## 國 立 交 通 大 學

### 應 用 數 學 研 究 所

### 博 士 論 文

### Vogan 圖之等價分類

### Equivalence Classes of Vogan Diagrams

### 研究生: 胡舉卿

### 指導教授: 蔡孟傑教授

### Vogan 圖之等價分類

### Equivalence Classes of Vogan Diagrams

### 學 生:胡舉卿 Student: Chu-Chin Hu

### 指導教授:蔡孟傑教授 Adviser: Pro. Meng-Kiat Chuah

### 國 立 交 通 大 學

### 應 用 數 學 研 究 所

### 博 士 論 文

### A Thesis

### Submitted to Department of Applied Mathematics

### College of Science National Chiao Tung University

### in Partial Fulfillment of Requirements

### for the Degree of

### in

### Applied Mathematics

### June 2006

### Hsinchu, Taiwan, Republic of China.

**Vogan 圖之等價分類 **

### 研究生： 胡舉卿

### 指導教授： 蔡孟傑教授

### 國 立 交 通 大 學

### 應 用 數 學 研 究 所 博 士 班

### 摘 要

### 在 Dynkin 圖上給定一個 involution, 並在該 involution 的固定點

### 上塗上黑或白色即為 Vogan 圖。一個 Vogan 圖則表示一個實數的簡單

### 李代數。若兩個 Vogan 圖所表示的李代數互相等價，我們稱這兩個圖

### 亦等價，而這篇論文將等價的 Vogan 圖作分類。在研究分類的方法

### 時，我們發現了等價的 Vogan 圖所對應的 Dynkin 圖在圖的塗色中有

### 特別的性質。這些組合學的性質簡化了大部分等價分類的證明。

### 本篇論文分為三個部分;在第一部份(第 1-6 節)簡單的介紹一些

### 李代數的基本概念及相關理論的概述。第二部分(第 7-10 節)為這篇

### 論文的主要研究結果。在第三部份(第 11 節)我們利用圖的塗色驗證

### 大部分Dynkin圖的可能性，但無法排除“E

### ＂。

**Equivalence Classes of Vogan Diagrams **

### Student : Chu-Chin Hu

### Advisor : Pro. Meng-Kiat Chuah

### Department of Applied Mathematics,

### National Chiao Tung University

**Abstract **

### A Vogan diagram is a Dynkin diagram with an involution, and the

### vertices fixed by the involution may be painted. They represent real simple

### Lie algebras, and two diagrams are said to be equivalent if they represent

### the same Lie algebra. Our purpose is to classify the equivalence classes of

### all Vogan diagrams. In doing so, we find that the underlying Dynkin

### diagrams have certain properties in graph painting. We show that this

### combinatorial property provides an easy classification for most of the

### simply-laced Dynkin diagrams.

### This thesis is divided into three parts. In the first part, consisting of

### Sections 1-6, we give a brief introduction of some fundamental concepts in

### Lie algebra and a survey of various results without proof. The second part,

### consisting of Sections 7-10, is the main results of this thesis. In the last part

### (Section 11), we use the method of graph painting to verify almost all

### simply-laced Dynkin diagrams except the "fake E_9".

### 誌 謝

### 在交大應用數學系待了六年的時間了，終於要畢業，心裡又是高興又

### 不捨。能順利畢業總是件開心的事，但要離開交大的點點滴滴，心中百感

### 交集。

### 能夠順利完成這篇論文，我最要感謝的是蔡孟傑老師對我的栽培及指

### 導。除了在學術方面的教導之外；更感謝蔡老師在生活上的關心及幫忙。

### 從老師身上我不僅學到做研究的態度更學到許多生活上寶貴的經驗。謝謝

### 老師一路來的栽培及支持。另外，感謝許義容、張麗萍老師在學業上的指

### 導；感謝于靖老師、潘樹衍老師以及莊重老師給我的指教及幫忙。

### 排球隊是我在交大的生活的另一個重心。打排球是我的興趣，也因

### 此結交了不少可貴的好友。感謝我的隊友兼死黨：珮蓉、怡純、彥妤及雅

### 欣，一起並肩作戰南征北討；在我心情不好時陪我徹夜聊天談心，謝謝妳

### 們的關心及付出。感謝雅文、乃心以及月偉在生活上給我的關懷及幫忙，

### 你們真的是很貼心的學妹！還有系排的學妹們除了忍受我在場上的咆哮

### 外，還得在打球之餘供我使喚，謝謝妳們囉！我常常覺得球場是人生的縮

### 影，在排球場上的練習，與隊友的合作，和他人相處，以及臨場的發揮…

### 種種的經驗讓我收穫不少。感謝鄉親張景榮給我的經驗分享，相互鼓勵。

### 我很慶幸能有你們這群好友，豐富並溫暖了我的生活。

### 最後，我要感謝我的家人：哥哥舉軍，妹妹舉芬及弟弟舉鳶相互的支

### 持及關懷，為家裡付出分憂解勞，讓我可以無後顧之憂的完成學業。能夠

### 有今天的成就，我更要感謝我的父親胡振光上校辛苦的養育及栽培，他不

### 求回報的付出造就了今天的我。在此將我的成就與榮耀與你們一起分享！

**Contents **

### 1 Introduction ...……… 1

### 2 Killing Form ……….. 3

### 3 Root Space Decomposition ………... 5

### 4 Cartan Involution and Cartan Subalgebra ………. 7

### 5 Dynkin Diagram ……… 9

### 6 Vogan Diagram ……… 11

### 7 Classifications of Vogan Diagrams ………. 12

### 8 Classical Diagrams ……….. 16

### 9 Exceptional Diagrams ……….…… 20

### 10 Nontrivial Involutions ……….…… 29

### 11 Graph Paintings ………..……. 30

### References ……….. 33

iv### 1

### Introduction

*An Lie algebra is a vector space g with an anti-symmetric bilinear map*
*[ , ] : g × g −→ g*

*which satisfies the Jacobi identity*

*[[X, Y ], Z] + [[Y, Z], X] + [[Z, X], Y ] = 0.*
*Let F be R or C. Then*

*sl(n, F ) = {A ∈ Mn(F ); trace(A) = 0} ;*

*u(n) = {A ∈ Mn(C); A + A∗* *= 0} (skew-Hermitian);*

*o(n) = u(n) ∩ Mn*(R) (skew-symmetric);

*su(n) = u(n) ∩ sl(n, C);*
*so(n) = o(n) ∩ sl(n, R);*

*are all simple Lie algebras with [A, B] = AB − BA for all A, B ∈ g.*
If a, b are subsets of g, we write

*[a, b] = span{[X, Y ]; X ∈ a, Y ∈ b}.*

*A Lie subalgebra h of g is a subspaces satisfying [h, h] ⊂ h; then h itself is a Lie*
*algebra. An ideal h in g is a subspaces satisfying [h, g] ⊂ g; an ideal is automatically*
*a subalgebra. A Lie algebra g is called abelian if [g, g] = 0. A finite-dimensional Lei*
*algebra is simple if it is nonabelian and has no nonzero proper ideas.*

*For any algebra g we get a linear map (adjoint map) ad : g −→ End(g) given by*
*(adX)Y = adXY = [X, Y ].*

*By the Jacobi identity, we see that for all X ∈ g, adX* is a derivation.

Let g1, g2 *be two Lie algebras. A Lei algebra homomorphism is linear map φ :*

g1 *−→ g*2 such that

Throughout this thesis, we assume all vector spaces to be finite dimensional, over R or C.

