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In this paper, we provide another non-linear example by using the method of [21] to compute a spine for SU(2, 1; Z[i]). Sections 2 and 3 set notation and define the exhaustion functions that are used to describe the pieces of the spine. In Section 4, we classify certain of configurations of isotropic line in C3. We show the spine has the structure of a cell complex with cells related to these configurations in Section 6. Explicit Γ-representatives of cells are fixed, and their stabilizers are computed in Section 8. After subdivision, we obtain a regular cell complex for D0 on which Γ acts cellularly. In Section 10, we recall some facts about orbifolds and develop machinery to investigate the cohomology of Γ. The results of Section 10 hold in more generality, and may be of independent interest. We apply these methods in Section 11 to Γ = SU(2, 1; Z[i]) to compute the cohomology of Γ with coefficients in various Γ-modules. I would like to thank my thesis advisor, Les Saper for his insight into this work. I would also like to thank Paul Gunnells for helpful conversations.
2. Preliminaries Let G be the identity component of the real points of the algebraic group G = SU(2, 1), realized explicitly as 0 0 i 0 0 i SU(2, 1) = g SL(3, C) g∗ 0 1 0 g = 0 1 0 . ∈ − − i 0 0 i 0 0 − − Alternatively, let be the (2,1)-quadratic form on C3 defined by Q 0 0 i (u, v)= u∗ 0 1 0 v. Q i −0 0 − Then G is the group of determinant 1 complex linear transformations of C3 that preserve . Let Γ be the arithmetic subgroup Γ = SU(2, 1) SL3(Z[i]). Let θ Qdenote the Cartan involution given by inverse conjugate∩ transpose and let K be the fixed points under θ. Let D = G/K be the associated Riemannian symmetric space of non-compact type. Let denote the set of (proper) rational parabolic subgroups of G. P Let P G be the rational parabolic subgroup of upper triangular matrices, and 0 ⊂ fix subgroups N0, A0, and M0: 2 2 yζ βζ− ζ r + i β /2 /y | | P = 0 ζ 2 iβζ/y ζ,β C, ζ =1, r R, y R , 0 − ∈ | | ∈ ∈ >0 0 0 ζ/y 1 β r + i β 2/2 | | N = 0 1 iβ β C, r R , 0 ∈ ∈ 00 1 y 0 0 A = 01 0 y R , 0 ∈ >0 0 0 1/y ζ 0 0 2 M = 0 ζ− 0 ζ C, ζ =1 . 0 ∈ | | 0 0 ζ
EXPLICIT REDUCTION FOR SU(2, 1; Z[i]) 3
P0 acts transitively on D, and every point z D can be written as p x0 for some p P . Using Langlands decomposition, there∈ exists u N ,a A ,· and m M ∈ 0 ∈ 0 ∈ 0 ∈ 0 such that p = uam. Since M0 K, z can be written as ua x0. Denote such a point z = (y,β,r). ⊂ · Zink showed that Γ has class number 1 [22]. Thus Γ G(Q)/P0(Q) consists of a single point, and all the parabolic subgroups of G are\ Γ-conjugate. The rational parabolic subgroups of G are parametrized by the maximal isotropic subspaces of C3 which they stabilize. These are 1-dimensional, and so to each P , there is 3 ∈t P 3 an associated reduced, isotropic vector vP Z[i] . (A vector (n,p,q) Z[i] is reduced if (n,p,q) generate Z[i] as an ideal.)∈ Similarly, given a reduced,∈ isotropic 3 vector v in Z[i] , there is an associated rational parabolic subgroup Pv. Notice, however, that vP is only well-defined up to scaling by Z[i]∗ = 1, i . Thus, the {± ± }γ vectors v and εv will be treated interchangeably for ε Z[i]∗. If P = Q for some γ Γ, then v = γv . ∈ ∈ P Q Unless explicitly mentioned otherwise, the vector vP will be written as vP = (n,p,q)t. The isotropic condition (v , v ) = 0 implies that Q P P (1) p 2 = 2 Im(nq). | | In particular, q = 0 for P = P0. Furthermore, since there are no isotropic 2-planes in C3, 6 6 (2) (v , v ) =0 for P = Q. Q P Q 6 6 Because these elements of Γ will be used frequently, set once and for all 0 0 1 1 1+ i i 1 i(1 + i) i w = 01− 0 , σ = 0 1 1+ i , σˇ = 0 1 i(1 + i) , 10 0 00 1 00− 1 1 0 1 i 0 0 τ = 0 1 0 ,ǫ = 0 1 0 , and ξ = τwτσwǫ3. 0 0 1 0− 0 i Note thatσ ˇ is contained in the group generated by ǫ,w,σ . In particular,σ ˇ = 1 { } wσǫwσ− w.
