The proof of Theorem 4.1.1

In this section, we give the proof of Theorem 4.1.1.

Proof. (I) The first step is to prove that if the Dynkin diagramDofM appears in one of the listed tables then the set of roots∆^[M]of Nichols algebraB(V) is finite.

Assume thatDappears in rowr of one of the Tables (5.1-5.8). ThenM ad-mits all reflections. Indeed, by Lemma 2.1.4 one obtains thatM isi-finite for all i ∈ I. For i ∈ I, one can determine the Dynkin diagram ofRi(M) by Lemma 2.1.7. One observes that it appears in the same row asD. Do the same for all the Dynkin diagrams in the same row ofD. It implies thatMadmits all reflections by Lemma 2.1.4. HenceC(M)is a well-defined Semi-Cartan graph by Theorem 2.1.9. Assume thatC =C(I,X, R,(A^X)X∈X)is the semi-Cartan graph attached toM. Notice that C(M)is connected. Now, we identify the objects ofC(M)with their Dynkin diagrams. The above calculations imply

4.2. The proof of Theorem 4.1.1 35

that the exchange graph ofC(M)appears in rowrof Table 5.10 and Table 5.11 for rank 2 and rank 3, respectively.

Ifθ = 2, then we calculate the smallest integern with(R₂R₁)ⁿ(D) = D and we observe that the integernappears in the third column andrth row of Table 5.9. Then we compute the characteristic sequence(c_k)k≥1with respect to the first Dynkin diagram in rowrand the label1. We observe that(c_k)k≥1

is the infinite power of the sequence in the fifth column andrth row of Ta-ble 5.9. Further, we get the numberl= 6n−Σ²ⁿ_i=1ci. It appears in the fourth column of Table 5.9. One checks thatl | 12 and(c₁, c₂, . . . , c_12n/l) ∈ A⁺ by Corollary 3.1.5(2). ThenC(M)is a finite Cartan graph by Theorem 3.1.14.

Hence the set of roots∆^[M]is finite by Corollary 2.1.11.

Ifθ= 3, then we obtain thatC(M)is the same as the Cartan graph obtained from the Dynkin diagrams appearing in rowsof Table 2 in [22], wheres ap-pears in the third column andrth row of Table 5.11. The detailed calculations are skipped here. Moreover, the Weyl groupoids of the above Cartan graphs from [22] are finite, see [22, Table 2]. Then∆^[M]is finite by Corollary 2.1.11.

(II) The second step is to prove that if the set of roots∆^[M] of Nichols algebra B(V)is finite then the Dynkin diagram ofM appears in the listed tables. If

∆^[M] is finite thenM admits all reflections by [28, Corollary 6.12]. Since

∆^[M]is finite, C(M) is a finite Cartan graph by Corollary 2.1.11. Since the braiding matrix ofMis indecomposable, then the generalized Cartan matrix A^X is indecomposable by the definition ofA^X and Lemma 2.1.4. SinceC(M) is connected, we obtain thatCis indecomposable by Proposition 1.5.9. We can apply Theorem 3.1.14 and Theorem 3.2.6 to rank 2 and rank 3 Cartan graph C(M), respectively. From the above argument I it is enough to prove that the Dynkin diagram of at least one point inC(M)is contained in the listed Tables (5.1-5.8.

We separate the proof intoθ= 2andθ= 3.

(IIa)Consider the caseθ= 2.

SetX = [M]₂andm =t^X₁₂. We may assume thati= 1, j = 2, but change the labels if necessary. Let (ck)k≥1 be the characteristic sequence of C(M) with respect toX and the labeli= 1. Then we have(c1, c2, . . . , cm) ∈ A⁺ and(c_k)k≥1 = (c₁, c₂, . . . , c_m)^∞by Theorem 3.1.14.

Ifm = 2then(c1, c2) = (0,0). Hence a^X₁₂ = a^X₂₁ = 0. Thenq12q21 = 1 and hence the braiding matrix ofX is decomposable, which is a contradic-tion. Hence m > 2. By Proposition 3.1.6 one of the subsequences (1,1), (1,2, a),(2,1, b),(1,3,1, b) or their transpose, where1 ≤ a ≤ 3 and3 ≤ b≤5, is a subsequence of(c1, c2, . . . , cm). Letnbe the smallest integer with

36 4. The classification result

(R2R1)ⁿ(X) = X. Thenn|m by Theorem 3.1.14. Sincecm+k = ck for all k∈N, we have the freedom to assume any position in(c_k)k≥1, where any of these subsequences is starting. Letc₀ :=c_mandq₀ :=q₁₂q₂₁.

We proceed case by case:

Step 1. Ifc0 =c1 = 1, thena^X₁₂=a^X₂₁=−1. Henceq0 6= 1. We distinguish four cases: 1aa,1ab,1ba and 1bb.

Case 1aa. Ifq11q0 = 1andq22q0 = 1, thenD=D₂₁.

Case 1ab. Ifq11q0 = 1,q22=−1, andq22q0 6= 1, thenD=D₃₁. Case 1ba. Ifq₁₁=−1,q₂₂q₀= 1, andq₁₁q₀ 6= 1, thenD=τD₃₁. Case 1bb. Ifq11=−1,q22=−1, andq0 6=−1, thenD=D₃₂.

Step 2. Assume that(c0, c1, c2) = (1,2, a⁰), wherea⁰ ∈ {1,2,3}. Then we obtain thata^X₂₁=−1,a^X₁₂=a^r₁₂^1(X)=−2, anda^r₂₁^1(X)=−a⁰. We distinguish four cases: 2aa, 2ab, 2ba and 2bb.

Case 2aa. Ifq112q0 = 1andq22q0 = 1, thenD=D₄₁.

Case 2ab. Ifq₁₁²q₀ = 1,q₂₂=−1, andq₂₂q₀ 6= 1, thenD=D₅₁.

Case 2ba. Assume that1 +q₁₁+q₁₁² = 0,q₂₂q₀= 1, andq₁₁²q₀ 6= 1. Ifp= 3 then1 +q11+q112 = 0yieldsq11= 1. Ifq22 6= −1thenD = D₆⁰_,1 and if q₂₂ =−1thenD =D₆⁰⁰⁰_,1. Assume thatp 6= 3. Setζ := q₁₁ andq := q₂₂. Thenq₀ = q⁻¹ ∈ {1, ζ/ ⁻¹}sincea^X₁₂ = −2, andq₀ 6= ζ sinceq₁₁²q₀ 6= 1. ThusD=D₆₁orD₆⁰⁰_,1,p6= 2.

