum _{i,j\in \{1,\cdots ,n\}}(a_i-a_j)(b_i-b_j)\\ & =2n\sum _{i=1}^n{a}_i{b}_i.\end{array}$$

Since

$$\begin{array}{rl}⟨\mathrm{diag}(a_1,\cdots ,a_n),x⟩& =\sum _{i\ne j}(a_i-a_j){\epsilon }_{ij}(x)\\ & =\sum _{i\ne j}(a_i-a_j)({\epsilon }_i-{\epsilon }_j)(x)\\ & =2n\sum _{i=1}^n{a}_i{\epsilon }_i(x),\end{array}$$

the identification $\mathcal{L}:h\to h^{\ast }$ is $\mathcal{L}(\mathrm{diag}(a_1,\cdots ,a_n))={\sum }_{i\ne j}(a_i-a_j){\epsilon }_{ij}=2n\sum a_j{\epsilon }_i$. And then the induced pairing on $h^{\ast }$ is

$$\begin{array}{rl}(\sum _{i=1}^n{\lambda }_i{\epsilon }_i,\sum _{j=1}^n{\mu }_j{ϵ}_j)& =⟨{\mathcal{L}}^{-1}(\sum _{i=1}^n{\lambda }_i{\epsilon }_i),{\mathcal{L}}^{-1}(\sum _{j=1}^n{\mu }_j{\epsilon }_j)⟩\\ & =⟨\frac{1}{2n}\mathrm{diag}({\lambda }_1,\cdots ,{\lambda }_n),\frac{1}{2n}\mathrm{diag}({\mu }_1,\cdots ,{\mu }_n)⟩\\ & =\frac{1}{2n}\sum _{i=1}^n{\lambda }_i{\mu }_i.\end{array}$$

View of $g$

Since $h\subseteq {\mathbb{C}}^n={\oplus }_{i=1}^n\mathbb{C}{E}_{ii}$ and $h^{\ast }\subseteq {\mathbb{C}}^{n\ast }={\oplus }_{i=1}^n\mathbb{C}{\epsilon }_i$, one can consider a pairing on ${\mathbb{C}}^{n\ast }$ induced from ${\mathbb{C}}^n$. We can choose suitable inner product of $\mathbb{C}$ and a map from ${\mathbb{C}}^n$ to ${\mathbb{C}}^{n\ast }$ s.t. the induced pairing on $h^{\ast }$ is compatible with the restriction of the induced pairing on ${\mathbb{C}}^{n\ast }$.

Define an inner product of ${\oplus }_{i=1}^n\mathbb{C}{E}_{ii}$ as $({\sum }_{i=1}^n{t}_i{E}_{ii},{\sum }_{j=1}^n{s}_j{E}_{jj})=2n{\sum }_{i=1}^n{t}_i{s}_i$, and define a map ${\mathbb{C}}^n\to {\mathbb{C}}^{n\ast },E_{ii}\mapsto 2n{\epsilon }_i$. Then it is easy to verify they are compatible.

Thus we may view the associated root system $\{{\epsilon }_{ij}:i\ne j\}$ is contained in the $(n-1)$-dim Euclidean space ${\mathbb{R}}^{n-1}=\{{\sum }_{i=1}^n\lambda {\epsilon }_i:{\sum }_{i=1}^n{\lambda }_i=0\}:=U_0$, which sits inside the natural Euclidean space, i.e., ${\mathbb{R}}^n=U_0\oplus \mathbb{R}{\sum }_{i=1}^n{\epsilon }_i$.

Remark. This example tells us for any orthonormal basis $\{e_1,\cdots ,e_n\}$, $\{e_i-e_j:i\ne j\}$ can be seen as a root system. So we can get some root systems for different $n$, which is called $A_{n-1}$ with $\dim V=n$.

Propositions

Proposition

Let $(V,R)$ be a reduced root system. Then $(V^{\ast },R^{\ast })$ is also a reduced root system, where $V^{\ast }$ is the dual space of $V$ equipped with the associated inner product via the identification

$$\begin{array}{rl}V& \stackrel{\sim }{\to }{V}^{\ast }\\ x& \mapsto (x,\cdot )\end{array}$$

and ${\alpha }^{\ast }:=(\frac{2\alpha }{(\alpha ,\alpha )},\cdot )$ for $\alpha \in R$.

\begin{proof} Recall for $\alpha ,\beta \in \Phi$, there is

$$S_{\alpha }(\beta )=\beta -2\frac{(\beta ,\alpha )}{(\alpha ,\alpha )}\alpha =\beta -{\alpha }^{\ast }(\beta )\alpha .$$

So $S_{\alpha }=1-{\alpha }^{\ast }\otimes \alpha \in \mathrm{Hom}(V,V)\cong V^{\ast }\otimes V$.

