"You are researching platypus."

数学

In this post, we are going to study the structure of the thing so called root system, and give a classification of it. The root system is a powerful tool to classify complex semisimple Lie algebras, but can be abstracted to a more general version.

Definition, Examples and Irreducibility

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 $s_\beta(\alpha)=\alpha-n_{\beta\alpha}\beta\in R$.

Its elements are called roots, and the rank of $E=\operatorname{Span}(R)$ is also said the rank of $R$. A thing shall be remarked is that the symbol $n_{\alpha\beta}$ may be written as $\langle\beta,\alpha\rangle$.

The condition 2 basically means the projection relationship between the roots are limited, which will be shown later. The condition 3 gives the reflection of $\alpha$ as the following image shows. Especially, if you take $\alpha=\beta$, this means $-\alpha\in R$, which overall implies $R=-R$.

We may operatorize $n_{\alpha\beta}$ by letting $\alpha^\vee(x)=\frac{2(\alpha,x)}{(\alpha,\alpha)}$. Hence $n_{\alpha\beta}$ now can be written as $\alpha^\vee(\beta)$. We call this functional, by the view of duality, the coroot of the root $\alpha$.

There are of course infinitely many root systems. However, we can add a restriction to make it a little bit simpler. A root system is called reduced, if there is no constant ratio between two roots, which formally means $\alpha,c\alpha\in R\Rightarrow c=\pm1$. Later, all the systems we are talking about are assumed to be reduced. (Actually, in some textbooks like Erdmann and Wildon’s Introduction to Lie Algebras, this is directly set as an axiom of root systems.)

Two roots determine a root system of rank $2$, therefore, it’s important to classify the root systems of rank $2$ at first.

For two different roots, let $\theta$ represent the angle between $\alpha$ and $\beta$. Then we have $(\alpha,\beta)=2\|\alpha\|\|\beta\|\cos\theta$ by Ancient Egyptian mathematics. Hence $n_{\alpha\beta}=2\frac{\|\alpha\|}{\|\beta\|}\cos\theta$, $n_{\alpha\beta}n_{\beta\alpha}=4\cos^2\theta\in[0,4]$ and finally only $0,1,2,3$. By analysis of each case, we have all the possibilities of the two roots:

1. (A1) $\theta=\pi/2$, $n_{\alpha\beta}=n_{\beta\alpha}=0$;
2. (A2) $\theta=\pi/3$, $n_{\alpha\beta}=n_{\beta\alpha}=1$, $\|\alpha\|^2=\|\beta\|^2$;
3. (A2’) $\theta=2\pi/3$, $n_{\alpha\beta}=n_{\beta\alpha}=-1$, $\|\alpha\|^2=\|\beta\|^2$;
4. (B2) $\theta=\pi/4$, $n_{\alpha\beta}=1$, $n_{\beta\alpha}=2$, $\|\alpha\|^2=2\|\beta\|^2$;
5. (B2’) $\theta=3\pi/4$, $n_{\alpha\beta}=-1$, $n_{\beta\alpha}=-2$, $\|\alpha\|^2=2\|\beta\|^2$;
6. (G2) $\theta=\pi/6$, $n_{\alpha\beta}=1$, $n_{\beta\alpha}=3$, $\|\alpha\|^2=3\|\beta\|^2$;
7. (G2’) $\theta=5\pi/6$, $n_{\alpha\beta}=-1$, $n_{\beta\alpha}=-3$, $\|\alpha\|^2=3\|\beta\|^2$.

The label of these situations, are actually from the classification of root systems. By applying the definition of root systems, we will get the following picture of root systems of rank $2$.

One should realize that the first picture is actually not A1. The root system of A1 is only one axis, which is the unique root system of dimension 1. However, when we study A2, we will easily find out that it can not be written as a union of two 1 dimensional root systems because of the restriction of rule (Root.C).

