In this section, we study the extended Vogan diagrams for A(2)n , Dn(2)+1, E6(2)and D4(3). Here the only possible nontrivial θ are in A(2)2n−1and D(2)n+1. We separate A(2)n into three cases n= 2, even n > 2, odd n > 3.
4.1. A(2)2
Label the vertices of A(2)2 as follows:
α0
α1
As mentioned before, type A(2)2 is not covered in (1.2). We now treat it separately.
Proposition 4.1. There are two inequivalent nontrivial diagrams of A(2)2 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.
Since α1is painted, c(α0) = c(α0+ α1). So Fα1 reverses the color of α0as claimed.
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(2)2n, n > 1
Next we consider Vogan diagrams of A(2)2n, 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(2)2n). 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(2)2n)in Table 1.
4.3. A(2)2n−1, n > 2
We label the vertices of A(2)2n−1as follows:
1
. . . n− 2
n− 1
n
0
We shall show that V (A(2)2n−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(2)2n−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(2)2n−1)in Table 1.
For A(2)2n−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(2)n+1, n > 1
Label the vertices of D(2)n+1as follows:
1
0
. . .n− 1
n
Proposition 4.5. Let v= (i1, . . . , ik)∈ V (D(2)n+1), n >1. Then
v∼
(φ(v))∼ (n − φ(v)) if k is odd;
(0, φ(v)) if k is even.
Furthermore,{(0), (n)} form an equivalence class.
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 that v∼ w ∈ V (E6(2)). Since vertex 4 is longer than vertex 3, v2= ∅ if and only if w2= ∅.
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(3)4
By the same arguments as in G(1)2 , 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(1)8 . 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 inter-changed and s= . . . , j, i, . . ..
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 proves the first part of (A.3). Similarly, suppose that Zk and Zk+1contain i, where i is 5
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
because Ym= ∅ and vertex 2 is unpainted in (1, 7). But this is a contradiction because
N (1)= {2} appears exactly once between 1 ∈ Ym−1and 1∈ Ym−2. So Ym−1cannot con-tain 2, and we have proved the first part of (A.9). Since Ym−2contains 2, and vertex 2 is unpainted in (1, 7), it forces 1∈ Ym−2, which proves the second part of (A.9). For the last part of (A.9), assume that Ym−3 = ∅. Then by (A.6), Ym−3= 3, 2, and so 2 ∈ Ym−2, Ym−3. By (A.3), 1∈ Ym−3, which is a contradiction. This proves the last part of (A.9).
Next we claim that
Zm−2= 7, 6, 5, Zm−3= 6, 5, Zm−4= 5, Zm−5= ∅. (A.10) We first check that
Zm−2 = ∅ ⇒ 5 ∈ Zm−2, Zm−3⇒ 6 ∈ Zm−2
Zm−4 = ∅ ⇒ 5 ∈ Zm−4, Zm−3⇒ 6 ∈ Zm−3
⇒ 7 ∈ Zm−2, 7 /∈ Zm−3. (A.11)
In (A.11), Zm−2, Zm−4 = ∅ follows from (A.5) and 5 ∈ Zm−3 because by (A.4) and by (A.9), Zm−3 = ∅. Other arguments follow from (A.3). Finally Zm−3cannot contain 7 be-cause otherwise N (7)= {6} appears exactly once between 7 ∈ Zm−3and 7∈ Zm−2. This explains (A.11). By (A.11), we have proved the first two parts of (A.10). If Zm−4contains 6, then (A.3) forces 7∈ Zm−3, a contradiction. So by (A.2) and (A.5), Zm−4= 5. For the last part of (A.10), if Zm−5 = ∅, then 5 ∈ Zm−5, Zm−4and (A.3) implies that 6∈ Zm−4, a contradiction. This completes the proof for (A.10).
We also claim that
Ym−4= 3, Ym−5= 3, 2, Ym−6= 3, 2, 1, Ym−7= ∅. (A.12) If 2∈ Ym−4, then together with (A.9), 2∈ Ym−4, Ym−2. We need N (2)= {1, 3} to appear even number of times between them, so 1∈ Ym−4, Ym−2. This is a contradiction, because N (1)= {2} appears exactly once between 1 ∈ Ym−4and 1∈ Ym−2. Therefore Ym−4cannot contain 2. Together with (A.5), we get Ym−4= 3. By (A.5) and (A.10), Zm−5= ∅, so Ym−5 = ∅. Together with (A.6), we get Ym−5= 3, 2. To prove that Ym−6= 3, 2, 1, we check that
Xm−6= ∅ ⇒ Ym−6, Zm−6 = ∅ ⇒ 3 ∈ Ym−6, Ym−5
⇒ 2 ∈ Ym−6, Ym−5 ⇒ 1 ∈ Ym−6. (A.13) The first part of (A.13) follows from (A.4) and (A.5). This implies that 3∈ Ym−6. The rest of (A.13) follows from (A.3). By (A.13), Ym−6= 3, 2, 1. For the last part of (A.12), as-sume that Ym−7 = ∅. By (A.6), it implies that Ym−7= 3, 2. But by (A.3), 2 ∈ Ym−7, Ym−6 implies 1∈ Ym−7, a contradiction. Hence Ym−7= ∅. This completes the proof for (A.12).