Define $f_x=(x,\cdot )$, then $V^{\ast }=\mathrm{span}\{f_x:x\in V\}$. With the induced pairing, there is

$$\begin{array}{rl}{S}_{{\alpha }^{\ast }}(f_x)& =f_x-2\frac{(f_x,{\alpha }^{\ast })}{({\alpha }^{\ast },{\alpha }^{\ast })}{\alpha }^{\ast }\\ & =f_x-f_x(\alpha ){\alpha }^{\ast }.\end{array}$$

Thus $S_{{\alpha }^{\ast }}=1-\alpha \otimes {\alpha }^{\ast }\in \mathrm{Hom}(V^{\ast },V^{\ast })\cong V\otimes V^{\ast }$. Since ${\alpha }^{\ast }$ is a reflection, $|\beta |=|{\alpha }^{\ast }(\beta )|$. Thus $(S_{\alpha }(\beta ){)}^{\ast }=S_{{\alpha }^{\ast }}({\beta }^{\ast })$ and so $S_{{\alpha }^{\ast }}(R^{\ast })=R^{\ast }$. Further, $({\alpha }^{\ast },{\beta }^{\ast })=(\frac{2\alpha }{(\alpha ,\alpha )},\frac{2\beta }{(\beta ,\beta )})\in \mathbb{Q}$ and $2({\alpha }^{\ast },{\beta }^{\ast })/({\alpha }^{\ast },{\alpha }^{\ast })=2(\alpha ,\beta )/(\beta ,\beta )\in \mathbb{Z}$. For any ${\alpha }^{\ast }\in R^{\ast }$, if there is an integer $k$ s.t. $k{\alpha }^{\ast }\in R^{\ast }$, then $1/k\alpha \in R$ as $(1/k\alpha {)}^{\ast }=k{\alpha }^{\ast }$. It contradicts with $R$ reduced root system. \end{proof}

Remark. In midterm, we have a similar exercise. See ^d044ea.

Proposition

Let $(V,R)$ be a reduced root system and $\alpha ,\beta \in R$. There are two integers

$$\begin{array}{rl}n(\alpha ,\beta )& =2\frac{(\alpha ,\beta )}{(\alpha ,\alpha )}=2\frac{|\beta |}{|\alpha |}\cos \theta \\ n(\beta ,\alpha )& =2\frac{(\alpha ,\beta )}{(\beta ,\beta )}=2\frac{|\alpha |}{|\beta |}\cos \theta .\end{array}$$

Then $4{\cos }^2\theta \in \mathbb{Z}$, ${\cos }^2\theta =0,1/4,2/4,3/4,1$. Thus the angle $\theta$ between $\alpha ,\beta$ satisfies

$$\theta \in \{\frac{\pi }{2},\frac{\pi }{3},\frac{2\pi }{3},\frac{\pi }{4},\frac{3\pi }{4},\frac{\pi }{6},\frac{5\pi }{6},0,\pi \}.$$

local picture

Let $(V,R)$ be a reduced root system and $\alpha ,\beta \in R$, $\beta \ne \pm \alpha$. Then:

• if the angle is $\pi /3$ or $2\pi /3$, then $|\alpha |=|\beta |$;
• if the angle is $\pi /4$ or $3\pi /4$, then the ratio of the length is $\sqrt{2}$;
• if the angle is $\pi /6$ or $5\pi /6$, then the ratio of the length is $\sqrt{3}$.

Proposition

Let $(V,R)$ be a reduced root system. Then there exists a subset $\Delta \subseteq R$ consisting of $\dim V$ elements, s.t. every element of $R$ is a linear combination of elements of $\Delta$ with coefficients either all in ${\mathbb{Z}}_{\ge 0}$ or all in ${\mathbb{Z}}_{\le 0}$. Such a subset is called a base or fundamental system or simple root system of $R$ and

$$W:=⟨S_{\alpha }:\alpha \in R⟩$$

is called the Weyl group acts transitively on the set of fundamental systems.

\begin{proof} The main idea is

\end{proof}