This gives us the idea of irreducible root system. More formally, we say a root system $R$ is irreducible if it cannot be expressed as a disjoint union of two non-empty subsets $R_1\sqcup R_2$ where $R_1$ and $R_2$ are two root systems and for all $\alpha\in R_1$ and $\beta\in R_2$, we have $(\alpha,\beta)=0$ (By this, we say $R_1$ and $R_2$ are orthogonal, denotes $R_1\perp R_2$).

For each root system, it can be written as a union of irreducible systems as

$$R=R_1\sqcup R_2\sqcup\cdots\sqcup R_k.$$

And the decomposition is unique modulo ordering. To see this, we define a relation between roots. For two roots $\alpha,\beta\in R$, we say $\alpha\sim\beta$ if there is a sequence

$$\alpha=\gamma_0,\gamma_1,\cdots,\gamma_n=\beta$$

where for each $i<n$, we have $\gamma_i\not\perp\gamma_{i+1}$. It’s easy to check that this relation is an equivalence relation. Each equivalence class forms an irreducible root system, and this is a decomposition of $R$. For arbitrary irreducible root system in arbitrary decomposition of $R$, we can find it in the decomposition we constructed, by applying the relation we defined to some roots, and vice versa. Therefore, such decomposition is unique.

The irreducible root systems are the simplest ones, and it is actually what is going to be classified. Each root system is corresponding to a complex simple Lie algebra, we will show this in another article.

Polarization, Basis and Recovery

From now on, we suppose $R$ is an irreducible root system.

Before classifying, we still need some preparations, which is used to simplify the irreducible root systems further. Firstly, we know that $R=-R$, therefore we can study a half part of the root system. Formally, we choose arbitrary hyperplane with codimension $1$ which doesn’t contain any root, This separates the root system into two parts. We can choose one side of it and say it is positive, while the other part of it is called negative. This is called a polarization of the root system.

Formally, we find a vector $t$ which is not orthogonal to any root, and calculate $(t,\alpha)$ for arbitrary root $\alpha$. If $(t,\alpha)>0$, we say $\alpha$ is positive, otherwise say negative. By Riesz-Fréchet representation theorem, this is equivalent to choose a $t\in E^*$ and calculate $t(\alpha)$. The positive roots are denoted by $R_+$ and negative ones by $R_-$.

Once a polarization of a root system is chosen, we can find simple roots on it. A positive root is called simple, if it can not be written as a sum of two other positive roots. We might see that every positive root can now be written as sum of simple roots. Given a positive root which is not simple, we have $\alpha=\beta+\gamma$, where $\beta,\gamma\in R_+$. Since $t\in E^*$, we have $0<t(\beta),t(\gamma)<t(\alpha)$. We know that $t$ has only finite values on $R_+$, therefore such process will terminate somewhere, and the decomposition is found.

By classification of root systems of rank $2$, it’s easy to find out that $\alpha+\beta\in R$ if $(\alpha,\beta)<0$. Therefore for two simple roots $\alpha,\beta\in R_+$, we have $(\alpha,\beta)<0$, since otherwise we have $(-\alpha,\beta)<0$ so $\gamma=\beta-\alpha$ is a root then make either $\beta$ or $\alpha$ not simple.

It’s obvious that simple roots span the space $E$. Actually, they forms a basis of $E$. Assume converse condition, then we can obtain a nontrivial relation that

$$\sum_{i\in I}c_iv_i=\sum_{j\in J}c_jv_j$$

where $I,J$ are disjoint and $c_i>0$. However,

$$0<\|\sum_{i\in I}c_iv_i\|^2=(\sum_{i\in I}c_iv_i,\sum_{i\in I}c_iv_i)=(\sum_{i\in I}c_iv_i,\sum_{j\in J}c_jv_j)\le0,$$

which is a contradiction. Hence, the simple roots are always a basis of $E$.

Actually, this leads to the following concept, which is the axiomatic version of simple roots and arranged as an alternative in some textbooks. A subset $B\subset R$ is called a base for $R$ if

1. (Base.A) $B$ is a basis of $E$.
2. (Base.B) every $\beta\in R$ can be written as $\beta=\sum_{\alpha\in B}k_\alpha \alpha$ with $k_\alpha\in\Z$, and $k_\alpha$ have the same sign.