Repeating the arguments for (A.9), (A.10) and (A.12), we have:
Zm−4k−1= ∅, Zm−4k−2= 7, 6, 5, Zm−4k−3= 6, 5, Zm−4k−4= 5, Ym−4k−1= 3, 2, Ym−4k−2= 3, 2, 1, Ym−4k−3= ∅, Ym−4k−4= 3, Xm−4k−1= 0, Xm−4k−2= ∅, Xm−4k−3= 0, Xm−4k−4= ∅.
Recall that m is odd. By the above conclusion for Xi, Yiand Zi, we have two possibilities to start the switching sequence s:
(i) m≡ 1 mod (4): s = 0, 41,3, 5, 42,0, 6, 5, 43,3, 2, 1, 7, 6, 5, 44, . . ., (ii) m≡ 3 mod (4): s = 0, 41,3, 2, 1, 7, 6, 5, 42, . . ..
In (i), N (2)= {1, 3} appears twice before 2 ∈ Y3. This is impossible, because after 3∈ Y3
appears, vertex 2 is unpainted. In (ii), 7 appears in Z1while it is still unpainted. Both (i) and (ii) lead to contradictions. Therefore, Case (I) with Zm= 5, 6, 7 is impossible. We next proceed with Case (II).
Case (II). Zm= 6, 7.
Vertex 0 is unpainted after 4m(also by (A.5)), so Xm= ∅, and therefore
Ym= 3, 2, 1. (A.14)
We claim that
Zm−1= ∅. (A.15)
Suppose otherwise, namely Zm−1 = ∅. By (A.2), 5 ∈ Zm−1. Then Zm−1 cannot contain 6, for otherwise N (6)= {5, 7} appears exactly once between 6 ∈ Zm−1 and 6∈ Zm via 5∈ Zm−1. On the other hand, (A.2) and (A.5) imply that 5∈ Zm−2. Further, by (A.3), 5∈ Zm−2, Zm−1implies 6∈ Zm−1. This is a contradiction. This proves (A.15).
By (A.2) and (A.15), 3∈ Ym−1. By (A.3) and Ym= 3, 2, 1,
3∈ Ym−1, Ym ⇒ 2 ∈ Ym−1, Ym ⇒ 1 ∈ Ym−1, Ym. (A.16) But the conclusion of (A.16) is impossible, because N (1)= {2} appears only once between 1∈ Ym−1and 1∈ Ym. By this contradiction, we have proved that Case (II) with Zm= 6, 7
is impossible.
To complete the proof of this proposition, we check that the final case is also impossible.
Case (III). Zm= 7.
As before, Ym= 3, 2, 1. If Ym−1 = ∅, then we obtain a contradiction by the same argument as (A.16). Together with (A.4), we get
Xm−1= 0, Ym−1= ∅, Zm−1 = ∅. (A.17) We check that
Xm−2= ∅ ⇒ 5 ∈ Zm−2, Zm−1 ⇒ 6 ∈ Zm−1
⇒ Zm−1= 6, 5 ⇒ Zm−2= 5, Zm−3= ∅. (A.18)
Here Xm−2= ∅ follows from (A.5). The next argument is due to (A.2) and (A.4). Also, 5∈ Zm−1 follows from (A.2) and (A.17). By (A.3), it leads to 6∈ Zm−1. Observe that 7 /∈ Zm−1, for otherwise N (7)= {6} appears once between 7 ∈ Zm−1and 7∈ Zm. It fol-lows that Zm−1= 6, 5. Then 6 /∈ Zm−2, for otherwise 7∈ Zm−1due to (A.3). So by (A.2) and (A.5), Zm−2= 5. Finally Zm−3= ∅, for otherwise if 5 ∈ Zm−3, then 6∈ Zm−2due to (A.3). This explains (A.18).