Case 2bb. Consider the last case 1 +q₁₁+q₁₁² = 0,q₂₂ = −1andq₀ ∈/ {1,−1, q₁₁, q₁₁² }.

Case 2bba. Ifp = 3thenq11 = 1. Setq := q0. By Lemma 2.1.7, the Dynkin diagrams ofr1(X)andXare withq∈k^∗\ {−1,1}. Thena^r₂₁^1(X)≤ −2since (−q²)q⁻¹ =−q 6= 1, and−q² 6=−1.

Case2bba1. Ifp= 3anda⁰=−a^r₂₁^1(X) = 2, then one gets(−q²)²q⁻¹ = 1or 1 + (−q²) + (−q²)² = 0. If(−q²)²q⁻¹ = 1thenq = 1, which is a contradic-tion. Hence−q² = 1from the second equation sincep= 3. ThenD=D₉⁰_,2. Case 2bba2. Ifp= 3anda⁰ =−a^r₂₁^1(X)= 3, then one has(−q²)³q⁻¹ = 1or 1 + (−q²) + (−q²)²+ (−q²)³ = 0. The first equation(−q²)³q⁻¹= 1yields (−q)⁵ = 1, henceD = D₁₆⁰_,2. If1 + (−q²) + (−q²)² + (−q²)³ = 0, then (1−q²)(1 +q⁴) = 0and henceq ∈G⁰₈. ThenD=D₁₃⁰_,2.

Case 2bbb. We now suppose thatp6= 3. Setζ :=q₁₁andq :=q₀. Hence the Dynkin diagram ofr1(X)is ^ζ_e^{(ζq)−1−ζq}_e²withζ ∈G⁰₃, q∈k^∗\{1,−1, ζ, ζ⁻¹}. Sincea⁰∈ {1,2,3}, we distinguish three cases:2bbb1,2bbb2and2bbb3.

Case 2bbb1. Consider thatp6= 3anda^r₂₁^1(X)=−1. then one gets(−ζq²)(ζq)⁻¹ = 1or 1 + (−ζq²) = 0. If(−ζq²)(ζq)⁻¹ = 1, thenq = −1, which is a

con-4.2. The proof of Theorem 4.1.1 37

tradiction. If1+(−ζq²) = 0thenζ² =q²and henceq =−ζ. ThenD=D₇₁. Case 2bbb2. If the conditionp6= 3anda^r₂₁^1(X)=−2hold, then(−ζq²)²(ζq)⁻¹ = 1orP2

i=0(−ζq²)ⁱ = 0.

Case 2bbb2a. Consider the equation(−ζq²)²(ζq)⁻¹ = 1. Thenζq³ = 1and henceq ∈G⁰₉sinceζ ∈G⁰₃andp6= 3. HenceD=D_10,2.

Case 2bbb2b. IfP2

i=0(−ζq²)ⁱ= 0, then−ζq² ∈ {ζ, ζ⁻¹}. Henceq² =−1or

−q²=ζ.

Case 2bbb2b1. Ifq² = −1, thenp 6= 2 and the Dynkin diagram of X is

e e

ζ q -1 withq∈G⁰₄,ζ ∈G⁰₃. Setη:=ζ²q⁻¹. Thenη∈G⁰₁₂,ζ =−η², and q=η³. HenceD=D₉₂.

Case 2bbb2b2. If−q² = ζ, thenq ∈ G⁰₁₂andp 6= 2sinceq 6= ζ⁻¹. Hence D=D₈₁.

Case 2bbb3. Consider thatp6= 3anda^r₂₁^1(X)=−3. Then(−ζq²)³(ζq)⁻¹= 1 orP3

i=0(−ζq²)ⁱ= 0,1−ζq² 6= 0.

Case 2bbb3a. Consider the equation(−ζq²)³(ζq)⁻¹ = 1, that is−q⁵ = ζ. Henceq =−ζ⁻¹or−q∈G⁰₁₅,p6= 5.

Case 2bbb3a1. If−q ∈G⁰₁₅,p6= 3,5, andζ =−q⁵, thenD=D_16,2.

Case 2bbb3a2. Ifq = −ζ⁻¹ thenp 6= 2sinceq 6= ζ⁻¹. Hence the Dynkin diagrams ofr₁(X),Xandr₂(X), respectively, are

e e

ζ -1 -ζ⁻¹

e e

ζ -ζ⁻¹ -1

e e

1 -ζ -1

withζ ∈ G⁰₃. Then one obtains thata^r₁₂^2(X) = 1−p. We distinguish four cases.

Case 2bbb3a2a. Ifp= 5, thenD=D₁₆⁰⁰₂. Case 2bbb3a2b. Ifp= 7, thenD=D_18,2.

Case 2bbb3a2c. Ifp = 6s+ 1 (s≥ 2),then the Dynkin diagrams ofr1(X), X,r₂(X),r₁r₂(X),r₂r₁r₂(X), and(r₁r₂)²(X), respectively, are

e e

ζ -1 -ζ⁻¹

e e

ζ -ζ⁻¹ -1

e e

1 -ζ -1

e e

1 -ζ⁻¹ -1

e e

ζ⁻¹ -ζ -1

e e

ζ⁻¹ -1 -ζ

withζ ∈ G⁰₃. Hencen = 6in Theorem 3.1.14 and(c_k)k≥0 = (2,3,2,1, p− 1,1,2,3,2,1, p−1,1)^∞. Thenl= 20−2p <0, which is a contradiction to Theorem 3.1.14.

Case 2bbb3a2d. Ifp = 6s+ 5, wheres ≥ 1, then the Dynkin diagrams of r1(X),X,r2(X)andr1r2(X), respectively, are

e e

ζ -1 -ζ⁻¹

e e

ζ -ζ⁻¹ -1

e e

1 -ζ -1

e e

1 -ζ⁻¹ -ζ

38 4. The classification result withζ ∈G⁰₃. Thenn= 4and(ck)k≥0 = (2,3,2,1, p−1,1, p−1,1)^∞. Hence l= 16−2p <0. Again, one gets a contradiction.

Case 2bbb3b. Consider the equation0 =P3

i=0(−ζq²)ⁱ= (1−ζq²)(1+ζ²q⁴), whereζq² 6= 1. One getsζ =−q⁴. Ifp = 2, thenζ⁴ =q⁴and henceζ =q, which is a contradiction toζq² 6= 1. Otherwiseq ∈G⁰₂₄andD=D_13,2.

Step 3. Now we change the label. It means that(c_k)k≥1is the characteristic sequence ofC_s(M)with respect toXand the label2.

