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Extended Vogan diagrams
✩Meng-Kiat Chuah, Chu-Chin Hu
∗Department of Applied Mathematics, National Chiao Tung University, Hsinchu, Taiwan
Received 13 April 2005 Available online 3 February 2006 Communicated by Peter Littelmann
Abstract
An extended Vogan diagram is an extended Dynkin diagram with a diagram involution, such that the vertices fixed by the involution can be painted or unpainted. Every extended Vogan diagram represents an almost compact real form of some affine Kac–Moody Lie algebra. Two diagrams may represent isomorphic algebras, and in this case we say that the diagrams are equivalent. In this paper, we classify the equivalence classes of extended Vogan diagrams, and provide a complete list of all diagrams within each class. It gives a combinatorial classification of the isomorphic classes of almost compact real forms of the affine Kac–Moody Lie algebras.
©2006 Elsevier Inc. All rights reserved.
Keywords: Extended Vogan diagram; Almost compact real form; Kac–Moody Lie algebra
1. Introduction
A Vogan diagram is a Dynkin diagram with a diagram involution, such that the vertices fixed by the involution are either painted or unpainted. This terminology first appeared in [7], and the Vogan diagrams represent the real forms of the complex simple Lie alge-bras. Similarly, given a complex affine Kac–Moody Lie algebra, we can represent it with
✩ This work is supported in part by the National Center for Theoretical Sciences, and the National Science
Council of Taiwan.
* Corresponding author.
E-mail addresses: [email protected] (M.K. Chuah), [email protected] (C.C. Hu).
0021-8693/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2005.12.022
a diagram, known as the extended Dynkin diagram [6, Chapter 4]. We define the extended Vogan diagrams as above, namely with an involution whose fixed points are painted or un-painted. The equivalence classes of extended Vogan diagrams correspond to the isomorphic classes of almost compact real forms [1,2]. In this paper, we classify all the equivalence classes of extended Vogan diagrams, and give a complete list of all the diagrams within each equivalence class. Consequently, this gives a combinatorial classification of the almost compact real forms of affine Kac–Moody Lie algebras, which is parallel to the algebraic classification given in [3].
Let gR be a real form of a complex affine Kac–Moody Lie algebra g. Fix an
isomor-phism from g to gR⊗ C, and let the Galois group Γ = Gal(C/R) act on g. We identify gR
with the fixed points of Γ . We say that gRis almost compact if the nontrivial element of Γ
transforms a Borel subalgebra of g to the Borel subalgebra of the opposite sign. Suppose that gR is an almost compact real form. By choosing a maximally compact Cartan
subal-gebra of gRwhich is stable under a Cartan involution, we can represent gRby an extended
Vogan diagram [1, Section 3].
In what follows, we recall the equivalence relation [1, (3.7.1)] on the extended Vogan diagrams v. It considers v whose edges may be single, double with one arrow, triple with one arrow. Namely we tentatively ignore A(11)(contains double edge with two arrows) and
A(22)(contains quadruple edge), and treat them separately later. If i is a painted vertex in v, let Fi be the algorithm which reverses the colors of all the vertices j adjacent to i, except
when j is a longer root joint to i by a double edge. Namely, define the neighborhood of vertex i by
N (i)= {vertices adjacent to i}, (1.1) excluding i itself. Then Fi(v)is the diagram given by
Fi: Reverse the colors of all j∈ N(i), except when j is a longer root joint to i
by a double edge or when j is not fixed by the involution. (1.2) The operation Ficorresponds to the reflection which sends the simple root i to−i. So v and
Fi(v)represent isomorphic Lie algebras. We say that two extended Vogan diagrams v and
ware equivalent if there is a sequence of operations v= v0→ v1→ · · · → vk= w, where
each va→ va+1is either some algorithm Fias given in (1.2), or a diagram automorphism.
This definition is justified by the following theorem.
Theorem 1.1 (Batra). Every extended Vogan diagram represents an almost compact real
form of an affine Kac–Moody Lie algebra. Two extended Vogan diagrams are equivalent if and only if their corresponding algebras are isomorphic.
Proof. The first statement follows from [2, Theorem 5.2], and the second statement
fol-lows from [1, Theorem 5.2]. 2
By this theorem, the equivalence classes of extended Vogan diagrams correspond to the isomorphic classes of almost compact real forms of affine Kac–Moody Lie algebras. It
al-lows us to use the diagrams to study the isomorphism of algebras. Nonequivalent diagrams of A(n1), Bn(1), Cn(1)and D4(1)are shown in [1]. In this paper, we give a complete list of all
the distinct equivalence classes, as well as all the diagrams within each class.
It is convenient to represent an equivalence class with a diagram with minimum number of painted vertices. So the following theorem will be useful.
Theorem 1.2 (Borel and de Siebenthal). Every equivalence class of extended Vogan
dia-grams has a representative with at most two vertices painted.
Proof. The Borel and de Siebenthal theorem [4] says that every real form of a complex
simple Lie algebra can be represented by a Vogan diagram with at most one painted vertex. In [5], we verify this theorem by using algorithms (1.2) and diagram automorphisms to explicitly reduce every painting on a Dynkin diagram D to another painting with at most one painted vertex.
Consider an extended Dynkin diagram given by a Dynkin diagram D, an extra vertex p, and some extra edges joint to p. Since a painting on D is equivalent to another one with at most one painted vertex, together with p, we obtain a painting with at most two painted vertices. 2
This theorem does not help to judge whether two diagrams are equivalent, or how to reduce a diagram to another one with at most two painted vertices. For instance two dia-grams, both with two painted vertices, could be nonequivalent to each other.
Clearly a diagram with trivial involution and no painted vertex is not equivalent to any other diagram. So once and for all, we ignore such diagrams. In Tables 1 and 2 below, we apply Theorem 1.2 and represent each equivalence class by a diagram with one or two painted vertices. Tables 1 and 2 handle the diagrams with trivial and nontrivial involutions, respectively. The tables give a complete list of all the diagrams within each equivalence class. We shall label the vertices, so that an extended Vogan diagram is denoted by
(i1, . . . , ik) or (θ; i1, . . . , ik), i1< i2<· · · < ik. (1.3)
Here (i1, . . . , ik) has trivial diagram involution and vertices i1, . . . , ik painted; while
(θ; i1, . . . , ik)has diagram involution θ and vertices i1, . . . , ikpainted. We also write (θ; ∅)
for the diagram with involution θ and no painted vertex.
In what follows, we explain the notations φ, c, B, M, ξ used in Table 1.
The notation φ shall be used very often. Given a diagram (i1, . . . , ik)where the painted
vertices are ordered by i1< i2<· · · < ik, we define
φ(i1, . . . , ik)= ik− ik−1+ · · · + (−1)k−1i1=
k
p=1
(−1)k−pip. (1.4)
For a vertex i of a given diagram v, let c(i) denote the color of i in v, which can be painted or unpainted.
Table 1
Trivial diagram involution
Extended Dynkin diagram Representative diagram Equivalent diagrams
A(11) 0 1 (0) (1). (0, 1) A(n1), n >1 . . . 0 n 1 (0, N ), 1 N n+12 (i1, . . . , ik), k is even and φ= N, n − N. (0) (i1, . . . , ik), k is odd. Bn(1), n >2 2 1 0 . . . n– 1n (1) c(0) = c(1). (N ), N 2 c(0)= c(1) and φ = N. (0, 1) (0, 1, 2), (k, k− 1), k 3. Cn(1), n >1 (0) c(0) = c(n). 1 0 . . . n– 1n (N ), Nn2 c(0)= c(n) = ◦ and φ = N, n − N. (0, n) c(0)= c(n) = •. Dn(1), n >4 2 . . . n– 2 n– 1 n 1 0 (0) c(0) = c(1), c(n − 1) = c(n) or c(0) = c(1), c(n − 1) = c(n). (0, n) c(0) = c(1) and c(n − 1) = c(n). (0, 1) (0, 1, 2), (n− 2, n − 1, n), (k − 1, k), 3 k n − 2. (N ),2 N n 2 v = (0, 1, 2); c(0) = c(1) and c(n − 1) = c(n), φ= N, n − N. E6(1) y1 y2 c0 x2 z2 z1 x1 (x1) Mis odd.
