複單李代數上的有限維對角的自同構

國

立

交

通

大

學

應用數學系

碩士論文

複單李代數上的有限維對角的自同構

Finite Order Diagonal Automorphisms

on

Complex Simple Lie Algebras

研究生：劉 筱 凡

複單李代數上的有限維對角的自同構

Finite Order Diagonal Automorphisms

on

Complex Simple Lie Algebras

研 究 生：劉筱凡 Student：Hsiao-Fan Liu

指導教授：蔡孟傑 Advisor：Dr. Meng-Kiat Chuah

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics

College of Science

National Chiao Tung University

In partial Fulfillment of the Requirements

For the Degree of Master

In

Applied Mathematics

June 2007

Hsinchu, Taiwan, Republic of China

複單李代數上的有限維對角的自同構

研究生：劉 筱 凡 指導教授：蔡 孟 傑 教授

國 立 交 通 大 學

應 用 數 學 系

摘要

複單李代數上的一自同構若保持住一個 Cartan 子李代數且以係數乘

積作用在相對應的根空間上，即稱為一對角的自同構。本篇論文中，

我們研究複單李代數上的有限維對角的自同構。特別地，我們可藉由

圖形來表示這些自同構，並且討論等價的圖形中彼此的組合性質。

關鍵字：對角的自同構，複單李代數。

中 華 民 國 九 十 六 年 六 月

Finite Order Diagonal Automorphisms

on Complex Simple Lie Algebras

Student: Hsiao-Fan Liu

Advisor: Meng-Kiat Chuah

Department of Applied Mathematics

National Chiao Tung University

Abstract

An automorphism on a complex simple Lie algebra is said to be diagonal if it preserves a Cartan subalgebra and acts as scalar multiples on the correspond-ing root spaces. In this thesis, we study finite order diagonal automorphisms on complex simple Lie algebras. In particular, we represent these automorphisms by some diagrams, and study the combinatorial properties of equivalent dia-grams.

誌 謝

能夠順利完成論文，最要感謝的就是我的指導老師蔡孟傑教授，

這段時間以來，老師不只悉心指導我課業及研究上的困難，也關心我

未來的學涯規劃。也要謝謝舉卿、采瑩和千砡三位學姊，雖然相處的

時間只有短短一年，無論生活上或是學業上，他們都不吝給我意見與

關心。

還有我的同學育慈和雁婷，我們所研究的領域完全不同，但她們

總是能對我面臨到的困難提出有建設性的意見，也讓我在撰寫論文的

過程中，面對電腦程式的問題都能迎刃而解。謝謝我的好朋友背影和

水母，總在電腦的一端為我加油打氣，一個人在研究室裡才不顯得孤

軍奮戰。謝謝志豪學長，在我遇到數學上的問題時，總是適時地提點

我。

謝謝惟揚任何時候都支持、陪伴在我身邊，在我論文遇到最大的

困難時，耐心地給我鼓勵並包容我的任性。最後要感謝的是始終關心

我的家人，給我絕對的支持與信任，是我的精神支柱。

最後，謹將本論文的成果獻給我的父母、師長及所有關心我的朋

友。

Contents

Abstract (in Chinese) I

Abstract (in English) II

Acknowledgement (in Chinese) III

1 Introduction 1

2 Vogan Diagrams 2

3 Equivalent Diagrams 5

1

Introduction

Let g be a finite dimensional complex simple Lie algebra. Given a Cartan subalgebra h ⊂ g, we have the root space decomposition

g = h +X

∆

gi,

where ∆ ⊂ h∗ _{are the root. We say that an automorphism σ : g −→ g is a diagonal}

automorphism if there exists a σ-stable Cartan subalgebra h such as σ preserves all the root spaces gi. The purpose of this thesis is to study the diagrammatic expression

of finite order diagonal automorphisms, and to classify the equivalent diagrams when g is a complex simple Lie algebra of type A.

Let σ be a diagonal automorphism with respect to h. So σ acts as scalar multiples on the root spaces gi. If in addition σ is of finite order m, then the eigenvalues are

some complex numbers ci such that cmi = 1. In Section 2, we introduce the Vogan

diagram of σ. It is a diagram which represents σ by indicating its eigenvalues on the vertices of the extended Dynkin diagram D1 _{of g. This is the natural generalization}

of the Vogan diagrams of involutions on g [7].

