Every simple lie, its root will be found, when winter comes.

数学

In this article, we will study Lie algebra and classify some of them. The central technique is given in the post of root systems, and the aim of this article is to find a root system of each Lie algebra and prove the correspondence of them.

Lie algebra

Axiomatically, A Lie algebra can be defined as a special algebra on a $\mathbb{F}$-vector space $\mathfrak{g}$, with a bilinear operation $[\cdot,\cdot]:\mathfrak{g}\rightarrow\mathfrak{g}$ called Lie bracket, which is demanded to have the following properties.

1. For all $x,y\in\mathfrak{g}$, $[x,y]=-[y,x]$.
2. For all $x,y,z\in\mathfrak{g}$, we have Jacobi identity,

$$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0.$$

In this article, we simply take $\mathbb{F}=\mathbb{C}$.

As it is an algebra, we can define the subalgebra and then the ideal of it. Basically, a subset $\mathfrak{h}$ of it is a subalgebra if it is closed under all the operations on $\mathfrak{g}$. And we say it is an ideal, if it is a subalgebra and for all $x\in\mathfrak{h}$ and $y\in\mathfrak{g}$, we have $[x,y]\in\mathfrak{h}$ (as what is defined in the case of a ring), denoted as $\mathfrak{h}\trianglelefteq\mathfrak{g}$.

The necessity of ideal is given by the calculation of quotient space. Given a subalgebra $\mathfrak{h}\subset\mathfrak{g}$, we can define $x+\mathfrak{h}\in\mathfrak{g}/\mathfrak{h}$. Then we can give the definition of commutator in quotient space by

$$[x+\mathfrak{h},y+\mathfrak{h}]=[x,y]+\mathfrak{h}.$$

Now, for arbitrary $h\in\mathfrak{h}$, we shall see $(x+h)+\mathfrak{h}= x+\mathfrak{h}$ since the equivalence relation of quotient space is defined as $x\sim y\Leftrightarrow x-y\in\mathfrak{h}$. Then we have

$$\begin{aligned} {} [x,y]+\mathfrak{h}=&[x+\mathfrak{h},y+\mathfrak{h}]\\ =&[x+h+\mathfrak{h},y+\mathfrak{h}]\\ =&[x+h,y]+\mathfrak{h}=[x,y]+[h,y]+\mathfrak{h}. \end{aligned}$$

We want this is well-defined, then we need $[h,y]\in\mathfrak{h}$. Considering the arbitrariness of $h\in\mathfrak{h}$ and $y\in\mathfrak{g}$, this is exactly the definition of an ideal.

Once we see the idea of quotient space, we can think of direct sum of Lie algebras.

Given two Lie algebras $\mathfrak a$ and $\mathfrak b$, we can define the direct sum of them as the direct sum of vector space $\mathfrak{a}\oplus\mathfrak b=\{(a,b);a\in\mathfrak a, b\in\mathfrak b\}$ with new Lie bracket

$$[(a_1,b_1),(a_2,b_2)]_{\mathfrak a \oplus\mathfrak b}=([a_1,a_2]_\mathfrak a,[b_1,b_2]_\mathfrak b).$$

And we can define the decomposition of direct sum for two ideals $\mathfrak{a,b}$ of $\mathfrak{g}$. If $\mathfrak{g}$ is the direct sum of $\mathfrak{a,b}$ in the sense of vector space (i.e, $\mathfrak a\cap\mathfrak b=0$), then we say $\mathfrak{a}\oplus\mathfrak{b}$ is the decomposition of $\mathfrak{g}$.

Actually, we can define the homomorphism on Lie algebra which is a linear map that preserves Lie bracket, and it is an isomorphism if it is a bijection. Then we shall see $\mathfrak{g}$ is isomorphic to $\mathfrak{a}\oplus\mathfrak{b}$.

And we shall see now, $\mathfrak{b}\cong \mathfrak{g}/\mathfrak{a}$ !!!!!!

Simple, Simple and Simple

Now, let’s consider the converse problem: given $\mathfrak a$ is an ideal of $\mathfrak g$, is $\mathfrak{a}\oplus\mathfrak{g/a}\cong \mathfrak{g}$?

At the first glance, it seems to be obvious, but actually, we can construct a counter example. Let’s think $\mathfrak a=\mathbb C$ and $\mathfrak{b}=\mathbb C^2$ are two trivial Lie algebras with

$$\forall a,b\in\mathbb C,[a,b]=0,\text{ and }\forall a,b\in\mathbb C^2,[a,b]=0.$$

Now, we shall see $\mathbb C^3$ can be a Lie algebra, and it can be a direct sum of $\mathbb C$ and $\mathbb C^2$.

However, we can give different Lie brackets between $\mathbb C$ and $\mathbb C^2$. We suppose $\mathfrak a=\mathbb C$ is an ideal of $\mathbb C^3$ and $\mathfrak b=\mathbb{C}^2$ is a subalgebra of it. Now, choose a base $\{e_1\}$ for $\mathfrak a$ and $\{e_2,e_3\}$ for $\mathfrak b$. Then we only have to find how $[e_1,e_2]$ and $[e_1,e_3]$ computes since the Lie bracket is a bilinear operation.