(x2) Mis even and B is odd.
(x1, y1) M, Bare even. E7(1) 1 2 3 4 7 5 6 0 (1) φis odd.
(2) φand ξ are even.
(0) φ= 0, 4 and ξ is odd. (1, 7) φ= 2, 6 and ξ is odd. E(81) 1 2 3 4 7 8 5 6 0 (1) (5), (0, N ), N= 4, 8. (7) (2), (3), (0, 6). (8) (0), (4), (6), (0, N ), N= 1, 2, 3, 5, 7. F4(1) 1 2 3 4 5 (1) (1 i1, . . . , ia 3, 4 ia+1, . . .), φ(i1, . . . , ia)is odd. (2) (1 i1, . . . , ia 3, 4 ia+1, . . .), φ(i1, . . . , ia)is even ( = 0). (4) (1 i1, . . . , ia 3, 4 ia+1, . . .), φ(i1, . . . , ia)= 0. G(21) 1 2 3 (1) φis odd. (2) φis even. A(22) 0 1 (0) (1) (0, 1). A(2n2), n >1 (0) (i1, . . . , ik), i1= 0. 1 0 . . . n– 1n (N ), 1 N n (i1, . . . , ik), i1 = 0 and φ = N. A(2n2)−1, n >2 1 . . . n– 2 n– 1 n 0 (0) c(0)= • and c(n − 1) = c(n). (n) c(0)= ◦ and c(n − 1) = c(n). (N ), 1 N n2 φ= N, n − N and c(0) = ◦, c(n − 1) = c(n). (0, n) c(0)= • and c(n − 1) = c(n).
Table 1 (continued)
Extended Dynkin diagram Representative diagram Equivalent diagrams
D(n2)+1, n >1 (0, N ), 1 N n (i1, . . . , ik), φ= N and k is even. 1 0 . . . n– 1n (N ), 0 N n 2 (i1, . . . , ik), φ= N, n − N and k is odd. E6(2) 1 2 3 4 5 (1) (i1, . . . , ik), ik 3 and φ is odd. (2) (i1, . . . , ik), ik 3 and φ is even. (4) (i1, . . . , ik), ik 4 and φ is even. (5) (i1, . . . , ik), ik 4 and φ is odd. D(43) 1 2 3 (1) φis odd. (2) φis even.
For a Vogan diagram v in E(61), let
B(v)= number of branches which contain painted vertices in v,
and
M(v)= number of painted odd vertices in v.
So 0 M(v) 4. More explanations for B(v) and M(v) are given in (3.1) and (3.2). For a Vogan diagram v in E(71), we write
v= (s, i1, . . . , ia, ia+1, . . . , ik), (1.5)
where 1 i1<· · · < ia 4 < ia+1<· · · < ik 7, and s ⊂ {0}. In this case, let
ξ =
a
p=1(−1)a−pip, if the vertex 0 is unpainted,
a
p=1(−1)a−pip+ 1, if the vertex 0 is painted.
(1.6)
For example, let v= (0, 1, 3, 4, 7) be a diagram for E7(1). Then φ(v)= 7−4+3−1+0 = 5 and ξ= 4 − 3 + 1 − 0 + 1 = 3 (the last +1 in the above equation is due to the vertex 0 being painted).
In Table 2, there are several cases where equivalent diagrams can be obtained by replac-ing θ with other σ via diagram automorphisms. For instance, consider the first diagram which deals with A(n1), n even. Here θ fixes 0 and θ (i)= n + 1 − i. If we rotate the indices
by one unit, we obtain σ which fixes n and σ (i)= n − 1 − i. But clearly the diagrams resulting from θ and σ can be identified. So we exclude such diagrams because they are obvious (but require messy notations). The same happen for other A(n1), Dn(1) (replacing
0↔ 1 with n − 1 ↔ n) and E6(1)(permuting xi, yi, zi).
The classification in Tables 1 and 2 is consistent with the classification of the almost compact real forms of affine Kac–Moody Lie algebras in [3, pp. 487–494]. For example, the equivalences classes for A(11) given in [3, p. 487] are τ0τ1, τ0 and ρ. And the corre-sponding classes are represented by (0, 1), (0) in Table 1 and (θ; ∅) in Table 2.
Table 2
Nontrivial diagram involution
Extended Dynkin diagram with nontrivial θ Representative diagram Equivalent diagrams
A(n1), neven (θ; ∅) 0 . . . n 1 . . . n+2 2 n 2
(θ; 0) A(n1), nodd 0 . . . n 1 . . . n+1 2 n+3 2 n−1 2 −→ ←− (θ; ∅) A(n1), nodd . . . 0 n . . . n+1 2 n−1 2
(θ; ∅) A(n1), nodd (θ; ∅) 0 . . . n 1 . . . n+1 2 n+3 2 n−1 2
(θ; 0) (θ;n+12 ). (θ; 0,n+1 2 ) Bn(1), n >4 (θ; ∅) 2 1 0 . . . n– 1n
(θ; N) N = 2, 3, . . . , n (θ; v), φ = N. Cn(1), n >3 (θ; ∅) 1 0 . . . n– 1 n (θ;n2), neven Dn(1), n >4 (θ; ∅) 2 . . .
←−−→
←−−→
n– 2 n– 1 n 1 0 (θ;n 2), neven Dn(1), n >4 (θ; ∅) 2 . . .n– 2 n– 1 n 1 0 (θ; N), 2 N n+12 or N= n (θ; v), φ = N. Dn(1), n >4 (θ; ∅) 2 . . .
n– 2
n– 1 n 1 0 (θ; N), 2 N n2 (θ; v), φ = N.
Table 2 (continued)
Extended Dynkin diagram with nontrivial θ Representative diagram Equivalent diagrams
E6(1) c0
x2 x1 z2 y2 z1 y1 (θ; ∅) (θ; x1) (θ; x2) E7(1) 4
0 1 7 3 5 2 6 (θ; ∅) (θ; 0) A(2n2)−1, n >2 (θ; ∅) 1 . . . n– 2 n– 1 n
0 (θ; N), 0 N n−12 (θ; v), φ = N. D(n2)+1, n >1 (θ; ∅) 1 0 . . . n– 1 n (θ;n 2), neven
Our arguments are divided into the following sections. In Section 2, we consider the classical nontwisted diagrams for A(n1), Bn(1), Cn(1) and Dn(1). In Section 3, we consider
the exceptional nontwisted diagrams for E6(1), E7(1), E8(1), F4(1)and G(21). In Section 4, we consider the twisted diagrams for A(n2), Dn(2), E6(2)and D
(3)
4 . There are two propositions for
E7(1) and E(81)which treat the Dynkin diagrams purely from a graph theoretic viewpoint. Their arguments are lengthy and less relevant, so we place them in Appendix A to keep the rest of the paper fluent.
2. Classical nontwisted diagrams
We consider the extended Vogan diagrams for A(n1), Bn(1), Cn(1)and Dn(1). Nonequivalent
diagrams of A(n1), Bn(1), C(n1)and D(41)are given in [1]. In this section, we show that the diagrams in [1] (as well as general Dn(1)) exhaust all the equivalence classes, and describe
the other diagrams which are equivalent to each of them.
2.1. A(11)
We start with A(11). Recall that the operation Fi in (1.2) does not cover the cases A(11)
and A(22). We now treat A(11), leaving A(22)for Section 4 later. Let
α0
α1
Proposition 2.1. There are three mutually nonequivalent nontrivial diagrams of A(11)given by{α0painted alone}, {α0, α1painted} and {involution α0↔ α1}.
Proof. Recall that the Cartan matrix of A(11)is−2 22 −2, and the positive roots are [6, p. 93]
Δ+=(k− 1)α0+ kα1, kα0+ (k − 1)α1, kα0+ kα1, where k= 1, 2, . . .
.
It implies that
α0, α0+ α1,2α0+ α1∈ Δ+. (2.1)
Suppose now that α0is painted. The operation Fα0corresponds to the effect on the diagram due to the Weyl reflection rα0 which sends α0to−α0. By [6, p. 86],
rα0(α1)= α1− (−2)α0= 2α0+ α1. (2.2)
The coefficient−2 in the above equation comes from the Cartan matrix.