### 2

### Killing Form

*The Killing form of a Lie algebra g is the symmetric bilinear form B : g × g −→ F*
defined by

*B(X, Y ) = trace(adX* *· adY).*

The Killing form is invariant, in the sense that

*B(adXY, Z) = −B(Y, adXZ).*

*Equivalently, B([X, Y ], Z) = −B(Y, [X, Z]).*

The Killing form can often be hard to compute, because it involves a careful
*choice of basis to avoid messy computation. If g is a subalgebra of Mn(F ), it is often*

convenient to use the trace form

*(A, B) = trace(AB), for all A, B ∈ g*

to substitute for the Killing form. If g is simple, then the Killing form and the trace form differs by a nonzero scalar.

Let g0 be a real Lie algebra. Define g = g0*⊗ C = g*0*+ ig*0. In this case we say that

*g is the complexification of g*0, and that g0 *is a real form of g. The complex linear*

*extension of [ , ] in g*0 is a natural complex Lie algebra structure on g. Notice that

non-isomorphic real Lie algebras may have the same complexification. For example,
*sl(n, R) and su(n) are both real forms of sl(n, C).*

*A Lie algebra is called semisimple if its Killing form is non-degenerate, e.g.*
*sl(n, F ), su(n), so(n) are all semisimple Lie algebras. Since the complex linear *
ex-tension preserves trace, we conclude that g0 is semisimple if and only if g0 *⊗ C is*

semisimple.

*We say a Lie algebra g is compact if its Killing form is negative definite. The*
following are some examples of complex Lie algebras and their compact real forms:

*Complex Lie algebra: Mn(C) so(n, C) sl(n, C) sp(n, C)*

Compact real form: *u(n)* *so(n)* *su(n)* *sp(n)*

### 3

### Root System and Root Space Decomposition

*Let (V, < , >) be an inner product space, and ∆ ⊂ V a finite set which spans V .*
*Each nonzero α ∈ V defines a reflection rα* *by α 7→ −α and fixing the hyperplane*

*orthogonal to α. We say that ∆ is a root system if it satisfies the following:*
*(1) If α ∈ ∆, the only multiples of α in ∆ are ±α.*

*(2) If α ∈ ∆, the reflection rα* leaves ∆ invariant.

*(3) If α, β ∈ ∆, then 2<α,β> _{<α,α>}*

*∈ Z.*

If ∆*i* *⊂ Ei* *are root systems for i = 1, 2, then ∆*1*× ∆*2 *⊂ E*1*× E*2 is also a root

*system. A root system is called reducible if it is a non-trivial product of two root*
*systems. Otherwise, it is called irreducible. A Lie algebra is simple if and only if its*
root system is irreducible.

*Let g be a complex semisimple Lie algebra. We say a subalgebra h ⊂ g is a Cartan*
*subalgebra if*

(1) h is maximal abelian, and

(2) adh is simultaneously diagonalizable. Namely, ad*X* *: g −→ g is diagonalizable*

*for all X ∈ h.*

*For example, h = {diagonal matrices} is a Cartan subalgebra of g = sl(n, C).*
From the Cartan subalgebra h of g, we can write

g = h + X

*α∈∆*

g*α,*

*where 0 6 ∈∆ ⊂ h∗* _{is a finite set and}

g*α* *= {Y ∈ g; adXY = α(X)Y, for all X ∈ h.}.*

(3.1)

*Consider V = h∗*_{, with inner product coming from the dual of the Killing form, then}

*∆ form a root system and (3.1) is called the root space decomposition of g. Elements*
*in ∆ are called roots and gα* *is called a root space.*

*Proposition 3.1 If α, β ∈ ∆ and α + β ∈ ∆, then [gα, gβ*] = g*α+β.*

*Given a root system ∆ ⊂ h∗ _{, we can choose a subset Π ⊂ ∆, called simple system,}*

such that

(1) Π is s basis for h*∗*_{, and}

(2) Every root is either a nonnegative linear combination of Π (positive root) or a nonpositive linear combination of Π (negative root).

Use the notation ∆*±* * _{for the positive and negative roots, so that Π ⊂ ∆}*+

_{⊂ ∆}and ∆ = ∆+_{∪ ∆}−_{.}

### 4

### Cartan Involution and Cartan Subalgebra

*Let g be a semisimple Lie algebra with the Killing form B. The sum g = k + p is*
*called a Cartan decomposition if it satisfies the following two conditions:*

*(1) [k, k] ⊂ k , [k, p] ⊂ p and [p, p] ⊂ k. (Hence k is a subalgebra of g.)*

*(2) The Killing form B of g is negative definite on k and positive definite on p.*
*A Cartan involution on g is a Lie algebra homomorphism θ : g −→ g with θ*2 _{= 1}

*such that −B(·, θ·) is an inner product on g.*

*Let V*+_{, V}−_{⊂ g denote the +1 and -1 eigenspaces of θ respectively, then g =}

*V*+_{+ V}−_{is a Cartan decomposition of g. Conversely, given a Cartan decomposition}

*g = k + p then we obtain a Cartan involution θ : g −→ g by*
*θ(X) =*
*X, if X ∈ k;*
*−X, if X ∈ p.*

Namely, there is a one-to-one correspondence between Cartan involutions and Cartan decompositions of g.

*Example: For g = sl(n, C), then θ(A) = −A∗* _{defines a Cartan involution with}

*the +1 eigenspace k = su(n) and the −1 eigenspace p = isu(n).*

*Let g be a real semisimple Lie algebra. We say that h ⊂ g is a Cartan subalgebra*
*if h ⊗ C is a Cartan subalgebra of g ⊗ C. The Cartan subalgebra is called θ-stable if*
*θ(h) = h. In this case*

*h = (h ∩ k) + (h ∩ p).*

*Write t = h ∩ k and a = h ∩ p, so that h = t + a. We call t and a the compact and*
*noncompact parts of h. Various Cartan subalgebras t + a ⊂ g may have different*
dimensions for t and a. But distinct Cartan subalgebras have the same dimension
*dim t + dim a. It is called the rank of g. It is a fact that we can always find a θ-stable*
maximally compact Cartan subalgebra of g.

*For example, in g = sl(2, R), we can take the diagonal matrices or so(2) as the*
Cartan subalgebra. In the first case dim t = 0 and dim a = 1; and in the second case
dim t = 1 and dim a = 0.

*A Cartan subalgebra is called maximally compact if the dimension of t is as large*
*as possible. It is called compact if h = t. For example so(2) is a compact Cartan*
*subalgebra of sl(2, R). For sl(3, R), which has rank 2, a maximally compact Cartan*
subalgebra is not compact because dim t = dim a = 1.

### 5

### Dynkin Diagram

Given a semisimple Lie algebra g with a Cartan subalgebra h and a choice of root
*system ∆ ⊂ h∗ _{. We shall draw a graph D = D(g) or D = D(∆) to represent g or}*

*root system ∆. It is called the Dynkin diagram.*

*Pick a choice of simple system Π ⊂ ∆. The vertices of D are the elements of Π.*
*Let αi, αj* *∈ Π. Write*

*aij* =

*2 < αj, αi* *>*

*< αi, αi* *>*

*,*

*here < , > comes from the Killing form. The matrix A = (aij) is called the Cartan*

*matrix.*

*The vertices of αi* *and αj* *in D are joint by aijaji* edges. Note in particular that

there is no edge joining them if and only if the roots are orthogonal. So the semisimple
Lie algebra is simple if and only if its Dynkin diagram is connected. It suffices to
*study the connected Dynkin diagrams. As a fact, the quantity aijaji* is 0,1,2 or 3.

*Suppose that the roots are not orthogonal, so that aijaji* *is 1, 2, 3. In the case that it*

is 2 or 3, draw an arrow pointing from the long root to the short root. The resulting diagram is the Dynkin diagram.