The existence of it is shown by constructing simple roots. The corresponding definition of positive roots can also be given after this proposition. A root is said to be positive if the coefficients in decomposition of (Base.B) are not negative.

The classification of irreducible root systems, are actually classifying bases of it. To show this, we left to show the unique rebuild of irreducible root systems.

Intuitively, we shall construct a root system by repeatedly acting $s_\alpha$ on the bases. We can see $s_\alpha$ are permutations on $R$. Therefore, it generates a group $W=\langle s_\alpha\rangle_{\alpha\in R}$ on $R$, and we call it the Weyl group. Naturally, we can define a subgroup of it as $W_B=\langle s_\gamma\rangle_{\gamma\in B}$. To obtain the rebuild, we only need to prove $W_B(B)=R$. More precisely, we need to prove that for each $\beta\in R$, there is some $\alpha\in B$ and $g\in W_B$ such that $\beta=g\alpha$.

The following is the proof.

We only need to consider $R^+\supset B$. Let $\beta\in R^+$, we have the decomposition $\beta=\sum_{\gamma\in B}k_\gamma\gamma$, where $\Z\ni k_\gamma \ge 0$. The following concept is important in the proof which gives an index to use induction on it.

$$\operatorname{ht}(\beta):=\sum_{\gamma\in B}k_\gamma.$$

This definition seems to be summoned from void, but is actually unavoidable. I don’t like this proof and I want to find a more general and more natural version, but has not been established.

We start from the situation when $\operatorname{ht}(\beta)=1$, then $\beta\in B$, we only take $\alpha=\beta$ and $g$ is the identity map.

Now we suppose the proposition holds for all $\beta$ that $\operatorname{ht}(\beta)<n$. Then for $\operatorname{ht}(\beta)=n\ge2$, at least two of $k_\gamma$ are strictly positive. There will always be some $\gamma_0\in B$ such that $(\beta,\gamma_0)>0$. If unfortunately otherwise, we have

$$(\beta,\beta)=(\beta,\sum_{\gamma\in B}k_\gamma\gamma)=\sum_{\gamma}k_\gamma(\beta,\gamma)\le0,$$

which contradicts to the positive definiteness of $\beta$. Now we consider $s_{\gamma_0}(\beta)$. Since $\beta\not=\gamma_0$, there is some $\alpha\in B$ such that $k_\alpha\not=0$ and $\alpha\not=\gamma_0$. Now by $s_{\gamma_0}(\beta)=\beta-n_{\beta\alpha}\gamma_0$, we shall see the coefficient of $\alpha$ in the decomposition of $s_{\gamma_0}(\beta)$ is also $k_\alpha$, then $s_{\gamma_0}(\beta)\in R^+.$ Now,

$$\operatorname{ht}(s_{\gamma_0}(\beta))=\operatorname{ht}(\beta)-n_{\beta\gamma_0}<\operatorname{ht}(\beta)=n.$$

Hence, we can find $\alpha\in B$ and $h\in W_B$ making $s_{\gamma_0}(\beta)=h\alpha$, then $\beta=(s_{\gamma_0}h)\alpha$.

Now we proved.

Dynkin Diagram and Restrictions

We now need to classify bases of simple root systems. However, they are still too abstract for us to see, although they are simple enough. A direct idea of realizing the classification, is find some invariants of bases, and here it is.

Given a root system $(\alpha_1,\cdots,\alpha_p)$, the Cartan matrix of it is written as $(n_{ij})_{1\le i,j\le p}$. We shall see this is a perfect encoding of bases.

Another encoding of bases is a diagram. We draw roots as vertices, and draw $d_{\alpha\beta}:=n_{\alpha\beta}n_{\beta\alpha}=:d_{\beta\alpha}$ lines between vertices $\alpha$ and $\beta$. Finally, we draw an arrow on edge from $\alpha$ to $\beta$ if $\|\alpha\|>\|\beta\|$. By the construction of irreducible root systems, we shall see that every Dynkin diagram of the bases is a connected graph.