A Vogan diagram represents a finite order diagonal automorphism σ under a choice of simple roots. However, different diagrams may represent the same σ due to different simple roots. In this case we say that the diagrams are equivalent. In Section 3, we introduce an algorithm which combinatorially describes how two diagrams are equivalent. As an application, in Section 4, we give a precise method to determine equivalent diagrams for type A Lie algebras.

2

Vogan Diagrams

In this section, we introduce the Vogan diagrams of finite order diagonal automor-phisms. Let g be a complex simple Lie algebra, with Cartan subalgebra h. Let Π denote a simple system, so Π ⊂ ∆ ⊂ h∗_{. The vertices of the extended Dynkin}

dia-gram are Π ∪ {ϕ}, where ϕ is the lowest root. The vertices i of D1 _{are equipped with}

canonical coefficients ai, where ai are positive integers without nontrivial common

factor andP_Π∪{ϕ}aii = 0.

Let σ be a diagonal automorphism on g of order m which preserves h, and let Zm

denote the abelian group of Z modulo mZ. Let ² = exp(2π√−1/m) ∈ C be the m-th

primitive root of unity. The eigenvalues ci of σ are exactly ²si, where si ∈ Zm. It

follows that if we assign si to vertex i of the Dynkin diagram D, then σX = ²siX for

all X ∈ gi. Moreover, Lie algebra homomorphism provides

σ[X, Y ] = [σX, σY ] = [²si_{X, ²}sj_{Y ] = ²}si+sj_{[X, Y ],}

where Y ∈ gj. This implies that the assignment to vertex ϕ is forced by Lie algebra

homomorphism while the other vertices are assigned. In other words, we can assign such si to each simple root and hence there is an assignment {si}Π∪ϕ on the vertices

of D1_.

Given si ∈ Zm, we may write si = [bi] for some bi ∈ Z. We say that the assignment

{si} is nontrivial if the set {bi}Π∪ϕ has no nontrivial common factor. To ensure that

the order of σ is m, {si} is required to be nontrivial. In this respect, we associate the

following diagram on g.

Definition 2.1. A Vogan diagram of order m on g is a nontrivial assignment of

si ∈ Zm to each vertex i of D1, such that

P

Π∪{ϕ}aisi is a positive multiple of m. In

particular, if P_Π∪{ϕ}aisi = m, we call it a standard Vogan diagram .

Obviously, these si indicate the behavior of σ on gi. Here, we may soon conclude

that every Vogan diagram represents an automorphism on the corresponding complex simple Lie algebra of order m.

Recall that a Vogan diagram defined in [7] is a Dynkin diagram with an involution

θ, such that the vertices fixed by θ are painted or not. For m = 2, σ is an involution

on g and the eigenvalues are ±1. By ignoring ϕ, we obtain a Dynkin diagram with

σ. We can let σ act on the roots i such that σ(gi) = gσ(i). Then σ(i) = i if and only

if i is imaginary, and hence we obtain an automorphism σ on D. So the imaginary simple roots are the vertices of D which are fixed by σ. Namely, the vertices which are painted represent the eigenvalue −1 of σ. It is easy to see that −1 = ²1_{. Therefore,}

we can assign 1 to the painted vertices and 0 to the others. This leads an assignment to be a Vogan diagram of order 2 on g and provides a natural way to define those Vogan diagrams of higher order m on g.

Throughout this paper, we shall always let V be the set of Vogan diagrams of order m on g and Vs be the set of standard Vogan diagrams.

Theorem 2.2 (Kac[4][6]). A Vogan diagram with {si} represents an order m

au-tomorphism σ on g by σXi = ²siXi on the root vector Xi of i. Conversely, every

diagonal automorphism is represented by a Vogan diagram. Each Vogan diagram is equivalent to a standard Vogan diagram. Up to conjugation, the automorphisms obtained this way exhaust all m-th order diagonal automorphisms on g.

This theorem follows from [4, Chap.X-5, Theorem 5.15]. Due to the choice of sim-ple systems, two different Vogan diagrams may represent conjugate automorphisms. In such case, we say that these Vogan diagrams are equivalent. We summarize it as follows.