*Notice that the Dynkin diagram D is independent of the choice of h or ∆. So*
*g determines D. Conversely, non-isomorphic algebras have distinct diagrams. So*
g1 *∼*= g2 *if and only if D(g*1*) ∼= D(g*2).

The classification of complex simple Lie algebras is done by E. Cartan in the end
*of the 19th century. They are gathered into 7 classes, denoted by A, B, C, D, E, F, G.*
Using the subscript to denote the dimension of h (or the number of vertices in the
*Dynkin diagram), they are An, Bn, Cn, Dn, E*6*, E*7*, E*8*, F*4*, G*2*. The classes A, B, C, D*

have infinite number of members, and are called the classical algebras. The classes
*E, F, G have finite number of members, and are called the exceptional algebras. Their*
Dynkin diagrams are given by:

*An* e *. . .* e
*Bn* e *. . .* e*i* e
*Cn* e *. . .* e*h* e
*Dn* e *. . .* e%
%%
e
e
e_{e e}
*E*6 e e e e e
e
*E*7 e e e e e e
e
*E*8 e e e e e e
e
e
*F*4 e e* _{i}* e e

*G*2 e

*e 10*

_{i}### 6

### Vogan Diagram

*We shall try to draw a picture to represent a real simple g. Since g ⊗ C is complex*
*simple, let D = D(g ⊗ C) denote its Dynkin diagram. The idea is to add extra*
*information to D to reveal the real form g. The resulting picture, namely D with*
*extra information, is called a Vogan diagram of g.*

*Start with a Cartan involution θ on g, and a θ-stable maximally compact Cartan*
*subalgebra h = t + a in g. Let ∆ be a root system. A root α ∈ ∆ is said to be a real*
*root or an imaginary root depends on α(h) ⊂ R or α(h) ⊂ iR. So α is imaginary if*
*and only if it annihilates a. We say that α is a complex root if it is neither real nor*
imaginary. Since h is the maximally compact Cartan of g, there is no real root, that
*is α(t) 6= 0.*

*Extend θ to be complex linear on g ⊗ C. We can let θ act on the roots such that*
*θ(g ⊗ C)α* *= (g ⊗ C)θα. Then θα = α if and only if α is imaginary. We can further*

*choose the simple system Π so that θ(Π) = Π. This leads to an automorphism θ on*
*the Dynkin diagram D = D(g ⊗ C). So the imaginary simple roots are the vertices*
*of D which are fixed by θ. The complex simple roots are the vertices on D which are*
*not fixed by θ; they form 2-element orbits.*

*Let α be an imaginary simple root. Define the 2-dimensional space gα* *= g ∩ ((g ⊗*

C)*α+ (g ⊗ C)−α). Since θ fixes α, it follows that θ(gα*) = g*α. So θ is 1 or −1 on gα*.

Equivalently, g*α* *⊂ k or gα* *⊂ p. We say that α is compact or noncompact accordingly.*

So there are three types of simple roots: complex, imaginary compact, imaginary
noncompact. Depending on whether the imaginary root is compact or noncompact,
*we color the vertex in D as white or black. The complex roots have no color.*

*A Vogan diagram is a Dynkin diagram with an involution θ, such that the vertices*
*fixed by θ are either white or black. We have used a real simple Lie algebra to*
construct a Vogan diagram. Conversely, every Vogan diagram represents a real simple
Lie algebra.

### 7

### Classifications of Vogan Diagrams

Each Vogan diagram corresponds to a real simple Lie algebra. Two diagrams are
said to be equivalent if they represent the same Lie algebra. We are interested in
equivalence classes of the Vogan diagrams. In this respect we can ignore once and
for all the diagrams with no painted vertex, as they represent Lie algebras without
noncompact imaginary root and so cannot be equivalent to any diagram with painted
vertices. Then the Borel-de Siebenthal theorem [3] says that every Vogan diagram is
equivalent to one with a single painted vertex. However, it does not give the explicit
equivalence. We shall develop algorithms which convert a diagram to an equivalent
one with fewer painted vertices. As a result, not only we have reproved the
Borel-de Siebenthal theorem, we give the equivalence classes explicitly. We shall label the
*vertices of the underlying Dynkin diagram with 1, ..., n. Then the Vogan diagram with*
*vertices i*1*, ..., ik* *painted, where i*1 *< ... < ik, is denoted by (i*1*, ..., ik*). For diagrams

*with θ = 1, the equivalence classes are listed in the following table.*

Dynkin Diagram Single Painted Vertex Equivalent Diagrams

*An*b
1
*. . .* b
n *(N ),* *1 ≤ N ≤*
*n+1*
2 *(i*1*, . . . , ik*),
P*k*
*p=1(−1)*
*k−p _{i}*

*p= N, n + 1 − N.*

*Bn*b 1

*. . .*b n-1

*i*b n

*(N ),*

*1 ≤ N ≤ n*

*(i*1

*, . . . , ik*), P

*k*

*p=1(−1)*

*k−p*

_{i}*p= N .*

*(n)*

*(i*1

*, . . . , ik, n)*

*Cn*b 1

*. . .*b n-1

*h*b n

_{(N ),}

_{1 ≤ N ≤}*n*2

*(i*1

*, . . . , ik), ik≤ n − 1 ,*P

*k*

*p=1(−1)k−pip= N, n − N.*

*(i*1

*, . . . , ik), ik≤ n − 2 ,*P

*k*

*p=1(−1)k−pip= N, n − N.*

*Dn*b 1

*. . .*b n-2 % % b e eb n-1 n

*(N ),*

*1 ≤ N ≤*

*n*2

*1*

_{(i}*, . . . , ik, n − 1, n),*P

*k*

*p=1(−1)*

*k−p*

_{i}*p= N − 1, n − N − 1.*

*(n)*

*(n − 1), (i*1

*, . . . , ik, n − 1), (i*1

*, . . . , ik, n).*Table 1

(continue next page)

Dynkin Diagram Single Painted Vertex Equivalent Diagrams
*(1), (5), (2, 4), (1, 3, 4), (1, ∗), (2, ∗), (4, ∗), (5, ∗), (3, 5, ∗)*
(1) *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
*2 − I and I + s is odd ,*
*4 − I and I + s is even ,*
*1 − I.*
*E*6 b
1
b
2
b
3
b
4
b
5
b
*∗*
*(i*1*, ..., ik, j*1*, s), J =*
(
*4 + I and I + s is odd ,*
*1 + I.*
*(2), (3), (4), (3, 5), (3, ∗), (2, 4, ∗), (1, 3, 4, ∗)*
*(∗)* *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
*2 − I and I + s is even ,*
*4 − I and I + s is odd ,*
*3 − I.*
*(i*1*, ..., ik, j*1*, s), j*1=
(
*4 + I and I + s is even ,*
*3 + I.*
*(2), (3), (5), (3, 5), (3, 4, 6), (3, 4, 5, 6), (4, ∗), (6, ∗), (2, 4, ∗), (1, 3, 4, ∗)*
(1) *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
(
*1 − I, 3 − I and I + s is odd ,*
*2 − I, 4 − I and I + s is even .*
*(i*1*, ..., ik, j*1*, s), j*1=
(
*1 + I, 2 + I, 3 + I, 5 + I and I + s is even ,*
*4 + I and I + s is odd .*
*(2, 4), (1, 3, 4), (1, ∗), (2, ∗), (5, ∗), (3, 5, ∗), (3, 4, 6, ∗), (3, 4, 5, 6, ∗)*
*E*7 b
1
b
2
b
3
b
4
b b
6
5
b
*∗*
(6) *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
(
*1 − I and I + s is even ,*
*2 − I and I + s is odd .*
*(i*1*, ..., ik, j*1*, s), j*1*= 1 + I, 2 + I, 5 + I and I + s is odd.*
*(4), (3, ∗)*
*(∗)* *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
(
*3 − I and I + s is even ,*
*4 − I and I + s is odd .*
*(i*1*, ..., ik, j*1*, s), j*1=
(
*3 + I and I + s is odd ,*
*4 + I and I + s is even .*
*(2), (3), (6), (1, ∗), (2, ∗), (5, ∗), (6, ∗)*
(7) *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
*1 − I, 5 − I and I + s is even ,*
*3 − I and I + s is odd ,*
*2 − I, 6 − I.*
*E*8 b
1
b
2
b
3
b
4
b b
6
5
b
*∗*
b
7
*(i*1*, ..., ik, j*1*, s), j*1=
*1 + I, 5 + I and I + s is odd ,*
*3 + I and I + s is even ,*
*2 + I, 6 + I.*
*(1), (4), (5), (3, ∗), (4, ∗), (7, ∗)*
*(∗)* *(i*1*, ..., ik, j*1*, ..., jl, s)*
*l 6= 1* *, J =*
*1 − I, 5 − I and I + s is odd ,*
*3 − I and I + s is even ,*
*4 − I.*
*(i*1*, ..., ik, j*1*, s), j*1=
*1 + I, 5 + I and I + s is even ,*
*3 + I and I + s is odd ,*
*4 + I.*
(1) *(i*1*, . . . , ik), {i*1*, . . . , ik} ∩ {1, 2} 6= ∅*
*F* b b*i* b b

