5.2 Representation theory of semi-simple Lie algebra

TLDR

By the representation theory of $s{l}_2(\mathbb{C})$, we define the positive weight lattice. And show that any irreducible finite dimensional representation of semi-simple Lie algebra can be uniquely determined by its maximal weight, which is in the positive weight lattice.

让我再用中文写一遍： 给定一个半单李代数 $g$ 和它的一个 cartan subalgebra，可以得到它的 cartan decomposition。取出这个分解中的 borel subalgebra，随便取一个 $h$ 上的线性函子 $\lambda$ 和一个向量 $v$，通过定义 $h(v)$ 和 $U_{+}(v)$，可以得到一个 borel subalgebra 上的一维表示 $\mathcal{V}:=\mathbb{C}v$，其中 $h(v)=\lambda (v)$, $U_{+}(v)=0$.

可以通过 tensor 形式地定义$g$上的乘法，通过商掉一些东西得到 universal developing algebra $U(g)$，这是一个 associate algebra 且它的 bracket 和 $g$ 上的 bracket 相容。此时 $g$ 上的 module 和 $U(g)$ 上的 module 是一一对应的。

$U(g)$和$\mathcal{V}$形式地做tensor，把其中的元素$(s,v_t)$射到$s{v}_t$。为了和borel subalgebra上的一维表示$\mathcal{V}$相容，商掉$I_{\lambda }$。这个商调的家伙就是上面那个映射自然的kernel。商掉$I_{\lambda }$之后的这个家伙叫做$M(\lambda )$. 这个$M(\lambda )$有一个唯一的不可约的quotient module，记作$M_{\lambda }$.

这时就需要weight lattice出场了——如果前面取的这个lambda在weight lattice中，那么$M_{\lambda }$是有限维的。此时$M_{\lambda }$是半单李代数的一个有限维不可约表示。对于随便某个highest weight是lambda的不可约表示$V^{\prime }$，可以证明$V^{\prime }$和$M_{\lambda }$是同构的。因此，$V^{\prime }$的highest weight是唯一的，在同构意义下就是最开始我们随便取出来的那个$v$。

另外某网友给出了另外一套思路，但是没找到ref：这里。

Suppose $g$ is a finite dimensional complex semi-simple Lie algebra and $h$ is a Cartan subalgebra of $g$. Write $g=h\oplus ({\oplus }_{\alpha \in \Phi }{L}_{\alpha })$. Choose a Weyl chamber and an according fundamental system $\Delta$.

Weight Lattice

Motivation

Observation 0. For $\alpha \in {\Phi }_{+}$, choose non-zero elements $E_{\alpha }\in L_{\alpha },E_{-\alpha }\in L_{-\alpha }$. Then $[E_{\alpha },E_{-\alpha }]=⟨E_{\alpha },E_{-\alpha }⟩h_{\alpha }^{\prime }$, where $h_{\alpha }^{\prime }\in h$ such that $⟨H,h_{\alpha }^{\prime }⟩=\alpha (H)$ for all $H\in h$.

Note that the standard $s{l}_2(\mathbb{C})=\mathbb{C}h\oplus \mathbb{C}e\oplus \mathbb{C}f$ with $[h,e]=2e$, $[e,f]=h$, $[h,f]=2f$. We can choose

$$\begin{array}{rl}{h}_{\alpha }& =2h_{\alpha }^{\prime }/⟨\alpha ,\alpha ⟩,\\ f_{\alpha }& =-E_{-\alpha },\\ e_{\alpha }& =2E_{\alpha }/⟨\alpha ,\alpha ⟩.\end{array}$$

It is easy to verify $s{l}_{2,\alpha }=\mathbb{C}{h}_{\alpha }\oplus \mathbb{C}{e}_{\alpha }\oplus \mathbb{C}{f}_{\alpha }\simeq s{l}_2(\mathbb{C})$. This is called the copy of $s{l}_2(\mathbb{C})$ associated to the positive root $\alpha$.

