4.5 Dynkin diagram

TLDR

Define Dynkin diagram and gives a classification of finite dimensional semi-simple Lie algebra.

What is Dynkin diagram?

Motivation. For a finite dimensional complex semi-simple Lie algebra, we can get a reduced root system w.r.t. a Cartan subalgebra. Suppose $|\Delta |=l$. Then define a Cartan matrix

$$(A_{ij}{)}_{l\times l},\text{ where }{A}_{ij}=\frac{2({\alpha }_i,{\alpha }_j)}{({\alpha }_i,{\alpha }_i)}=2\frac{|{\alpha }_j|}{|{\alpha }_i|}\cos {\varphi }_{ij},$$

and so define its Dynkin diagram. Note that any root system is uniquely determined by a Cartan matrix up to scalar. Furthermore, we can prove that there is a 1-1 correspondence between finite dimensional complex simple Lie algebras and Dynkin diagrams.

Definition

The Dynkin diagram is determined by the Cartan matrix. It is a graph with vertices labelled $1,\cdots ,l$. If $i\ne j$, the vertices $i,j$ are joined by $n_{ij}$ edges, where

$$n_{ij}=A_{ij}{A}_{ji}=4{\cos }^2{\theta }_{ij}\in \mathbb{Z},$$

by ^58ihkp, and the arrow points from the longer root to the shorter root, that is, $i\to j$ if $|A_{ji}|\ge |A_{ij}|$.

Remark. The associated quadratic form

$$2\sum _{i=1}^l{x}_i^2-\sum _{1\le i,j\le l,i\ne j}\sqrt{n_{ij}}{x}_i{x}_j$$

is positive definite with the pairing on the Euclidean space, because

$$2\sum _{i=1}^l{x}_i^2-\sum _{1\le i,j\le l,i\ne j}\sqrt{n_{ij}}{x}_i{x}_j=2(\sum _{i=1}^l{x}_i\frac{{\alpha }_i}{|{\alpha }_i|},\sum _{j=1}^l{x}_j\frac{{\alpha }_j}{|{\alpha }_j|}).$$

Lemma

Let $(V,\Phi )$ be a root system. If $V=V_1\oplus V_2$ and $\Phi \subseteq V_1\cup V_2$, then:

• $V_1\perp V_2$;
• $(V_1,V_1\cap \Phi )$ and $(V_2,V_2\cap \Phi )$ are root systems.

\begin{proof} i) For any $\alpha \in V_1$ and $\beta \in V_2$, since $S_{\alpha }(\beta )=\beta -2\frac{(\alpha ,\beta )}{(\alpha ,\alpha )}\alpha \in \Phi \subseteq V_1\cup V_2$ and $V_1\cap V_2=0$, $S_{\alpha }(\beta )\in V_2$ and $2\frac{(\alpha ,\beta )}{(\alpha ,\alpha )}=0$. Then $V_1\perp V_2$ by $\mathrm{span}\Phi =V$.

ii) Define ${\Phi }_1=V_1\cap \Phi$ and ${\Phi }_2=V_2\cap \Phi$. It is obvious that $S_{\alpha }({\Phi }_i)={\Phi }_i$ for any $\alpha \in \Phi$, and $\mathrm{span}{\Phi }_i=V_i$. Hence $(V_1,{\Phi }_1)$ and $(V_2,{\Phi }_2)$ are root systems. \end{proof}

Remark. In this case, we say $(V,\Phi )$ is the direct sum of $(V_1,V_1\cap \Phi )$ and $(V_2,V_2\cap \Phi )$. If a root system cannot be written as the direct sum of two non-empty root systems, then we say $(V,\Phi )$ is irreducible.

Proposition

The followings hold.

• Let $g$ be a finite dimensional complex semi-simple Lie algebra. Then $g$ is simple iff its associated root system is irreducible.
• Let $(V,\Phi )$ be a reduced root system. Then it is irreducible iff its associated Dynkin diagram (with arrows) is connected.

Remark. A Cartan matrix is called indecomposable if its associated Dynkin diagram is connected.

uniqueness and existence theorem

i) (the uniqueness part) Let $g_1,g_2$ be two finite dimensional simple complex Lie algebras. Then the followings are equivalent:

• $g_1\simeq g_2$.
• the root systems associated to $g_1,g_2$ are isomorphic.
• the associated Cartan matrices $(A_{ij}^{(1)})$ and $(A_{ij}^{(2)})$ are equivalent, i.e. they are of the same size $l\times l$ and $\exists \sigma \in S_l$ s.t. $A_{\sigma (i)\sigma (j)}^{(1)}=A_{ij}^{(2)}$.
• the associated Dynkin diagrams with arrows are the same.

ii) (the existence part) A connected Dynkin diagrams with arrows and whose associated quadratic form is positive definite can only be the following forms: and each of them is associated with a finite dimensional complex simple Lie algebra. Also, you can see here.