By Xm−2= ∅ in (A.18), we have Ym−2 = ∅. Since 3 ∈ Ym−2, Ym, and since Ym−1= ∅, it follows that 2 ∈Ym−2, because N (3)= {2, 4} already appears twice between 3 ∈ Ym−2
and 3∈ Ymvia 4m−1and 4m. Therefore,
Ym−2= 3. (A.19)
Since Zm−3= ∅, by (A.4) and by (A.6),
Xm−3= 0, Ym−3= 3, 2. (A.20)
By arguments similar to (A.16),
Xm−4= ∅ ⇒ 3 ∈ Ym−4, Ym−3 ⇒ 2 ∈ Ym−4, Ym−3
⇒ 1 ∈ Ym−4 ⇒ Ym−4= 3, 2, 1. (A.21) If Ym−5= 3, 2, then 2 ∈ Ym−5, Ym−4and (A.3) imply 1∈ Ym−5, a contradiction. So Ym−5 = 3, 2. By (A.6),
Ym−5= ∅. (A.22)
From (A.17) through (A.22), it follows that
Zm−4k−1= 6, 5, Zm−4k−2= 5, Zm−4k−3= ∅, Zm−4k−4= 7, 6, 5, Ym−4k−1= ∅, Ym−4k−2= 3, Ym−4k−3= 3, 2, Ym−4k−4= 3, 2, 1.
So there are two possibilities, depending on the odd integer m:
(i) m≡ 1 mod (4): s = 0, 41,3, 2, 1, 7, 6, 5, . . .,
(ii) m≡ 3 mod (4): s = 0, 41,3, 5, 42,0, 6, 5, 43,3, 2, 1, . . ..
In (i), the entry 7∈ Z1is impossible because vertex 7 is unpainted at that time. In (ii), the entry 2∈ Y3is impossible since vertex 2 is unpainted at that time (because its neighborhood {1, 3} appears exactly twice before it). This shows that Case (III) leads to a contradiction too.
We have shown that each of the Cases (I)–(III) on Zmleads to a contradiction. There-fore, there is no switching sequence between (0) and (1, 7). This proves the proposi-tion. 2
Proposition A.2 completes the proof of Proposition 3.6. Finally, we want to show that the E(1)8 diagrams (1) and (7) are not equivalent. To achieve this, we consider the more general diagram
1
2
3
n
. . . 0
Proposition A.3. In the above diagram,
(1)∼ (n − 1) ⇔ n ≡ 2 (mod 4).
Proof. Suppose that there is a minimum switching sequence s for ((1), (n− 1)). We want to show that n≡ 2 (mod 4). We will follow the spirit of Proposition A.2. By Lemma 3.3, it is easy to see that
(a) tiis even for odd i n − 1, (b) t2, t0+ t4, tn−2+ tnare odd, (c) ti−1+ ti+1is even for i = 3, n − 1, (d) tn−1is even.
(A.23)
In particular
m= t3
is even, and we write
s= 1, 2, 31, X1, Y1, Z1,32, X2, . . . ,3m, Xm, Ym, Zm, Xi⊂ {0}, Yi⊂ {1, 2}, Zi⊂ {4, . . . , n}.
We proceed with arguments similar to (A.2). If Yistarts with 1, then by Lemma A.1(b), s= . . . , 3i, Xi,1, . . . = . . . , 1, 3i, Xi, . . . contradicts the assumption that s is minimum (A.1). So
Yi= 2, 1, 2, ∅ for all i = 1, . . . , m. (A.24) If Ziends with some p > 4, then by Lemma A.1(b), s= . . . , p, 3i+1, . . . = . . . , 3i+1, p, . . . again contradicts the assumption that s is minimum. So
Zi= ki, ki− 1, ki− 2, . . . , 4, ∅ for all i = 1, . . . , m − 1. (A.25) For nonempty Zi in (A.25), the need for consecutive decreasing integers comes from the fact that s is minimum.
From s= . . . , 3i, Xi, Yi, Zi,3i+1, . . . and N(3) = {0, 2, 4}, we again see that for i= 1, . . . , m − 1, exactly one of Xi, Yi, Zi is empty. (A.26) After1, 2, 31, X1, . . . in the beginning of s, vertex 2 is unpainted. So Y1= ∅ and X1, Z1 = ∅. Since N(0) = {3} and X1= 0, it follows that X2, X4, . . . are all empty.