*The left column labels the vertices with 1, 2, 3, ... and so on. The middle column*
lists the diagrams with single painted vertex, for example (2) corresponds to the
diagram with vertex 2 painted. The right column provides all the Vogan diagrams in
*their equivalence classes. For example if we consider (1, 3, 4) in A*5, then the formula

*i*3 *− i*2 *+ i*1 *= 4 − 3 + 1 = 2 says that it is equivalent to the diagram with vertex 2*

painted.

*It turns out that En*are the most complicated ones. The following methods explain

*how to use Table 1 for Vogan diagrams of En*:

(1) Diagrams in the following special cases:

*(2, 4), (1, 3, 4), (3, 5), (2, 4, ∗), (1, 3, 4, ∗), (3, 5, ∗),*

*(3, 4, 6), (3, 4, 5, 6), (3, 4, 6, ∗), (3, 4, 5, 6, ∗)* *in E*6 *and E*7*.*
(7.1)

*Obviously we disregard the second row of (7.1) in E*6 because there is no vertex

6. Their equivalence classes can be found directly in Table 1. (2) Diagrams not in (7.1):

Write it in the form

*(i*1*, ..., ik, j*1*, ..., jl, s),*

(7.2)

*where 1 ≤ i*1 *< ... < ik* *≤ 3 < j*1 *< ... < jl* *≤ n − 1, and s is either ∗ or empty*

*depending on whether ∗ is painted or not. In this case, let*
*I =*P*k*

*p=1(−1)k−pip, (I = 0 if no i appears)*

*J =*P*l*

*p=1(−1)l−pjp. (J = 0 if no j appears)*

(7.3)

*Set l = 0 if no j appears. In computing the sign of I +s, we make the convention*
*that s = ∗ is odd and s = ∅ is even. Then find the equivalence class in Table 1.*
Note that method (2) cannot be used against the diagrams in (7.1), because that
would lead to the wrong equivalence classes. The significance of (7.1) will be explained
in Proposition 9.2 later.

*For example, consider (1, 2, 3, 5, ∗) in E*7*, which is not in (7.1). We see that l = 1,*

*I = i*3*− i*2*+ i*1 *= 3 − 2 + 1 = 2, and J = j*1 *= 5 = 3 + I.*

*Here s = ∗, and so I + s = 2 + ∗ is odd. By Table 1, (1, 2, 3, 5, ∗) ∼ (∗) in E*7.

*We shall prove Table 1, for the classical diagrams in §2, and the exceptional*
*diagrams in §3. We shall only prove the equivalence of each grouping in Table 1.*
We need not prove inequivalence of different groupings, since this is done in [4]. For
*instance [4, p.355] says that in A*4*, (1) is su(1, 4), and (2) is su(2, 3), so (1) and (2)*

are not equivalent.

*Next we consider the Vogan diagrams with nontrivial involutions θ. Here θ imposes*
a symmetry requirement on the underlying Dynkin diagrams, and the only vertices
*fixed by θ may be painted. Therefore such Vogan diagrams are limited. They are listed*
*in the following table, together with their equivalence classes, where “↔” indicates*
*the two-element-orbits of θ. Once again we ignore the ones without painted vertex,*
which are obviously not equivalent to any other diagram.

Dynkin Diagram Single Painted Vertex Equivalent Diagrams

*An*
*n odd* b
n + 1
2 %%
b
eeb*l*
*. . .* b
*. . .* *l*b
n
1
(*n+1*
2 )
*Dn* b
1
*. . .* b
n-2
%
%
b
eeb*l*
n-1
n
*(N ), N ≤* *n−1*
2
*(i*1*, ..., ik), ik≤ n − 2,*
P*k*
*p=1(−1)k−pip= N, n − N − 1.*
*E*6 b b
*∗* 3
2
4
%
%
b
eeb*l*
b
b
*l*
5
1
*(∗)* *(3), (3, ∗)*
Table 2

*We shall prove Table 2 in §4. Tables 1 and 2 confirm the Borel-de Siebenthal*
*theorem. Their proofs use some algorithms F [i] (see (8.1) below) which reduce the*
*number of painted vertices to one. In §5, we show that these algorithms lead to a*
necessary condition for a graph to be Dynkin (Corollary 1). We shall see that this
necessary condition is almost sufficient, thereby providing a very easy classification
for almost all simply-laced Dynkin diagrams.

### 8

### Classical Diagrams

*In this section we consider Vogan diagrams of types A, B, C, D in Table 1, with θ = 1.*
*We label their vertices with 1, ...., n as in Table 1. A Vogan diagram with painted*
*vertices i*1*, ..., ik, where 1 ≤ i*1 *< ... < ik* *≤ n, is denoted by (i*1*, ..., ik*). Suppose that

*i ∈ {i*1*, ..., ik}, so that i is a painted vertex. We introduce an operation F [i] on the*

*Vogan diagram as follows. Let F [i] act on the root system by reflection corresponding*
*to the noncompact simple root i. As a result, it leads an equivalent Vogan diagram.*
*The effect of F [i] on the Vogan diagram is as follows (developed in [1], see also [2,*
p.89]):
*F [i] :*

*• The colors of i and all vertices not adjacent to i remain*
unchanged.

*• If j is joined to i by a double edge and j is long, the color*
*of j remains unchanged.*

*• Apart from the above exceptions, reverse the colors of all*
*vertices adjacent to i.*

(8.1)

*For instance, if we apply F [4] to (1, 3, 4, 7), then we reverse the colors of 3, 5 and get*
*(1, 4, 5, 7). Thus (1, 3, 4, 7) is equivalent to (1, 4, 5, 7).*

*Using the operation F [i], the next lemma shows that a pair of painted vertices*
can be shifted leftward or rightward.

*Lemma 8.1 Let i*1 *< ... < ik.*

*(a) (i*1*, ..., ik) ∼ (i*1*, ..., ir−1, ir− c, ir+1− c, ir+2, ..., ik) whenever ir−1* *< ir− c.*

*(b) (i*1*, ..., ik) ∼ (i*1*, ..., ir−1, ir+ c, ir+1+ c, ir+2, ..., ik) whenever ir+1+ c < ir+2. We*

*require ir+1+ c ≤ n − 1 in Cn* *and ir+1+ c ≤ n − 2 in Dn.*

*Proof: We now prove (a). Suppose we want to move ir, ir+1* *leftward c steps, where*

*ir−1* *< ir− c. It is equivalent to moving them 1 step for c times, namely it suffices to*

show that

*(i*1*, ..., ik) ∼ (i*1*, ..., ir−1, ir− 1, ir+1− 1, ir+2, ..., ik).*

(8.2)

*By applying F [ir+ 1], F [ir+ 2], ..., F [ir+1− 1] consecutively to (i*1*, ..., ik*), we get (8.2),

and (a) follows.