Definition 2.3. Two Vogan diagrams v and w are equivalent if and only if they represent two conjugate automorphisms.

We denote it by v ∼ w. By Theorem 2.2, it allows us to use the diagrams to study finite order automorphisms on g. Since each automorphism of order m must give one assignment {si} with

P

Π∪{ϕ}aisi = m, we derive that every Vogan diagram v ∈ V is

equivalent to a standard Vogan diagram v0 _{∈ V} s.

Theorem 2.4 (Borel-de Siebenthal [3]). Every real form of a complex simple Lie

algebra can be represented by a Vogan diagram with at most one painted vertex.

For the case of m = 2, Theorem 2.2 says that each involution σ determines an assignment {si} on D1 where

P

Π∪{ϕ}aisi is even and si is either 1 or 0. Suppose

there is only one black vertex of D. By Theorem 2.4, the vertex ϕ must be black. Namely, every Vogan diagram is equivalent to a diagram with P_Π∪{ϕ}aisi = 2.

3

Equivalent Diagrams

In the previous section, we have constructed the Vogan diagram v of the automor-phism σ with respect to a simple system. If we change the simple system, we obtain a different diagram w which also represents σ, and in this case we say that v and

w are equivalent. In this section, we give a combinatorial description for equivalent

diagrams.

Let aut(∆) denote the automorphism group of the root system ∆. It acts on the simple system, and so it also acts of the Vogan diagrams V . Since it acts transitively on all the ordered simple system, the orbits of its action on V are precisely the equivalent classes of Vogan diagrams. Namely, v ∼ w if and only if there exists some

f ∈ aut(∆) such that f (v) = w. We shall give a combinatorial description for such f ∈ aut(∆).

Consider two possible ways to change the choice of simple system. Let W be the Weyl group generated by simple reflections and aut(D) be the set of all automorphisms on the Dynkin diagram D. Evidently, aut(D) is a subgroup of aut(∆). Moreover, W is a normal subgroup [5, Lemma 9.2]. Exactly, a possible change of simple system is provided as follows.

Theorem 3.1. There is a semi-direct product

aut(∆) = W × aut(D).

This theorem follows from [4, Chap.X-3, Theorem 3.29] or [5, Chap.12.2].

Due to aut(∆) = W × aut(D), we shall consider an operation which acts on ∆ by reflection corresponding to the simple root i. As a result, it leads to an equivalent Vogan diagram. In what follows, we introduce and generalize Fi developed in [1] and

[2]. Let v be a Vogan diagram of an order m automorphism, with assignment of {sj}

to the vertices j. Given a vertex i, we define Fi(v) to be another Vogan diagram with

(3.1) Fi(v) :                      ti = −si,

tj = sj+ si if j is an adjacent root of equal or shorter length,

tj = sj+ 2si if j is a longer root joint to i by a double edge,

tj = sj+ 3si if j is a longer root joint to i by a triple edge,

tj = sj if j and i are not adjacent.

Indeed, Fi corresponds to the reflection defined by the simple root i. Consider

the effect of the reflection defined by i. There is no effect to vertices j which are not adjacent to i, since it means that the simple roots j and i are mutually orthogonal. So, we obtain tj = sj. For adjacent vertices, we consider just locally on the plane,

namely A2, B2 and G2. There are five conditions for adjacent vertices. In each case,

one can draw the roots of i and j on the plane and justify the condition of Fi visually.

Such Fi provides a way to judge whether two Vogan diagrams are equivalent, and

they generate the Weyl group W . Definition 2.3 therefore can be restated as follows. Proposition 3.2. Two Vogan diagrams v and w are equivalent if and only if there

exists a sequence of Vogan diagrams va with

(3.2) v = v0 7→ v1 7→ ... 7→ vk= w,

such that each va→ va+1 is given by some Fi of (3.1) or a diagram automorphism.