Let c(·) denotes “the color of,” which could be painted or unpainted. The almost com-pact real form determines the colors of all real roots, though only the colors of simple roots are indicated on the extended Vogan diagrams. Suppose that we regard the two colors as the two element group with “unpainted” being the identity. Then whenever i, j, i+ j are roots, they satisfy c(i)+ c(j) = c(i + j). For example, the sum of two painted roots is unpainted, and so on.
Since α0is painted, by (2.1), c(α0+ α1) = c(α1), and also c(2α0+ α1) = c(α0+ α1). So c(2α0+ α1)= c(α1). Together with (2.2), we conclude that Fα0 does not change the color of α1. By symmetry of the diagram, clearly the diagram with α0 painted alone is equivalent to the one with α1painted alone. The proposition follows. 2
By the above proposition, we have proved the information for A(11)in Tables 1 and 2. Let
Xbe a type of complex simple or affine Kac–Moody Lie algebra. Let V (X) and V (θ; X)
respectively denote the diagrams with trivial diagram involution (with at least one painted vertex) and with diagram involution θ . We write
(i1, . . . , ik)∈ V (X) and (θ; i1, . . . , ik)∈ V (θ; X), i1< i2<· · · < ik,
where i1, . . . , ik are the painted vertices. Here we label the vertices as in Tables 1 and 2.
Define the function φ on V (X) by (1.4).
2.2. A(n1), n > 1
The diagram of A(n1)is a loop with vertices 0, 1, . . . , n in this order. Proposition 2.2. Let v= (i1, . . . , ik)∈ V (A(n1)). Then
v∼
(
0, N )∼ (0, n + 1 − N) if k is even and φ(v) = N or n + 1 − N;
Proof. By Theorem 1.2, v is equivalent to another diagram with at most two painted
ver-tices. Further, each Fi preserves the parity of the number of painted vertices in v. So if v
has odd number of painted vertices, it is equivalent to a diagram with one painted vertex. By rotating the diagram of A(n1), it is clear that all the diagrams with one painted vertex are
equivalent to one another.
Next we consider the case where v has even number of painted vertices. For the proof of this proposition (only), we modify the requirement for the notation (i1, . . . , ik)in (1.3)
by allowing i1 i2 · · · ik. In this notation, a vertex appears odd number of times if
and only if it is painted. So for instance (1, 1, 2) and (2, 2, 2) refer to the same diagram with vertex 2 painted. Observe that φ of (1.4) remains well defined in this convention. As we shall see, it allows us to express Fi easily. Note that k is even.
For ir = 0, n, φ· Fir(v)= φ(i1, . . . , ir−1, ir− 1, ir, ir+ 1, ir+1, . . . , ik) = φ(ir+1, . . . , ik)+ (−1)k−r (ir+ 1) − ir+ (ir− 1) − φ(i1, . . . , ir−1) = φ(v). (2.3)
If we can apply Fnto v= (i1, . . . , ik), then ik= n and so
φ· Fn(v)= φ(0, i1, . . . , ik−1, n− 1, n) = n + 1 − φ(v). (2.4)
If we can apply F0to v= (i1, . . . , ik), then i1= 0 and so
φ· F0(v)= φ(0, 1, i2, . . . , ik, n)= n + 1 − φ(v). (2.5)
The last equation uses the fact that k is even. We conclude from (2.3)–(2.5) that
v∼ w ⇔
φ(v)= φ(w) or
φ(v)= n + 1 − φ(w). (2.6)
By Theorem 1.2, v is equivalent to some diagram with two painted vertices i and j . By (2.6),|j − i| is either φ(v) or n + 1 − φ(v). But both cases represent equivalent dia-grams, via diagram automorphisms. For instance the diagrams v= (1, 3) and w = (1, n) are equivalent, with φ(w)= n + 1 − φ(v). This proves the proposition. 2
As explained in the proof, the diagrams with odd number of painted vertices form an equivalence class. Proposition 2.2, together with (2.6), show that two Vogan diagrams v and w with even number of painted vertices are equivalent if and only if φ(v)= φ(w) or
φ(v)= n + 1 − φ(w). This leads to all the information for V (A(n1))in Table 1.
We next consider V (θ; A(n1)). If n is even (i.e. odd number of vertices), then up to
diagram automorphisms, θ has only one possibility 1↔ n, 2 ↔ n − 1, . . . where 0 is fixed by θ . So there are two equivalence classes, given by vertex 0 being painted or unpainted. If
nis odd (i.e. even number of vertices), then up to diagram automorphisms, there are three cases for θ :
0 ... n 1 . . . n+1 2 n+3 2 n−1 2 −→ ←− . . . 0 n . . . n+12 n−1 2
0 ... n 1 . . . n+1 2 n+3 2 n−1 2
(a) (b) (c)
In cases (a) and (b), θ has no fixed point. In case (c), θ has fixed points 0 and n+12 . So case (c) has three equivalence classes represented by (θ; ∅), (θ; 0) and (θ; 0,n+12 ). This completes the discussion for V (θ; A(1)
n )in Table 2.
In our labeling for X= A, B, C, D, if we omit vertex 0 and its adjacent edges in X(n1),
then we obtain the Dynkin diagram for Xn. This idea allows us to apply the results of [5]
in the following manner. Suppose that S is a collection of extended Vogan diagrams, and
Sis closed under each Fi. To study S, we shall often omit one or two vertices (especially
vertices 0 and n) from each diagram in S, and denote the resulting Vogan diagrams by T . The bijection π : S→ T is an isomorphism in the sense that Fi· π(v) = π · Fi(v)and
φ(v)= φ · π(v) for all v ∈ S. In this way, we can apply the results of [5] on T to S. We
first recall some results of [5].
Proposition 2.3. (a) In Anand Bn, (i1, . . . , ik)∼ ( k p=1(−1)k−pip). (b) In Cn, if ik= n, then (i1, . . . , ik)∼ (n). (c) In Dn, (i1, . . . , ik, n− 1) ∼ (n − 1), (i1, . . . , ik n − 2) ∼ ( k p=1(−1)k−pip) and (i1, . . . , ik, n− 1, n) ∼ (1 + k p=1(−1)k−pip).
Proof. Part (a) follows from [5, Proposition 2.3], part (b) follows from [5, Proposition 2.4],
and part (c) follows from [5, Proposition 2.5]. 2
2.3. Bn(1), n > 2
Given a Vogan diagram, recall that c(i) denote the color of vertex i in that diagram. The vertices of Bn(1)are labeled as follows:
2 1 0 . . . n− 1
nProposition 2.4. Let v∈ V (Bn(1)). Then
v∼ ⎧ ⎨ ⎩ (φ(v)) if c(0)= c(1) and φ(v) = 1; (a) (0, 1) if c(0)= c(1) and φ(v) = 1; (b) (1) if c(0) = c(1). (c)
Proof. We prove parts (a) and (b) simultaneously. Let S⊂ V (Bn(1))be the diagrams with
vertices 0, 1 having the same color. It is preserved by all the Fi. By ignoring vertex 0, we
obtain an isomorphism π : S→ V (Bn). Recall from [5] that in Bn, the distinct equivalence
classes are represented by (1), (2), . . . , (n), where v∼ (φ(v)). We conclude that in S, the equivalence classes are (0, 1), (2), (3), . . . , (n), with
v∼
(φ(v)) if φ(v) > 1,
(0, 1) if φ(v)= 1.
We next consider part (c), where vertices 0 and 1 have opposite colors. If we ignore vertex n and think of the diagram as in V (Dn), then Proposition 2.3(c) says that the colors
of 2, 3, . . . , n− 1 are irrelevant. Namely all the diagrams in {v ∈ V (Dn); c(0) = c(1)} are
equivalent to one another. In particular if we let vertex n− 1 be painted and apply Fn−1,
then the color of vertex n is irrelevant too. We conclude that all the diagrams in part (c) are equivalent to one another. This completes the proof. 2
By Proposition 2.4, to prove all the information for V (Bn(1))in Table 1, it remains only
to show that the diagrams (0, 1), (1), (2), . . . , (n) are not equivalent to one another. The diagram (1) is obvious, because the colors of vertices 0 and 1 remain different under all the Fi. For the other diagrams, we use the function φ of (1.4). The computation similar
to (2.3) shows that φ· Fi(v)= φ(v). Since the values of φ on (0, 1), (2), (3), . . . , (n) are
different, they are not equivalent to one another. This proves all the cases for V (Bn(1))in
Table 1.