*The proof of (b) is similar. The restrictions on Cn, Dnare added because F [n − 1]*

*does not change the color of n in Cn, and F [n − 2] changes the colors of n − 1, n in*

*Dn*. *2*

*For example, in (1, 5, 7, 9), we can move the pair 5, 7 leftward three steps and get*
*(1, 5, 7, 9) ∼ (1, 2, 4, 9). The following lemma provides a way to reduce the number of*
painted vertices.

*Lemma 8.2 In An, Bn, (i*1*, ..., ik) ∼ (i*2*− i*1*, i*3*, ..., ik). If i*2 *≤ n − 1, this is true in*

*Cn. If i*2 *≤ n − 2, this is true in Dn.*

*Proof: We divide the arguments for (i*1*, ..., ik*) into two cases.

*Case 1:* *i*1 *= 1. If i*2 *= 2 then F [1](1, 2, i*3*, ..., ik) = (1, i*3*, ..., ik*) and we are

*done. So suppose that i*2 *> 2. Apply F [1], F [2], ..., F [i*2 *− 1] to (1, i*2*, ..., ik*), we get

*(1, i*2*, ..., ik) ∼ (i*2*− 1, i*3*, ..., ik*). This solves Case 1.

*Case 2: i*1 *> 1. By Lemma 8.1(a), (i*1*, ..., ik) ∼ (1, i*2*− i*1*+ 1, i*3*, ..., ik*). This is

*reduced to Case 1, so we get (1, i*2*− i*1*+ 1, i*3*, ..., ik) ∼ (i*2*− i*1*, i*3*, ..., ik*). This solves

Case 2.

*The extra conditions are imposed to deal with the special cases of F [n − 1] in*
*Cn* *and F [n − 2], F [n − 1], F [n] in Dn*, as explained in Lemma 8.1. This proves the

lemma. *2*

*Proposition 8.3 In An* *and Bn, (i*1*, ..., ik) ∼ (*

P_{k}

*p=1(−1)k−pip).*

*Proof: Consider (i*1*, ..., ik) in An* *or Bn*. By Lemma 8.2,

*(i*1*, ..., ik) ∼ (i*2*− i*1*, i*3*, ..., ik*)

*∼ (i*3*− i*2*− i*1*, i*4*, ..., ik*)

*∼ ... ∼ (*P*k*

*p=1(−1)k−pip).*

This proves the proposition. *2*
*Obviously (N) ∼ (n + 1 − N) in An*, by symmetry of the diagram. Therefore by

*Proposition 8.3, we have verified the equivalence classes of diagrams of types A and*
*B in Table 1. The next proposition considers the type C diagrams of Table 1. The*
*argument is similar unless the vertex n is painted.*

*Proposition 8.4 Consider (i*1*, ..., ik) in Cn.*

*(a) If ik* *< n, then (i*1*, ..., ik) ∼ (*

P_{k}

*p=1(−1)k−pip).*

*(b) If ik* *= n, then (i*1*, ..., ik) ∼ (n).*

*Proof: If ik* *< n, we can repeat the argument as in (8.3) and get the desired result.*

*We now consider the case ik* *= n, namely (i*1*, ..., ik−1, n). Let c = n − 1 − ik−1*. Then

*(i*1*, ..., ik−1, n) ∼ (i*1*, ..., ik−3, ik−2+ c, n − 1, n)* *by Lemma 8.1(b)*

*∼ (i*1*, ..., ik−3, ik−2+ n − 1 − ik−1, n).* *by F [n]*

(8.4)

*Thus the number of entries has gone from k to k − 1. Repeat the applications of*
*Lemma 8.1(b) and F [n] as in (8.4), we end up with (n).* *2*
*Most of Cn* in Table 1 follow from Proposition 8.4. It remains only to check that

*if (i*1*, ..., ik) with ik* *< n satisfies*

P_{k}

*p=1(−1)k−pip* *= n − N, then (i*1*, ..., ik) ∼ (N). This*

*can be done by modifying (8.3) to (i*1*, ..., ik) ∼ (i*1*, ..., ik−2, n − (ik− ik−1)) ∼ ... ∼*

*(n −*P*k _{p=1}(−1)k−p_{i}*

*p) and proceed with similar arguments, or by observing that (N)*

*and (n − N) correspond to the Lie algebras sp(N, n − N) ∼= sp(n − N, N ) [4, p.355].*
*This proves Table 1 for Cn*.

*For Dn*, the following proposition considers the various situations based on the

*colors of n − 1 and n.*
*Proposition 8.5 In Dn:*
*(a) If ik* *≤ n − 2, then (i*1*, ..., ik) ∼ (*
P_{k}*p=1(−1)k−pip).*
*(b) (i*1*, ..., ik, n − 1, n) ∼ (1 +*
P_{k}*p=1(−1)k−pip).*
*(c) (i*1*, ..., ik, n − 1) ∼ (n − 1).*
18

*Proof:* *The argument for (a) is similar to An*; we simply move pairs of painted

vertices to the left by Lemma 8.1(a). We perform this operation in (b), and get
*(i*1*, ..., ik, n − 1, n) ∼ (*
P_{k}*p=1(−1)k−pip, n − 1, n). By F [n − 1] followed by F [n − 2], we*
get (P*k*
*p=1(−1)k−pip, n−1, n) ∼ (*
P_{k}

*p=1(−1)k−pip, n−3, n−2). This reduces to (a), and*

simple operations show that the last expression is equivalent to (1 +P*k*

*p=1(−1)k−pip*).

This proves (b).

*Now consider (i*1*, ..., ik, n − 1) in (c). The first k painted vertices can be dealt*

*with as before, leaving (i*1*, ..., ik, n − 1) ∼ (I, n − 1), where I =*

P_{k}

*p=1(−1)k−pip*. By

*performing F [n − 1], F [n − 2], ..., F [I + 1] to (I, n − 1), we get (I, n − 1) ∼ (I + 1, n) ∼*
*(I+1, n−1). Repeating this method gives (I, n−1) ∼ (I+1, n−1) ∼ ... ∼ (n−2, n−1).*
*Then F [n − 1](n − 2, n − 1) = (n − 1) and we are done.* *2*
*Most equivalence classes of type D in Table 1 are covered by Proposition 8.5. The*
*remaining cases follow from two simple observations. Firstly, (N) ∼ (n − N) because*
*they correspond to Lie algebras so(2N, 2n − 2N) ∼= so(2n − 2N, 2N). Secondly, if*
*exactly one of n − 1, n is painted, obviously it does not matter which of them is*
painted due to symmetry of the diagram.