Observation 1. Let $M$ be a finite dimensional $g$-module. Then $M={\oplus }_{\mu \in \Lambda }{M}_{\mu }$, where $\Lambda$ is the set of weights of $h\curvearrowright M$.

\begin{proof} For each $\alpha \in \Delta$, $s{l}_{2,\alpha }$ acts on $M$. By the representation theory of $s{l}_2(\mathbb{C})$, $M$ is the direct sum of eigenspaces of $h_{\alpha }$. Since $h=\mathrm{span}\{h_{\alpha }:\alpha \in \Delta \}$ and $h$ is abelian, one can write $M$ as the direct sum of weight spaces w.r.t the action of $h$. Now we finish the proof. \end{proof}

In addition, there is $L_{\alpha }(M_{\mu })\subseteq M_{\alpha +\mu }$. For any $H\in h$, $X_{\alpha }\in L_{\alpha }$ and $m\in M_{\mu }$,

$$\begin{array}{rl}H{X}_{\alpha }(m)& =[H,X_{\alpha }]m+X_{\alpha }Hm\\ & =\alpha (H)X_{\alpha }m+X_{\alpha }\mu (H)m\\ & =(\alpha (H)+\mu (H))X_{\alpha }m.\end{array}$$

Furthermore, $\mu (h_{\alpha })\in \mathbb{Z}$ by the definition of ${\pi }_m$. On the other hand, $\mu (h_{\alpha })=2⟨\mu ,\alpha ⟩/⟨\alpha ,\alpha ⟩$ because $\mu \in h^{\ast }=\mathrm{span}\{\alpha :\alpha \in \Phi \}$.

Put ${\Phi }^{\vee }=\{{\alpha }^{\vee }:\alpha \in \Phi \}$, then ${\Phi }^{\vee }$ is a root system. ${\alpha }^{\vee }$ is called the coroot and is defined by ${\alpha }^{\vee }=h_{\alpha }=2h_{\alpha }^{\prime }/⟨\alpha ,\alpha ⟩$. It is easy to verify ${\Phi }^{\vee }\subseteq h$ is a root system, because

$$\begin{array}{r}\frac{({\alpha }^{\vee },{\beta }^{\vee })}{({\alpha }^{\vee },{\alpha }^{\vee })}=\frac{(\alpha ,\beta )}{(\beta ,\beta )},\\ S_{{\alpha }^{\vee }}({\beta }^{\vee })={S_{\alpha }(\beta )}^{\vee }.\end{array}$$

Observation 2. When $M$ is irreducible as a $s{l}_{2,\alpha }$-module for some $\alpha$, there is a unique weight $\mu$ such that $\pi (L_{\alpha })M_{\mu }=0$ for all $\alpha \in {\Phi }_{+}$, and such $M_{\mu }$ is $1$-dimensional by the definition of ${\pi }_m$. At this moment, the basis vector of $M_{\mu }$ is the highest weight w.r.t. the action of ${\mathrm{sl}}_{2,\alpha }$. Hence $\mu (h_{\alpha })\in {\mathbb{Z}}_{\ge 0}$. Specially, $\mu (h_{\alpha })\in {\mathbb{Z}}_{\ge 0}$ for all $\alpha \in \Delta$. Note that $M_{\mu }$ is a $1$-dimensional representation of $g$.

Rmk. It is not easy to show an irreducible representation has the unique highest weight, and this conclusion is proved by the construction of $M_{\lambda }$, as this proposition says:

If $V$ is an irreducible $g$-module with a maximal weight $\lambda$, then $V\cong M_{\lambda }$. In particular, a maximal weight of an irreducible module, if it exists, it is unique.

Link to original

Define weight lattice

In Observation 2, the $1$-dimensional representation $M_{\mu }$ can be described by the weight $\mu$. Naturally, we want to list all possibilities of $\mu$. Since $\mu (h_{\alpha })\in {\mathbb{Z}}_{\ge 0}$, we have the following definition.

weight lattice

Suppose $l=\dim h$. Define the weight lattice $X$ to be the subset of $h_{\mathbb{R}}^{\ast }$ taking integer values on $\{h_{\alpha }:\alpha \in \Phi \}$. The elements of the dual basis $w_1,\dots w_l$ of $\Delta =\{h_1,\dots ,h_l\}$ are called the fundamental weights, which satisfy $w_i(h_j)={\delta }_{ij}$. Put $X^{+}={\sum }_{i=1}^l{\mathbb{Z}}_{\ge 0}{w}_i$.