Let int(g) be the subgroup of aut(g) generated by {exp(adX) ; X ∈ g}, where

exp : end(g) −→ aut(g) is the exponential map. The members of int(g) are called

inner automorphisms. The theorem of Kac [4, Chap.X-5, Theorem 5.16] says that

each diagonal automorphism of finite order on g is an inner automorphism. The inner automorphism corresponds only to Fi, without diagram automorphisms. Thus, we

may impose a stricter notion on the Vogan diagrams. Namely, two equivalent Vogan diagrams v and w are said to be inner equivalent if and only if each va → va+1 in

4

Equivalence Classes of Type A Diagrams

Recall that Kac’s theorem says that a Vogan diagram is equivalent to a standard diagram, but it does not say which standard diagram. In this section, we provide a method to find the standard diagram explicitly. We also show the method to judge whether two Vogan diagrams are equivalent.

Let V (A1

n) denote the set of all Vogan diagrams of order m on A1n. Label the

vertices of A1

n naturally by 0, 1, ..., n, where 0 is the extra vertex and express a Vogan

diagram on A1

n by

(4.1) (i1, i2, ..., ik) ∈ V (A1n), 0 ≤ i1 ≤ ... ≤ ik ≤ n and k is a multiple of m.

Recall that each vertex i of a Vogan diagram is assigned by si ∈ Zm, and thus we allow

an index to appear s (or s + mZ) times in (i1, ..., ik) if and only if the corresponding

vertex is assigned by s. We also allow 0 to appear, and we may ignore it since the assignment to 0 is forced. So for example (2, 2, 3) and (1, 1, 1, 2, 2, 3) both refer to the following Vogan diagram of order 3 on A1

3. e 0 e 2 e 1 e 0 ©©© HHH

Define the standard Vogan diagrams by setting k = m in (4.1), namely (i1, ..., im) ∈ V (A1n), 0 = i1 ≤ i2 ≤ ... ≤ im ≤ n.

From now on, we shall let Vs(A1n) denote the set of all standard Vogan diagrams of

order m on A1

n. It is easily seen that Vs(A1n) ⊂ V (A1n).

Recall that we want to find a standard Vogan diagram which is equivalent to a given diagram v ∈ V (A1

n). This will be done in Proposition 4.4, where we construct

a map τ : V (A1

n) −→ Vs(A1n) which satisfies v ∼ τ (v). The next few lemmas study a

function φ : V (A1

Let C be the set of all complex numbers. Define (4.2) φ : V (A1 n) −→ C, φ(i1, ..., ik) = k−1 X p=0 ²p_i k−p,

where ² is the m-th primitive root of unity as before. By this definition of φ, we can check that

(4.3) φ(i1, ..., ir, ..., ik) = φ(ir+1, ..., ik) + ²k−rφ(i1, ..., ir).

For example, φ(2, 2, 3, 3, 6, 6, 6) = (1 + ² + ²2_{)6 + ²}3_{φ(2, 2, 3, 3)} = (1 + ² + ²2_{)6 + ²}3_{((1 + ²)3 + ²}2_{φ(2, 2))} = (1 + ² + ²2_{)6 + ²}3_{((1 + ²)3 + ²}2_{(1 + ²)2)} = (1 + ² + ²2_{)6 + (²}3_{+ ²}4_{)3 + (²}5_{+ ²}6_)2.

Two important properties of ² are

(4.4) ²m _{= 1 , 1 + ² + ... + ²}m−1 _{= 0.}

The second equation is proved by (² − 1)(1 + ² + ... + ²m−1_{) = 0, and hence we know}

that 1, ², ²2_{, ..., ²}m−2 _{are linearly independent. Observe that φ is well-defined because}

of the second equation in (4.4). Namely, if the indices ir appear several times for a

same diagram as allowed in (4.1), then the value of φ remains the same. Further, the second equation in (4.4) yields

(4.5) 1 + ² + ... + ²s−1 = −²s(1 + ² + ... + ²m−s−1), for s ∈ {0, 1, ..., m − 1}. For the following proposition and lemma, it is convenient for us to rewrite (4.1) to be

(4.6) (0s0_{, 1}s1_{, ..., n}sn_{) ∈ V (A}1

to denote the Vogan diagram with value si at vertex i. For example, if m = 4,

(2, 3, 3, 6, 6, 6) = (21, 32, 63) = (25, 32, 63) = (10, 25, 32, 63). Here vertex 2 is assigned s2 = 1 or s2 = 5, and so on.