3. Construction of the spine In this section we briefly describe the construction of a Γ-invariant, 3-dimensional cell complex which is a deformation retract of D. This construction is described for the general Q-rank 1 case in [21]. We first define an exhaustion function fP for every rational parabolic subgroup P G. These exhaustion functions are then used to define a decomposition of D into⊆ sets D( ) for . I I⊂P 3.1. Exhaustion functions. Let z = (y,β,r) D and P a rational parabolic subgroup of G with associated isotropic vector∈ (n,p,q)t. Then the exhaustion function fP can be written as (3) f (z) f (z)= y 0 ≡ P0 y (4) fP (z)= 1/2 . n βp + i β 2/2 r q 2 + y2 p iβq 2 + y4 q 2 | − | | − | | − | | | 4 DAN YASAKI
The family of exhaustion functions defined above is Γ-invariant in the sense that 1 (5) fγ (z)= f (γ− z) for γ Γ. P P · ∈ 3.2. Admissible sets. For a parabolic P , define D(P ) D to be the set of z D such that f (z) f (z) for every Q P . More generally,⊂ for a subset ∈ , P ≥ Q ∈P\{ } I⊆P (6) E( )= z D f (z)= f (z) for every pair P,Q I { ∈ | P Q ∈ I} (7) D( )= D(P ) I P \∈I (8) D′( )= D( ) D( ′). I I \ I ′) I[I It follows that D′( ) D( ) E( ) and D( )= ˜ D′(˜). Let f denote the I ⊆ I ⊂ I I I I restriction to E( ) of f for P . I⊇I I P ∈ I ` Definition 3.1. Let , P , and z E( ). Then z is called a first contact for if f (z) is a globalI⊆P maximum∈ I of f on∈E( I). I I I I Definition 3.2. A subset is called admissible if D( ) is non-empty and I ⊂P I strongly admissible if D′( ) is non-empty. I Let D D denote the subset 0 ⊂ (9) D = D′( ). 0 I >1 |I|a The deformation is defined separately on each D(P ) for P . For z D(P ), we ∈P ∈ use the (negative) gradient flow of fP to flow z to a point on D0. This corresponds to using the geodesic action [9] of AP on z [21].
Let fD0 denote the function on D0 given by
(10) fD0 (z)= f (z) for z D( ). I ∈ I 3.3. First contact points. Given the explicit description of the exhaustion func- tions in coordinates, one can readily describe the set E( P0, P ). Writing z = (y,β,r) and using (3), { } 2 f (z) 2 2 (11) 0 = n βp + i β 2/2 r q + y2 p iβq + y4 q 2. f (z) − | | − − | | P Proposition 3.3. Let P be a rational parabolic subgroup of G with associated isotropic vector (n,p,q)t. Every z = (y,β,r) E( P , P ) satisfies ∈ { 0 } 1 p 2 1 n βp 2 y2 = iβ¯ + Re − r −2 q − s q 2 − q − | | 2 f0(z) Proof. Note that for z E( P0 , P ), = 1. Then the result follows from ∈ { } fP (z) (11) using the quadratic formula to solve fory2, and simplifying the result using the isotropic condition (1).
Proposition 3.3 allows us to easily calculate the first contact point for P0, P . The Γ-invariance of the exhaustion functions allows us to translate this for{ general} P,Q . { } EXPLICIT REDUCTION FOR SU(2, 1; Z[i]) 5
Proposition 3.4. Let = P,Q . Let z be a first contact point for . Then I { }⊂P I 1 fP (z)= fQ(z)= . (v , v ) |Q P Q | In particular, the first contact for P , P pis z = 1/ q ,i(p/q), Re(n/q) . { 0 } | | p 4. Configurations of vectors Definition 4.1. Let be a subset of vectors in C3. Then is said to be c-bounded if J J (u, v) 2 c for every u and v in . |Q | ≤ J Set once and for all the following sets i of isotropic vectors in C3. Jj 1 0 1 i 2 = 0 , 0 2 = 0 , 1+ i J1 J2 0 1 0 1+ i 1 i 1+ i 3 = 2 0 3 = 2 1+ i 3 = 2 1+ i J1 J1 ∪ J2 J1 ∪ J3 J1 ∪ 1 1 1 i 1 − 4 = 3 1+ i 4 = 3 1+ i J1 J1 ∪ J2 J3 ∪ − 1 1+ i 1+ i 5 = 4 1+ i J J1 ∪ 1 1 0 1 1+ i 1+i i 2i i − − 8 = 0 , 0 , 1+ i , 1+ i , 1 i , 1+ i , 2 , 2 . J − 0 1 1+ i 1 1 1+ i 1 2 8 One easily checks that