Since $\mathfrak a$ is an ideal, then $[e_1,e_2]$ and $[e_1,e_3]$ are all resides in $\mathfrak a$. Then, we can assume $[e_1,e_2]=\lambda e_1$ and $[e_1,e_3]=\mu e_1$. For each pair of $(\lambda,\mu)$, it defines a unique Lie algebra $\mathbb C^3_{(\lambda,\mu)}$. For $(a_1,b_1,c_1),(a_2,b_2,c_2)\in \mathbb C^3_{(\lambda,\mu)}$, we have

$$\begin{aligned} {}[(a_1,b_1,c_1),(a_2,b_2,c_2)]=&[a_1e_1+b_1e_2+c_1e_3,a_2e_1+b_2e_2+c_2e_3]\\ =&[a_1e_1,a_2e_1]+[a_1e_1,b_2e_2]+[a_1e_1,c_2e_3]\\ +&[b_1e_2,a_2e_1]+[b_1e_2,b_2e_2]+[b_1e_2,c_2e_3]\\ +&[c_1e_3,a_2e_1]+[c_1e_3,b_2e_2]+[c_1e_3,c_2e_3]\\ =&(\lambda a_1b_2+\mu a_1c_2-\lambda a_2b_1-\mu a_2c_1)e_1\\ =&[\lambda(a_1b_2-a_2b_1)+\mu(a_1c_2-a_2c_1)]e_1\\ =&(\lambda(a_1b_2-a_2b_1)+\mu(a_1c_2-a_2c_1),0,0). \end{aligned}$$

If $\lambda,\mu\not=0$, then we shall see $\mathfrak{g/a}\cong\mathfrak b$, but $\mathfrak a \oplus\mathfrak{b}\not=\mathfrak g$, which means $\mathfrak{a}\oplus\mathfrak{g/a}\not\cong \mathfrak{g}$, and this is a “wrong decomposition”!

Now, we can think of a kind of “good” Lie algebras, which has the property that for any ideal of it, we have a decomposition of direct sum. Why? Because our aim is to classify Lie algebras, thus we need them to be structured and simple enough.

Let’s think the simplest situation, where a Lie algebra doesn’t have any nontrivial ideal. Moreover, it’s easy to see that $[\mathfrak{g},\mathfrak{g}]:=\{[a,b];a,b\in\mathfrak{g}\}$ is an ideal of $\mathfrak g$, then $[\mathfrak g,\mathfrak g]$ equals either $0$ or $\mathfrak g$. The former is too trivial and we are not interested in it, we only want the latter. This leads to the definition of simple Lie algebra:

A Lie algebra $\mathfrak{g}$ is called simple if it has no non-trivial ideal and there exists $a,b\in\mathfrak g$ such that $[a,b]\not=0$.

If we can classify simple ones, the direct sum of them is naturally classified. Such objects, are called semisimple Lie algebras.

Formally, a Lie algebra $\mathfrak g$ is called semisimple, if and only if

$$\mathfrak {g\cong s_1\oplus\cdots\oplus s_n},$$

where all $\mathfrak s_i$ are simple Lie algebra. (Semi.A)

Now, consider any ideal $\mathfrak a$ of $\mathfrak{g}=\bigoplus_{i\in I}\mathfrak{s}_i$. For each $i\in I$, $[\mathfrak{a},\mathfrak{s}_i]\subset\mathfrak{a}\cap\mathfrak{s}_i$, which is an ideal of $\mathfrak{s}_i$. By the simplicity of $\mathfrak s_i$, we shall see $[\mathfrak a,\mathfrak s_i]$ is either $\mathfrak{s}_i$ or $0$. Assume it is $0$ for all $\mathfrak{s}_i$, then for each $i\in I$, $\pi_{\mathfrak{s}_i}(\mathfrak{a})\subset Z(\mathfrak{s}_i)=0$, where $\pi_{\mathfrak{s}_i}(\mathfrak{a})$ is the project on $\mathfrak{s}_i$ from $\mathfrak{a}$ and $Z(\mathfrak s_i)$ is the center of $\mathfrak s_i$s. Hence $\mathfrak{a}=0$ in such case. Suppose $\mathfrak{a}$ is not $0$, then we have $\mathfrak{a}=\bigoplus_{i\in J\subset I}\mathfrak{s}_i$, we choose $\mathfrak{b}=\bigoplus_{i\not\in J}\mathfrak{s}_i$, then we know $\mathfrak{g}=\mathfrak{a}\oplus\mathfrak{b}$. Since $\mathfrak{b}\cong\mathfrak{g/a}$, then

$$\mathfrak{g}\cong\mathfrak{a}\oplus\mathfrak{g/a}.$$

Here we know, semisimple Lie algebras are the ideal objects we want.

Characterization of Semisimple Lie Algebras

Consider $[\mathfrak{g},\mathfrak{g}]$. It’s clear that it is an ideal of $\mathfrak{g}$, and it gives a decomposition of $\mathfrak{g}$. If this process keeps, we may find it down to $0$. To formalize this idea, we define $\mathfrak{g}^{(0)}=\mathfrak{g}$, and for each $n\in\mathbb{N}$, we write $\mathfrak{g}^{(n+1)}=[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]$. Then we have something called a derived sequence of $\mathfrak{g}$:

$$\mathfrak{g}=\mathfrak{g}^{(0)}\trianglerighteq \cdots\trianglerighteq \mathfrak{g}^{(n)}\trianglerighteq\cdots.$$

We say $\mathfrak{g}$ is solvable if there is some $n$ such that $\mathfrak{g}^{(n)}=0$. It’s clear that the direct sum of two solvable Lie algebra is still solvable. Then we can define the thing called radical of a Lie algebra $\mathfrak{g}$, which is the biggest solvable ideal, denoted as $\operatorname{rad}(\mathfrak{g})$.

Since $\mathfrak{r}=\operatorname{rad}(\mathfrak{g})$ is an ideal of $\mathfrak{g}$, we have $\mathfrak{r}=\bigoplus_{i\in I}\mathfrak{s}_i$ where $\mathfrak{s}_i$ are simple algebras. However for every $i\in I$, $[\mathfrak{s}_i,\mathfrak{s}_i]=0$. Hence, we know that $\operatorname{rad}(\mathfrak{g})=0$.

Conversely, we suppose $\operatorname{rad}(\mathfrak{g})=0$. We want to prove that $\mathfrak{g}$ is semisimple, which means it can be decomposed into direct sum of simple Lie algebras. This will give a characterization of semisimplicity.