The function φ is useful, because it is invariant under F1, ..., Fn−1, as shown by

the following lemma. Let Rc denote counter clockwise rotation of assignment {si}

by c steps. For example, in the following diagram, the right diagram is obtained by applying R2 to the left diagram.

e 1 e 0 e 2 e 0 e 0 ´´´ QQQ _e 0 e 0 e 1 e 0 e 2 ´´´ QQQ

Two Vogan diagrams on A1

4 with m = 3.

Lemma 4.1. Let v ∈ V (A1

n) be a Vogan diagram. Then

(a) φFr(v) = φ(v) for all r = 1, 2, ..., n − 1.

(b) φFn(v) = φR1(v).

(c) φF0(v) = φR−1(v).

Proof. To prove this lemma, we use the notation in (4.6) to express v. For part (a),

the identity (4.3) says that

(4.7) φFr(0s0, ..., nsn) = φ(0s0, ..., (r − 1)sr−1+sr, rm−sr, (r + 1)sr+1+sr, ..., nsn) = φ((r + 2)sr+2 _{+ ... + n}sn₎ +²sr+2+...+snφ((r − 1)sr−1+sr, rm−sr, (r + 1)sr+1+sr) +²sr−1+...+sn_φ(0s0_{, ..., (r − 2)}sr−2_).

A direct computation shows that

(4.8) φ((r − 1)sr−1+sr_{, r}m−sr_{, (r + 1)}sr+1+sr_{) = φ(r − 1, r, r + 1).}

Therefore, when we substitute (4.8) into the middle term of the last expression of (4.7), we get

φFr(0s0, ..., nsn)

= φ((r + 2)sr+2_{+ ... + n}sn₎

+²sr+2+...+snφ(r − 1, r, r + 1)

+²sr−1+...+sn_φ(0s0_{, ..., (r − 2)}sr−2₎

= φ(0s0, ..., nsn).

This proves part (a) of the lemma.

To show part (b), the definition (4.2) yields

(4.9)

φR1(v)

= φ(0sn_{, 1}s0_{, ..., n}sn−1₎

= (1 + ² + · · · + ²sn−1−1) · n + ²sn−1(1 + ² + · · · + ²sn−2−1)(n − 1)

+... + ²sn−1+...+s1_{(1 + ² + · · · + ²}s0−1_).

Subtracting (1 + ² + · · · + ²sn−1+...+s1−1) before adding it in (4.9) gives

φR1(v) = φ(1s1, 2s2, ..., (n − 1)sn−1) +(1 + ² + · · · + ²sn−1+...+s1+s0−1). Note that φFn(v) = φ(0s0+sn, 1s1, ..., (n − 1)sn−1+sn, nm−sn) = φ((n − 1)sn, nm−sn) + ²(m−sn)+snφ(1s1, ..., (n − 1)sn−1) = φ(1s1, 2s2, ..., (n − 1)sn−1) + (1 + ² + · · · + ²m−sn−1).

The last expression is proved by using (4.5). Since Pn_i=0si is a multiple of m, it follows that

This verifies part (b) of the lemma and (c) follows from the same argument in (b), completing the proof.

Lemma 4.2. FnFn−1· · · F1 = R−1 and F1F2· · · Fn = R1.

Proof. Given a Vogan diagram v, let vi be its value at vertex i. We first claim that

(FiFi−1· · · F1(v))i = −v1− v2− ... − vi and

(4a)

(FiFi−1· · · F1(v))i+1= v1+ v2 + ... + vi+1 for all i = 1, 2, ..., n.

(4b)

We prove (4a) and (4b) by induction on i. It is clear that (F1(v))1 = −v1 and

(F1(v))2 = v1+ v2, so (4a) and (4b) hold for i = 1.

Now suppose that (4a) and (4b) hold for i, and we want to show that they therefore hold for i + 1 as well. It is obvious that

(4.10) (FiFi−1· · · F1(v))i+2= vi+2.

Then

(4.11) (Fi+1· · · F1(v))i+1= −(Fi· · · F1(v))i+1

= −v1− v2− ... − vi+1 by (4b).