Dynkin diagram: simplification

Now, by classifying Dynkin diagrams, we can give a classification of bases. To make it simpler, we can ignore the arrows on it, by normalizing the roots.

Given a base $A$ with dimension $n$, we consider the set

$$\bar A=\{\bar \alpha:=\frac{\alpha}{\sqrt{(\alpha,\alpha)}};\alpha\in A\}.$$

Then we shall see for each $\alpha,\beta\in \bar A$,

$$\begin{aligned} d_{\alpha\beta}:=4(\bar\alpha,\bar\beta)^2 &=4(\frac{\alpha}{\sqrt{(\alpha,\alpha)}},\frac{\beta}{\sqrt{(\beta,\beta)}})^2\\ &=4(\frac{1}{\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}})^2(\alpha,\beta)^2\\ &=\frac{4}{(\alpha,\alpha)(\beta,\beta)}(\alpha,\beta)^2\\ &=\frac{2(\alpha,\beta)}{(\alpha,\alpha)}\frac{2(\alpha,\beta)}{(\beta,\beta)}=n_{\alpha\beta}n_{\beta\alpha}. \end{aligned}$$

Then the Dynkin diagram of $\bar A$ is actually a graph forgetting arrows on the Dynkin diagram of $A$. And the vertices are corresponding.

We can give an axiomatic description of such sets. A set $\bar A$ is said to be admissible if

1. $(v_i,v_i)=1$ for all $i$ and $(v_i,v_j)\le 0$ if $i\not= j$;
2. If $i\not=j$, then $4(v_i,v_j)^2\in\{0,1,2,3\}$.

Later, a lot of lemmas will be derived only from these two conditions. We use the notation $\bar A$ to denote the simplified set a base $A$ of a root system, which is admissible and has $|\bar A|=|A|=n$.

Dynkin diagram: admissibility

In this section, all we suppose is that the set is admissible. We now study the restrictions an admissible set can give.

Firstly, the numbers of pairs of vertices joined by at least one edge is at most $n-1$. (Res.A) To prove this, we consider $v=v_1+\cdots+v_n$, then

$$\begin{aligned} 0<(v,v)&=(v_1+v_2+\cdots+v_n,v_1+v_2+\cdots+v_n)\\ &=(v_1,v_1+v_2+\cdots+v_n)+\cdots+(v_n,v_1+v_2+\cdots+v_n)\\ &=(v_1,v_1)+(v_1,v_2)+\cdots+(v_1,v_n)\\ &+(v_2,v_1)+(v_2,v_2)+\cdots+(v_2,v_n)\\ &+\cdots\\ &+(v_n,v_1)+(v_n,v_2)+\cdots+(v_n,v_n)\\ &=n+2\sum_{i<j}(v_i,v_j). \end{aligned}$$

Therefore, $n>-2\sum_{i<j}(v_i,v_j)=\sum_{i<j}\sqrt{d_{ij}}$.

Now, the number of pairs of vertices $\{v_i,v_j\}$ such that $d_{ij}\ge1$ is less or equal to $\sum_{i<j}\sqrt{d_{ij}}$, which is smaller than $n$. Hence the claim proved.

By this, we can easily see that there will be no cycle in a diagram. (Res.B)

What’s more, no vertex of $\Gamma$ is incident to four or more edges. (Res.C) We take $v$ is an arbitrary vertex and there are $k$ vertices $v_1,\cdots,v_k$ incident to it. Since there is no cycles, we have $(v_i,v_j)=0$ for $0\le i<j\le k$. Now, consider the subspace $\operatorname{Span}(v,v_1,\cdots,v_k)=:U$. Then there must be an orthogonal basis of $U$ like $v_0,v_1,\cdots,v_k$. Where $\|v_0\|=0$ and $(v,v_0)\not=0$.