The only possible nontrivial diagram involution for Bn(1)is given by 0↔ 1, fixing the
other vertices. In this case the arguments are similar to Proposition 2.4(a), and the equiv-alence classes are represented by diagrams with only one vertex painted from 2, 3, . . . , n, respectively.
2.4. Cn(1), n > 1
The vertices of Cn(1), n >1, are labeled as follows:
1 0
. . . n− 1 nProposition 2.5. Let v∈ V (Cn(1)). Then
v∼
⎧ ⎨ ⎩
(φ(v))∼ (n − φ(v)) if 0, n are unpainted; (a)
(0)∼ (n) if exactly one of 0, n is painted; (b)
(0, n) if 0, n are painted. (c)
Proof. We first consider part (a), namely the diagrams v with vertices 0, n unpainted. Since
0, n are long, they remain unpainted under any Fi. So by ignoring vertices 0 and n, such
diagrams are isomorphic to An−1. By Proposition 2.3(a), v∼ (φ(v)) ∼ (n − φ(v)). This
Next we consider part (b), where exactly one of 0, n is painted. Without loss of gener-ality, let S be the diagrams with 0 unpainted and n painted. Once again the colors of 0 and
nremain unchanged under any Fi. Let
T =v∈ V (Cn); vertex n of v is painted
. (2.7)
By ignoring vertex 0, we obtain an isomorphism between S and T . By Proposition 2.3(b), the diagrams in T are all equivalent to (n). Therefore, the diagrams in S are all equivalent to (n). By symmetry of the diagram, (0)∼ (n) in V (Cn(1)).
The argument for part (c) is similar to part (b). Namely by ignoring the painted vertex 0, the diagrams with 0, n painted can be identified with T of (2.7). By applying Proposi-tion 2.3(b) again, it follows that the diagrams in part (c) are all equivalent to (0, n). The proof follows. 2
By Proposition 2.5, to prove the information for V (Cn(1))in Table 1, we only have to
show that the diagrams (0), (0, n) and{(N); 1 N n2} are not equivalent to one another. The colors of vertices 0 and n remain the same under all the Fi, so (0) and (0, n) are not
equivalent to the other diagrams in this list. As for{(N); 1 N n2}, apply the function φ of (1.4) to them. Similar to the computation in (2.3), if v is a diagram with vertices 0 and n unpainted, then φ· Fi(v)equals φ(v) or n− φ(v) for all i = 1, . . . , n − 1. So the diagrams
in{(N); 1 N n2} are not equivalent to one another. This proves the information for
V (Cn(1))in Table 1.
In Cn(1), the only nontrivial diagram involution is the reflection 0↔ n, 1 ↔ n − 1, . . . .
If n is odd (i.e. even number of vertices), then the involution has no fixed point and so all vertices remain unpainted. If n is even (i.e. odd number of vertices), then the involution has exactly one fixed point at vertex n2. In this case there are two equivalence classes, given by n2 painted or unpainted.
2.5. Dn(1), n > 4
As before, c(i) denotes the color of vertex i. The vertices of Dn(1)are labeled as follows:
2 . . . n− 2 n− 1 n 1 0
Proposition 2.6. Let v∈ V (Dn(1)). Then
v∼ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ (φ(v))∼ (n − φ(v)) if c(0) = c(1), c(n − 1) = c(n), φ(v) = 1; (a) (0, 1) if c(0)= c(1), c(n − 1) = c(n), φ(v) = 1;
(0) if c(0)= c(1), c(n − 1) = c(n) (or vice versa); (b)
Proof. We first prove part (a). Let S denote the diagrams v in which c(0)= c(1) and
c(n− 1) = c(n). By ignoring vertices 0 and n, we see that S is isomorphic to V (An−1).
By Proposition 2.3(a),
v∼
(φ(v))∼ (n − φ(v)) if φ(v) = 1,
(0, 1) if φ(v)= 1.
By symmetry of the diagram, (0, 1)∼ (n − 1, n). This proves (a).
We next prove part (b). Without loss of generality, we may consider the Vogan diagrams
Sin which c(0)= c(1) and c(n − 1) = c(n). So S is closed under each Fi. Let T ⊂ V (Dn)
be the diagrams where c(n− 1) = c(n). By ignoring vertex 0, we obtain an isomorphism
S→ T . By Proposition 2.3(c), the diagrams in T are all equivalent to (1) ∈ V (Dn).
There-fore all the diagrams in S are equivalent to (1). By diagram automorphisms, they are also equivalent to (0), (n− 1) and (n). This completes the proof for (b).
Next we prove (c). Let S be the diagrams with c(0) = c(1) and c(n − 1) = c(n). Then
Sis closed under each Fi. The argument for Proposition 2.3(c) can be used to show that
each v∈ S is equivalent to some w ∈ S whose vertices 2, 3, . . . , n − 2 are unpainted. For instance if v= (0, 3, 4, n), we may perform F3, F2, F1and obtain w= (1, n). We conclude that all the diagrams in S are equivalent to (1, n). By diagram automorphisms, they are also equivalent to (0, n− 1), (0, n) and (1, n − 1). This proves (c). 2
We now prove the information for V (Dn(1))in Table 1. If v and w belong to different
parts of Proposition 2.6(a), (b) and (c) (for example if vertices 0 and 1 have the same color in v but different colors in w), then they are inequivalent. Since each of parts (b) and (c) consists of a single equivalence class, it suffices to show that in (a), the diagrams in
(0, 1)∪ (N ); 2 N n 2 (2.8) are mutually not equivalent. We modify φ of (1.4) by ignoring vertex n, so for instance
φ(4, 6, n)= 6 − 4 = 2. By a computation similar to (2.3), φ · Fi(v)= φ(v) or n − φ(v). If
vand w are distinct diagrams chosen from (2.8), then φ(w) is neither φ(v) nor n− φ(v). So the diagrams in (2.8) are mutually not equivalent. This proves all the information for
V (D(n1))in Table 1.
Next we consider V (θ; D(n1))in Table 2. Up to diagram automorphisms, there are three
cases for θ , 2 . . .
←−−→
←−−→
n− 2 n− 1 n 1 0 2 . . . n− 2 n− 1 n 1 02 . . . n− 2 n− 1 n 1 0
(a) (b) (c)
In (a), if n is odd, then there is no fixed point, so all vertices are unpainted. If n is even, there is one fixed pointn2, so there are two classes represented by (θ; ∅) and (θ;n2).
In (b), the diagram obtained by ignoring vertices 0 and 1 is simply Dn−1, so the distinct
equivalence classes are represented by{(θ; N); 2 N n+12 } ∪ {(θ; n)} [5].
In (c), the diagram obtained by ignoring vertices 0, 1, n− 1, n is An−3, so the distinct
equivalence classes are represented by{(θ; N); 2 N n2} [5].
3. Exceptional nontwisted diagrams
In this section, we study the extended Vogan diagrams for E6(1), E(71), E8(1), F4(1) and G(21). Observe that if σ is a diagram automorphism, then σ· Fi= Fσ (i)· σ . So given
a sequence of mixed Fi and σj, we can move the σj over the Fi and gather them. This
proves the following proposition.
Proposition 3.1. If diagrams v and w are equivalent, then there exist some Fi1, . . . , Fik
and diagram automorphisms σj1, . . . , σjl such that σjl· . . . · σj1· Fik· . . . · Fi1(v)= w.
3.1. E6(1)
Label the vertices of E6(1)as follows:
y1 y2 c0 x2 z2 z1 x1
Given a Vogan diagram v, let
B(v)= number of branches which contain painted vertices in v. (3.1) In this definition we ignore vertex c0, except that B= 1 if c0is the only painted vertex. For example, B(c0)= B(c0, x1, x2)= 1, while B(x1, y1)= 2.
We say that a vertex is odd or even depending on whether there are odd or even number of edges joined to it. Given a Vogan diagram v, let
M(v)= number of painted odd vertices in v. (3.2)
So 0 M(v) 4. For example, M(c0, x1, x2)= 2, due to the odd vertices c0and x1.