Also,

(4.12)

(Fi+1· · · F1(v))i+2

= (Fi· · · F1(v))i+2+ (Fi· · · F1(v))i+1

= (v1+ v2+ ... + vi+1) + vi+2 by (4b) and (4.10).

By (4.11) and (4.12), we have shown that (4a) and (4b) are also true for i + 1. This completes the induction, and so (4a) and (4b) hold for all i = 1, 2, ..., n.

To prove the lemma, we want to show that

By (4a) and (4b),

(Fi+1· · · F1(v))i = (Fi· · · F1(v))i+ (Fi· · · F1(v))i+1

= (−v1− v2− ... − vi) + (v1+ v2+ ... + vi+1)

= vi+1.

Since Fi+2, Fi+3, · · · , Fn has no effect on vertex i, it leads to (4c). This proves the

lemma.

By similar arguments, we obtain F1F2· · · Fn= R1.

Lemma 4.3. Let v ∈ V (A1

n). Then there exists v0 ∈ Vs(A1n) such that v0 ∼ v and

φ(v0_{) = φ(v).}

Proof. Let v ∈ V (A1

n). We claim that there exists i1, ..., iN ∈ {1, ..., n} such that

(4.13) FiNFiN −1· · · Fi1(v) ∈ Vs(A

1

n).

By Theorem 2.2, there exists w ∈ Vs(A1n) such that v ∼ w. Since the simple reflections

Fi ∈ W and aut(D) generate aut(∆), there exists a sequence {va} such that

(4.14) v = v0 7→ v1 7→ ... 7→ vr−1 7→ vr = w,

where each va 7→ va+1 is given by some Fi (where i ∈ {1, ..., n}) or the diagram

reflection γ ∈ aut(D). Since W is a normal subgroup of aut(∆), for each Fi, there

exists some Fj such that Fiγ = γFj. Therefore, using another sequence in (4.14) if

necessary, we may move the γ0_{s so that they appear only at the end of the sequence.}

Further, since γ2 _{= 1, we may assume that γ appears at most once in (4.14). We then}

let either vr−1 or vr be FiNFiN −1· · · Fi1(v) in (4.13). This proves (4.13) as claimed.

We now prove the lemma for v ∈ V (A1

n) by induction on N of (4.13). When N = 1,

and hence φ(v0_{) = φ(v). If otherwise, let w = F}

n(v) and choose v0 = R1(w) since

φ(v) = φFn(w)

= φR1(w) by Lemma 4.1(b).

= φ(v0_).

(4.15)

Suppose that the assertion is true for N − 1. For the case of N, let w =

FiNFiN −1· · · Fi1(v). The case of iN < n is obvious by Lemma 4.1 and the assumption.

Now, we shall consider the case w = FnFiN −1· · · Fi1(v) only.

If iN −1 = n, then w = FiN −2· · · Fi1(v) and hence there is nothing to prove.

Con-versely, if iN −1is not adjacent to n, namely 0 < iN −1 < n−1, then we can interchange

FiN −1 and Fn to obtain w = FiN −1Fn· · · Fi1(v). Then

φ(w) = φFiN −1FnFiN −2· · · Fi1(v)

= φFnFiN −2· · · Fi1(v).

Using the assumption again, the result is followed. Consequently, it suffices to deal with

(4.16) w = FnFn−1Fn−2· · · Fn−j(v), j ∈ {1, ..., n − 1}.

Note that Fn(w) = Fn−1Fn−2· · · Fn−j(v) and hence

φFn(w) = φ(v) by Lemma 4.1(a).

Therefore, the similar arguments in (4.15) imply that there is v0 _{= R}

1(w) ∈ Vs(A1n)

such that φ(v0_{) = φ(v). By induction, this proof is completed .}

We are now ready to construct the map τ : V (A1

n) −→ Vs(A1n) which satisfies

v ∼ τ (v). Define P ⊂ C by

Note that the image of φ is contained in P and we therefore define the useful τ as follows. Let v ∈ V (A1

n) and write φ(v) = b0+ b1² + ... + bm−2²m−2 ∈ P . Define

(4.17) τ : V (A

1

n) −→ Vs(A1n),

τ (v) = (0, bm−2, ..., b1, b0) ∈ Vs(A1n).