The projection decomposition is $v=\sum_{0\le i\le k}(v,v_i)v_i$. Then we have

$$\begin{aligned} 1=(v,v)&=(\sum_{0\le i\le k}(v,v_i)v_i,\sum_{0\le j\le k}(v,v_j)v_j)\\ &=\sum_{0\le i\le k}\sum_{0\le j\le k}(v,v_i)(v,v_j)(v_i,v_j)\\ &=\sum_{0\le i\le k}(v,v_i)^2. \end{aligned}$$

Hence $\sum_{1\le i\le k}(v,v_i)^2<1$. By the previous knowledge of bases, $4(v,v_i)^2>1$, then $(v,v_i)^2>\frac{1}{4}$. It’s so obvious now that $k$ should be no more than $3$.

Then obviously we know that, if a graph has a triple edge, then it will be only two vertices like below. (Res.D)

An interesting and useful lemma can be given here, which is called the Shrinking Lemma. Given a subset $\bar A'\subset \bar A$, and the graph $\Gamma'$ of $\bar A'$ is a subgraph of $\Gamma$, where $\Gamma$ is the graph of $A$. The lemma claims that, if $\Gamma'$ is a line, then $(\bar A\setminus \bar A')\cup\{\sum_{v\in A'}v\}$ is still an admissible set. That is to say, intuitively, we can shrink a line in a graph to a point.

We only have to verify the conditions of admissible sets. We enumerate $\bar A'$ as $\bar A'=\{v_i\}_{i=1}^k$ and define $u=\sum_{i=1}^kv_i$.

$$\begin{aligned} (u,u)&=(\sum_{i=1}^kv_i,\sum_{i=1}^kv_i)\\&=k+2\sum_{i=1}^{k-1}(v_i,v_{i+1})=k-(k-1)=1. \end{aligned}$$

Choose some $v\in \bar A\setminus\bar A'$, since $v$ can only be incident to at most one $v_i$, then either $(v,u)=0$ or $(v,v_i)<0$. The same thing holds for $4(v,u)^2$. Now we proved.

A simple conclusion (Res.E) can now be given that a graph $\Gamma$ has

1. no more than one double edge;
2. no more than one branch vertex;
3. not both a double edge and a branch vertex.

A vertex is said to be a branch vertex if, it is incident to three or more vertices, and it’s always $3$ in the context of Dynkin diagrams. We only apply the Shrinking Lemma, then by (Res.C), it holds.

Now, we can study the situation of double edge and branch vertex respectively.

We know that if a graph $\Gamma$ has a double edge, then it should have the form below.

And without loss of generality, $p\ge q$. Now, we let $v=\sum_{i=1}^piv_i$, then

$$\begin{aligned} (v,v)&=(\sum_{i=1}^piv_i,\sum_{i=1}^piv_i)\\ &=\sum_{i=1}^p\sum_{j=1}^pij(v_i,v_j)\\ &=\sum_{i=1}^pi^2-\sum_{i=1}^{p-1}i(i+1)\\ &=\frac{p(p+1)(2p+1)}{6}-\frac{p(p+1)(p-1)}{3}\\ &=\frac{p(p+1)}{2}. \end{aligned}$$

Similarly, we let $w=\sum_{i=1}^qiw_i$, then $(w,w)=q(q+1)/2$. Now,

$$(v,w)^2=(pq(v_p,w_q))^2=\frac{p^2q^2}{2}.$$

By the Cauchy-Schwarz inequality, $(v,w)^2<(v,v)(w,w)$, then $2pq<(p+1)(q+1)$, which implies $pq<p+q+1$. Hence,

$$\begin{aligned} (p-1)(q-1)&=pq-p-q+1\\ &<p+q+1-p-1+1=2. \end{aligned}$$

Either $q=1$ or $p=q=2$. Hence, if $\Gamma$ has a double edge, it can only be one of below. (Res.F)

Now, we can study the situation where there is a branch vertex.

As you know, without loss of generality, we take $p\ge q\ge r$.