Proposition 3.2. There are three equivalence classes of V (E6(1)), namely
Z1=
M(v) is odd, Z2=
M(v) is even and B(v) is odd, Z3=
Proof. Observe that in E6(1), any vertex i has even number of adjacent odd vertices. There-fore Fi preserves the parity of M. We conclude that Z1and Z2∪ Z3are both unions of
equivalence classes.
Direct manipulation with the various Fi shows that Z1is indeed one equivalence class
by itself. We next check that each of Z2and Z3is preserved by the various Fi. Recall that
c0 is the central vertex. Clearly B is preserved by all the Fi except possibly Fc0. So we only need to consider Fc0(v)for diagrams v which contain c0.
First, consider Z2. Here M(v) is even and B(v)= 1, 3. If B(v) = 1, then up to dia-gram automorphisms, either v= (c0, x1)or v= (c0, x1, x2). In either case B(Fc0(v))= 3. If B(v) = 3, then up to diagram automorphisms v has three possibilities, namely
(c0, x1, y2, z2), (c0, x1, x2, y2, z2)and (c0, x1, y1, z1, s), where s⊂ {x2, y2, z2}. In the first two possibilities B(Fc0(v))= 1, and in the third B(Fc0(v))= 3. We conclude that Z2 is preserved by all the Fi.
In Z3, M(v)= B(v) = 2; and in particular if c0 is painted, then v= (c0, x1, y2)or
v= (c0, x1, x2, y2)up to diagram automorphisms. It follows that B(Fc0(v))= 2.
We conclude that each of Z2 and Z3 is preserved by all the Fi and so is a union of
equivalence classes. Direct manipulation with the various Fi shows that each of them is
indeed one equivalence class. The proposition is proved. 2
The above proposition proves the information for V (E6(1))in Table 1. The case with nontrivial diagram involution θ is easy. Up to diagram automorphisms, θ is given by {y1↔ z1and y2↔ z2}. The fixed points of θ are x1, x2and c0. From A3, we know that there are three equivalence classes. They are represented by (θ; ∅), (θ; x1)and (θ; x2).
3.2. E7(1)
Label the vertices of E7(1)as follows:
1 2 3 4 7 5 6 0
If v, w are equivalent diagrams,
a switching sequencei1, . . . , ik (3.3)
for (v, w) is a sequence of Fi1, . . . , Fik such that Fik · . . . · Fi1(v)= w. For example 1, 2, 3 is a switching sequence for ((1), (3, 4)). There is only one nontrivial diagram automorphism on V (E7(1))given by the reflection r(i1, . . . , ik)= (8 − ik, . . . ,8− i1). So
by Proposition 3.1, if v, w∈ V (E7(1))are equivalent, then either s(v)= w or r · s(v) = w, where s is a switching sequence. For a switching sequence s, let
Lemma 3.3. (For diagrams with single edges only.) Let s be a switching sequence for
(v, w). Then vertex i has the same color in v and w if and only ifj∈N(i)tj is even.
Here N (i) is the neighborhood of vertex i as defined in (1.1). The lemma is obvious and we omit the proof. Lemma 3.3 will be useful when proving inequivalence of some diagrams. Recall that φ and ξ are as defined in (1.4) and (1.6).
Lemma 3.4. Each Fi preserves the parities of φ and ξ .
Proof. As in (1.5), write v∈ V (E7(1))in the form v= (s, i1, . . . , ia, ia+1, . . . , ik), where
1 i1<· · · < ia 4 < ia+1<· · · < ik 7, and s ⊂ {0}. First we show that each Fi
preserves the parity of φ. By using arguments similar to (2.3) and (2.4), it is clear that this is true for 1 i 7. It remains to show that F0also preserves the parity of φ. Suppose that 0 is painted. Since φ· F0(v)= φ(0, i1, . . . , ia,4, ia+1, . . . , ik) = k r=a+1 (−1)k−r(ir)+ (−1)k−a 4− a p=1 (−1)a−pip = φ(v) + (−1)k−a 4− 2 a p=1 (−1)a−pip ,
F0preserves the parity of φ.
Next we show that each Fi preserves the parity of ξ . For i = 4, Fi preserves the value
a
p=1(−1)a−pip and the color of vertex 0, and hence preserves the parity of ξ . Since F4
changes the colors of vertices 3 and 0, it follows that F4also preserves the parity of ξ . This proves the lemma. 2
We shall show that V (E(71))consists of the following four equivalence classes,
Z1= {φ is odd},
Z2= {φ and ξ are even},
Z3= {φ = 0, 4 and ξ is odd},
Z4= {φ = 2, 6 and ξ is odd}. (3.5)
This will be proved using the next two propositions.
Proposition 3.5. Let v∈ V (E7(1)). Then
(a) v∈ Z1⇒ v ∼ (1), (b) v∈ Z2⇒ v ∼ (2), (c) v∈ Z3⇒ v ∼ (0), (d) v∈ Z4⇒ v ∼ (1, 7).
Proof. By Theorem 1.2, v is equivalent to a diagram of the form (k) or (0, k).
We first prove part (a). It suffices to show that the diagrams in {(k), (0, k); k is odd} are mutually equivalent. Further, by symmetry of the diagram, it suffices to show that
(0, 1)∼ (7) ∼ (0, 3) ∼ (5). We do this by giving their switching sequences:
(0, 1)∼ (7) by 1, 2, 3, 4, 5, 6, 7,
(7)∼ (0, 3) by 7, 6, 5, 4, 0,
(0, 3)∼ (5) by 3, 2, 1, 4, 3, 2, 5, 4, 3, 0, 4, 5. This proves part (a) as claimed.
For v∈ Z2, v∼ (k) where k is even. Since (2) ∼ (6) by symmetry of the diagram and
(2)∼ (4) by 2, 3, 4, 0, 1, 2, 3, 4, we have (2) ∼ (4) ∼ (6). This proves part (b).
The only possibilities of the form (k) and (0, k) are (0), (0, 4) in Z3and (0, 2), (0, 6) in Z4. Clearly each pair of above diagrams are equivalent and since (0, 6)∼ (1, 7) by 6, 5, 4, 3, 2, 1, this proves parts (c) and (d). Hence we complete the proof. 2
Proposition 3.6. Each Zi of (3.5) is a union of equivalence classes.
Proof. The diagram reflection preserves the parity of φ. So together with Lemma 3.4, we
have
{φ is even} {φ is odd}.
It follows that Z1and Z2∪ Z3∪ Z4are unions of equivalence classes.
By Theorem 1.2 and Lemma 3.4, for any v∈ Zi, there exists an even k = 0 such that
v∈ Z2 ⇒ v ∼ (k),
v∈ Z3∪ Z4 ⇒ v ∼ (0, k). (3.6)
We claim that
k1, k2∈ {2, 4, 6} ⇒ (k1) (0, k2). (3.7)
Suppose otherwise, namely (k1)∼ (0, k2)for some k1, k2∈ {2, 4, 6}. By Proposition 3.1, there exists a switching sequence s and the reflection r such that
s(k1)= (0, k2) (3.8)
or r· s(k1)= (0, k2). In the latter case, we may replace k2 by 8− k2to eliminate r and again obtain (3.8). Apply Lemma 3.3 to (3.8), we see that t4is odd because vertex 0 changes color, then t2is odd because vertex 3 does not change color. This is a contradiction because vertex 1 does not change color. This proves (3.7) as claimed.
In (3.7), ξ(k1)and ξ(0, k2)have different parities. So by (3.6) and (3.7), it follows that Z2and Z3∪ Z4are both unions of equivalence classes. To complete the proof of the proposition, it remains to prove that Z3and Z4are both unions of equivalence classes. By
Proposition 3.5(c), (d), this will follow from (0) (1, 7). Unfortunately, and surprisingly, its argument is much harder than other inequivalences, so we accept it for now, leaving
(0) (1, 7) for Appendix A. So Z3, Z4are both equivalence classes. This completes the proof. 2
It follows from Propositions 3.5 and 3.6 that V (E7(1))consists of the four equivalence classes given in (3.5).
For E7(1), the nontrivial involution θ fixes vertices 0 and 4, with i↔ 8 − i. Regarding vertices 0 and 4 as type A2, there are two distinct classes represented by (θ; ∅) and (θ; 0).