Assume that(c0, c1, c2) = (2,1, b⁰), whereb⁰∈ {3,4,5}. Then we obtain that a^X₁₂ = −2,a^X₂₁ = −1anda^r₁₂^2(X) = −b⁰. Ifq₁₁²q₀ = 1andq₂₂ = −1then a^r₁₂^2(X)=−2, which is a contradiction. Ifq0q22= 1thena^r₁₂^2(X)=a^X₁₂=−2, which is again a contradiction. Hence we may assume that1 +q₁₁+q₁₁² = 0, q22 =−1andq0 ∈ {1,/ −1, q⁻²₁₁}.Sincea^X₁₂ = −2, we also obtain thatq0 6=

q⁻¹₁₁.

Case 3a. If p = 3, then by settingq := q₀, the Dynkin diagrams ofX and r2(X), respectively, are

e e

1 q -1

e e

-q q⁻¹ -1

q∈k^∗\ {−1,1}. (4.1)

Case 3a1. Ifp= 3andb⁰ = 3, then one gets(−q)³q⁻¹ = 1orP3

i=0(−q)ⁱ = 0. Sinceq6= 1, both equations imply thatq² =−1. HenceD=D₉⁰_,2.

Case 3a2. Ifp= 3andb⁰ = 4, then one has(−q)⁴q⁻¹ = 1orP4

i=0(−q)ⁱ = 0.

If(−q)⁴q⁻¹= 1thenq = 1, which is a contradiction to (4.1). IfP4

i=0(−q)ⁱ= 0thenq⁵=−1andD=D₁₆⁰_,2.

Case 3a3. Ifp= 3andb⁰ = 5, then one obtains(−q)⁵q⁻¹= 1orP5

i=0(−q)ⁱ= 0. If (−q)⁵q⁻¹ = 1 then q ∈ G⁰₈ andD = D₁₃⁰_,2. Consider the equation 0 =P5

i=0(−q)ⁱ = (1−q)(1 +q²+q⁴). Thenq² = 1sincep= 3andq6= 1, which is a contradiction to (4.1).

Case 3b. We now consider the cases in which the conditionp6= 3holds. Set ζ :=q₁₁andq:=q₀. The Dynkin diagram ofr₂(X)is

e e

-ζq q⁻¹ -1 ζ ∈G⁰₃ q∈k^∗\ {1,−1, ζ, ζ⁻¹}. (4.2)

Case 3b1. Ifp6= 3andb⁰= 3, then one gets(−ζq)³q⁻¹ = 1orP3

i=0(−ζq)ⁱ= 0. If(−ζq)³q⁻¹= 1thenq ∈G⁰₄,p6= 2andD=D₉₂. If0 =P3

i=0(−ζq)ⁱ= (1−ζq)(1 + (ζq)²), then(ζq)² = −1and p 6= 2sinceq 6= ζ⁻¹. Hence D=D₈₁.

Case 3b2. Ifp6= 3andb⁰= 4, then one gets(−ζq)⁴q⁻¹ = 1orP4

i=0(−ζq)ⁱ=

4.2. The proof of Theorem 4.1.1 39

0. If(−ζq)⁴q⁻¹ = 1thenζ = q⁻³. Sinceq /∈ G⁰₃, one obtainsq ∈ G⁰₉ and D = D_10,2. The equationP₄

i=0(−ζq)ⁱ = 0givesζ = −q⁵. Sinceζ ∈ G⁰₃, one gets−q ∈G⁰₃,ζ =−q⁻¹,p= 5or−q∈G⁰₁₅,p6= 3,5. If−q ∈G⁰₃ then D=D₁₆⁰⁰_,2and if−q ∈G⁰₁₅thenD=D_16,2.

Case 3b3. Ifp6= 3andb⁰ = 5, then one gets(−ζq)⁵q⁻¹ = 1orP5

i=0(−ζq)ⁱ = 0, −ζq 6= 1. If(−ζq)⁵q⁻¹ = 1 and p = 2 then qζ⁻¹ = 1, which is a contradiction to (4.2). If (−ζq)⁵q⁻¹ = 1 and p 6= 2 then ζ = −q⁴ and D =D_13,2. Consider0 =P5

i=0(−ζq)ⁱ = (1−ζq)(1 + (−ζq)²+ (−ζq)⁴). Sinceq /∈ {1,−1, ζ, ζ⁻¹}, one getsp 6= 2andq³ =−1. Since−ζq6= 1, one getsq =−ζanda^r₁₂^2(X)=−2, which is a contradiction.

Step 4. Again we use the same labeling as in steps 1 and 2. Assume that (c₀, c₁, c₂, c₃) = (1,3,1, c⁰), where c⁰ ∈ {3,4,5}. Then a^X₂₁ = −1 and a^X₁₂=−3. We distinguish four cases: 4aa, 4ab, 4ba and 4bb.

Case 4aa. Ifq113q0 = 1andq22q0 = 1, thenD=D_11,1.

Case 4ab. Setq :=q₁₁. Ifq₁₁³q₀= 1andq₂₂=−1, then the Dynkin diagrams ofX,r₁(X)andr₂r₁(X), respectively, are

e e

q q⁻³ -1

e e

q q⁻³ -1

e e

-q⁻² q³ -1 q ∈k^∗\ {1,−1}, q /∈G⁰₃.

Then−a^r₁₂^2r1(X) = c⁰ ∈ {3,4,5}. Hence we distinguish three cases: 4ab1, 4ab2 and 4ab3.

Case 4ab1. Ifc⁰ = 3, then(−q⁻²)³q³ = 1orP3

i=0(−q⁻²)ⁱ = 0,1−q⁻² 6= 0. If(−q⁻²)³q³ = 1then D = D_11,1, whereq³ = −1. IfP3

i=0(−q⁻²)ⁱ = 0, thenD=D_12,1.

Case 4ab2. Ifc⁰ = 4, then(−q⁻²)⁴q³ = 1orP4

i=0(−q⁻²)ⁱ = 0.

Case 4ab2a. The equation(−q⁻²)⁴q³ = 1gives q⁵ = 1andp 6= 5, since q6= 1. HenceD=D_14,1.

Case 4ab2b. Consider the equationP4

i=0(−q⁻²)ⁱ = 0. One getsq¹⁰=−1. If p = 2thenD= D_14,1. Ifp = 5thenq² =−1andD =D₁₅⁰_,1. Ifp 6= 2,5, thenq ∈G⁰₂₀andD=D_15,1.

Case 4ab3. Ifc⁰ = 5, then(−q⁻²)⁵q³ = 1orP5

i=0(−q⁻²)ⁱ = 0.

Case 4ab3a. Consider the equation(−q⁻²)⁵q³ = 1, which gives−q⁷ = 1.