However, the direct proof is hard. It’s because we don’t have enough tools to decompose a Lie algebra. Given an ideal $\mathfrak{i}\trianglelefteq \mathfrak{g}$, how can we find a standard complement of it? The traditional way (like what we do in linear algebra) to do this, is finding an orthogonal complement

$$\mathfrak{i}^{\perp}:=\{x\in\mathfrak{g};B(x,y)=0,y\in\mathfrak{i}\},$$

where $B$ is a bilinear form (Kill.A). Moreover, we need $\mathfrak{i}^\perp$ to be an ideal of $\mathfrak{g}$, therefore for $x\in\mathfrak{i}^\perp,y\in\mathfrak{i},z\in\mathfrak{g}$, $[z,x]\in\mathfrak{i}^\perp$, then $B([z,x],y)=0$. Since $B(x,[z,y])$ is always $0$, hence ideally, we need $B$ to have the property (Kill.B) that

$$B([z,x],y)=-B(x,[z,y]),\text{ for all } x,y,z\in\mathfrak{g}.$$

Therefore, we need to find such a bilinear form $B:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathbb{F}$ at first.

Since $\mathfrak{g}$ is too far from numbers, we first need to give a representation $\rho:\mathfrak{g}\rightarrow\operatorname{End}(\mathfrak{g})$. Naturally, we can use adjoint representation here,

$$\operatorname{ad}:\mathfrak{g}\rightarrow\operatorname{End}(\mathfrak{g}),\operatorname{ad}(x)=[x,\cdot].$$

It’s quite clear that such map is a good representation of Lie algebras, which is given by the Jacobi’s identity that

$$\begin{aligned} &[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0\\ \Rightarrow\quad&[x,[y,z]]+[y,[z,x]]=-[z,[x,y]]\\ \Rightarrow\quad&[x,[y,z]]-[y,[x,z]]=[[x,y],z]\\ \Rightarrow\quad&(\operatorname{ad}(x)\circ\operatorname{ad}(y))(z)-(\operatorname{ad}(y)\circ\operatorname{ad}(x))(z)=\operatorname{ad}([x,y])(z)\\ &\text{here we define}[A,B]=AB-BA\\ \Rightarrow\quad&[\operatorname{ad}(x),\operatorname{ad}(y)](z)=\operatorname{ad}([x,y])(z)\\ \Rightarrow\quad&[\operatorname{ad}(x),\operatorname{ad}(y)]=\operatorname{ad}([x,y]). \end{aligned}$$

Now, the problem becomes finding a bilinear form $C:\operatorname{End}(\mathfrak{g})\times\operatorname{End}(\mathfrak{g})\rightarrow\mathbb{C}$.

By some representation theory, we have $\operatorname{End}(\mathfrak{g})\cong V\otimes V^*$, and the problem is equivalent to find a linear map

$$D:\underset{e_i}V\otimes\underset{e^j}V^*\otimes\underset{e_k}V\otimes\underset{e^l}V^*\rightarrow\mathbb{C}.$$

For $A,B\in\operatorname{End}(\mathfrak{g})$, we now can write $A=\sum_{i,j}A_{ij}e_i\otimes e^j$ and $B=\sum_{k,l}B_{kl}e_k\otimes e^l$. We have two ways to compute such linear map by contraction. If we let $e^j$ acts on $e_i$ and $e^l$ acts on $e_k$, then

$$D_1(A,B)=\sum_{i,j,k,l}A_{ij}B_{kl}\delta_i^j\delta_k^l=\sum_{i,k}A_{ii}B_{jj}=\operatorname{tr}(A)\operatorname{tr}(B).$$

If we let $e^l$ acts on $e_i$ and $e^j$ acts on $e_k$, then

$$D_2(A,B)=\sum_{i,j,k,l}A_{ij}B_{kl}\delta_i^l\delta_k^j=\sum_{i,j}A_{ij}B_{ij}=\operatorname{tr}(AB).$$

Now, we have two candidates for the bilinear form. For $x,y\in\mathfrak{g}$,

$$B_1(x,y)=\operatorname{tr}(\operatorname{ad}(x))\operatorname{tr}(\operatorname{ad}(y)),$$ $$B_2(x,y)=\operatorname{tr}(\operatorname{ad}(x)\circ\operatorname{ad}(y)).$$

However, in semisimple Lie algebra $\mathfrak{g}$, $[\mathfrak{g},\mathfrak{g}]=\bigoplus_i[\mathfrak{s}_i,\mathfrak{s}_i]=\bigoplus_i\mathfrak{s}_i=\mathfrak{g}$. Then for each $x\in\mathfrak{g}$, it can be written as $x=\sum_i[y_i,z_i]$. Then

$$\begin{aligned} \operatorname{tr}(\operatorname{ad}(x))&=\sum_i\operatorname{tr}(\operatorname{ad}([x_i,y_i]))\\ &=\sum_i\operatorname{tr}(\operatorname{ad}(x_i)\operatorname{ad}(y_i)-\operatorname{ad}(y_i)\operatorname{ad}(x_i))\\ &=\sum_i\operatorname{tr}(\operatorname{ad}(x_i)\operatorname{ad}(y_i))-\operatorname{tr}(\operatorname{ad}(x_i)\operatorname{ad}(y_i))=0. \end{aligned}$$

Hence, $B_1$ is disused here because $B_1(x,y)\equiv0$.

For $B_2$, we let $\mathfrak{z}=\{x\in\mathfrak{g};B_2(x,y)=0,\forall y\in\mathfrak{g}\}$. It’s easy to verify that $\mathfrak{z}\trianglelefteq\mathfrak{g}$. A machinery can be given here called Cartan’s criterion that a Lie algebra $\mathfrak{g}$ over a field $\mathbb{k}$ characteristic zero is solvable if and only if $[\mathfrak{g},\mathfrak{g}]\subset\{x\in\mathfrak{g}; B_2(x,y)=0,\forall y\in\mathfrak{g}\}$ (See [Etingof, Theorem 16.18], and we won’t prove it since it’s too technical). Then, we see $\mathfrak{z}$ is solvable. Since $\operatorname{rad}(\mathfrak{g})=0$, then $\mathfrak{z}=0$.

Now, we know $B_2$ is non-degenerate on semisimple Lie algebras, and it is exactly the bilinear form we want to construct. Historically, this is originally given by Wilhelm Killing, thus we call it Killing form, denoted as $\kappa$.