3.3. E8(1)
Next we study the equivalence classes of E(81). Label the vertices of E(81)as follows:
1 2 3 4 7 8 5 6 0
Here the only diagram involution is the trivial one. We shall show that there are three nontrivial equivalence classes represented by
(1), (7), (8). (3.9)
Proposition 3.7. The three diagrams (1), (7), (8)∈ V (E8(1)) are mutually not equivalent.
Proof. To prove this, we assume that a switching sequence exists between two diagrams
and derive a contradiction. By Lemma 3.3, we can prove inequivalence for most of them: Suppose that there is a switching sequence for ((1), (8)). The colors of vertices 0, 4, 6 are unchanged. So t3is even, which implies that t5is even, which implies that t7is even. This is a contradiction because vertex 8 changes colors. The same arguments show that there is no switching sequence for ((7), (8)). So we have
(1) (8), (7) (8).
Unfortunately, the remaining arguments for (1) (7) require a lengthy proposition, and we leave it for Appendix A. For the time being, we accept the fact that (1) (7). 2
In the next three propositions, we shall show that every other diagram in V (E8(1))is equivalent to one of (3.9).
Lemma 3.8.
(a) For q 4 and p = 2, 3, we get (p, q) ∼ (0, p−1, q −1) and (0, p, q) ∼ (p−1, q −1). (b) For q 4, (1, q) ∼ (0, q − 1) and (0, 1, q) ∼ (q − 1).
Proof. This follows directly from [5, Lemma 3.1]. Note that vertex 0 is denoted by the
notation∗ in [5]. 2
The next proposition simplifies a diagram of the form (α) or (0, α) to (3.9).
Proposition 3.9. The equivalence classes of (α) and (0, α) in E8(1)are given by
(a) (1)∼ (5) ∼ (0, 4) ∼ (0, 8), (b) (7)∼ (2) ∼ (3) ∼ (0, 6),
(c) (8)∼ (0) ∼ (4) ∼ (6) ∼ (0, N), for N = 1, 2, 3, 5, 7.
Proof. We prove this proposition by Lemma 3.8 and switching sequences. For part (a), (5) ∼ (0, 1, 6) by Lemma 3.8(b) ∼ (1, 5) by0, 1, 3, 4, 5 ∼ (0, 4) by Lemma 3.8(b) ∼ (1) by0, 3, 2, 1 ∼ (0, 8) by1, 2, 3, 4, 5, 6, 7, 8. For part (b), (0, 6) ∼ (1, 7) by Lemma 3.8(b) ∼ (0, 2, 8) by Lemma 3.8(a) ∼ (7) by0, 3, 4, 5, 6, 7 ∼ (0, 1, 8) by Lemma 3.8(b) ∼ (2) by8, 7, 6, 5, 4, 3, 2 ∼ (0, 1, 4) by 2, 3, 0 ∼ (3) by Lemma 3.8(b). For part (c), (0, 5)∼ (1, 6) by Lemma 3.8(b) ∼ (0, 2, 7) by Lemma 3.8(a) ∼ (6) by0, 3, 4, 5, 6 ∼ (0, 1, 7) by Lemma 3.8(b) ∼ (2, 8) by Lemma 3.8(a) ∼ (0) by8, 7, 6, 5, 4, 3, 0 ∼ (0, 3) by0 ∼ (1, 4) by Lemma 3.8(b) ∼ (0, 2, 5) by Lemma 3.8(a) ∼ (4) by0, 3, 4
∼ (0, 1, 5) by Lemma 3.8(b) ∼ (2, 6) by Lemma 3.8(a) ∼ (0, 7) by6, 5, 4, 3, 0 ∼ (1, 8) by Lemma 3.8(b) ∼ (0, 2) by8, 7, 6, 5, 4, 3, 2 ∼ (0, 1) by0, 2, 1 ∼ (8) by1, 2, 3, 4, 5, 6, 7, 8. 2
By the above proposition, the only remaining problem is to simplify a diagram to (α) or (0, α). We divide the diagrams into the following two cases:
(a)
(s,2, 4), (s, 1, 3, 4), (s, 2, 3, 8), (s, 1, 2, 8), (s, 1, 8),
(s,3, 4 j1, . . . , jl)where φ(j1, . . . , jl)= 5,
(b) diagrams which do not belong to (a).
(3.10)
In (3.10)(a), s⊂ {0} depending on whether vertex 0 is painted. The reason for this division is that we shall apply [5, Proposition 3.2] which is not valid for the special cases (3.10)(a).
Proposition 3.10. The diagrams (s, v) in (3.10)(a) are equivalent to (t, 7), where t⊂ {0}.
In the first row of (3.10)(a), s = t. In the second row of (3.10)(a), s = t.
Proof. Observe that the diagrams in the first and the second row of (3.10)(a) are equivalent
to (s, 2, 4) and (s, 3, 5), respectively. By [5, Proposition 3.5], (s, 2, 4)∼ (t, 3, 5) ∼ (t, 7) where s = t. This completes the proof. 2
By Propositions 3.9 and 3.10, we have solved the diagrams in (3.10)(a). We now consider (3.10)(b). Denote a diagram v by
v= (s, i1, . . . , ia, ia+1, . . . , ik),
where 1 i1<· · · < ia 3 < ia+1<· · · < ik 8 and s ⊂ {0}. Let I , J be defined by
I= a p=1 (−1)a−pip and J= k p=a+1 (−1)k−pip (3.11) and let α= J− I if J > 4 or J= 4, k − a = 1, 9− J − I if J < 4 or J = 4, k − a = 1. (3.12) In fact, there are only two cases for J = 4: v = (i1, . . . , ia,4) (with k− a = 1) and
v= (i1, . . . , ia,4, 8) (with k− a = 2). Using I and α defined above, the next proposition
Proposition 3.11. Let v= (s, i1, . . . , ik) be in(3.10)(b). Then v∼ (t, α), where s = t if I
is even, and s = t if I is odd.
Proof. The proof follows from [5, Proposition 3.2]. 2
By Propositions 3.9 and 3.11, we have solved the diagrams in (3.10)(b).
Propositions 3.7, 3.9, 3.10, 3.11 explain all the information for V (E8(1))in Table 1 as follows. Proposition 3.7 shows that there are at least three distinct classes, represented by
(1), (7), (8). The remaining propositions explain how any v∈ V (E(81))is equivalent to one of them. Namely, if v belongs to (3.10)(a), we apply Propositions 3.9 and 3.10. And if v belongs to (3.10)(b), we apply Propositions 3.9 and 3.11.
3.4. F4(1)
Label the vertices of F4(1)as follows: 1 2 3 4 5
The only diagram involution is the trivial one. In the next proposition, we write a typical
v∈ V (F4(1))as
v= (v1, v2), v1⊂ {1, 2, 3}, v2⊂ {4, 5}. So v1is a Vogan diagram of A3.
Proposition 3.12. Let v, w∈ V (F4(1)) both contain painted vertices. Then v∼ w if and
only if v1∼ w1.
Proof. In what follows, “F4” refers to the algorithm (1.2) on vertex 4, rather than the
diagram of type F4. Since there is no risk of confusion, we do not create extra notation to distinguish them. Since (4)∼ (4, 5) ∼ (5), and since F4, F5do not change the colors of vertices 1, 2 and 3, the proposition obviously holds when v1= ∅ or w1= ∅. Therefore, in what follows, we may assume that v1 = ∅ and w1 = ∅. By applying F4and F5, obviously
(v1,4)∼ (v1,4, 5)∼ (v1,5), (3.13)
for any v1. So we may assume that v2= (5). We first claim that
(v1, v2)∼ (v1,∅). (3.14)
That is equivalent to prove that
Since v1 = ∅, there exists a sequence s of operations involving F1and F2such that vertex 3 is painted in sv1(if 3 is already painted in v1, we may take s= 1). Hence
(v1,5)∼ (sv1,5). (3.16)
By applying F3, F4, F3in that order,
(sv1,5)∼ (sv1,∅). (3.17)
Apply s−1to (sv1,∅), we get
(sv1,∅) ∼ (v1,∅). (3.18)
Then (3.16), (3.17) and (3.18) lead to (3.15). This proves (3.14) as claimed.