Since τ is defined from φ, τ is well-defined clearly. For arbitrary v ∈ V (A1

n), the following proposition gives us the method to express

one form of standard Vogan diagrams which are equivalent to v. Proposition 4.4. Let v ∈ V (A1

n). Then v and τ (v) are equivalent.

Proof. By Lemma 4.3, we can choose a standard Vogan diagram v0 _{∈ V}

s(A1n) such that φ(v) = φ(v0_{). Write v}0 _{= (i} 1, i2, ..., im), and we obtain φ(v) = (im− i1) + (im−1 − i1)² + ... + (i2− i1)²m−2. By (4.17), we have that τ (v) = (0, i2− i1, ..., im−1− i1, im− i1).

It is clear that v0 _{∼ τ (v), which implies v ∼ τ (v). This completes the proof.}

By the above proposition, we are able to find a standard Vogan diagram τ (v) which is equivalent to a given diagram v. Together with the following proposition of Kac, we are also able to judge whether two given diagrams are equivalent.

Proposition 4.5 (Kac). Two standard Vogan diagrams v, w ∈ Vs(A1n) are equivalent

if and only if there exists a diagram automorphism which maps v to w.

Proof. Recall that a standard Vogan diagram which represents an automorphism is

an assignment {si} with

P

si = m. Therefore, v and w ∈ Vs(A1n) are equivalent if

and only if their corresponding automorphisms, σ and σ0_{, are conjugate. The theorem}

of Kac [4, Chap.X-5, Theorem 5.16] says that two automorphisms on g are conjugate if and only if {si} can be transformed to {s0i} by a diagram automorphism. The

For example, consider the standard Vogan diagram (0, 1, 1) of A1

3. We obtain

(4.18) (0, 1, 1) 7→ (1, 2, 2) 7→ (2, 2, 3).

Obviously, the first step is achieved by rotation, while the second step by reflection. By Proposition 4.5, it implies immediately that (0, 1, 1) and (2, 2, 3) are equivalent.

We summarize our results in the following theorem. Theorem 4.6. Let v and w ∈ V (A1

n) be Vogan diagrams. Then v and w are equivalent

if and only if τ (v) and τ (w) are related by a diagram automorphism. Proof. By Proposition 4.4 and Proposition 4.5, we derive this theorem.

Example 4.7. Consider the following three Vogan diagrams v1, v2, v3 with m = 4.

e 3 e 0 e 3 e 0 e 2 ´´´ QQQ v1 e 1 e 0 e 3 e 2 e 2 ´´´ QQQ v2 e 3 e 0 e 2 e 1 e 2 ´´´ QQQ v3 Write v1 = (0, 0, 1, 1, 1, 3, 3, 3), v2 = (0, 0, 1, 3, 3, 3, 4, 4), and v3 = (0, 0, 1, 1, 1, 3, 3, 4).

By Theorem 4.6, we obtain standard diagrams τ (v1), τ (v2) and τ (v3) equivalent to

v1, v2and v3, respectively. That is, τ (v1) = (0, 2, 3, 3), τ (v2) = (0, 0, 2, 4), and τ (v3) =

(0, 2, 3, 4) are as follows. e 0 e 1 e 2 e 0 e 1 ´´´ QQQ τ (v1) e 0 e 1 e 0 e 1 e 2 ´´´ QQQ τ (v2) e 0 e 1 e 1 e 1 e 1 ´´´ QQQ τ (v3)

Note that by diagram automorphism, τ (v1) ∼ τ (v2), but τ (v2)  τ (v3). Therefore,

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] S. Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces, Gradu-ate Studies in Math. vol. 34, Amer. Math. Soc., Providence 2001.

[5] J. E. Humphreys, Introduction to Lie Algebras and Representation Theory, Grad-uate Text in Math. vol. 9, Springer-Verlag, New York 1972.

[6] V. Kac, Automorphisms of finite order of semisimple Lie algebras, Funkcional

Anal. i Prilozen 3 (1969), 94-96.

[7] A. Knapp, Lie Groups Beyond an Introduction, 2nd. ed., Progr. Math. vol. 140, Birkh¨auser, Boston 2002.