Big picture

Define $U_{+}={\oplus }_{\alpha \in {\Phi }_{+}}{L}_{\alpha }$ and $U_{-}={\oplus }_{\alpha \in {\Phi }_{-}}{L}_{\alpha }$. Then the Borel subalgebra is defined by

$$b=h\oplus U_{+}$$

and $g=b\oplus U_{-}$. Let $(\pi ,V)$ be a finite dimensional irreducible representation of $g$.

Philosophy. Representation of $b$ induces representation of $g$.

Recall $b$ has a $1$-dimensional representation. So the question is, how does this representation induce a $g$-module? We will see that a $b$-module $\mathcal{V}$ induces a $g$-module $U_g{\otimes }_{U_b}\mathcal{V}$, which is called the Verma module.

$U_g$ is defined here. Roughly speaking, let $g$ be a finite dimensional complex Lie algebra and $U_g$ its universal enveloping algebra. Then

$$\text{the category of representations of }g\simeq \text{the category of }{U}_g\text{ -modules}.$$

What does $U_g{\otimes }_{U_b}\mathcal{V}$ exactly mean? You can see here.

Recall: existence of $1$-dim $b$-submodule

Observation. There exists a vector $v_t$ such that $U_{+}\cdot v_t=0$.

\begin{proof}Since $V={\oplus }_{\lambda \in \Lambda }{V}_{\lambda }$ where $\Lambda$ is the set of the weights w.r.t. $h\curvearrowright V$. Since there are only finitely many weights, there exists a weight $\lambda$ such that

$$\lambda +\text{any positive root}\notin \Lambda .$$

Then a non-zero vector $v_t\in V_{\lambda }$ meets the propose. \end{proof}

Universal enveloping algebra

Let $g$ be a Lie algebra and $(\pi ,V)$ a finite dimensional representation of $g$. Since $\pi$ maps $g$ to $\mathrm{End}(V)$ where $\mathrm{End}(V)$ is an associative algebra, we should have a multiplication operation on "$g$".

Consider the tensor algebra over $g$

$$T(g)={\oplus }_{n\ge 0}{T}^n(g).$$

Let $I$ be the two side ideal generated by $X\otimes Y-Y\otimes X-[X,Y]$. Then $U_g=T(g)/I$ is called the universal enveloping algebra of $g$.

Define the natural map

$$\begin{array}{rl}g& \to U_g,\\ X& \mapsto \overline{X}.\end{array}$$

Lemma

There is a natural equivalence between the category of $g$-module and $U_g$-modules.

PBW

Let $\{x_1,\cdots ,x_n\}$ be a basis of $g$. Then

$$\{x_1^{i_1}\cdots x_n^{i_n}:i_1,\cdots ,i_n\in {\mathbb{Z}}_{\ge 0}\}$$

is a basis of $U_g$.

\begin{proof}See textbook. First show that such vectors span $U_g$. Then show that they are linearly independent. \end{proof}

What is $U_g{\otimes }_{U_b}\mathcal{V}$ ?

Let $(\pi ,\mathcal{V})$ be a finite dimensional $b$-module and $v_t$ be a highest weight vector of weight $\lambda$, i.e. $U_{+}{v}_t=0$, $\pi (H)v_t=\lambda (H)v_t$.

Define

$$\begin{array}{rl}\phi :U_g& \twoheadrightarrow U_g{v}_t,\\ s& \mapsto s{v}_t.\end{array}$$

Then the left ideal $I_{\lambda }$ generated by $x\in U_{+}$ and $H-\lambda (H)$, $H\in h$ belongs to the kernel of $\phi$. Thus $U_g{\otimes }_{U_b}\mathcal{V}=U_g/I_{\lambda }$. Also you can see here.