Back to the waist line, we now use the Killing form to decompose a given semisimple Lie algebra $\mathfrak{g}$. For any ideal $\mathfrak{i}\trianglelefteq\mathfrak{a}$, let

$$\mathfrak{i}^\perp=\{x\in\mathfrak{g};\kappa(x,y)=0,\forall y\in\mathfrak{i}\}.$$

For arbitrary $z\in\mathfrak{g},i\in\mathfrak{i}$ and $a\in\mathfrak{i}^\perp$, $\kappa([z,a],i)=-\kappa(a,[z,i])=0$ (This is by (Kill.B)). Then $\mathfrak{i}^\perp$ is an ideal. And again by Cartan’s criterion, we have $\mathfrak{i}\cap\mathfrak{i}^\perp=0$. Since $\kappa$ is non-degenerate, we have $\mathfrak{g}=\mathfrak{i}\oplus\mathfrak{i}^\perp$.

Then we do induction on the dimension of $\mathfrak{g}$. Suppose when $\dim\mathfrak{g}<n$, it can be decomposed into direct sum of simple Lie algebras. When $\dim\mathfrak{g}=n$, take $\mathfrak{m}$ to be the minimal non-zero ideal of $\mathfrak{g}$, then $\mathfrak{g}=\mathfrak{m}\oplus\mathfrak{m}^\perp$. Assume $\mathfrak{t}\trianglelefteq\mathfrak{m}$ is a non-trivial ideal of $\mathfrak{m}$, then

$$[\mathfrak{g},\mathfrak{t}]=[\mathfrak{m,t}]+[\mathfrak{m}^\perp,\mathfrak{t}]=\mathfrak{t}+0=\mathfrak{t}.$$

Hence $\mathfrak{n}\trianglelefteq\mathfrak{g}$, which contradicts to the minimality of $\mathfrak{m}$. And $[\mathfrak{m},\mathfrak{m}]$ by $\operatorname{rad}(\mathfrak{g})=0$, then $\mathfrak{m}$ is a simple Lie algebra. And $\mathfrak{m}^\perp$ is semisimple, because $\operatorname{rad}(\mathfrak{g})\trianglelefteq\operatorname{rad}(\mathfrak{g})=0$. Then by assumption, $\mathfrak{m}^\perp$ can be written as direct sum of simple Lie algebras since $\dim\mathfrak{m}^\perp<n$. Therefore, $\mathfrak{g}$ can be decomposed as direct sum of simple Lie algebras.

Here, we proved the characterization of semisimplicity. Formally, the following 3 are equivalent to each other,

1. $\mathfrak{g}=\bigoplus_{i}\mathfrak{s}_i$, where $\mathfrak{s}_i$ are simple; (Semi.A)
2. $\operatorname{rad}(\mathfrak{g})=0$; (Semi.B)
3. the Killing form $\kappa$ is non-degenerate on $\mathfrak{g}$. (Semi.C)

The equivalence of (Semi.C) is given by sandwiching it between (Semi.A) and (Semi.B). In some textbooks, (Semi.B) is used as the standard definition of semisimplicity, but here we use (Semi.A) because of the clarity of it.

Jordan Decomposition of Simple Lie Algebras

Now, we have found the right object to be classified. That is the complex simple Lie algebras.

How to classify something linear? Maybe we need to recall the Jordan canonical form, which is a good way to see through things.

Given a square matrix $A$, it can be written as

$$A= \begin{bmatrix} A_{1} &&\\ &\ddots&\\ &&A_{n}\\ \end{bmatrix}$$

under some basis, where

$$A_i= \begin{bmatrix} \lambda_i &1&\\ &\lambda_i&1\\ &&\ddots&\ddots\\ &&&\lambda_i&1\\ &&&&\lambda_i \end{bmatrix}= \begin{bmatrix} \lambda_i\\ &\ddots\\ &&\lambda_i \end{bmatrix} +\begin{bmatrix} 0&1\\ &\ddots&\ddots\\ &&0&1\\ &&&0 \end{bmatrix}$$

and $\lambda_i$ is the eigenvalue of $A_i$.

Then, the linear operator $\mathscr{A}$ corresponding to $A$ can be decomposed as

$$\mathscr{A}=\mathscr{A}_d+\mathscr{A}_n,$$

where $\mathscr{A}_d$ is diagonalizable and $\mathscr{A}_n$ is nilpotent (means some $m\in\mathbb{N}$ be such that $\mathscr{A}_n^m=0$). Moreover, we always have $\mathscr{A}_d\mathscr{A}_n=\mathscr{A}_n\mathscr{A}_d$.

Here, we can do the same thing on Lie algebras. Given a $x\in\mathfrak{g}$, $\operatorname{ad}x$ is a complex matrix, then it can be decomposed into $(\operatorname{ad}x)=(\operatorname{ad}x)_d+(\operatorname{ad}x)_n$. Then, we hope we can find $x=x_d+x_n$ where $x_d,x_n\in\mathfrak{g}$ and $(\operatorname{ad}x)_d=\operatorname{ad}x_d,(\operatorname{ad}x)_n=\operatorname{ad}x_n$.

To do such thing, we need some knowledge of derivation. Watch, start from the Jacobi’s identity,

$$\begin{aligned} [x,[y,z]]+[y,[z,x]]+[z,[x,y]]&=0\\ [x,[y,z]]&=[[x,y],z]+[y,[x,z]]\\ (\operatorname{ad}x)[y,z]&=[(\operatorname{ad}x)y,z]+[y,(\operatorname{ad}x)z].\\ \end{aligned}$$

If something has such identity, we call it a derivation. Formally, an operator $\delta$ is called a derivation if and only if

$$\delta[y,z]=[\delta y,z]+[y,\delta z].$$

All derivations on $\mathfrak{g}$, is written as $\operatorname{der}\mathfrak{g}$. Apparently, $\operatorname{ad}\mathfrak{g}\subset\operatorname{der}\mathfrak{g}$.