We now let $u=\sum_{i=1}^piu_i$, $v=\sum_{i=1}^qiv_i$, and $w=\sum_{i=1}^riw_i$. We easily see that $u,v,w$ are pairwisely orthogonal. Let $\hat u=u/\|u\|$, $\hat v=v/\|v\|$ and $\hat w=w/\|w\|$. By the Gram–Schmidt process, we have an orthogonal basis of $\operatorname{Span}(\hat u,\hat v,\hat w,z)$ as $\{\hat u,\hat v,\hat w,z_0\}$ and $(z,z_0)>0$. Therefore we can decompose $z$ as

$$z=(z,\hat u)\hat u+(z,\hat v)\hat v+(z,\hat w)\hat w+(z,z_0)z_0,$$

So,

$$\begin{aligned} &\quad(z,\hat u)\hat u+(z,\hat v)\hat v+(z,\hat w)\hat w=z-(z,z_0)z_0\\ \Rightarrow&\quad(z,\hat u)^2+(z,\hat v)^2+(z,\hat w)^2<1\\ \Leftrightarrow&\quad\frac{1}{\|u\|^2}(z,u)^2+\frac{1}{\|v\|^2}(z,v)^2+\frac{1}{\|w\|^2}(z,w)^2<1\\ \Leftrightarrow&\quad\frac{2p^2}{4p(p+1)}+\frac{2q^2}{4q(q+1)}+\frac{2r^2}{4r(r+1)}<1\\ \Leftrightarrow&\quad\frac{p}{2(p+1)}+\frac{q}{2(q+1)}+\frac{r}{2(r+1)}<1\\ \Leftrightarrow&\quad p(q+1)(r+1)+q(p+1)(r+1)+r(p+1)(q+1)\\ &\qquad<2(p+1)(q+1)(r+1)\\ \Leftrightarrow&\quad pqr<p+q+r+2\\ \Leftrightarrow&\quad pqr+[pq+pr+qr]+[p+q+r]+1\\ &\qquad<[pq+pr+qr]+2(p+q+r)+3\\ \Leftrightarrow&\quad (p+1)(q+1)(r+1)\\ &\qquad<(p+1)(q+1)+(p+1)(r+1)+(q+1)(r+1)\\ \Leftrightarrow&\quad\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}>1. \end{aligned}$$

A thing has to be said here is that the way I calculated this is way more complicated than it could have been. Actually, we only have to notice that

$$\frac{p}{2(p+1)}=\frac{1}{2}(1-\frac{1}{p+1}),$$

then we will have

$$\begin{aligned} &\quad\frac{p}{2(p+1)}+\frac{q}{2(q+1)}+\frac{r}{2(r+1)}<1\\ \Leftrightarrow&\quad\frac{1}{2}\left[3-\left(\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}\right)\right]<1\\ \Leftrightarrow&\quad\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}>1. \end{aligned}$$

This is given by Claude opus. THANKS TO IT.

Since $\frac{1}{p+1}\le\frac{1}{q+1}\le\frac{1}{r+1}\le\frac{1}{2}$, we have $\frac{1}{r+1}>\frac{1}{3}$ and hence $r<2$, so we must have $r=1$. Now, $\frac{1}{p+1}+\frac{1}{q+1}>\frac{1}{2}$, then $\frac{1}{q+1}>\frac{1}{4}$. Hence $q<3$.

1. If $q=2$, we have $\frac{1}{p+1}>\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. Hence $p<5$.
2. If $q=1$, we only have $\frac{1}{p+1}>0$. Then it seems to have no constraints.

Conclude the study of branch vertex, we have shown that either $q=r=1$ or $q=2$, $r=1$ and $p<5$. (Res.G)

Dynkin diagram: constraints

We have studied the restriction on admissible sets, and it is indeed restrictions on bases. We now return to the study of the Dynkin diagram of bases, which may have arrows on it.