Sinceq6=−1, one getsp6= 7and−q ∈G⁰₇. HenceD=D_17,1. Case 4ab3b. Consider the equation0 =P5

i=0(−q⁻²)ⁱ = (1−q⁻²)(1 +q⁻⁴+ q⁻⁸). Sinceq² 6= 1, one gets 1 +q⁻⁴ +q⁻⁸ = 0. If p = 3 thenq ∈ G⁰₄ sinceq² 6= 1. Hencea^r₁₂^2r1(X) = −2, which is a contradiction. Thenp 6= 3 andq⁻⁴ ∈ G⁰₃. Sincec⁰ = 5, one getsq ∈ G⁰₆ or q ∈ G⁰₁₂. Ifq ∈ G⁰₆, then a^r₁₂^2r1(X) = −3, which is a contradiction. Ifq ∈ G⁰₁₂, thena^r₁₂^2r1(X) = −2,

40 4. The classification result which is again a contradiction.

Case 4ba. The conditions1 +q₁₁+q₁₁²+q₁₁³ = 0,q₁₁ 6=−1andq₂₂q₀ = 1 hold. Then q₁₁ ∈ G⁰₄ andp 6= 2 since a^X₁₂ = −3. Set ζ := q₁₁ and q = q₂₂. The Dynkin diagram of r₁(X) is ^ζ_e ^{−q ζq}_e⁻²withζ ∈ G⁰₄ and q ∈k^∗\ {1,−1, ζ, ζ⁻¹}. Since−a^r₂₁^1(X) =c₂ = 1, one gets(ζq⁻²)(−q) = 1 orζq⁻² =−1. If(ζq⁻²)(−q) = 1thenq =−ζ, which is a contradiction. If ζq⁻² =−1thenq∈G⁰₈andD=D_12,3.

Case 4bb. Consider the last case: 1 +q₁₁+q²₁₁+q³₁₁ = 0,q₂₂ = −1and q116=−1. Thenq₁₁² =−1andp 6= 2. Setζ := q11andq =q0. The Dynkin diagram of r1(X) is ^ζ_e^-q−1-ζq_e³withq ∈ k^∗ \ {1,−1, ζ, ζ⁻¹} andζ ∈ G⁰₄. Since−a^r₂₁^1(X)=c2 = 1, one has(−q⁻¹)(−ζq³) = 1orζq³ = 1.

Case 4bb1. Ifζq³ = 1, thenζ =q⁻³ ∈ G⁰₄. Ifp= 3, thenζ =q, which is a contradiction. Hencep6= 3andD=τD₈₅.

Case 4bb2. The condition(−q⁻¹)(−ζq³) = 1holds. Thenζ = q⁻² ∈ G⁰₄. Henceq∈G⁰₈andD=D_12,2.

By checking all cases in Proposition 3.1.6, the proof of Theorem 4.1.1 is com-pleted.

(IIb)Consider the caseθ= 3.

Since the matrix(qij)i,j∈I is indecomposable, the finite Cartan graphC(M) is indecomposable. Set X = [M]3, q1 = q12q21, q2 = q23q32, andq3 = q₃₁q₁₃. We write the Cartan matrixA^X := (a_ij)i,j∈I. SinceC(M)is a finite indecomposable Cartan graph, by Theorem 3.2.6 we are free to assume that Cartan matrixA^X is one of the following cases:(a),(b)and(c).

Case(a). Assume that eitherXhas a goodA₃neighborhood orCis standard, A^X =A3for allX∈ X. Let

(a, b, c, d) := (−a^r₂₃^1(X),−a^r₁₃^2(X),−a^r₃₁^2(X),−a^r₂₁^3(X))

be the sequence in Definition 3.2.2. By applying Lemma 2.1.4 to a₁₃ = 0, a12 = −1, anda23 = −1, we getq3 = 1,q1 6= 1andq2 6= 1, respectively.

SinceA^X is ofA3 type, we distinguish subcasesa1,a2,a3,a4,a5,a6anda7

by Lemma 2.1.4.

Casea1. Consider the case(2)q11 = (2)q22 = (2)q33 = 0. Setq1 := q and q2:=r. Then we getq11=q22=q33=−1forp6= 2orq11=q22=q33= 1 forp= 2. Hence we distinguish subcasesa_1aanda_1b.

casea1a. Consider the caseq11 = q22 = q33 = −1,p 6= 2. Ifqr = 1and q 6= −1, thenD = D₈₂,p 6= 2. Ifqr = 1 andq = −1, thenCis standard

4.2. The proof of Theorem 4.1.1 41

andD = D₁₁, whereq = −1, andp 6= 2. Ifqr 6= 1, thenXhas a goodA3

neighborhood and by Lemma 2.1.7 we get the reflectionr₂(X)ofX

e u e

−1 q −1 r −1

⇒ ^{e e}

e

A AA

q

−1

r q⁻¹ r⁻¹

qr

withq6= 1,r6= 1,qr6= 1, andp6= 2. SinceXhas a goodA₃neighborhood, we obtain thata^r₁₃^2(X) = −1anda^r₃₁^2(X) ∈ {−1,−2}. Then we distinguish subcasesa1a1anda1a2.

Casea_1a1. Consider the casea^r₁₃^2(X) =a^r₃₁^2(X) = −1. We obtain(2)_q(q²r− 1) = 0and(2)_r(qr²−1) = 0by Lemma 2.1.4. Ifq =−1, thenr 6= −1by qr 6= 1. Ifq = −1andqr² −1 = 0, thenr ∈G⁰₄ and henceD =τ321D₆₂, whereq ∈ G⁰₄ andp 6= 2. If q²r = 1andr = −1, then D = D₆₂, where q ∈ G⁰₄ andp 6= 2. Ifq²r = qr² = 1, thenq = r ∈ G⁰₃,p 6= 3and hence D=D_15,2,p6= 2,3.