For some $\delta\in\operatorname{der}\mathfrak{g}$, it can be decomposed into $\delta=\delta_d+\delta_n$. Here, we can define generalized eigenspace of eigenvalue $\lambda$ of $\delta$ as (here $I$ is the identity)

$$\mathfrak{g}_\lambda:=\{x\in\mathfrak{g};\delta_dx=\lambda x\}.$$

This is equivalent to say that

$$\mathfrak{g}_\lambda=\{x\in\mathfrak{g};(\delta-\lambda I)^N=0,\text{ for some big enough }N\}.$$

It’s clear that there is a decomposition $\mathfrak{g}=\bigoplus_\lambda\mathfrak{g}_\lambda$ (in the sense of vector space).

Now for $x\in\mathfrak{g}_\lambda,y\in\mathfrak{g}_\mu$, there must be some $p,q\in\mathbb{N}$ such that $(\delta-\lambda I)^px=0,(\delta-\mu I)^qx=0$. Now, compute

$$\begin{aligned} (\delta-(\lambda+\mu)I)[x,y]&=\delta[x,y]-(\lambda+\mu)[x,y]\\ &=[\delta x,y]+[x,\delta y]-[\lambda x,y]-[x,\mu y]\\ &=[(\delta-\lambda I)x,y]+[x,(\delta-\mu I)y]. \end{aligned}$$

Continue such procedure, by the parody of binomial theorem, we have

$$(\delta-(\lambda+\mu)I)^n[x,y]=\sum_{k=0}^n\binom{n}{k}[(\delta-\lambda)^kx,(\delta-\mu)^{n-k}y].$$

We take $n=p+q$ and check every term in this formula, then $(\delta-(\lambda+\mu)I)^n[x,y]=0$. Hence, we have $[x,y]\in\mathfrak{g}_{\lambda+\mu}$. (Deco.A)

Now, for arbitrary $\delta\in\operatorname{der}\mathfrak{g}$,

$$\delta_d[x,y]=(\lambda+\mu)[x,y]=[\lambda x,y]+[x,\delta y]=[\delta_d x,y]+[x,\delta_d y].$$

Hence, we know that $\delta_d$ is still a derivation, and naturally follows that, $\delta_n$ is still a derivation.

Now, we use a functional analytical method to prove the fact that $\operatorname{ad} \mathfrak{g}=\operatorname{der}\mathfrak{g}$ for $\mathfrak{g}$ is semisimple. (Deco.B)

For arbitrary $\delta\in\operatorname{der}\mathfrak{g},y\in\mathfrak{g}$, we define a linear functional that $f(y)=\operatorname{tr}(\delta\circ\operatorname{ad} y)$. Since $\kappa$ is non-degenerate on $\mathfrak{g}$, then there exists unique $z\in\mathfrak{g}$ such that $f(y)=\kappa(z,y)$. Let $\delta'=\delta-\operatorname{ad}z$, then $\delta'$ is a derivation and

$$\begin{aligned} \operatorname{tr}(\delta'\circ\operatorname{ad}y)&=\operatorname{tr}((\delta-\operatorname{ad}z)\circ\operatorname{ad}y)\\ &=\operatorname{tr}(\delta\circ\operatorname{ad}y)-\operatorname{tr}(\operatorname{ad}z\circ\operatorname{ad}y)\\ &=f(y)-\kappa(z,y)=0. \end{aligned}$$

For any $x,y\in\mathfrak{g}$,

Notice that

$$\begin{aligned} &\delta'[x,y]=[\delta' x,y]+[x,\delta' y]\\ \Rightarrow\quad&(\delta\circ\operatorname{ad}x)y=\operatorname{ad}(\delta' x)(y)+\operatorname{ad}x(\delta' y)\\ \Rightarrow\quad&\operatorname{ad}(\delta' x)(y)=(\delta'\circ\operatorname{ad}x)y-(\operatorname{ad}x\circ\delta')y\\ \Rightarrow\quad&\operatorname{ad}(\delta' x)=\delta'\circ\operatorname{ad}x-\operatorname{ad}x\circ\delta', \end{aligned}$$

therefore,

$$\begin{aligned} \kappa(\delta'x,y)&=\operatorname{tr}(\operatorname{ad}(\delta'x)\circ\operatorname{ad}y)\\ &=\operatorname{tr}(\delta'\circ\operatorname{ad}x\circ\operatorname{ad}y)-\operatorname{tr}(\operatorname{ad}x\circ\delta'\circ\operatorname{ad}y)\\ &=0. \end{aligned}$$

Hence, $\delta'=0$, so we now know $\operatorname{ad}z=\delta$. This is equivalent to say that $\operatorname{ad}\mathfrak{g}=\operatorname{der}\mathfrak{g}$. Since $(\operatorname{ad}x)_d\in\operatorname{der}\mathfrak{g}$, then there is some $x_d\in\mathfrak{g}$ such that $(\operatorname{ad}x)_d=\operatorname{ad}(x_d)$. Now, we can do the Jordan decomposition on the simple Lie algebra (actually by the process, we see it is the same on semisimple ones) that

$$x=x_d+x_n.$$

Cartan Subalgebra

We say an element $x\in\mathfrak{g}$ is diagonalizable if $x_d=x$, and we say it is nilpotent if $x_n=x$. The toral subalgebra is defined as a subalgebra $\mathfrak{t}\subset\mathfrak{g}$ that every $x\in\mathfrak{t}$ is diagonalizable. And we can define a thing called Cartan subalgebra, which is named after Élie Cartan, that $\mathfrak{h}$ is a maximal toral subalgebra.

A thing interesting can be given here is that $\mathfrak{h}$ can be diagonalized simultaneously. First, we prove that $\mathfrak{h}$ is abelian, i.e. for any $x,y\in\mathfrak{h}$, $\operatorname{ad}x\circ\operatorname{ad}y=\operatorname{ad}y\circ\operatorname{ad}x$. To prove this, is equivalent to prove that $[x,y]=0$.