these sets are not Γ-conjugate, and is 4-bounded, while the other sets are 2-bounded. J Proposition 4.2. A 2-bounded set of reduced, integral, isotropic vectors is Γ- conjugate to exactly one of 2, 2, 3, 3, 3, 4, 4, or 5. J1 J2 J1 J2 J3 J1 J2 J Proof. Let denote such a subset of order 2. Since G has class number 1, we can assume thatJ one of the vectors of is (1, 0, 0)t. Let v = (n,p,q)t be the other vector of . Let σ, σ,ˇ τ, and ǫ be definedJ as in Section 2. These elements preserve t J (1, 0, 0) (up to scaling by Z[i]∗). By applying powers of σ andσ ˇ to v, we can add any Z-linear combination of (1 + i)q and (1 + i)iq to p to force it into the square in the complex plane with vertices q, iq, q, and iq. By applying powers of iǫ to v, one can force p to lie in the triangle with− q, iq,− and 0 as vertices while leaving− q fixed. Then by applying powers of τ to v, one can add a Z-scalar multiple of q to n so that n now has the form dq + ciq, where 1 < d 1 for some c R. Since − 2 ≤ 2 ∈ v is isotropic, p 2 = 2 Im(nq)=2c q 2. Since is 2-bounded, q 2 2, so that in | | 2| | 2 J | | ≤ particular, is Γ-equivalent to 1 or 2 . J J J 2 2 For > 2, we can arrange that 1 or 2 . The 2-bounded condition in each|J of | these two cases has onlyJ finitely ⊃ J manyJ ⊃solutions. J The proposition then follows from listing the solutions and identifying Γ-conjugate sets. 6 DAN YASAKI
5. Reduction theory
Proposition 5.1. Every point z D is conjugate under ΓP0 to a point (y,β,r), 1 1 ∈ where 2 < r 2 and β lies in the square in the complex plane with vertices 1+i − ≤1+i 0, 2 , i, and − 2 .
Proof. Consider the elements σ, σ,ˇ τ, ǫ ΓP0 defined in Section 2. The action of σ, σ,ˇ τ, ǫ leaves D(P ) {D stable,}⊂ and is given explicitly by { } 0 ∩ 0 σ (y,β,r) = (y,β +(1+ i), r Re(β) + Im(β)), · − σˇ (y,β,r) = (y,β + i(1 + i), r Re(β) Im(β)), (12) · − − τ (y,β,r) = (y,β,r + 1), and · ǫ (y,β,r) = (y, iβ,r). · − Thus, by applying powers of σ andσ ˇ, β can be put in the square in the complex plane with vertices 1, i, 1, and i. By applying a power of ǫ, β can be put in the − − 1+i 1+i square in the complex plane with vertices 0, 2 , i, and − 2 . Then by applying 1 1 powers of τ, it can be arranged that 2 < r 2 without changing the value of β. − ≤
Proposition 5.2. For each z D , ∈ 0 1
Note that on F , f0(z) fP (z) for all P . By Proposition 3.3, this is equivalent to the condition that ≥ ∈P
2 2 2 2 1 p 1 n βp (14) f0(z) = y iβ + Re − r . ≥−2 q − s q 2 − q − | | w Consider the rational parabolic subgroups P0, P , and Q corresponding to the t t t vectors vw = (0, 0, 1) , vP = (i, 1+ i, 1+ i) and vQ = ( 1, 1+ i, 1+ i) . Divide F into the following regions: − 9 3 1 F = x = (y,β,r) F β > , < r P ∈ | | 10 10 ≤ 2 n 9 1 o3 F = x = (y,β,r) F β > , r< Q ∈ | | 10 −2 ≤ −10 n o Fw = F F1 F2 . \{ ∪ } EXPLICIT REDUCTION FOR SU(2, 1; Z[i]) 7
Since β is constrained to lie in the square indicated in Proposition 5.1, one may calculate that on FP and FQ (10 √62) 10 √62 (15) − < Re(β) < − − 20 20 81 10√62 (16) β i 2 < − . | − | 100 Comparing f0 to fP on FP , f0 to fQ on FQ, and f0 to fw on Fw, (14) gives the desired bound on each piece. Since F = F F F , this proves the result. P ∪ Q ∪ w 6. Admissible sets