Casea1a2. Consider the casea^r₁₃^2(X)=−1anda^r₃₁^2(X)=−2. Then(2)q(q²r− 1) = 0,(3)r(qr³−1) = 0and(2)r(qr²−1)6= 0. Hencer 6=−1andqr² 6= 1. Ifq = −1,r ∈ G⁰₃ andp 6= 3, then D = D_17,1, p 6= 2,3. Ifq = −1and qr³ = 1, thenr ∈ G⁰₆,p 6= 3 and henceD = τ321D₇₂, whereq ∈ G⁰₆ and p 6= 2,3. Ifq²r = r³ = 1, then−q = r ∈ G⁰₃,p 6= 3 byqr² 6= 1. Hence D=τ₃₂₁D_17,3,p6= 2,3. Ifq²r=qr³ = 1, thenr² =q, wherer∈G⁰₅,p6= 5. Hence by Lemmas 2.1.4 and 2.1.7 we get the sequence(a, b, c, d) = (2,1,2,3). Furthera^r₂₁^3r1(X)=−4, which is a contradiction.

casea_1b. Consider the caseq11 = q22 = q33 = 1,p = 2. Ifqr = 1, then D=D₈₂,p= 2. Ifqr 6= 1, then the Dynkin diagrams ofXandr₂(X)are

e u e

1 q 1 r 1

⇒ ^{e e}

e

A AA

q 1

r q⁻¹ r⁻¹

qr

withq6= 1,r6= 1,qr6= 1, andp= 2. SinceXhas a goodA3neighborhood, we obtain thata^r₁₃^2(X)=−1anda^r₃₁^2(X)∈ {−1,−2}. Then we have subcases a_1b1 anda_1b2.

Casea_1b1. Consider the casea^r₁₃^2(X) =a^r₃₁^2(X) =−1. Sinceq 6= 1andr 6= 1, we obtain q²r = qr² = 1by Lemma 2.1.4. Thenq = r ∈ G⁰₃ and hence D=D_15,2,p= 2.

Casea_1b2. Consider the case a^r₁₃^2(X) = −1 anda^r₃₁^2(X) = −2. Then we get

42 4. The classification result

q²r = 1,(3)r(qr³−1) = 0, andqr² 6= 1. Ifq²r =r³ = 1, thenq =r ∈G⁰₃ and henceqr² = 1, which is a contradiction. Then we obtainq²r=qr³ = 1, thenr² =q, wherer ∈G⁰₅,p= 2. Hence by Lemmas 2.1.4 and 2.1.7 we get the sequence(a, b, c, d) = (2,1,2,3)anda^r₂₁^3r1(X) =−4in Definition 3.2.2, which is again a contradiction.

Casea₂. Consider the case(2)_q22 = (2)_q33 = 0,q₁₁q₁−1 = 0, and(2)_q11 6= 0. Set q11 := q andq2 := r. Then we getq 6= −1for p 6= 2 andq 6= 1 for p = 2. Hence we obtainq₂₂ = q₃₃ = −1,qq₁ = 1,q 6= −1for p 6= 2or q₂₂ = q₃₃ = 1,qq₁ = 1,q 6= 1forp = 2. We distinguish two subcases a_2a anda2b.

Casea_2aConsider the caseq₂₂ =q₃₃ =−1,qq₁ = 1,q 6=−1, andp 6= 2. If r = q 6=−1, thenCis standard andD =τ₃₂₁D₄₂,p 6= 2. Ifr =−1,q 6=r, andq /∈ G⁰₄, thenD =D₉₁, wherer = −1,q /∈ G⁰₄ andp 6= 2. Ifr = −1, q 6= r, andq ∈ G⁰₄, then D = D_10,2, whereq ∈ G⁰₄ andp 6= 2. Now we consider the caser 6= −1andq 6= r. ThenX has a goodA₃ neighborhood and by Lemma 2.1.7 we get the following reflectionr2(X)ofX

e u e

q q⁻¹ −1 r −1

⇒ ^{e e}

e

A AA

−1

−1

r q r⁻¹

rq⁻¹

withq6= 1,r /∈ {−1,1, q}, andp6= 2. SinceXhas a goodA₃neighborhood, we get a^r₃₁^2(X) ∈ {−1,−2}. Then we get either (2)r(r²q⁻¹ − 1) = 0 or (3)r(r³q⁻¹−1) = 0,(2)r(r²q⁻¹−1)6= 0. Hence we can distinguish three subcasesa_2a1,a_2a2 anda_2a3.

Casea2a1. Consider the case(2)r(r²q⁻¹−1) = 0. Sincer 6=−1andp 6= 2, we obtain thatr²q⁻¹ = 1and henceD=τ321D₆₂,p6= 2.

Casea_2a2. Consider the caser³q⁻¹ = 1and(2)_r(r²q⁻¹−1)6= 0. Then we getq=r³6= 1andr /∈ {1,−1}. HenceD=τ₃₂₁D₇₂,p6= 2.

Casea2a3. Consider the caser ∈G⁰₃,(2)r(r²q⁻¹−1)6= 0, andp 6= 3. Then qr6= 1. By Lemma 2.1.7 we get the following reflectionr₃(X)ofX

e e u

q q⁻¹ −1 r −1

⇒ ^q_e ^{q−1 r}_e ^{r−1 −1}_e

withq 6= 1,r /∈ {−1,1, q, q⁻¹}. By Lemmas 2.1.4 and 2.1.7 we obtain that (a, b, c, d) = (1,1,2,2), which is a contradiction.

Casea2b. Consider the caseq22=q33= 1,qq1= 1,q 6= 1forp= 2. Ifr=q, thenD =τ321D₄₂,p = 2. Ifr 6= q, then we geta^r₃₁^2(X) ∈ {−1,−2}. Hence

4.2. The proof of Theorem 4.1.1 43

we distinguish subcasesa2b1,a2b2, anda2b3.

Casea2b1. Ifa^r₃₁^2(X)=−1, thenr²q⁻¹= 1and henceD=τ321D₆₂,p= 2.

Casea_2b2. Consider the casea^r₃₁^2(X)=−2. Ifr∈G⁰₃, then we get(a, b, c, d) = (1,1,2,2), which is a contradiction. Ifr³q⁻¹ = 1, thenq =r³6= 1and hence D=τ₃₂₁D₇₂,p= 2.

Casea₃. Consider the case(2)_q11 = (2)_q22 = 0,q₃₃q₂−1 = 0, and(2)_q33 6= 0. Letq33 := q andq1 := r. Then by(2)q 6= 0we getq 6= −1 forp 6= 2and q6= 1forp= 2. Hence we obtainq₁₁=q₂₂=−1,qq₂= 1,q6=−1forp6= 2 orq₁₁=q₂₂= 1,qq₂ = 1,q6= 1forp= 2. We distinguish two subcasesa_3a anda3b.

Casea_3aConsider the caseq₁₁ =q₂₂ =−1,qq₂ = 1,q 6=−1, andp 6= 2. If r =q 6=−1, thenCis standard andD =D₄₂,p6= 2. Ifr =−1,q 6=r, and q /∈ G⁰₄, thenD =τ321D₉₁, wherer = −1,q /∈ G⁰₄ andp 6= 2. If r = −1, q 6=r, andq ∈G⁰₄, thenD =τ321D_10,2, whereq ∈ G⁰₄andp 6= 2. Now we consider the caser 6=−1andq 6= r. ThenXhas a goodA₃ neighborhood and by Lemma 2.1.7 we get the following reflectionr2(X)ofX

e u e

−1 r −1 q⁻¹ q ⇒ ^{e e}

e A AA

r

−1

−1 r⁻¹ q

rq⁻¹

withq 6= 1,r /∈ {−1,1, q}, andp6= 2. SinceXhas a goodA3neighborhood, we geta^r₃₁^2(X)=−1and hencer²q⁻¹ = 1. ThenD=D₆₂,p6= 2.