We constrain $x$ on $\mathfrak{h}$, then $\operatorname{ad}x|_\mathfrak{h}$ is still diagonalizable. Suppose $\{\lambda_i\}_{i\in I}$ are eigenvalues of it and $\{h_i\}_{i\in I}$ are corresponding eigenvectors. Then $[x,h_i]=\lambda_i h_i$. Now,

$$\begin{aligned} (\operatorname{ad} h_i)^2x&=(\operatorname{ad}h_i)[h_i,x]\\ &=(\operatorname{ad}h_i)(-[h_i,x])\\ &=(\operatorname{ad}h_i)(\lambda_ih_i)\\ &=-\lambda_i[h_i,h_i]=0. \end{aligned}$$

Hence, we see $x$ resides in the generalized eigenspace of $h_i$ where the eigenvalue is $0$. However, we know that $h_i$ is diagonalized. Hence $x$ resides in the eigenspace of $h_i$ where eigenvalue is $0$. Therefore, $[h_i,x]=0\cdot x=0$. Thus we know $[x,h_i]=0$ for all $h_i\in\mathfrak{h}$. So $[x,y]=0$.

Now, since $\operatorname{ad}\mathfrak{h}$ is abelian, while everything inside can be diagonalized, then we see $\operatorname{ad}\mathfrak{h}$ can be diagonalized by simple linear algebra.

Now, for each $x\in\mathfrak{h}$, it gives a decomposition of $\mathfrak{g}=\bigoplus_{\lambda_i}\mathfrak{g}_{\lambda_i}$ in the sense of vector space. Such decomposition is the same for all $x\in\mathfrak{h}$, hence we have a simultaneous eigenspace decomposition that

$$\mathfrak{g}=\bigoplus_{\alpha\in\mathfrak{h}^*}\mathfrak{g}_\alpha$$

where

$$\mathfrak{g}_\alpha=\{x\in\mathfrak{g};[h,x]=\alpha(h)x,\forall h\in\mathfrak{h}\}.$$

For $x\in\mathfrak{g},h\in\mathfrak{h}$, by the maximality of Cartan subalgebra $\mathfrak{h}$, we know $[h,x]=0\Leftrightarrow x\in\mathfrak{h}$. Then we can write $\Phi=\{\alpha\in \mathfrak{h}^*\setminus 0;\mathfrak{g}_\alpha\not=0\}$, and

$$\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_\alpha.$$

$\Phi$

In this chapter, we will dive into $\Phi$ to study its structure, and we will see it is a root system which has been studied throughoutly in the previous post.

$\Phi$, Canto I, the real Hilbert space

We first prove that $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is still non-degenerate. For arbitrary $h\in\mathfrak{h},x\in\mathfrak{g}_\alpha$, by (Deco.A),

$$\operatorname{ad}(h)\circ\operatorname{ad}(x):\mathfrak{g}_\beta\rightarrow\mathfrak{g}_{\alpha+\beta}.$$

Hence, $\operatorname{tr}(\operatorname{ad}(h)\circ\operatorname{ad}(x))=0$. Then we see if $\kappa(h,\mathfrak{h})=0$, then $\kappa(h,\mathfrak{g})=0$. Then by non-degenerateness of $\kappa$ on $\mathfrak{g}$, we have $h=0$.

Now, we can see $\kappa$ gives a linear isomorphism that

$$t:\mathfrak{h}\rightarrow\mathfrak{h}^*,t(h)=\kappa(h,\cdot).$$

Then for any $\alpha\in\Phi$, there is a $t^{-1}(\alpha)\in\mathfrak{h}$. Now, we define

$$(\alpha,\beta):=\kappa(t^{-1}(\alpha),t^{-1}(\beta)).$$

It is bilinear and non-degenerate.

Now we take $E=\operatorname{span}_\mathbb{R}(\Phi)$, then

$$\begin{aligned} (\alpha,\alpha)&=\kappa(t^{-1}(\alpha),t^{-1}(\alpha))\\ &=\operatorname{tr}(\operatorname{ad}(t^{-1}(\alpha))^2)\\ &=\sum_{\beta\in\Phi}\dim(\mathfrak{g}_\beta)\beta(t^{-1}(\alpha))^2\\ &=\sum_{\beta\in\Phi}\beta(t^{-1}(\alpha))^2\\ &=\sum_{\beta\in\Phi}\kappa(t^{-1}(\beta),t^{-1}(\alpha))^2\\ &=\sum_{\beta\in\Phi}(\beta,\alpha)^2, \end{aligned}$$

which is a sum of square of real numbers. Hence, we see $(\cdot,\cdot)$ is an inner product on $E$.

$\Phi$, Canto II, $\mathfrak{sl}(2,\mathbb{C})$-triple and some prerequisites

For each $\alpha\in\Phi$, assume $-\alpha\not\in\Phi$, then consider non-zero $x\in\mathfrak{g}_\alpha$, when

1. $y\in\mathfrak{g}_\beta$ where $\beta\in\Phi$, then $\alpha+\beta\not=0$. We then see $\operatorname{ad}x\circ\operatorname{ad}y:\mathfrak{g}_{\gamma}\rightarrow\mathfrak{g}_{\gamma+\alpha+\beta}$. Since they are different components, $\operatorname{tr}(\operatorname{ad}x\circ\operatorname{ad}y)=0$. Hence $\kappa(x,y)=0$.
2. $y\in \mathfrak{h}$, then $\operatorname{ad}x\circ\operatorname{ad}y:\mathfrak{g}_{\gamma}\rightarrow\mathfrak{g}_{\gamma+\alpha}$. Still different components, so $\kappa(x,y)=0$.

Since $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\beta\in\Phi}\mathfrak{g}_\beta$, we have $\kappa(x,\mathfrak{g})=0$. However, we know $x\not=0$, which is a contradiction. Hence, we know $-\alpha\in\Phi$.

