### A Quick Proof on the Equivalence Classes

### of Extended Vogan Diagrams

Meng-Kiat Chuah and Chu-Chin Hu Department of Mathematics, National Tsing Hua University,

Hsinchu, Taiwan.

chuah@math.nthu.edu.tw cchu@math.nthu.edu.tw

Abstract

An extended Vogan diagram is an extended Dynkin diagram together with a diagram involution, such that the vertices fixed by the involution are colored white or black. Every extended Vogan diagram represents an almost compact real form of the affine Kac-Moody Lie algebra. Two extended diagrams are said to be equivalent if they represent isomorphic real forms. The equivalence classes of extended Vogan diagrams have earlier been classified by the authors. In this paper, we present a much shorter and instructive argument.

2000 Mathematics Subject Classification: 17B67, 17B05.

Keywords: Dynkin diagram, extended Vogan diagram, simple Lie algebra.

This work is supported in part by the National Center for Theoretical Sciences, and the National Science Council of Taiwan.

### 1

### Introduction

*Let X be a Dynkin diagram. Let X*(1)* _{, A}*(2)

*n* *, Dn*(2)*, E*

(2)

6 *, D*4(3) be the extended or affine

*diagram [6, Chap.X-5]. For example if X = An, which is a line with n vertices and*
*n − 1 edges, then X*(1) _{is a loop with n + 1 vertices. A Vogan diagram on X or X}(k)

is a diagram involution, such that the vertices fixed by the involution are colored
*white or black. Every Vogan diagram on X represents a real simple Lie algebra, and*
*every Vogan diagram on X(k)* _{represents an almost compact real form of the affine}

Kac-Moody Lie algebra [1],[2],[3]. If two diagrams represent isomorphic Lie algebras,
we say that they are equivalent. Thus the classification of the Lie algebras amounts
*to the classification of equivalence classes of Vogan diagrams on X [4] and on X(k)*

*[5]. Some combinatorial arguments for X*(1) _{in [5] are messy and not instructive. For}

example, the lengthy arguments [5, p.138-147] merely show that two particular pairs
*of diagrams in E*_{7}(1) *and E*_{8}(1) are not equivalent. This paper provides a much shorter
*and instructive proof on the equivalence classes of X*(1)_{, thereby greatly simplifies the}

### 2

### Induced Diagrams

*Let V (X) and V (X*(1)* _{) be the Vogan diagrams of X and X}*(1)

_{with trivial diagram}

*involution. Given any labelling 1, ..., n on the vertices of X, we let 1, ..., n, e be the*
*corresponding labelling on the vertices of X*(1)* _{, where e is the extra vertex in X}*(1)

_{.}

*An element of V (X) or V (X*(1)_{) is denoted by (i}

1*, ..., ik), where i*1*, ..., ik* are the black

*vertices. Each vertex i of X*(1) _{is assigned a positive integer m}

*i* [6, p.503]. Define
*I : V (X) −→ V (X*(1)_{),}*I(i*1*, ..., ik*) =
*(i*1*, ..., ik*) *if mi*1 *+ ... + mik* is even;
*(i*1*, ..., ik, e) if mi*1 *+ ... + mik* *is odd.*
(2.1)

*We call I(v) ∈ V (X*(1)* _{) the induced diagram of v ∈ V (X). An element of V (X}*(1)

_{) not}

*belonging to the image of I is said to be non-induced.*

*For example, label the vertices of Cn* *by 1, 2, ..., n, where n is the unique long*

*root. Here m*1 *= ... = mn−1* *= 2 and mn* *= 1. Consider (1, 2) ∈ V (Cn*), namely the

*Vogan diagram with vertices 1 and 2 colored black. We have I(1, 2) = (1, 2) because*

*m*1*+ m*2 *= 4 is even. Similarly, I(1, n) = (1, n, e) because m*1*+ mn*= 3 is odd.