Casea_3b. Consider the caseq₁₁=q₂₂= 1,qq₂ = 1,q6= 1forp= 2. Ifr=q, thenD =D₄₂,p= 2. Ifr 6=q, then we geta^r₃₁^2(X) =−1. Hencer²q⁻¹ = 1 andD=D₆₂,p= 2.

Casea₄. Consider the case(2)_q11 = (2)_q33 = 0,q₂₂q₁−1 = q₂₂q₂−1 = 0, and(2)q22 6= 0. Letq22:=q. Then by(2)q 6= 0we getq6=−1forp6= 2and q 6= 1forp= 2. Hence we obtainq11 =q33 =−1,qq1 = qq2 = 1,q 6=−1 forp6= 2orq₁₁=q₃₃= 1,qq₁ =qq₂ = 1,q 6= 1forp= 2. Ifp6= 2, thenC is standard andD=D₈₂,p6= 2. Ifp= 2, thenCis standard andD =D₈₂, p= 2.

Casea₅. Consider the case(2)_q11 = 0,q₂₂q₁−1 =q₂₂q₂−1 =q₃₃q₂−1 = 0, (2)q22 6= 0, and(2)q33 6= 0. Setq33 := q. If q11 = 1andp = 2, thenC is standard andD = D₄₁,p = 2. Ifq11 = −1andp 6= 2, thenq 6= −1,C is standard, andD=D₄₁,p6= 2.

Casea6. Consider the case(2)q33 = 0,q11q1−1 =q22q1−1 =q22q2−1 = 0, (2)q11 6= 0, and(2)q22 6= 0. Setq11:=q. Thenq6=−1. Ifq33= 1andp= 2,

44 4. The classification result thenC is standard andD = τ321D₄₁,p = 2. Ifq33 = −1andp 6= 2, then q 6=−1andD=τ₃₂₁D₄₁,p6= 2.

Casea₇. Consider the case(2)_q22 = 0,q₁₁q₁ = q₃₃q₂ = 1,(2)_q11 6= 0, and (2)q33 6= 0. Setq11:=qandq33:=r. We distinguish subcasesa7aanda7b. Casea7a. Consider the caseq22=−1,qq1 =rq2 = 1, andp6= 2. Ifqr= 1, thenq 6= −1andD= D₈₁,p 6= 2. Ifqr 6= 1,q = r ∈ G⁰₃, andp 6= 3, then D = D_11,1,p 6= 2,3. Ifqr 6= 1,q = r /∈ G⁰₃ andp 6= 3, thenD = D_10,1, p 6= 2,3. If qr 6= 1,q 6= r,qr² 6= 1, andrq² 6= 1, then D = D₉₁,p 6= 2. Ifqr 6= 1,q 6= r, and rq² = 1, thenq /∈ G⁰₃ for p 6= 3. HenceD = D_10,2, p 6= 2. Ifqr 6= 1,q 6=r, andqr² = 1, thenr /∈G⁰₃forp6= 3,r² 6= 1. Hence D=τ321D_10,2,p6= 2.

Casea_7b. Consider the caseq₂₂ = 1,qq₁ = rq₂ = 1, andp = 2. Ifqr = 1, thenD=D₈₁,p = 2. Ifqr 6= 1andq =r ∈G⁰₃, thenD=D_11,1,p = 2. If qr 6= 1andq =r /∈ G⁰₃, thenD =D_10,1,p = 2. Ifqr 6= 1,q 6=r,qr² 6= 1, andrq² 6= 1, then D = D₉₁,p = 2. Ifqr 6= 1,q 6= r, andrq² = 1, then q /∈G⁰₃andD=D_10,2,p = 2. Ifqr 6= 1,q 6=r, andqr² = 1, thenr /∈G⁰₃, andD=τ321D_10,2,p= 2.

Casea₈. Consider the caseq₁₁q₁ =q₂₂q₁ =q₂₂q₂ =q₃₃q₂ = 1,(2)_q11 6= 0, (2)q22 6= 0, and(2)q33 6= 0. ThenD=D₁₁,q6=−1.

Case(b). Assume that eitherXhas a goodB3neighborhood orCis standard, A^X = B₃ for allX ∈ X. SinceA^X is ofB₃ type, we obtain thatq₃ = 1, q₁ 6= 1and q₂ 6= 1 by Lemma 2.1.4. By Lemma 2.1.7 we get q₁q₂ = 1 if (2)_q22 = 0, sincea^r₁₃^2(X) = 0. Further, we distinguish subcasesb₁,b₂,b₃,b₄, b5,b6,b7andb8.

Caseb₁. Consider the case(2)_q11 = (2)_q22 = (3)_q33 = 0and(2)_q33(q₃₃q₂− 1) 6= 0. Letq₁ := q. Thenq₂ = q⁻¹ andq₃₃ 6= q. We distinguish subcases b1a,b1bandb1c.

caseb_1a. Consider the caseq₁₁ =q₂₂ =−1,q₃₃:=ζ ∈G⁰₃, andp6= 2,3. By Lemma 2.1.7 we get the reflectionr₃(X)ofX

e e u

−1 q −1 q⁻¹ ζ

⇒ ⁻¹_e ^{q -ζq}_e^{−2qζ−1 ζ}_e

withζ ∈ G⁰₃,q 6= ζ, andp 6= 2,3. Sincea^r₂₃^3(X) = −1by Definition 3.2.3, we get((−ζq⁻²)qζ⁻¹−1)(ζq⁻²−1) = 0and henceq = −1,q = ζ⁻¹, or q = −ζ⁻¹. Ifq = ζ⁻¹, then C is standard and D = D₅₂, where q ∈ G⁰₃, p6= 2,3. Ifq=−ζ⁻¹, thenCis standard andD=D_14,2,p6= 2,3. Ifq=−1, thena^r₂₁^3(X) =−3, which is a contradiction.

Caseb_1b. Consider the case q11 = q22 = −1, q33 = 1, andp = 3. Then

4.2. The proof of Theorem 4.1.1 45

the Dynkin diagram ofr3(X)is ⁻¹_e ^{q -q−2}_e ^{q 1}_e. Bya^r₂₃^3(X) =−1we get q=−1. HenceCis standard andD=D₁₂⁰_,1.