We are now ready to prove the proposition. Suppose that v1∼ w1. So there is a sequence
t of operations involving F1, F2, F3, such that t (v1)= w1. Since F3may affect the color of vertex 4, we have
t (v1, v2)= (w1, x),
for some x. By (3.14), (w1, x)∼ (w1,∅) ∼ (w1, w2). It follows that v∼ w.
Conversely, suppose that v∼ w. Let r be a sequence of Fi such that r(v)= w. Let r1
be the subsequence of r obtained by removing all the F4and F5in r. Then r1(v1)= w1, so v1∼ w1. This completes the proof of the proposition. 2
The equivalence classes for v1∈ V (A3)and v2∈ V (A2)are well known. So Proposi-tion 3.12 proves the informaProposi-tion for V (F4(1))in Table 1.
3.5. G(21)
Since the effect of Fi on the Vogan diagrams of G(21) is the same as that of A3, the
equivalence classes of G(21) is that of A3 given in [5, Table 1]. That is, if we label the vertices of G(21)by 1 2 3
then there are two equivalence classes in V (G(21)), namely
(1)∼ (1, 2) ∼ (2, 3) ∼ (3) and (2) ∼ (1, 3) ∼ (1, 2, 3). Since the diagram is not symmetric, there is no nontrivial diagram involution.
4. Twisted diagrams
In this section, we study the extended Vogan diagrams for A(n2), Dn(2)+1, E6(2)and D4(3).
Here the only possible nontrivial θ are in A(2n2)−1and D(n2)+1. We separate A(n2)into three
cases n= 2, even n > 2, odd n > 3.
4.1. A(22)
Label the vertices of A(22)as follows:
α0
α1
As mentioned before, type A(22)is not covered in (1.2). We now treat it separately.
Proposition 4.1. There are two inequivalent nontrivial diagrams of A(22)given by
{α0painted alone} and {α1painted alone}.
Proof. Similar to Proposition 2.1, we want to consider the effects of the Weyl reflections
rα0, rα1 on the diagram. Recall that the Cartan matrix of A
(2) 2 is
2 −4 −1 2
, and the positive roots are [6, p. 94]
Δ+=4kα0+ (2k − 1)α1,4(k− 1)α0+ (2k − 1)α1, (2k− 1)α0+ kα1, (2k− 1)α0+ (k − 1)α1,2kα0+ kα1; where k = 1, 2, . . ..
It implies that
α1, α0+ α1,2α0+ α1,3α0+ α1,4α0+ α1∈ Δ+. (4.1) Suppose that α0is painted. We claim that Fα0 does not change the color of α1. Since the upper right entry of the Cartan matrix is−4, by [6, p. 86],
rα0(α1)= α1− (−4)α0= 4α0+ α1. (4.2)
Let c(·) denote “the color of,” as in Proposition 2.1. Since α0 is painted, by (4.1),
c(kα0+α1) = c((k +1)α0+α1)for all k= 0, 1, 2, 3. Hence c(α1)= c(4α0+α1). By (4.2), we conclude that Fα0does not change the color of α1, as claimed.
Next, suppose that α1is painted. We claim that Fα1 reverses the color of α0. Since the lower left entry of the Cartan matrix is−1, by [6, p. 86],
rα1(α0)= α0− (−1)α0= α0+ α1.
We conclude that there are two nontrivial equivalence classes, represented by{α0painted alone} and {α1painted alone}. Note that {α0and α1painted} is equivalent to {α1painted alone}, via Fα1. This proves the proposition. 2
4.2. A(2n2), n > 1
Next we consider Vogan diagrams of A(2n2), n > 1. Label the vertices as follows:
1 0
. . . n− 1 n Throughout this section, φ denotes the function defined in (1.4).Proposition 4.2. Let v∈ V (A(2n2)). Then
(a) v∼ (0), if the vertex 0 is painted in v,
(b) if the vertex 0 is not painted in v, then v∼ (φ(v)), 1 φ(v) n.
Proof. Since the vertex 0 represents the longest root, part (a) follows from
Proposi-tion 2.3(b). Next suppose that the vertex 0 is not painted in v. Since vertex 0 remains unpainted under any Fi, we can ignore it and regard the remaining diagram as a diagram
of Bn. Hence (b) follows from Proposition 2.3(a). This completes the proof. 2
Similar to the argument in (2.3), F1, . . . , Fn preserve φ. Therefore, the diagrams
{(N); 1 N n} in Proposition 4.2(b) are mutually not equivalent. This proves the infor-mation for V (A(2n2))in Table 1.
4.3. A(2n2)−1, n > 2
We label the vertices of A(2n2)−1as follows:
1 . . . n− 2 n− 1 n 0
We shall show that V (A(2n2)−1)consists of the following four equivalence classes,
Z1= c(n− 1) = c(n) and 0 is painted, Z2= c(n− 1) = c(n) and 0 is painted, Z3= c(n− 1) = c(n) and 0 is unpainted, Z4= c(n− 1) = c(n) and 0 is unpainted.
Proposition 4.3. If v∈ Zi, w∈ Zj and i = j, then v w.
Proof. Notice that Fi preserves the color of the long root 0. Moreover, if a diagram v
satisfies c(n− 1) = c(n) or c(n − 1) = c(n), then the same property is satisfied by all the diagrams equivalent to v. 2
Proposition 4.4. Let v∈ V (A(2n2)−1). Then
(a) v∈ Z1⇒ v ∼ (0), (b) v∈ Z2⇒ v ∼ (0, n),
(c) v∈ Z3⇒ v ∼ (φ(v)) ∼ (n − φ(v)), (d) v∈ Z4⇒ v ∼ (n).
Proof. Consider parts (a) and (b), where 0 is painted in v. By Theorem 1.2, v is equivalent
to a diagram w with at most two painted vertices. In part (a), w∈ Z1by Proposition 4.3, so
w= (0) or w = (0, k) for some k n − 2. Using the arguments in [5, Proposition 2.4(b)],
we see that (0)∼ (0, k) for k n − 2. This proves (a). In part (b), w ∈ Z2 by Proposi-tion 4.3, so w= (0, n − 1) or w = (0, n). This proves (b).
Next we prove (c), (d) simultaneously. Since the vertex 0 is long, the color of 0 does not change under any Fi. So we can ignore vertex 0 and its adjacent edges and regard it as a
diagram of Dn. Hence (c), (d) follow from Proposition 2.3(c) and we are done. 2
By Proposition 4.3, each Zi is a union of equivalence classes. Proposition 4.4 says
that in addition, each of Z1, Z2 and Z4 is an equivalence class. In Z3, we see that
φ· Fi(v)equals φ(v) or n− φ(v), so the distinct equivalence classes in Z3are represented
by{(N); 1 N n2}. This proves all the information for V (A(2n2)−1)in Table 1.
For A(2n2)−1, the only nontrivial involution is given by θ (n− 1) = n. Regarding vertices 0, . . . , n− 2 as type Cn−1, there are n+32 distinct classes represented by (θ; ∅) and (θ; N),
0 N n−12 .
4.4. D(n2)+1, n > 1
Label the vertices of D(n2)+1as follows:
1 0
. . . n− 1 nProposition 4.5. Let v= (i1, . . . , ik)∈ V (D(n2)+1), n >1. Then
v∼
(φ(v))∼ (n − φ(v)) if k is odd;
(0, φ(v)) if k is even.
Proof. We first claim that φ· Fi(v)= φ(v) for all i. By the same argument as in (2.3), we
have φ· Fi(v)= φ(v) for i = 0, n. Since 0 and n are short, F0and Fndo not change the
colors of their neighborhoods. So φ· Fi(v)= φ(v) for all i = 0, . . . , n as claimed.
By Theorem 1.2, v is equivalent to some diagram w with one or two painted vertices. Further, each Fipreserves the parity of the number of painted vertices in v. So if v has odd
(respectively even) number of painted vertices, then w has one (respectively two) painted vertex. Also, φ(v)= φ(w). The equivalence of (φ(v)) and (n − φ(v)) follows from the symmetry of the diagram. And the last statement is obvious since vertices 0 and n are short. So we complete the proof. 2
Regarding the subdiagram with 1, . . . , n− 1 as type An−1, it follows that the diagrams
in{(N); 0 N n2} ∪ {(0, N); 1 N n} are mutually not equivalent. So together with Proposition 4.5, this proves the information for V (Dn(2)+1)in Table 1.