*We have checked the equivalence classes of Vogan diagrams of types A, B, C, D*
*given in Table 1. The next section considers the diagrams of types E, F, G.*

### 9

### Exceptional Diagrams

*In this section, we consider the Vogan diagrams of types E, F , G in Table 1 with*
*θ = 1. We first treat the diagrams of En*. Label the vertices as follows.

e
1
e
2
e
3
*. . .* e
*n − 1*
e
*∗*
Lemma 9.1

*(a) For q ≥ 4 and p = 2, 3, we get (p, q) ∼ (p − 1, q − 1, ∗) and (p, q, ∗) ∼ (p − 1, q − 1).*
*(b) For q ≥ 4, (1, q) ∼ (q − 1, ∗) and (1, q, ∗) ∼ (q − 1).*

*Proof: For (p, q) or (p, q, ∗), where q ≥ 4, apply F [p],...,F [q − 1] to it and we get the*

desired results. *2*

*The next proposition simplifies a Vogan diagram to one of the form (α) or (α, ∗).*
However it excludes the special cases in (7.1) because they are not valid in argument
(9.7) below. We will deal with them seperately in Proposition 9.5. Although argument
*(9.7) also cannot be applied to (7.1) of E*8, Propositions 9.3 and 9.5 show that they all

*happen to be equivalent to (7) in E*8, which coincides with the formulae in Proposition

*9.2. Therefore, we need not exclude (7.1) of E*8 in Proposition 9.2.

*As in (7.2), the Vogan diagrams are denoted by (i*1*, . . . , ik, j*1*, . . . , jl, s), where*

*1 ≤ i*1 *< ... < ik* *≤ 3 < j*1 *< ... < jl* *≤ n − 1 and s is ∗ or empty. Throughout this*

*section, let I, J be defined as in (7.3), and let*
*α =*
*J − I* *if J ≥ 4*
*n − J − I* *if J < 4.*
(9.1)

*The next proposition simplifies (i*1*, ...ik, j*1*, ..., jl, s) to (α, t), where t is ∗ or empty.*

*Proposition 9.2 Consider (i*1*, ..., ik, j*1*, ..., jl) or (i*1*, ..., ik, j*1*, ..., jl, ∗) other than in*

*(7.1). Then (i*1*, ..., ik, j*1*, ..., jl, s) ∼ (α, t), where s = t if I is even, and s 6= t if I is*

*odd.*

*Proof: For the case (i*1*, j*1*) = (3, 4), by F [3], F [2], F [1], we get (1, ∗). Now consider*

*the case (i*1*, . . . , ik, j*1*, . . . , jl), we may regard (i*1*, . . . , ik) and (j*1*, . . . , jl*) as painted

*diagrams of A*3 *and An−4* respectively. By Proposition 8.3, we have

*(i*1*, ..., ik) ∼ (*
*k*
X
*p=1*
*(−1)k−p _{i}*

*p) = (I) in A*3 (9.2) and

*(j*1

*, ..., jl) ∼ (*

*l*X

*p=1*

*(−1)l−p*

_{j}*p) = (J) in An−4.*(9.3)

*Notice that J ≥ 4 if and only if l = 1, this implies that there is a single painted*
*vertex in {ji, . . . , jl}; and if J < 4, then the corresponding single painted vertex of*

*En* *is n − J. Let β denote the single painted vertex of En* reduced from the painted

*vertices {j*1*, . . . , jl}, then*
*β =*
*n − J,* *if J < 4*
*J,* *if J ≥ 4* *and α = β − I.*
(9.4)

*In reducing the diagrams (9.2) and (9.3), we did not use the operation F [3]. So ∗*
does not occur and

*(i*1*, ..., ik, j*1*, ..., jl) ∼ (I, β).*

(9.5)

*Since β ≥ 4 and by Lemma 9.1(a), we see that*
*(I, β) ∼ (I − 1, β − 1, ∗)*
*∼ (I − 2, β − 2)*
...
*∼*
*(1, β − I + 1, ∗)* *if I − 1 is odd*
*or (1, β − I + 1)* *if I − 1 is even* *.*
(9.6)

Hence we have
*(i*1*, ..., ik, j*1*, ..., jl) ∼ (I, β)* *by (9.5)*
*∼*
*(1, β − I + 1)* *if I − 1 is even*
*(1, β − I + 1, ∗)* *if I − 1 is odd* *by (9.6)*
*∼*
*(β − I, ∗)* *if I is odd*
*(β − I)* *if I is even* *by Lemma 9.1(b)*
*∼*
*(α, ∗)* *if I is odd*
*(α)* *if I is even .* *by (9.4)*
(9.7)

*The use of Lemma 9.1(b) in (9.7) requires β − I + 1 ≥ 4, which is not valid for the*
diagrams in (7.1). This is the reason which excludes them from this proposition.

*By (9.7), we solve the case (i*1*, ..., ik, j*1*, ..., jl). The case of (i*1*, . . . , ik, j*1*, . . . , jl, ∗)*

follows from similar argument. This completes the proof. *2*
The above proposition shows how a Vogan diagram is equivalent to one of the
*form (α) or (α, ∗). The next two propositions deal with (α, ∗) and (α) respectively.*
*Proposition 9.3 The Vogan diagrams of the form (α, ∗) are equivalent to diagrams*
*with single painted vertex in the following table.*

*(α, ∗)* *E*6 *E*7 *E*8
*(1, ∗)*
*(2, ∗)* (5) (6) (7)
*(3, ∗)* *(∗)*
*(4, ∗)* (1)
*(5, ∗)* (5) (6) (7)
*(6, ∗)* *—* (1) (7)
*(7, ∗)* *—* *—* (1)
Table 3
22

*Proof: For (1, ∗), we apply F [1], F [2], ..., F [n − 1] consecutively and get*
*(1, ∗) ∼ (1, 2, ∗) ∼ (2, 3, ∗) ∼ (3, 4) ∼ ... ∼ (n − 1).*
(9.8)

*For (2, ∗), we apply F [∗] to it and get (2, 3, ∗), then proceed as in (9.8). Clearly*
*(3, ∗) ∼ (3). For (4, ∗), we apply F [∗], F [3], F [2], F [1] to it and get (4, ∗) ∼ (1).*

*We next show that (5, ∗) ∼ (2, ∗), so that we can proceed with (2, ∗) as above.*
*By F [∗], F [3], F [4], we get (5, ∗) ∼ (2, 4). By Lemma 9.1(a), (2, 4) ∼ (1, 3, ∗). Then*
*apply F [1], F [2] to (1, 3, ∗) we get (2, ∗). This solves (5, ∗).*

*We now consider (6, ∗) in E*7 *and E*8*. In E*7*, apply F [6], ..., F [1] consecutively to*

*(6, ∗) and we get (1). In E*8*, by Lemma 9.1(b), (6, ∗) ∼ (1, 7). Apply F [7], F [6], ..., F [2]*

*to (1, 7) and we get (2, ∗). This solves (6, ∗).*

*Finally for (7, ∗) in E*8*, we apply F [7], ..., F [1] to it and get (1). This proves the*

proposition. *2*

*By Propositions 9.2 and 9.3, we have simplified all type E diagrams to single*
painted vertex diagrams. We consider these single painted vertex diagrams in the
following proposition.

Proposition 9.4

*(a) E*6 *has two equivalence classes (1) ∼ (5) and (2) ∼ (3) ∼ (4) ∼ (∗).*

*(b) E*7 *has three equivalence classes (6), (1) ∼ (2) ∼ (3) ∼ (5) and (4) ∼ (∗).*

*(c) E*8 *has two equivalence classes (1) ∼ (4) ∼ (5) ∼ (∗) and (2) ∼ (3) ∼ (6) ∼ (7).*

*Proof: We only have to prove the equivalence claimed in this proposition. The *
in-equivalence of different groupings follows from [4]. For example, [4, p533-534] says
*that (1) and (∗) in E*6 are not equivalent.

*We first claim that (∗) ∼ (4) in all En*:

*(∗) ∼ (3, ∗)* *apply F [∗]*

*∼ (2, 5, ∗)* *by Proposition 9.2*
*∼ (4).* *apply F [3], F [4]*
(9.9)

*Hence (∗) ∼ (4) as claimed. In the following we consider E*6*, E*7*, E*8 seperately.