Properties

Proposition

$M(\lambda ):=U_g/I_{\lambda }$ has a unique irreducible non-zero quotient module, which is denoted by $M_{\lambda }$, i.e. it has a unique maximal proper submodule.

\begin{proof}Recall $g=U_{-}\oplus h\oplus U_{+}$. It follows that $M(\lambda )$ is a free $U_{U_{-}}$-module generated by $[1]$. On the other words, numerate the positive roots by ${\alpha }_1,\cdots ,{\alpha }_n$ and choose a non-zero vector $E_{{\alpha }_i}$ in $L_{{\alpha }_i}$ and $E_{-{\alpha }_i}$ in $L_{-{\alpha }_i}$. Then

$$\{E_{-{\alpha }_1}^{i_1}\cdots E_{-{\alpha }_l}^{i_l}[1]:i_1,\cdots ,i_l\ge 0\}$$

is a basis of $M(\lambda )$. Note that the weight of $E_{-{\alpha }_1}^{i_1}\cdots E_{-{\alpha }_l}^{i_l}[1]$ is $\lambda -i_1{\alpha }_1-\cdots -i_l{\alpha }_l$. At this moment,

$$M(\lambda )={\oplus }_{\mu \in \Lambda }M(\lambda )[\mu ]$$

where $\Lambda$ is the set of weights w.r.t. the action of $h$ and $M(\lambda )[\mu ]$ is the weight space of $\mu$. Furthermore, $M(\mu )$ can be written as

$$M(\mu )=M(\lambda )[\lambda ]\oplus M(\lambda {)}^{\prime }$$

where $M(\lambda {)}^{\prime }$ is the direct sum of $M(\lambda )[\mu ]$ with all $\mu \in \Lambda$, $\mu \ne \lambda$.

If $N$ is a proper $g$-submodule of $M[\lambda ]$, then it’s contained the second summand $M(\lambda {)}^{\prime }$. Thus the sum of proper $g$-submodules of $M(\lambda )$ is still contained in $M(\lambda {)}^{\prime }$ and it’s the unique proper $g$-submodule of $M(\lambda )$. \end{proof}

Proposition

$M_{\lambda }$ is finite dimensional iff $\lambda \in X^{+}$.

\begin{proof}If $M_{\lambda }$ is finite dimensional, then $M_{\lambda }$ has a maximal weight $\lambda$. Also by the representation theory of $sl(2,\mathbb{C})$, $\lambda (h_{\alpha })\in {\mathbb{Z}}_{\ge 0}$ for all $\alpha \in \Delta$. Therefore, $\lambda \in X^{+}$.

Conversely, suppose $\lambda \in X^{+}$. Define $M_{\lambda }=M(\lambda )/J(\lambda )$ where $J(\lambda )$ is the unique maximal proper $g$-module of $M(\lambda )$ and $M_{\lambda ,\mu }$ is the weight $\mu$-space of $M_{\lambda }$. Then

$$M_{\lambda }={\oplus }_{\mu \in \Lambda }{M}_{\lambda ,\mu },$$

where $\Lambda$ is the collection of allowable weights. Notice $\Lambda \subseteq X\cap (\lambda -Y^{+})$, where $X={\oplus }_{i=1}^l\mathbb{Z}{w}_i$ is the weight lattice and $Y={\mathrm{span}}_{\mathbb{Z}}\Phi$ is the root lattice. And $\dim M_{\lambda ,\mu }$ is finite, because $\mu =\lambda -i_1{\alpha }_1-\cdots -i_l{\alpha }_l$ has only finitely many solutions $(i_1,\cdots ,i_l)$.

Key observation: When $\lambda \in X^{+}$, $\Lambda$ is $W$-invariant where $W$ is the Weyl group. Whence for any $\mu \in \Lambda$, there is a $w\in W$ such that $w\mu \in X^{+}\cap (\lambda -Y^{+})$ and $X^{+}\cap (\lambda -Y^{+})$ is a finite set. Then

$$\dim M_{\lambda }\le \sum _{\mu \in X^{+}\cap (\lambda -Y^{+})}\sum _{w\in W/W_{\mu }}\dim M_{\lambda ,w\mu }$$

is finite, where $W_{\mu }$ is the stabilizer of $\mu$ in $W$.