Now consider two incident root $\alpha$ and $\beta$, we shall see they have the same length if and only if they have a single edge. Actually, $n_{\alpha\beta}n_{\beta\alpha}=\frac{2(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}=1$, if and only if $n_{\alpha\beta}=n_{\beta\alpha}=-1$ (they are all integers and obtuse). This implies $(\alpha,\alpha)=(\beta,\beta)$. Conversely, if we have $\|\alpha\|=\|\beta\|$, by the study of the root system of rank $2$, we have $n_{\alpha\beta}=n_{\beta\alpha}=-1$.

Now, we can give the classification of Dynkin diagrams. There are only these Dynkin diagrams which are possible that the corresponding root system exists. The letters of the families are given by Cartan.

A thing can be remarked here, is that $B_1=A_1$, $C_2=B_2$ and $D_3=A_3$. Therefore, we always start $B_n$ from $n=2$, $C_n$ from $n=3$ and $D_n$ from $n=4$ to make the classification without repetition.

Recovery of Bases

We have already given all the possibilities of Dynkin diagrams. But, is there really a root system exists? We still have to show the bases corresponding to all the graphs. The derivation of these results are quite complicated, and I don’t really think it’s necessary, so they will be omitted here.

In this chapter, we write $\{e_i\}_{i=1}^n$ as the orthonormal basis of the space $\mathbb{R}^n$.

$A_n(n\ge1)$

Here, $\alpha_i=e_i-e_{i+1}$.

$B_n(n\ge2)$

Here, $\alpha_i=e_i-e_{i+1}$ for $0\le i\le n-1$, and $\alpha_n=e_n$.

$C_n(n\ge3)$

Here, $\alpha_i=e_i-e_{i+1}$ for $0\le i\le n-1$, and $\alpha_n=2e_n$.

$D_n(n\ge4)$

Here, $\alpha_i=e_i-e_{i+1}$ for $0\le i\le n-1$, and $\alpha_n=e_{n-1}+e_n$.

$E_6$

Here,

$$\begin{aligned} &\alpha_1=\frac{1}{2}(e_1+e_8-e_2-e_3-e_4-e_5-e_6-e_7),\\ &\alpha_2=e_1+e_2,\\ &\alpha_i=e_{i-1}-e_{i-2}(3\le i\le6). \end{aligned}$$

$E_7$

Here,

$$\begin{aligned} &\alpha_1=\frac{1}{2}(e_1+e_8-e_2-e_3-e_4-e_5-e_6-e_7),\\ &\alpha_2=e_1+e_2,\\ &\alpha_i=e_{i-1}-e_{i-2}(3\le i\le7). \end{aligned}$$

$E_8$

Here,

$$\begin{aligned} &\alpha_1=\frac{1}{2}(e_1+e_8-e_2-e_3-e_4-e_5-e_6-e_7),\\ &\alpha_2=e_1+e_2,\\ &\alpha_i=e_{i-1}-e_{i-2}(3\le i\le8). \end{aligned}$$

$F_4$

$$\begin{aligned} &\alpha_1=e_2-e_3,\\ &\alpha_2=e_3-e_4,\\ &\alpha_3=e_4,\\ &\alpha_4=\frac{1}{2}(e_1-e_2-e_3-e_4). \end{aligned}$$

$G_2$

Here, $\alpha_1=e_1-e_2$, and $\alpha_2=e_2+e_3-2e_1$.

Wait…

Wait, is everything done?

Everything seems to be perfect. We have simplified the problem and found an object simple enough to be classified, and we have found a good way to encoding them. We have discussed all the possibilities, and shown that there actually are corresponding objects of them.

However, what if an irreducible root system has two different bases, while each of them has a Dynkin diagram respectively? If such thing happens, they will live in different classes. Although this is no biggie, we still do not want such thing happens, since such mathematics is not “beautiful”, just like the classification of Platypus.

Therefore, it remains to prove that, two bases of one root system, have the same Dynkin diagram, which means, they are geometrically equivalent.

Previous contents basically refer to [Erdmann–Wildon] and [Etingof], and the contents following is based on [Humphreys].