*Theorem 1 Two Vogan diagrams v, w ∈ V (X) are equivalent if and only if their*

*induced diagrams I(v), I(w) ∈ V (X*(1)_{) are equivalent.}

*Proof. A way to view the coefficients mi* *is as follows: The vertices of X represent*

the simple roots Π of a finite dimensional complex simple Lie algebra, and the extra
*vertex e represents the lowest root. The coefficients mi*are introduced by Kac while he

studies finite order automorphisms [7]. Namely the coefficients of Π are the coefficients
*of the highest root with respect to Π, and e has coefficient 1. Hence the linear*
*combination of Π ∪ {e} over the coefficients {mi} is 0.*

Recall that the white (resp. black) vertices of a Vogan diagram represent the
compact (resp. noncompact) roots of a real simple Lie algebra; namely their root
*spaces are in the 1 (resp. −1) eigenspace of a Cartan involution of the Lie algebra. If*

*α, β and α+β are roots, then the compactness (i.e. whether compact or noncompact)*

*of α, β, α + β are related by the law*

*c + c = c , c + n = n , n + n = c ,*

(2.2)

*where c denotes compact root and n denotes noncompact root. In this way, the*
*compactness of e is determined by the compactness of the simple roots. The vertex*

*e of an induced diagram I(i*1*, ..., ik*) satisfies

*e is white ⇐⇒ mi*1 *+ ... + mik* is even by (2.1)

*⇐⇒ e is compact.* by (2.2)

*Hence in an induced diagram in V (X*(1)_{), the vertices receive an “over-determined”}

*coloring, where the color of e correctly represents its compactness. Similarly, in a*
*non-induced diagram, the color of e does not correctly represent its compactness.*

*We conclude that every induced diagram I(v) represents a finite dimensional real*
*simple Lie algebra, as given by v. The equivalence relation, as defined by the *
*algo-rithms Fi* *in [4, (2.1)] and [5, (1.2)], satisfies v ∼ w if and only if I(v) ∼ I(w). This*

### 3

### Examples

*With the aid of Theorem 1, we can use the equivalence classes of V (X) [4] to solve*
*many equivalence classes of V (X*(1)_{) [5]. It greatly reduces the messy computations}

in [5]. We illustrate this with the following two examples.

*We have used lengthy computations in [5, p.138-147] to show that the two E*_{7}(1)
diagrams in Picture 1 are not equivalent,

h h h h h h h

x

x h h h h h x

h

Picture 1

*and that the two E*_{8}(1) diagrams in Picture 2 are not equivalent.

x h h h h h h h

h

h h h h h h x h

h

Picture 2

Theorem 1 helps to simplify their arguments. By the coefficients of the roots [6,
*p.503], we see that all diagrams in Pictures 1 and 2 are induced diagrams. Namely, I*
*of (2.1) maps the following E*7 diagrams

h h h h h h

x

h h h h h x

h

Picture 3

to the diagrams in Picture 1. The diagrams in Picture 3 are not equivalent [4]. Con-sequently, by Theorem 1, the diagrams in Picture 1 are also not equivalent.

*Similarly, I maps the following E*8 diagrams

x h h h h h h

h

h h h h h h x

h

Picture 4

to the diagrams in Picture 2. The diagrams in Picture 4 are not equivalent [4], so by Theorem 1, the diagrams in Picture 2 are also not equivalent.

*More generally, consider [5, Table 1]. For each V (X*(1)_{), the equivalence classes of}

the induced diagrams can be checked by Theorem 1 and [4, Table 1]. 5

The above technique does not cover the non-induced diagrams, or diagrams with
*nontrivial involution, or diagrams of A*(2)

*n* *, D*(2)*n* *, E*

(2)

6 *, D*4(3) in [5]. However, the

argu-ments in [5] for these diagrams are quite straight forward. The more messy arguargu-ments
*occur in the induced diagrams of V (X*(1)_{), and they are handled by the above }

### 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] H. Ben Messaoud and G. Rousseau, Classification des formes r´eelles presque
*compactes des alg´ebres de Kac-Moody affines, J. Algebra 267 (2003), 443-513.*
*[4] M. K. Chuah and C. C. Hu, Equivalence classes of Vogan diagrams, J. Algebra*

279 (2004), 22-37.

*[5] M. K. Chuah and C. C. Hu, Extended Vogan diagrams, J. Algebra 301 (2006),*
112-147.

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

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

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