To show $\Lambda$ is $W$-invariant, it is enough to show if $\mu$ is a weight then $w\mu$ is also a weight. Since $W=⟨S_{\alpha }:\alpha \in \Delta ⟩$, it remains to show $S_{\alpha }(\mu )$ is a weight for $\mu \in \Lambda$ and $\alpha \in \Delta$.

Note

$$S_{\alpha }(\mu )=\mu -2\frac{(\mu ,\alpha )}{(\alpha ,\alpha )}\alpha =\mu -\mu (h_{\alpha })\alpha .$$

Assume $\mu (h_{\alpha })=m$. Since $M_{\lambda }$ is a $U_g$-module, $M_{\lambda }$ is also a $s{l}_{2,\alpha }$-module by observation 1. By the representation theory of $s{l}_2(\mathbb{C})$, $\mu (h_{\alpha })=m\in \mathbb{Z}$. Thus there is an irreducible $s{l}_{2,\alpha }$-submodule $M_1$ which has a non-empty intersection with $M_{\lambda ,\mu }$. Suppose $M_1=\mathrm{span}\{v,fv,f^{2}v,\cdots ,f^{k}v\}$. Take $w\in M_1\cap M_{\lambda ,\mu }$, then $h_{\alpha }w=\mu (h_{\alpha })w=mw$ and there exists an $i$ such that $f^{i}v=w$. Whence $k=2i+m$.

Case 1. If $m\ge 0$, $f^m{M}_{\lambda ,\mu }\subseteq M_{\lambda ,\mu -m\alpha }$ because $f=-E_{-\alpha }\in L_{-\alpha }$. Then $f^{m}w=f^m{f}^{i}v=f^{m+i}v\in M_{\lambda ,\mu -m\alpha }$ and $k=2i+m\ge m+i$ yields $f^{m+i}v$ is not zero. Thus $M_{\lambda ,\mu -m\alpha }$ is not empty and so $S_{\alpha }(\mu )\in \Lambda$.

Case 2. If $m\le 0$, $e^{-m}{M}_{\lambda ,\mu }\subseteq M_{\lambda ,\mu +(-m)\alpha }=M_{\lambda ,\mu -m\alpha }$ because $e=2E_{\alpha }/⟨\alpha ,\alpha ⟩\in L_{\alpha }$,. Since $e^{-m}w=e^{-m}{f}^{j}v=c{f}^{j+m}v$ with $c$ a non-zero constant, $e^{-m}w$ is not zero. Thus $M_{\lambda ,\mu -m\alpha }$ is not empty and so $S_{\alpha }(\mu )\in \Lambda$.

Therefore, $\Lambda$ is $W$-invariant and we finish the proof. \end{proof}

Rmk. In fact, for any $w\in M_{\lambda ,\mu }$, either $f^{m}w$ or $e^{-m}w$ is in $M_{\lambda ,S_{\alpha }(\mu )}$. So $\dim M_{\lambda ,\mu }\ge \dim M_{\lambda ,S_{\alpha }(\mu )}$. Conversely, $\dim M_{\lambda ,S_{\alpha }(\mu )}\ge \dim M_{\lambda ,S_{-\alpha }(S_{\alpha }(\mu ))}=\dim S_{\lambda ,\mu }$. Whence $\dim M_{\lambda ,\mu }=\dim M_{\lambda ,S_{\alpha }(\mu )}$. Furthermore, the dimension can be computed by Weyl character formula.