Since vP and vQ are integral vectors, (vP , vQ) Z[i]. Proposition 3.4 and Proposition 5.2 imply the following. Q ∈ Proposition 6.1. Let be an admissible set. Then I (v , v ) 2 4 for every P and Q in . |Q P Q | ≤ I In particular, the set of vectors associated to an admissible set is 4-bounded. Proposition 6.2. Let = P,Q be an admissible subset of . If (v , v ) 2 = I { } P |Q P Q | 4, then there exists a strongly admissible set ˜ of order 8 such that D( ) = I⊂P I D(˜)= z , where z is the first contact for . Let ˜ be the set of isotropic vectors I { } I J associated to ˜. Then ˜ is 4-bounded and is Γ-equivalent to 8. A set of isotropic vectors that isI not 2-bounded,J but is associated to a stronglyJ admissible set, is Γ- equivalent to 8. J Proof. Suppose = P,Q is admissible and (v , v ) 2 = 4. Since G has class I { } |Q P Q | number 1, is Γ-conjugate to a set of the form P0, P . By Γ-action, one only I t { } t needs to consider the cases when vP = (i, 2, 2) and vP = (1, 0, 2) . In the case that v = (1, 0, 2)t we claim that P , P is not admissible, and { 0 v} hence D( P0, Pv ) is empty. To see this, it suffices to show that fQ(x) >f0(x) on E( P , P { ) for some} Q. Proposition 3.3 implies that E( P , P ) is the set where { 0 v} { 0 v} β 2 (17) y2 = | | + r r2. − 2 − In particular, 0
Consider the rational parabolic subgroups Q1,Q2,Q3, and Q4 with associated isotropic vectors (0, 0, 1)t, (1+i, 1 i, 1)t, (2i, 2, 1)t, and ( 1+i, 1+i, 1)t respectively. Explicit computation shows that− on S, −
f (x) f (x) f (x)= 0 f (x)= 0 Q1 β 2 Q2 β ( 1+ i) 2 | | | − − | f (x) f (x) f (x)= 0 f (x)= 0 . Q3 β 2i 2 Q4 β (1 + i) 2 | − | | − |
2 Since β i < 1 and fQj (x) f0(x) on D( ), we see that β is forced to equal i. Thus | − | ≤ I
1 D( )= D( P0, Pv, P1, P2,Q1,Q2,Q3,Q4 )= , i, 0 . I { } √2
One checks that 8 is not properly contained in any 4-bounded set to complete the proof. J
Corollary 6.3. The set of isotropic vectors associated to a strongly admissible set of order less than eight is 2-bounded. Furthermore, there are no strongly admissible sets of order greater than eight.
Proof. Propositions 6.1 and 6.2 imply the first statement. The second follows from explicit calculation that shows there are no 4-bounded sets that properly contain 8. J
Proposition 6.4. Let i denote the set of rational parabolic subgroups associated Ij i as defined in Section 4. Then Jj
2 = P , wP 2 = P , ξP I1 { 0 0} I2 { 0 0} 3 2 τw 3 2 σw 3 2 τσw 1 = 1 P0 2 = 1 P0 3 = 1 P0 I I ∪{ } I I ∪{ } I−1 I ∪{ } 4 = 3 σwP 4 = 3 w τσwˇ P I1 I1 ∪{ 0} I2 I3 ∪{ 0} 5 4 τσw = 1 P0 I−1 I ∪{ } 2 4 8 = P , wP , wτσwP , τ σwP , τσwˇ P , τwτσwP , τ σσwˇ P , ǫwξ wP I { 0 0 0 0 0 0 0 0}
Using (3) and Propositions 3.4 and 6.4, one calculates the following first contact points z( ). I EXPLICIT REDUCTION FOR SU(2, 1; Z[i]) 9
1+√5 Proposition 6.5. Let φ = 2 . Then 1 1 z( 2)=(1, 0, 0) z( 2)= , i, I1 I2 √4 2 2 4 3 1 √3 1+ i z( 3)= , 0, z( 3)= , , 0 I1 4 2 I2 2 2 r 1 1 1 z( 3)= φ2 φ 2, (φ2 φ + i(1 φ + φ)), (1 φ + φ) I3 √2 − 2 − − 2 − q p p p p 3+ √3+ √2+ √6 1+ √3 √2 1 z( 4)= − , − (1 + √3i), I1 2 4 2 p 3+ i z( 4)= φ 1, (1 √2), 0 I2 − 2 − 1+2√p3 1+ i 1 1 z( 5)= − , , z( 8)= , i, 0 . I 2 2 2 I √2 p Theorem 6.6. Up to Γ-conjugacy, the strongly admissible sets are exactly 2, 2, 3, 3, 3, 4, 4, 5, 8. I1 I2 I1 I2 I3 I1 I2 I I Proof. Propositions 4.2 and 6.2 and Corollary 6.3 imply that every strongly admis- sible set must be one of the ones listed. Thus, it suffices to show that for each set listed, D′( ) is non-empty. In particular, it suffices to show that D′( ) contains Iits first contactI point. One uses (3) and Proposition 6.5 to show that forIP , 2 ∈P\I fP (z( )) 7. Pieces of the spine Theorem 7.1. Let (y,β,r) denote a point in D. Then 2 2 2 β (i) E( )= y = | | + √1 r2 I1 − 2 − n 2 o 2 (ii) E( 2)= y2 = 1 1 iβ¯ + 1 1 Re(β) r I2 − 2 − 2 − 2 − − q (iii) D( 2) E( 2) 1