We know that, every base is a simple roots live in some corresponding polarization of a root system. Although a base lives in many polarizations, one polarization has only one base. Therefore, we can establish an equivalence relation between polarizations, and corresponding equivalence classes appear. Actually, if we take the perspective of geometry, all the $t$ define the polarizations in an equivalence class are residing in a cone, and this is exactly what we call a Weyl Chamber. Therefore, strictly speaking, each base is corresponding to a Weyl Chamber uniquely.

Then, for sake of convenience, we define some notations. Given a root system $R$, $B$ is a base of it. Then we use $C_B$ to denote the Weyl Chamber of it. Apparently, there are many polarizations in $C_B$, then we write $C_B^+$ for the equivalence class of positive sides, and conversely, we say $C_B^-$.

Given a root $r$, we say $r\in C_B^+$ abusively, if and only if for any $R^+\in C_B^+$, $r\in R^+$.

Now, let’s consider two bases $A$ and $B$ of a root system $R$. We then show that there always be some $\omega\in W$ makes $\omega A=B$, where $W$ is the Weyl group of $R$.

Define $\delta_A=\frac{1}{2}\sum_{r\in C_A^+}r$ and $\delta_B=\frac{1}{2}\sum_{r\in C_B^+}r$. Now, for each $\alpha\in A$, we have

$$\begin{aligned} s_\alpha(\delta_A)&=\frac{1}{2}\sum_{r\in C_A^+}s_\alpha(r)\\ &=\frac{1}{2}\left(s_\alpha(\alpha)+\sum_{r\in C_A^+\setminus\{\alpha\}}s_\alpha(r)\right)\\ &=\frac{1}{2}\left(-\alpha+\sum_{r\in C_A^+\setminus\{\alpha\}}r\right)\\ &=\frac{1}{2}\left(\sum_{r\in C_A^+}r-2\alpha\right) =\delta_A-\alpha. \end{aligned}$$

The third equals sign is given because the simple roots permute $C_A^+\setminus\{\alpha\}$, which has been proved in the recovery of root system from a base.

Now, we iterate the elements in the group $W$, and then choose some $\omega\in W$ to make $(\omega(\delta_A),\delta_B)$ maximal. Then we compute $(\omega(s_\alpha(\delta_A)),\delta_B)$,

$$(\omega(s_\alpha(\delta_A)),\delta_B)=(\omega(\delta_A),\delta_B)-(\omega(\alpha),\delta_B).$$

By the maximality of $(\omega(\delta_A),\delta_B)$, we have $(\omega(\alpha),\delta_B)\ge0$. Now, if $\omega(\alpha)\in C_B^-$, we have $(\omega(\alpha),\delta_B)<0$, this is a contradiction. Hence, we have $\omega(\alpha)\in C_B^+$. Since $\omega$ keeps addition, we know that $\omega(\alpha)$ is a simple root. This implies $\omega(A)=B$ by iterating $\alpha\in A$.

Finally, let’s compute

$$\begin{aligned} &(s_\alpha(\alpha_1),s_\alpha(\alpha_2))\\=&(\alpha_1-\frac{2(\alpha_1,\alpha)}{(\alpha,\alpha)}\alpha,\alpha_2-\frac{2(\alpha_2,\alpha)}{(\alpha,\alpha)}\alpha)\\ =&(\alpha_1,\alpha_2)-\frac{2(\alpha_1,\alpha)}{(\alpha,\alpha)}(\alpha,\alpha_2)-\frac{2(\alpha_2,\alpha)}{(\alpha,\alpha)}(\alpha_1,\alpha)\\ &+\frac{4(\alpha_1,\alpha)(\alpha_2,\alpha)}{(\alpha,\alpha)^2}(\alpha,\alpha)\\ =&(\alpha_1,\alpha_2). \end{aligned}$$

This implies elements in the Weyl group preserve inner product. And hence it is an isometric map. Therefore, their Dynkin diagram shall be the same.

Now, everything we want has been proved, and this is the end of this article.