Similarly, we have for $x\in\mathfrak{g}_\alpha$, $\kappa(x,\mathfrak{g}_{-\alpha})\not=0$. We then can choose $e\in\mathfrak{g}_\alpha,f'\in\mathfrak{g}_{-\alpha}$ such that $\kappa(e,f')\not=0$ and take

$$f=\frac{2}{(\alpha,\alpha)\kappa(e,f')}f'.$$

Then for any $h\in\mathfrak{h}$,

$$\begin{aligned} \kappa([e,f],h)&=\kappa(e,[f,h])\\ &=\kappa(e,-[h,f])\\ &=\kappa(e,\alpha(h)f)\\ &=\alpha(h)\kappa(e,f)\\ &=\kappa(t^{-1}(\alpha),h)\kappa(e,f)\\ &=\kappa(\kappa(e,f)t^{-1}(\alpha),h). \end{aligned}$$

By the non-degenerateness of $\kappa$ on $\mathfrak{h}$, we have $[e,f]=\kappa(e,f)t^{-1}\alpha$.

Now,

$$[e,f]=\kappa\left(e,\frac{2}{(\alpha,\alpha)\kappa(e_\alpha,f')}f\right)t^{-1}(\alpha)=\frac{2}{(\alpha,\alpha)}t^{-1}\alpha=:h$$

While $e\in \mathfrak{g}_\alpha$, then $[h,e]=\alpha(h)e=\frac{2\alpha(t^{-1}\alpha)e}{(\alpha,\alpha)}=\frac{2(\alpha,\alpha)e}{(\alpha,\alpha)}=2e$. Similarly, we have $[h,f]=-2f$. Therefore, we have $\{e,f,h\}$ satisfies

$$[h,e]=2e,[h,f]=-2f,[e,f]=h.$$

This give a basis for a special Lie algebra, which is called $\mathfrak{sl}(2,\mathbb{C})$.

Here, we introduce some results of representations of $\mathfrak{sl}(2,\mathbb{C})$. Since these are totally irrelevant algebraic results, we won’t prove them here, but only give as machineries. For more information, refer to [Erdmann–Wildon, Chapter 8].

Every finite dimensional module of $\mathfrak{sl}(2,\mathbb{C})$ is totally reducible. And every irreducible module of $\mathfrak{sl}(2,\mathbb{C})$ is determined by a non-negative integer $d$, written by $V_d$, with dimension $d+1$, and there is a basis $\{v_0,v_1,\cdots,v_d\}$ such that

$$h v_k=(d-2k)v_k,\quad e v_k=(d-k+1)v_{k-1},\quad fv_k=(k+1)v_{k+1},$$

where $v_{-1}=v_{d+1}:=0$.

Therefore, we know that

1. Every eigenvalue of $h$ is integer, and is symmetric about zero; (Repr.A)
2. In any finite dimensional $\mathfrak{sl}(2,\mathbb{C})$-module, if $v$ is an eigenvector of $h$ with eigenvalue greater than $0$, then $fv\not=0$. (Repr.B)

$\Phi$, Canto III, verification of root system

Recall, the knowledge of root systems. In this chapter, we will give the fact that $\Phi$ is actually a root system.

We suppose $E\cong \mathbb{R}^r$, and $(\cdot,\cdot)$ is a positive definite inner product on it. A finite subset $R\subset E\setminus\{0\}$ is called an abstract root system, if

1. (Root.A) $E$ is spanned by $R$;
2. (Root.B) $\forall\alpha,\beta\in R$, $n_{\alpha\beta}=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}$ is an integer;
3. (Root.C) If $\alpha,\beta\in R$, then $s_\beta(\alpha)=\alpha-n_{\beta\alpha}\beta\in R$.

(Root.A) is vacuous.

To verify (Root.B), define the $\alpha$-string through $\beta$ as

$$M_\beta^\alpha=\bigoplus_{k\in\mathbb{Z}}\mathfrak{g}_{\beta+k\alpha}.$$

By (Deco.A), it’s easy to verify that this is a module of $\mathfrak{sl}(2,\mathbb{C})$. Then the eigenvalue of $h$ is

$$(\beta+k\alpha)h=\beta(h)+2k,$$

(here $\alpha(h)=2$ because $[h,e]=2e$ and $e\in\mathfrak{g}_\alpha$).

Since we know the eigenvalue of $h$ is always integer, then $\beta(h)$ is integer.

Now,

$$\beta(h)=\beta(\frac{2t^{-1}\alpha}{(\alpha,\alpha)})=\frac{2\kappa(t^{-1}\beta,t^{-1}\alpha)}{(\alpha,\alpha)}=\frac{2(\alpha,\beta)}{(\alpha,\alpha)}=n_{\alpha\beta},$$

which gives the verification of (Root.B).

To verify (Root.C), we need to compute $\beta-n_{\alpha\beta}\alpha$. The eigenvalue of $h$ is $\beta(h)+2k$, while in $V(d)$, the eigenvalue of $h$ is $d-2k$. Then we see the set $K=\{k;\mathfrak{g}_{\beta+k\alpha}\}\not=0$ is continuous, i.e. $K=\{-p,-p+1,\cdots,q\}$. By (Repr.A), we see $\beta(h)+2q=-(\beta(h)-2p)$, so $n_{\alpha\beta}=p-q$. Since $-q\le q-p\le q$, then $s_\alpha(\beta)=\beta-(p-q)\alpha\in\Phi$.

$\Phi$, Canto IV, reducedness and irreduciblity

In the post of root system, what we have proved is the reduced and irreducible root systems. Hence we now have to prove that $\Phi$ is reduced and irreducible.

We now see $\mathfrak{g}$ as an $\mathfrak{sl}(2,\mathbb{C})=\{e,f,h\}$ deduced $\alpha\in\Phi$. Assume that $\dim\mathfrak{g}_\alpha\ge2$. $\kappa(f,\cdot):\mathfrak{g}_\alpha\rightarrow\mathbb{C}$ then is a linear functional. Since $\dim\mathfrak{g}_\alpha\ge 2$, $\ker \kappa(f,\cdot)\cap\mathfrak{g}_\alpha\not=0$, take $0\not= x\in\mathfrak{g}_\alpha$ such that $\kappa(f,x)=0$. Use the same procedure we used in Canto II, we can replace $e$ by $x$ and get $[f,x]=0$. However, $x$ is the eigenvector of $h$, $\alpha(h)=2$ and $f(x)=[f,x]=0$, which contradicts to (Repr.B).