Caseb_1c. Consider the caseq₁₁ =q₂₂ = 1,q₃₃:= ζ ∈G⁰₃, andp= 2. Then the Dynkin diagram ofr3(X)is ¹_e ^{q ζq−2}_e^{qζ−1 ζ}_ewithζ ∈G⁰₃,q 6= ζ, and p = 2. The conditiona^r₂₃^3(X) =−1givesq = ζ⁻¹. HenceCis standard and D=D₅₂, whereq ∈G⁰₃andp= 2.

Caseb₂. Consider the case(2)_q11 = (2)_q22 = 0,q₃₃² q₂−1 = 0,(3)_q33 6= 0, and(2)_q33(q₃₃q₂−1)6= 0. Letq₃₃:= q. Thenq₂ =q⁻² and henceq₁ =q² byq1q2 = 1. Hence we distinguish two casesb2aandb2b.

caseb2a. Consider the caseq11=q22=−1andp6= 2. Ifq² 6=−1andp6= 3, thenC is standard andD = D₅₂, where q /∈ G⁰₃,p 6= 2,3. Ifq² 6= −1and p = 3, thenD= D₅₂,p = 3. Ifq² =−1, thenCis standard andD =D₂₁, whereq∈G⁰₄,p6= 2.

Caseb_2b. Consider the caseq₁₁=q₂₂= 1andp= 2. ThenCis standard and D=D₅₂, whereq /∈G⁰₃,p= 2.

Caseb3. Consider the case(2)q11 = (3)q33 = 0,q22q1−1 = 0,q22q2−1 = 0, (2)_q22 6= 0, and(2)_q33(q₃₃q₂ −1) 6= 0. Letq₂₂ := q andq₃₃ := ζ. Then q1 =q2 =q⁻¹andq 6=ζ. We distinguish subcasesb3a,b3b, andb3c.

Caseb3a. Consider the caseq11 =−1,ζ ∈G⁰₃ andp6= 2,3. By Lemma 2.1.7 we get the Dynkin diagrams ofXandr₃(X)

e e u

−1 q⁻¹ q q⁻¹ ζ ⇒ ⁻¹_e ^q−1ζq−1_e^{qζ−1 ζ}_e

withζ ∈ G⁰₃,q /∈ {−1,1, ζ}, andp 6= 2,3. We geta^r₂₁^3(X) ∈ {−1,−2}by Definition 3.2.3. Hence((ζq⁻¹) + 1)(ζq⁻²−1)(q⁻³−1)(ζ²q⁻³−1) = 0. Then we obtainq ∈ {ζ⁻¹,−ζ⁻¹,−ζ}or q⁻³ = ζ. If q = ζ⁻¹, then D = D₅₁, where q ∈ G⁰₃,p 6= 2,3. Ifq = −ζ⁻¹, thenC is standard and D = D_14,1,p 6= 2,3. Ifq = −ζ, then by Lemmas 2.1.4 and 2.1.7 we geta^r₂₃^1r3(X) a^r₂₃^1r3(X)=−3∈ {−1,/ −2}, which is a contradiction. Ifq⁻³ =ζ ∈G⁰₃, then a^r₂₃^1r3(X)=−8∈ {−1,/ −2}, which is again a contradiction.

Caseb_3b. Consider the caseq₁₁= 1,ζ ∈G⁰₃andp = 2. By Lemma 2.1.7 we get the reflectionr3(X)ofXis

e e u

1 q⁻¹ q q⁻¹ ζ ⇒ ¹_e ^q−1ζq−1_e^{qζ−1 ζ}_e

46 4. The classification result

withζ ∈ G⁰₃,q 6= ζ, and p = 2. We getq = ζ⁻¹ orq⁻³ = ζ by a^r₂₁^3(X) ∈ {−1,−2}. Ifq=ζ⁻¹, thenCis standard andD=D₅₁, whereq∈G⁰₃,p= 2. Ifq⁻³ = ζ ∈ G⁰₃, then by Lemmas 2.1.4 and 2.1.7 we geta^r₂₃^1r3(X) = −8 ∈/ {−1,−2}, which is a contradiction.

Caseb_3c. Consider the caseq₁₁=−1,ζ = 1andp = 3. By Lemma 2.1.7 the Dynkin diagrams ofr3(X)is ⁻¹_e ^{q−1 q−1}_e ^{q 1}_e withq /∈ G⁰₂, andp = 3. Thena^r₂₁^3(X)∈ {−1,/ −2}, which is a contradiction.

Caseb₄. Consider the case(2)_q22 = (3)_q33 = 0,q₁₁q₁−1 = 0,(2)_q11 6= 0, and(2)q33(q33q2 −1)6= 0. Letq11 := qandq33 := ζ. Thenq1 = q⁻¹ and q2 =qbyq1q2 = 1. Henceq 6=ζ⁻¹. We distinguish three subcasesb4a,b_4b, andb_4c.

caseb4a. Consider the caseq22=−1,ζ ∈G⁰₃,qq1−1 = 0, andp6= 2,3. The Dynkin diagrams ofXandr3(X)are

e e u

q q⁻¹ −1 q ζ ⇒ ^q_e ^q−1-ζq_e^2(qζ)−1ζ_e

with ζ ∈ G⁰₃, q /∈ {1,−1, ζ⁻¹}, and p 6= 2,3. We get a^r₂₃^3(X) = −1 by Definition 3.2.3. Then (ζq² −1)(ζq²(ζq)⁻¹ + 1) = 0. Hence q = −ζ or q =ζ. Ifq =−ζthenCis standard andD=D_14,3,p6= 2,3. Ifq=ζthenC is standard andD=D₅₃, whereq :=−ζ⁻¹,ζ ∈G⁰₃, andp6= 2,3.

Caseb_4b. Consider the caseq₂₂ = −1,ζ = 1, andp = 3. Then the Dynkin diagram ofr3(X)is ^q_e ^{q−1 -q}_e^{2 q−1 1}_ewithq /∈G⁰₂,p = 3. Thena^r₂₃^3(X) 6=

−1, which is a contradiction.

Caseb4c. Consider the caseq22 = 1,q33 := ζ ∈ G⁰₃, and p = 2. Then the Dynkin diagram ofr3(X)is ^q_e ^{q−1 ζq}_e^{2qζ−1 ζ}_e. The conditiona^r₂₃^3(X) =−1 gives thatq=ζ. ThenCis standard andD=D₅₃, whereq :=ζ⁻¹,ζ ∈G⁰₃, andp= 2.