In Dn(2)+1, the only nontrivial diagram involution is the reflection 0↔ n, 1 ↔ n − 1, . . . . If n is odd (i.e. even number of vertices), then the involution has no fixed point and so all vertices remain unpainted. If n is even (i.e. odd number of vertices), then the involution has exactly one fixed point at vertex n2. In this case there are two equivalence classes, given by n2 painted or unpainted.
4.5. E6(2)
Label the vertices of E6(2)as follows: 1 2 3 4 5 In the next proposition, we write a typical v∈ V (E6(2))as
v= (v1, v2), v1⊂ {1, 2, 3}, v2⊂ {4, 5}. So v1is a Vogan diagram of A3.
Proposition 4.6. Let v= (v1, v2)∈ V (E6(2)). Then there are four equivalence classes of
V (E6(2)), given by
(1)∈ {v2= ∅ and φ(v) is odd},
(2)∈ {v2= ∅ and φ(v) is even},
(4)∈ {v2 = ∅ and φ(v) is even},
(5)∈ {v2 = ∅ and φ(v) is odd}.
Proof. By direct computations, we see that each Fi preserves the parity of φ(v). Suppose
So each of the four subsets in the proposition is a union of equivalence classes. Direct manipulations show that their diagrams are equivalent to (1), (2), (4) and (5), respectively. Hence the proposition follows. 2
4.6. D(43)
By the same arguments as in G(21), if we label the vertices of D4(3)by 1 2 3
then the equivalence classes are given by
(1)∼ (1, 2) ∼ (2, 3) ∼ (3) and (2) ∼ (1, 3) ∼ (1, 2, 3).
Acknowledgments
The authors thank the referee for helping to improve the presentation of this article. Also, G.J. Chang showed us the method of switching sequences as used in Appendix A.
Appendix A
In this section, we show that (0) (1, 7) in E7(1), and that (1) (7) in E(81). This will complete the proofs of Propositions 3.6 and 3.7. We have isolated these remaining steps into two propositions here, because their arguments are very lengthy and purely computa-tional. It would be nice to replace them with more concise and instructive arguments.
Recall that we define the switching sequence i1, . . . , ik in (3.3). Define the
lexico-graphic ordering on the set of all switching sequences as follows. Given s= i1, . . . , ik
and u= j1, . . . , jl, we declare that s < u by
s < u≡
k < l, or
k= l, i1= j1, . . . , ia= jaand ia+1< ja+1for some a. (A.1)
The following lemma on switching sequences will be useful. We omit the proof, which is obvious. Recall that N (i) is the neighborhood of vertex i, as defined in (1.1).
Lemma A.1.
(a) If s= . . . , i, j1, . . . , jr, i, . . . and ja = i for all a, then N(i) appears even number of
times in j1, . . . , jr.
(b) If s= . . . , i, j, . . . and the vertices i, j are not adjacent, then they can be
Proposition A.2. In E7(1), (0) is not equivalent to (1, 7).
Proof. Since the diagram reflection fixes (0) and (1, 7), by Proposition 3.1, it suffices to
show that there is no switching sequence for ((0), (1, 7)). Suppose otherwise, let s be a switching sequence for ((0), (1, 7)) which is minimum in the sense of (A.1). We now start our series of arguments to obtain a contradiction. By Lemma 3.3,
j∈N(i) tj= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ t4is odd for i= 0; t2is odd for i= 1; t6is odd for i= 7; t0+ t3+ t5is even for i= 4;
ti−1+ ti+1is even for other i.
We will often make use of t4. So denote it by
m= t4
and note that m is odd. Write
s= 0, 41, X1, Y1, Z1,42, X2, . . . ,4m, Xm, Ym, Zm,
Xi⊂ {0}, Yi ⊂ {1, 2, 3}, Zi⊂ {5, 6, 7}.
Here 4i denotes the ith time entry 4 appears in s. For example if s starts with
0, 4, 3, 2, 5, 4, . . ., then the ordered sets satisfy X1= ∅, Y1= {3, 2}, Z1= {5} and so on. We claim that
Yi= 3, 2, 1, 3, 2, 3, ∅ for all i = 1, . . . , m,
Zi= 7, 6, 5, 6, 5, 5, ∅ for all i = 1, . . . , m − 1. (A.2)
Note that nonempty Yi has to start with 3. This is because if Yi starts with q < 3, then
by Lemma A.1(b), s= . . . , 4i, Xi, q, . . . = . . . , q, 4i, Xi, . . .. This contradicts the
as-sumption that s is a minimum switching sequence (A.1). Similarly, if Zi ends with p > 5,
then by Lemma A.1(b), s= . . . , p, 4i+1, . . . = . . . , 4i+1, p, . . . again contradicts the
assumption that s is a minimum switching sequence. The need for consecutive decreasing integers in Yi and Zi comes from the fact that s is minimum. This proves (A.2).
We also claim that
i∈ Yk, Yk+1 ⇒ i − 1 ∈ Yk for i= 2, 3,
i∈ Zk, Zk+1 ⇒ i + 1 ∈ Zk+1 for i= 5, 6. (A.3)
Suppose that Ykand Yk+1contain i, where i is 2 or 3. By Lemma A.1(a), we need N (i)=
{i − 1, i + 1} to appear even number of times between i ∈ Yk and i∈ Yk+1. By (A.2)
or 4k+1, we know that i+ 1 ∈ Yk+1definitely appears, so it forces Ykto contain i− 1. This
or 6. By Lemma A.1(a), we need N (i)= {i − 1, i + 1} to appear even number of times between i∈ Zkand i∈ Zk+1. By (A.2) or 4k, we know that i− 1 ∈ Zk definitely appears,
so it forces Zk+1to contain i+ 1. This completes the proof for (A.3) as claimed. We shall
repeatedly apply (A.3) in future arguments.
Consider s= . . . , 4i, Xi, Yi, Zi,4i+1, . . . for i = 1, . . . , m − 1. Since each Xi, Yi, Zi
contains exactly one element of N (4)= {0, 3, 5}, it follows from Lemma A.1(a) that for i= 1, . . . , m − 1, exactly one of Xi, Yi, Zi is empty. (A.4)
It is clear that X1= ∅. So by (A.4), Y1and Z1are nonempty. Applying Lemma A.1(a) to N (0)= {4}, we conclude that
for odd i= 1, . . . , m, Xi= ∅,
for odd i= 1, . . . , m − 2, Yi, Zi = ∅. (A.5)
We claim that
for even i m − 3, Yi = 3,
for even i m − 1, Yi = 3, 2, 1. (A.6)
Suppose that Yi= 3 for some even i m − 3. By (A.2) and (A.5), 3 ∈ Yi+1. By (A.3),
2∈ Yi, which is a contradiction. This proves the first part of (A.6). Next suppose that
Yi= 3, 2, 1 for some even i m − 1. By (A.2) and (A.5), 3 ∈ Yi−1. So by (A.3),
3∈ Yi−1, Yi ⇒ 2 ∈ Yi−1, Yi ⇒ 1 ∈ Yi−1, Yi. (A.7)
The conclusion in (A.7) is impossible, because N (1)= {2} appears exactly once between 1∈ Yi−1and 1∈ Yi. This completes the proof for (A.6).
There are three cases for Zm, namely5, 6, 7, 6, 7 and 7. We shall show that each
case leads to a contradiction.
Case (I). Zm= 5, 6, 7.
By (A.5), Xm= ∅. So Ym cannot contain 3 because vertex 4 is unpainted in (1, 7).
By (A.2), Ym = ∅. Then Zm−1= ∅, for otherwise 5 ∈ Zm−1, Zm, which contradicts
Lemma A.1(a). Therefore, by (A.4), we have
Xm−1, Ym−1 = ∅, Zm−1= ∅. (A.8)
We claim that
Ym−1= 3, Ym−2= 3, 2, 1, Ym−3= ∅. (A.9)
By (A.2), (A.5) and (A.8), 3∈ Ym−2, Ym−1and so 2∈ Ym−2. If 2∈ Ym−1, then 1∈ Ym−1