*In E*6*, clearly (1) ∼ (5) and (2) ∼ (4) by symmetry of the diagram. So by (9.9)*

*it suffices to show that (3) ∼ (∗). By applying F [3], F [4], F [5] to (3) we get (2, 5, ∗),*
*and by Proposition 9.2, (2, 5, ∗) ∼ (3, ∗). Clearly (3, ∗) ∼ (∗). This proves (a).*

*We next consider E*7 in (b):

*(3) ∼ (3, 6, ∗)* *by Proposition 9.2*
*∼ (4, ∗)* *apply F [6], F [5], F [4]*
*∼ (1).* *by Proposition 9.3*
*We conclude that (3) ∼ (1). Next we claim that (2) ∼ (3):*
*(2) ∼ (1, 3)* *apply F [2], F [1]*

*∼ (2, 4, ∗)* *by Lemma 9.1(a)*
*∼ (3).* *apply F [∗], F [3]*
*Hence (2) ∼ (3) as claimed. Finally, we prove that (5) ∼ (2):*

*(5) ∼ (1, 6, ∗)* *by Proposition 9.2*

*∼ (2).* *apply F [6], F [5], . . . , F [2]*
Together with (9.9), this proves (b).

*Finally, we consider E*8 in (c):

*(5) ∼ (2, 7)* *by Proposition 9.2*
*∼ (3, ∗)* *apply F [7], F [6], ..., F [3]*
*∼ (∗).* *apply F [∗]*

On the other hand,

*(4) ∼ (3, 7, ∗)* *by Proposition 9.2*
*∼ (4, ∗)* *apply F [7], F [6], ..., F [4]*
*∼ (1).* *by Proposition 9.3*

*Recall that (4) ∼ (∗) by (9.9), so we conclude that (1) ∼ (4) ∼ (5) ∼ (∗). We next*
*check the other equivalence class (2) ∼ (3) ∼ (6) ∼ (7):*

*(6) ∼ (1, 7, ∗)* *by Proposition 9.2*
*∼ (2)* *apply F [7], F [6], ..., F [2]*
*∼ (1, 4, ∗)* *apply F [2], F [3], F [∗]*
*∼ (3)* *by Proposition 9.2*
*∼ (3, 6, ∗)* *by Proposition 9.2*
*∼ (6, ∗)* *apply F [∗]*
*∼ (7).* *by Proposition 9.3*

*That is, (6) ∼ (2) ∼ (3) ∼ (7). This completes the proof.* *2*
The next proposition deals with the Vogan diagrams in (7.1). They have been
excluded by Proposition 9.2.

*Proposition 9.5 The equivalence classes of the Vogan diagrams in (7.1) are given*
*in Table 4 below. In particular, each of them is equivalent to (n − 2, ∗) or (n − 2).*
*Proof: In all En*,

*(3, 5, ∗) ∼ (2, 4)* *by Lemma 9.1(a)*
*∼ (1, 3, 4)* *apply F [2], F [1]*

*∼ (1, n − 1)* *apply F [4], ..., F [n − 1]*
*∼ (n − 2, ∗).* *by Lemma 9.1(b)*

*The equivalence class of (n − 2, ∗) is given by Proposition 9.3. By similar arguments*
*we have (3, 5) ∼ (2, 4, ∗) ∼ (1, 3, 4, ∗) ∼ (n − 2). The equivalence class of (n − 2)*
*is given by Proposition 9.4. And clearly, in E*7*, (3, 4, 6) ∼ (3, 4, 5, 6) ∼ (3, 5) and*

*(3, 4, 6, ∗) ∼ (3, 4, 5, 6, ∗) ∼ (3, 5, ∗). This completes the proof.* *2*
By Propositions 9.2, 9.3, 9.4 and 9.5, we have completely characterized all the
*equivalence classes of Vogan diagrams of type E. We summarize these results in the*

*following table. Recall that α is defined in (9.1).*

Dynkin Diagram Single Painted Vertex Equivalent Diagrams
*(5),(2, 4),(1, 3, 4)*
*E*6 b
1
b
2
b
3
b
4
b
5
b
*∗*
(1)
*(3, 5, ∗),(α, ∗), α = 1, 2, 4, 5*
*(2),(3),(4),(3, 5)*
*(∗)*
*(3, ∗),(2, 4, ∗),(1, 3, 4, ∗)*
*(2),(3),(5),(3, 5),(3, 4, 6),(3, 4, 5, 6)*
(1)
*(2, 4, ∗),(1, 3, 4, ∗),(α, ∗), α = 4, 6*
*E*7 b
1
b
2
b
3
b
4
b b
6
5
b
*∗*
(6) *(2, 4),(1, 3, 4)*
*(3, 5, ∗), (3, 4, 6, ∗), (3, 4, 5, 6, ∗), (α, ∗), α = 1, 2, 5*
*(∗)* *(4), (3, ∗)*
(2),(3),(6)
(7)
*(α, ∗), α = 1, 2, 5, 6*
*E*8 b
1
b
2
b
3
b
4
b b
6
5
b
*∗*
b
7
(1),(4),(5)
*(∗)*
*(α, ∗), α = 3, 4, 7*
Table 4

Table 4 summarizes the following method to determine the equivalence class of a
*Vogan diagram of En*:

(1) Diagrams belong to the special cases (7.1):

*Use Proposition 9.5 to reduce it to the form (n − 2, ∗) or (n − 2), then use*
Proposition 9.3 or 9.4 to find the equivalence class. The result is in Table 4.
(2) Diagrams not in (7.1):

*Write it as (i*1*, ..., ik, j*1*, ..., jl, s), then use (7.3) and (9.1) to compute I, J, α.*

*Use Proposition 9.2 to reduce it to (α, ∗) or (α). The equivalence classes of*
*(α, ∗) and (α) are given in Propositions 9.3 and 9.4, and are summarized in*
Table 4.

Methods (1) and (2) here correspond to methods (1) and (2) for Table 1. The methods for Table 4 have been simplified to the various cases of Table 1.

*For example, consider (1, 2, 3, 5, ∗) in E*7. It does not belong to (7.1), so we

compute

*I = i*3*− i*2*+ i*1 *= 3 − 2 + 1 = 2, and J = j*1 *= 5 > 4,*

*hence α = J − I = 3. Since I is even, by Proposition 9.2, (1, 2, 3, 5, ∗) ∼ (3, ∗).*
*By Proposition 9.3, (3, ∗) ∼ (∗). So Table 4 shows that (1, 2, 3, 5, ∗) ∼ (∗) in E*7.

*Alternatively, from j*1 *= 3 + I and I + s = 2 + ∗ is odd, we find (1, 2, 3, 5, ∗) ∼ (∗) in*

*E*7 of Table 1.

*We next consider the Vogan diagrams of F*4. We label the vertices as follows.

e
1
e
2
e
3
*i* e
4

*Proposition 9.6 In F*4*, (i*1*, . . . , ik) ∼ (1) if and only if {i*1*, . . . , ik} ∩ {1, 2} 6= ∅.*

*Proof: Suppose that (i*1*, . . . , ik) does not contain 1 or 2. That is (3, 4) ∼ (4) or (3).*

*It follows either from [4, p541-542] or Theorem 1 later that (3) 6∼ (2) and (3) 6∼ (1).*
*By applying F [4], F [3] on (4) we get (3).*
Conversely,
*(1) ∼ (1, 2)* *by F [1]*
*∼ (2, 3)* *by F [2]*
*∼ (2, 3, 4)* *by F [3]*
*∼ (2, 4)* *by F [4]*
*∼ (1, 2, 3, 4)* *by F [2]*
*∼ (1, 2, 3)* *by F [3]*
*∼ (2).* *by F [2]*
(9.10)
And clearly,
*(1, 2, 3, 4) ∼ (1, 3, 4) ∼ (1, 4) ∼ (1, 2, 4) and (1, 2, 3) ∼ (1, 3).*
(9.11)

*All cases of {i*1*, . . . , ik}∩{1, 2} 6= ∅ are considered in (9.10) and (9.11), this completes*

the proof. *2*

Proposition 9.6 shows that there are only two equivalence classes of Vogan
*dia-grams of F*4 as listed in Table 1.