Caseb₅. Consider the case(2)_q11 = 0,q₂₂q₁ = 1,q₂₂q₂ = 1,q₃₃² q₂ = 1, (2)q22 6= 0, and(3)q33 6= 0. Letq33:=q. Thenq2 =q⁻²,q22=q²and hence q1 =q⁻². We obtain thatq² 6= 1,−1, andq³ 6= 1. Ifp = 3andq11 = −1, then C is standard andD = D₅₁,p = 3. Ifp 6= 2,3andq₁₁ = −1, then D =D₅₁, whereq /∈G⁰₃andp 6= 2,3. Ifp = 2andq11= 1, thenD=D₅₁, whereq /∈G⁰₃andp= 2.

Caseb₆. Consider the case(2)_q22 = 0,q₁₁q₁ = 1,q₃₃² q₂ = 1,(2)_q11 6= 0, and (3)q33 6= 0. Letq33 := q. Then q2 = q⁻² andq1 = q² byq1q2 = 1. Hence q11 = q⁻². Ifp = 3andq22 = −1, thenD = D₅₃,p = 3. Ifp 6= 2,3and

4.2. The proof of Theorem 4.1.1 47

q22 =−1, thenD=D₅₃, whereq /∈G⁰₃andp 6= 2,3. Ifp= 2andq22= 1, thenD=D₅₃, whereq /∈G⁰₃ andp= 2.

Caseb₇. Consider the case (3)_q33 = 0, q₁₁q₁ = 1, q₂₂q₁ = 1, q₂₂q₂ = 1, (2)q11 6= 0,(2)q22 6= 0and(2)q33(q33q2−1)6= 0. Letq22:= qandq33:= ζ.

Thenq2 =q1 =q⁻¹,q11=q, andq 6=ζ. Hence we distinguish two subcases b_7aandb_7b.

Caseb7a. Consider the caseζ ∈ G⁰₃ andp 6= 3. Then by Lemma 2.1.7 the reflectionr3(X)ofXis

e e u

q q⁻¹ q q⁻¹ ζ

⇒ ^q_e ^q−1ζq−1_e^{qζ−1 ζ}_e

withζ ∈ G⁰₃,q /∈ {1,−1, ζ}, andp 6= 3. We geta^r₂₁^3(X) ∈ {−1,−2} from Definition3.2.3. Hence((ζq⁻¹)+1)(ζq⁻²−1)(q⁻³−1)(ζ²q⁻³−1) = 0. Then q∈ {ζ⁻¹,−ζ⁻¹,−ζ}orq⁻³ =ζ. Ifq=ζ⁻¹, thenCis standard andD=D₂₁, whereq ∈ G⁰₃,p 6= 3. Ifq = −ζ⁻¹, thenC is standard and D = D_12,1. If q=−ζ, thenXhas a goodB3neighborhood andD=τ321D_16,5. Ifq⁻³ =ζ, thenD=D_18,1.

Caseb7b. Consider the casep = 3andζ = 1. Since(2)q 6= 0andq 6= 1, we geta^r₂₁^3(X)∈ {−1,/ −2}, which is a contradiction.

Caseb8. Consider the caseq11q1 =q22q1 =q22q2 =q²₃₃q2 = 1,(2)q11 6= 0, (2)_q22 6= 0, and(3)_q33 6= 0. If p 6= 2,3, then C is standard andD = D₂₁, whereq /∈G⁰₃∪G⁰₄andp6= 2,3. Ifp= 3, thenD=D₂₁, whereq /∈G⁰₄and p= 3. Ifp= 2, thenD=D₂₁, whereq /∈G⁰₃andp= 2.

Case(c). Assume that eitherA^Xhas a goodC₃neighborhood orCis standard, A^X = C₃ for allX ∈ X. Since A^X is ofC₃ type, we obtain that q₃ = 1, q1 6= 1andq2 6= 1by Lemma 2.1.4. Sincea23 = −2, we get(2)q22 6= 0and henceq22q1 = 0by a21 = −1. Henceq22 6= 1. Sincea^r₂₃^1(X) = −2, we get (2)q11 6= 0and henceq11q1−1 = 0bya12=−1. If(3)q22 = 0, thenq1q2= 1 by Lemma 2.1.4. Hence, by Lemma 2.1.4 we distinguish the following cases c1,c2,c3, andc4.

Casec1. Consider the caseq11q1 =q22q1 =q33q2 =q₂₂² q2 = 1. Letq22:=q. Thenq²6= 1andCis standard and henceD=D₃₁.

Casec2. Consider the caseq11q1−1 =q22q1−1 =q33q2−1 = 0,(3)q22 = 0, andq222q2−1 6= 0. Sinceq22 6= 1, we getq22 :=ζ ∈ G⁰₃ andp 6= 3. Then we getq₁=ζ⁻¹,q₂=ζand henceq₂₂²q₂−1 = 0, which is a contradiction.

Casec3. Consider the caseq11q1−1 =q22q1−1 =q₂₂² q2−1 = 0,(2)q33 = 0, andq33q2 −1 6= 0. Letq11 := q. Ifp 6= 2, then by Lemma 2.1.7 we get the

48 4. The classification result

reflection ofX

e e u

q q⁻¹ q q⁻² −1

⇒ ^q_e ^q−1-q−1_e ^{q2 −1}_e

withq /∈G⁰₂∪G⁰₄. We obtaina^r₂₁^3(X)∈ {−1,−2}from Definition 3.2.4. Since q² ∈/ G⁰₂∪G⁰₄, we geta^r₂₁^3(X) = −2and hence q⁻³ = −1or q⁻³ = 1. If q⁻³ = −1andp 6= 3, thenC is standard andD = D_13,1, withζ ∈ G⁰₆. If q⁻³ = 1andp 6= 3, thenD = D_13,1, wíthζ ∈ G⁰₃. Ifp = 2, then we get q⁻³ = 1bya^r₂₁^3(X)∈ {−1,−2}and thenD=D₁₃⁰_,1.

Casec4. Consider the caseq11q1−1 = q22q1−1 = 0,(2)q33 = (3)q22 = 0, q²₂₂q₂ −1 6= 0, andq₃₃q₂−1 6= 0. Since (3)_q22 = 0andq₂₂ 6= 1, we get q22 := ζ ∈ G⁰₃, q2 = ζ, andp 6= 3. Henceq²₂₂q2 −1 = 0, which is a contradiction.

Chapter 5

Generalized Dynkin diagrams

In this section, we give all the Nichols algebras of diagonal type of rank 2 and rank 3 with finite root systems as well as their exchange graphs.

Im Dokument Finite dimensional Nichols algebras of diagonal type over fields of positive characteristic (Seite 50-65)