Proposition

If $V$ is an irreducible $g$-module with a maximal weight $\lambda$, then $V\cong M_{\lambda }$. In particular, a maximal weight of an irreducible module, if it exists, it is unique.

\begin{proof}There is a non-zero homomorphism of modules $\Pi :M(\lambda )\to V$ from the universal property of $M(\lambda )$. Since $V$ is irreducible, $\Pi$ is surjective. So $V$ is isomorphic to a quotient of $M(\lambda )$. By Proposition 1, $V\cong M_{\lambda }$. \end{proof}

Rmk. About the universal property, see here. In fact, the universal property is the reason why we define the Verma module.

from wiki: Our goal is to construct a representation $W_{\lambda }$ of $g$ with highest weight $\lambda$ that is generated by a single nonzero vector $v$ with weight $\lambda$. The Verma module is one particular such highest-weight module, one that is maximal in the sense that every other highest-weight module with highest weight $\lambda$ is a quotient of the Verma module.

Main theorems

theorem of the highest weight

Let $g$ be a finite dimensional complex semi-simple Lie algebra. Then the map

\begin{align} X^+ &\to \{\text{ finite dimensional irreducible representations }\}, \\

\mu &\mapsto \text{the irreducible representation whose highest weight is }\mu

\end{align}

$$isbijective{.}^{7}cae7d$$

\begin{proof} For each finite dimensional irreducible representation $V$, there is $M_{\lambda }\cong V$ by ^0c8949. Then by ^7cae7d, $\lambda$ belongs to $X^{+}$. If $M_{\lambda }\cong M_{\mu }$, then the highest weights of $M_{\lambda }$ and $M_{\mu }$ are equal, that is, $\lambda =\mu$.

Conversely, for a fixed $\lambda \in X^{+}$, suppose $\lambda ={\sum }_{i=1}^l{a}_i{w}_i$. Take a vector $v$ and define $h_{{\alpha }_i}v=a_{i}v$, where $\Delta =\{{\alpha }_1,\cdots ,{\alpha }_l\}$. And also suppose $L_{\alpha }v=0$ for all $\alpha \in {\Phi }_{+}$. Then $\mathbb{C}v$ is a $1$-dimensional $U_b$-module and we get Verma module $M(\lambda )$. Whence $M_{\lambda }$ is an finite dimensional irreducible representation of $g$. \end{proof}

Weyl's theorem on complete reducibility

Every finite dimensional representation of a finite dimensional complex semi-simple Lie algebra is completely reducible.

\begin{proof}It needs the construction of the Casimir element $\Omega$ and the computation of its action on the highest weight vector of a non-trivial finite dimensional of a non-trivial finite dimensional irreducible representation ${\pi }_{\lambda }$ of $g$.

Choose a basis $\{x_1,\cdots ,x_n\}$ of $g$ and let $\{x^1,\cdots ,x^n\}$ be the dual basis w.r.t the killing pairing. Note that

$$\begin{array}{rl}\Omega z-z\Omega & =\sum _{i=1}^n(x_i{x}^{i}z-z{x}_i{x}^i)\\ & =\sum _{i=1}^n{x}_i[x^i,z]-\sum _{i=1}^n[z,x_i]x^i\\ & =\sum _{i=1}^n{x}_i(\sum _{j=1}^n{a}^{ji}{x}^j)-\sum _{i=1}^n(\sum _{j=1}^n{a}_{ji}{x}_j)x^i\\ & =0,\end{array}$$

because $a^{ji}=([x^i,z],x_j)$ and $a_{ji}=([z,x_i],x^j)$. Thus $\Omega ={\sum }_{i=1}^n{x}_i{x}^i$ is in the center of $U_g$.

For any $g$-module $M$, to show $M$ is completely reducible, it is enough to show for any irreducible submodule $N$ with $\dim M/N=1$, there is $M\simeq N\oplus \mathbb{C}$. By Schur’s lemma, $\Omega$ acts on $N$ as a scalar. Since $N$ is a $g$-module, $N={\oplus }_{\lambda \in \Lambda }{N}_{\lambda }$ where $\Lambda$ is the set of weights w.r.t $h\curvearrowright N$. Suppose $\rho :g\to \mathrm{End}(M)$ is the representation corresponding $M$. Then $\mathrm{tr}(\rho (\Omega ))=\sum \mathrm{Tr}(\rho (x_i)\rho (x^i))=\dim M\ne 0$. Then see here to finish the proof, which is similar to case 1 in the proof of this theorem. \end{proof}