*It is clear that all paintings on G*2 (unless we keep all vertices unpainted) are

*equivalent to one another. This can be checked by the performing various F [i], or by*
looking at its painted root system.

### 10

### Nontrivial Involutions

In this section we study the equivalence classes of the Vogan diagrams with nontrivial involutions, and prove the informations in Table 2.

*The condition θ 6= 1 restricts the underlying Dynkin diagrams to An, Dn* *and E*6.

We also ignore the diagrams without painted vertex, since they cannot be equivalent
*to one with painted vertices. So the possibilities for θ 6= 1 and with painted vertices*
*are limited to An* *(n odd), Dn* *and E*6. We may not paint vertices that are not fixed

*by θ (since compactness of roots makes sense only on the imaginary ones). We label*
*the vertices as in Table 2. The only way to paint An(n odd) is by painting the vertex*

*n + 1*

2 , so it is not equivalent to any other diagram.

*Next we consider Dn* *with vertex N painted, where N ≤ n − 2. In the previous*

*case where θ = 1, we have shown in Proposition 8.5(a) that*
*(N) ∼ (i*1*, ..., ik) for N =*
*k*
X
*p=1*
*(−1)k−p _{i}*

*p*

*and ik*

*≤ n − 2.*(10.1)

*This argument uses F [i] for i ≤ n − 3. In general, F [i] differs in the cases θ = 1*
*and θ 6= 1 only if a vertex adjacent to i is not fixed by θ. Therefore, since n − 1*
*and n are the only vertices not fixed by θ here, the arguments in Proposition 8.5(a)*
*is still valid in our present situation. Namely we also have (10.1) for θ 6= 1. Also,*
*(N) ∼ (n − N − 1) because they represent the Lie algebras so(2N + 1, 2n − 2N − 1) ∼*=
*so(2n − 2N − 1, 2N + 1). This proves Dn* in Table 2.

*Finally in E*6*, there is only one equivalence class with θ 6= 1 and with some vertices*

painted [4, p532-535]. Therefore all such cases are equivalent to one another. This completely verifies Table 2.

### 11

### Graph Paintings

This section develops an idea in the opposite direction: The Vogan diagrams can
classify almost all the simply-laced Dynkin diagrams. Since we are interested in the
*underlying Dynkin diagrams, we may consider only the Vogan diagrams with θ = 1*
in this section.

*Recall that the algorithm F [i] in (8.1) is used to reduce the number of painted*
vertices within an equivalence class of Vogan diagrams until we end up with a single
painted vertex. This is not so surprising, by the following theorem.

*Theorem 1 Two Vogan diagrams with θ = 1 are equivalent if and only if one can be*
*transformed to the other by a sequence of F [i] operations.*

*Proof:* *The “if” part of the theorem is obvious, since F [i] preserves equivalence*
*classes. The converse has been verified explicitly for each Dynkin diagram in §§2, 3*
when we check Table 1. We now give a more intrinsic argument which does not
take into account the shapes of the Dynkin diagrams. Recall that two equivalent
Vogan diagrams correspond to the same Lie algebra under different choices of Weyl
*chambers. The Weyl group W acts transitively on the chambers, and so it acts*
*transitively on each equivalence class of Vogan diagrams. Since θ = 1, all roots are*
*imaginary, and they are either compact or noncompact. Let Wc* *and Wn* denote the

subgroups generated by reflections about the compact and noncompact simple roots
*respectively. Clearly W is generated by Wc* *and Wn. Further, since Wc* acts trivially

*on the Vogan diagrams, it follows that Wn* acts transitively on each equivalence class

*of Vogan diagrams. Since F [i] corresponds to reflection about the noncompact simple*
*root labelled i, this proves the theorem.* *2*
The proof of this theorem does not make use of knowledge on the shapes of the
Dynkin diagrams. Therefore, if we accept the Borel-de Siebenthal theorem, then it
gives a necessary condition for a connected graph to be a Dynkin diagram:

*Corollary 1 If a connected graph Γ is a Dynkin diagram, then*

(a) *Every painting on Γ can be simplified via a sequence of F [i] to a painting with*
*single painted vertex;*

*(b) Every connected subgraph of Γ satisfies property (a).*

*Proof:* *To prove (a), let Γ be a Dynkin diagram. Suppose that p is a painting on*
*Γ. By the Borel-de Siebenthal theorem, (Γ, p) ∼ (Γ, s), where s paints just a single*
*vertex of Γ. By Theorem 1, (Γ, p) can be transformed to (Γ, s) with some sequence*
*of F [i] operations. This proves (a). Since connected subgraphs of a Dynkin diagram*
correspond to simple subalgebras, condition (b) is trivial. The corollary follows. *2*
The corollary provides an obstruction for a graph to be Dynkin via conditions
(a) and (b). We shall see that they come close to being sufficient conditions. The
simply-laced Dynkin diagrams are classified by showing that they cannot contain the
following subgraphs:
c
c¢¢
c _{p pp}
pp
...
p ppp
AA c p
c%%
e
e
c c
e
e c
%
%
c
c
%
%
e
e
c
c
*. . .* c%%
c
e
e c
c c c c c
c
c
c c c c c c c
c
c c c c c c
c
c c
(11.1)

In the top row of (11.1), the first two diagrams say that a Dynkin diagram has
no loop and no node (branch point) with more than three edges. The third diagram
says that there is at most one node. In this case we can topologically think of the
*node as being joined to three “lines” l*1*, l*2*, l*3 whose lengths are defined in the obvious

*manner. The fourth diagram of the top row says that one of the li, say l*1, is of length

*1. Then the remaining diagrams put some restrictions based on the lengths of l*2 and

Corollary 1(b) says that a connected subgraph of a Dynkin diagram is again
Dynkin; so it suffices to show that the six graphs in (11.1) are not Dynkin. We
attempt to use Corollary 1(a) to achieve this; namely we find a painting which cannot
*be simplified to a graph with single painted vertex via the algorithms F [i]. Such*
attempt is successful for all but one of them:

s
s¢¢
c _{p pp}
pp
...
p ppp
AA c
p
s%%
e
e
s c
e
e c
%
%
s
s
%
%
e
e
s
s
*. . .* c%%
c
e
e c
c s c c c
s
c
s c c c c c s
c
(11.2)

*For instance in the loop in (11.2), no matter how we apply F [i], we always end*
up with a loop with two painted vertices. So we conclude that every Dynkin diagram
cannot contain any loop. Unfortunately, in the last graph of (11.1), any painting can
*be reduced to a diagram with a single painted vertex. This “fake E*9” is the only

structure which does not exist in Dynkin diagrams but cannot be dismissed by the
*algorithms F [i].*

### References

*[1] P. Batra, Invariants of real forms of affine Kac-Moody Lie algebras, J. Algebra*
223 (2000), 208-236.

*[2] P. Batra, Vogan diagrams of real forms of affine Kac-Moody Lie algebras, J.*
*Algebra 251 (2002), 80-97.*

[3] A. Borel and J. de Siebenthal, Les sous-groupes ferm´es de rang maximum des
*groupes de Lie clos, Comment. Math. Helv. 23 (1949), 200-221.*

*[4] A. W. Knapp, Lie Groups Beyond an Introduction, Progress in Mathematics*
140, Birkh¨auser 1996.