Together with (A.26), we conclude that
Xi= ∅ for even i < m,
Yi, Zi = ∅ for even i m − 2. (A.27) Consider some odd i m − 3. If Yi = 2, then by (A.27), N(2) = {3} appears ex-actly once between 2∈ Yi and 2∈ Yi+1. This is a contradiction. If Yi= 2, 1, then since 2∈ Yi−1, Yi, by the argument similar to (A.3), 1∈ Yi−1. This is a contradiction, because N (1)= {2} appears exactly once between 1 ∈ Yi−1and 1∈ Yi. We conclude that
for odd i m − 3, Yi= ∅. (A.28)
Since s is a switching sequence for ((1), (n− 1)), we can directly check the end of s that Xmand Ymare empty, and that Zm= 4, 5, . . . , n − 1. If Zm−1 = ∅, (A.25) implies 4∈ Zm−1, then N (4)= {3, 5} appears exactly once between 4 ∈ Zm−1and 4∈ Zmvia 3m, which is impossible. So together with (A.26), we conclude that
Xm−1, Ym−1 = ∅, Zm−1= ∅,
Xm= Ym= ∅, Zm= 4, 5, . . . , n − 1. (A.29) By (A.27), (A.28) and (A.29), we see that Yi is nonempty exactly for even i m − 2, or for i= m − 1. Recall that m is even. So t2 is the sum of m−22 (from the nonempty Y2, Y4, . . . , Ym−2) and 2 (entry 2 appears right before 31and in Ym−1). Namely t2=m2+ 1.
By (A.23)(b), t2is odd. We conclude that
m= 4k for some integer k. (A.30)
By (A.27), (A.28) and (A.29), we see that Xiis empty if and only if i is even. It follows that
t0=m
2 is even.
Therefore, by (A.23)(b) and by (A.23)(c), t4= m − 1 is odd, and hence ti is odd for all even 2 i n − 1. Together with (A.23)(a), we have
ti is odd ⇔ i is even and 2 i n − 1. (A.31) By (A.23)(a), (d), n− 1 is odd, so n is even.
By direct inspection at the beginning of s, we see that Z1 does not contain 5. Hence Z1= 4. Recall that Zi = ki, ki − 1, . . . , 4 from (A.25). We claim that k1, k2, . . . is strictly increasing, namely
ki−1< ki (A.32)
for all 2 i m − 2. Suppose otherwise, namely ki ki−1for some i. Then N (ki)= {ki ± 1} appears only once between ki ∈ Zi−1 and ki ∈ Zi, which is impossible. This proves (A.32) as claimed.
We claim further that|Zi| = i, namely
Zi= i + 3, i + 2, . . . , 4. (A.33) This is proved inductively on i. We have seen from above that Z1= 4. By (A.32), we know that 5 k2. But if 5 < k2, then 5, 6 are contained in all of Z2, Z3, . . . ,this implies that t5= t6, which contradicts (A.31). So 5= k2and Z2= 5, 4. We continue this argu-ment and see that if i+3 < ki, then i+3, i +4 are contained in all of Zi, Zi+1, . . . ,leading to the impossible ti+3= ti+4. Hence i+ 3 = ki. This proves (A.33) as claimed.
Recall that n is even. By (A.31), tn−2is odd and hence (A.23)(b) implies that tnis even.
We claim that in fact tn= 0. Since n ∈ Zm and Zm−1= ∅ (by (A.29)), n appears even number of times in Z1, . . . , Zm−2.This is impossible, because (A.32) implies that n can appear exactly once in Zm−2if tn = 0. So tn= 0 as claimed.
By (A.23)(d), tn−1is even, since n− 1 ∈ Zm, (A.32) implies that
Zm−2= n − 1, . . . , 4. (A.34)
By (A.33) and (A.34), m+ 2 = n, together with (A.30) we have n≡ 2 (mod 4).
We have proved the first part of Proposition A.3; namely if (1)∼ (n − 1), then n ≡ 2 (mod 4).
Conversely, suppose that n≡ 2 (mod 4). Then the minimum switching sequence for ((1), (n− 1)) is given by
1, 2, 31, X1, Y1, Z1, . . . ,3n−3, Xn−3, Yn−3, Zn−3,3n−2, Zn−2, where
Xi=
0, for odd i,
∅, for even i,
Yi= 2, for even i n − 4,
2, 1, for i = n − 3,
∅, for odd i n − 5,
Zi= i + 3,i + 2,...,4, for 1 i n − 4,
∅, for i= n − 3,
4, 5, . . . , n − 1, for i= n − 2.
For example if n = 6, then the minimum switching sequence for ((1), (5)) is
1, 2, 3, 0, 4, 3, 2, 1, 5, 4, 3, 0, 2, 3, 4, 5. This completes the proof of Proposition A.3. 2
We remark that Proposition A.3 can be extended to
(1)∼ (k) ⇔ k ≡ 1 (mod 4), 1 k n.
But for the purpose of this paper, we only need the weaker version presented by Propo-sition A.3. By this propoPropo-sition, it follows that (1) is not equivalent to (7) in E(1)8 . This completes the proof of Proposition 3.7.
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