Therefore, we see $\dim\mathfrak{g}_\alpha=1$.

And assume $\mathfrak{g}_{2\alpha}\not=0$, take $0\not=y\in\mathfrak{g}_{2\alpha}$, then $h(y)=2\alpha(h)y=4y$. We now decompose $\mathfrak{g}$ into direct sum irreducible $\mathfrak{sl}(2,\mathbb{C})$-modules, then $y$ is in some $D_d$, and there must be an eigenvalue $4$, then there must be some eigenvalue $2$, which contributes a component of $\mathfrak{g}_\alpha$. Meanwhile, we know that $\operatorname{span}\{e,f,h\}$ is a submodule of $\mathfrak{g}$ with eigenvalues $\{-2,0,2\}$. However, $e\in\mathfrak{g}_\alpha$, therefore $\dim\mathfrak{g}_\alpha\ge 2$. But this is a contradiction. Hence $2\alpha\not\in\Phi$. Similar for $k\alpha\not\in\Phi$. Hence $\Phi$ is a reduced root system.

Assume that $\Phi$ is not irreducible. Then there is a decomposition that

$$\Phi=\Phi_1\sqcup\Phi_2,\text{ where }\alpha\in\Phi_1,\beta\in\Phi_2,(\alpha,\beta)=0.$$

Assume $\alpha+\beta\in\Phi$ where $\alpha\in\Phi_1,\beta\in\Phi_2$, W.L.O.G, we have $\alpha+\beta\in\Phi_1$. Then $(\alpha+\beta,\beta)=0$. However, $(\alpha,\beta)+(\beta,\beta)>0$, this is a contradiction. Hence $[\mathfrak{g}_\alpha,\mathfrak{g}_\beta]\subset\mathfrak{g}_{\alpha+\beta}=0$. Similarly, $\alpha-\beta\not\in\Phi$.

We define $h_\alpha=\frac{2t^{-1}\alpha}{(\alpha,\alpha)}$, and similarly for $e_\alpha,f_\alpha$. Let $\mathfrak{h}_1=\operatorname{span}\{h_\alpha;\alpha\in\Phi_1\}$, $\mathfrak{h}_2=\operatorname{span}\{h_\alpha;\alpha\in\Phi_2\}$, then $\mathfrak{h}=\mathfrak{h}_1\oplus\mathfrak{h}_2$. Define $\mathfrak{i}=\mathfrak{h}_1\oplus\bigoplus_{\alpha\in\Phi_1}\mathfrak{g}_\alpha\not=0$. Since $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_\alpha$, we only need to check every component of $\mathfrak{g}$,

1. $[\mathfrak{h}_1,\mathfrak{i}]\subset\mathfrak{i}$ is obvious.

2. Take $h_\beta\in\mathfrak{h}_2$ and $\alpha\in\Phi_1$, $[h_\beta,e_\alpha]=\alpha(h_\beta)e_\alpha=n_{\alpha\beta}e_\alpha=0$. Hence $[\mathfrak{h}_2,\mathfrak{g}_\alpha]=0$. Meanwhile, $[\mathfrak{h}_2,\mathfrak{h}_1]=0$ by $\mathfrak{h}$ is abelian. Therefore, $[\mathfrak{h}_2,\mathfrak{i}]=0\subset\mathfrak{i}$.

3. Take $\beta\in\Phi_2,\alpha\in\Phi_1$, $\mathfrak{h}_\beta$, $[\mathfrak{g}_\beta,h_\alpha]=-\beta(h_\alpha)\mathfrak{g}_\beta=-n_{\beta\alpha}\mathfrak{g}_\beta=0$. $[\mathfrak{g}_\beta,\mathfrak{g}_\alpha]=0$ is already proved. Hence $[\mathfrak{g}_\beta,\mathfrak{i}]=0\subset\mathfrak{i}$.

4. $[\mathfrak{g}_\alpha,\mathfrak{i}]\subset\mathfrak{i}$ is obvious.

Hence, $[\mathfrak{g},\mathfrak{i}]=\mathfrak{i}$. Therefore $\mathfrak{i}$ is an ideal of $\mathfrak{g}$. However, this is a contradiction since $\mathfrak{g}$ is simple. Hence $\Phi$ is irreducible.

Consequence

Now, we have given a way to classify the complex simple Lie algebras.

But, we still have two things to prove, which are the equivalence between different Cartan subalgebra of a given Lie algebra (i.e. they give the same root system), and the recovery of simple Lie algebra from a given root system. Formally speaking, they are the following two theorems.

(Cons.A) Let $\mathfrak{g}$ be a complex simple Lie algebra and $\mathfrak{h}_1$ and $\mathfrak{h}_2$ are two Cartan subalgebra of $\mathfrak{g}$, then there is some $\varphi\in\operatorname{Inn}(\mathfrak{g})$ such that $\varphi(\mathfrak h_1)=\mathfrak{h_2}$.

(Cons.B, Serre’s Theorem) Let $C = (c_{ij})$ be the Cartan matrix of a root system, of rank $\ell$. Let $L$ be the complex Lie algebra generated by elements $e_i, f_i, h_i$ for $1 \leq i \leq \ell$, subject to the following relations:

1. $[h_i, h_j] = 0$ for all $i, j$;

2. $[h_i, e_j] = c_{ji}, e_j$ and $[h_i, f_j] = -c_{ji}, f_j$ for all $i, j$;

3. $[e_i, f_i] = h_i$ for each $i$, and $[e_i, f_j] = 0$ if $i \neq j$;

4. $(\mathrm{ad}; e_i)^{1-c_{ji}}(e_j) = 0$ and $(\mathrm{ad}; f_i)^{1-c_{ji}}(f_j) = 0$ if $i \neq j$.

Then $L$ is finite-dimensional and semisimple, with Cartan subalgebra $H = \mathrm{span}\{h_1, \dots, h_\ell\}$, and its root system has Cartan matrix $C$.

However, the proof of them won’t be given in the post. Because I don’t really care about the techniques used in these proof, and these two theorems are so intuitive that I don’t really think them should be proved again.

Hence, this is the end of this article.