On the cohomology groups of certain quotients of products of upper half planes and upper half spaces

Transcription

1 On he cohomolog groups of cerain quoiens of producs of upper half planes and upper half spaces Amod Agashe and Ldia Eldredge Absrac A heorem of Masushima-Shimura shows ha he he space of harmonic differenial forms on he quoien of producs of upper half planes under he acion of cerain groups, when he quoien is compac, is he direc sum of wo subspaces called he universal and cuspidal subspaces. We generalize his resul o compac quoiens of producs of upper half planes and upper half spaces (hperbolic hree spaces under he acion of cerain groups o obain a similar decomposiion. 1 Inroducion and main resul Le n be a posiive ineger. For i = 1,..., n, le H i be eiher he upper half plane H 2 = {z C : I(z > 0} or upper half space H 3 = C R >0 (a model for hperbolic hree space, and le C i = R if H i is he upper half plane and C i = C if H i is upper half space. The group SL 2 (R acs on H 2 as usual b fracional linear ransformaions and he acion of SL 2 (C on H 3 is given b [ ] a b (z, = 1 c d δ ((az + b(cz + d + ac2, ad bc, where δ = cz + d 2 + c 2. Le Γ be a subgroup of n i=1 SL 2(C i, which hus acs on n i=1 H i. For i = 1,..., n, if H i is he upper half plane, hen le x i and i denoe he x and coordinaes on H i (i.e., for which he corresponding poin is x i + 1 i, and if H i is he upper half space, hen le x i, i and i denoe he x,, and coordinaes on H i (i.e., for which he corresponding poin is (x i + 1 i, i. The sandard Riemannian merics on he upper half plane and upper half space induce a meric (he produc meric on n i=1 H i. For i = 1,..., n, le η i denoe he volume form (also called fundamenal form on H i, i.e., if H i is he upper half plane, hen η i = dx i / i d i / i, and if H i is he upper half space, hen η i = dx i / i d i / i d i / i. If C {1,..., n}, hen le η C = i C η i. Noe ha η C is invarian under n i=1 SL 2(C i. For i = 1,..., n, le d i = 2 if H i is he upper half plane, hen le d i = 3 if H i is he upper half space. Define { H m C {1,...,n}: univ = Rη C, if m = 2i + 3j for some inegers i and j i C d i =m 0 oherwise Noe ha if H i is he upper half plane for all i, hen H m univ C is isomorphic o he usual universal cohomolog group associaed o Γ of degree m (e.g., see [Fre90, III.1]. Le H m (Γ denoe he group of harmonic differenial forms of degree m on n i=1 H i ha are invarian wih respec o Γ. Since he volume form on an H i is harmonic, H m univ H m (Γ. The firs auhor was pariall suppored b he Collaboraion Grans for Mahemaicians from he Simons Foundaion (award number

2 Now assume ha Γ is discree in n i=1 SL 2(C i, has no elemen of finie order oher han he ideni, and is such ha he quoien b is acion on n i=1 H i is compac. For i = 1,..., n, if H i is he upper half plane, hen le B i = {dx i / i, d i / i }, and if H i is he upper half space, hen le B i = {dx i / i, d i / i, d i / i }. For an i, if A i B i, hen le ω Ai = ω Ai ω. If a 1,..., a n are inegers, hen we sa ha a differenial form is of pe a 1,..., a n if i can be wrien as a linear combinaion (wih coefficiens being smooh funcions of some ω A1 ω An, such ha for each i, A i B i and A i = a i. Le H m cusp denoe he subgroup of H m (Γ generaed b elemens ha are of pe a 1,..., a n such ha none of he a i are equal o 0 or d i (noe ha i a i = m necessaril. Assume now ha he image of Γ under he projecion o an parial produc of n i=1 SL 2(C i is dense; his is called an irreducibili condiion in [Fre90, I.2.13] (cf. p of [MS63], and is ofen rue for arihmeic subgroups if n > 1. Theorem 1.1. We have H m (Γ = H m univ + H m cusp and H m univ H m cusp = {0}. Hence H m (Γ, R = H m( Γ\( n i=1 H i, R ( = H m dr Γ\( n i=1 H i, R = H m (Γ = H m univ Hcusp, m where he firs cohomolog is group cohomolog, he second one is Bei cohomolog, and he hird one is derham cohomolog. Noe ha he firs hree isomorphisms in he second saemen are sandard, and for his reason, his saemen follows from he firs. The proof of he firs saemen is given in he nex secion. If H i is he upper half plane for all i, hen he saemen of he heorem, when ensored wih C, follows from [MS63] (cf. Theorem 1.6 in Chaper III of [Fre90]. The proof given in loc. ci. (see also [Fre90, III.1] for an exposiion uses cruciall he fac ha here is a basis of complex differenial forms on he upper half plane for which he acion of SL 2 (R is diagonal (e.g., one such basis is {dz, dz}, where z denoes he coordinae on he upper half plane; his does no seem o be rue for upper half space (as far as we know, and so he proof of [MS63] does no generalize in an obvious wa o our siuaion. While our proof does borrow ideas from loc. ci., even in he case where H i is he upper half plane for all i, i differs from he one in loc. ci. and is more elemear, in he sense ha he proof in loc. ci. uses he maximum principle for harmonic funcions, while we jus use he fac ha harmonic forms on compac manifolds are closed. Moreover, we ge a resul wih coefficiens in R, as opposed o in C, so our resul is more general as well. The moivaion for he auhors o generalize he resul of Masushima-Shimura (o include upper half spaces is for a poenial applicaion o he consrucion of Sark-Heegner poins (also called Darmon poins on ellipic curves over arbirar number fields (he case of oall real fields is considered in [AT]; for an example of he applicaion of he original heorem of Masushima- Shimura in his conex, see he proof of Theorem 8.9 on p. 92 of [Dar04]. Acknowledgemen: We are ver graeful o E. Aldrovandi for several discussions and advice on work ha led o his aricle. 2 Proof of Theorem 1.1 We coninue o use he noaion inroduced in he previous secion. If ω is a differenial form, hen i can be wrien uniquel as ω = α S ω α where S is some finie indexing se and he ω α s are non-rivial forms of disinc pes. The Laplacian operaor and an elemen of Γ ake a form of a given pe o a form of he same pe. Thus if ω as above is an invarian harmonic form, hen each ω α is also an invarian harmonic form (here and henceforh, when we sa ha a form is 2

3 invarian, we shall mean wih respec o he full group Γ. Thus i suffices o show ha a nonzero harmonic invarian differenial form of a given pe is eiher in H m univ or in H m cusp, bu no in boh. Proposiion 2.1. Le ω be a nonzero invarian harmonic form of pe a 1,..., a n for some a 1,..., a n. Suppose here is an i such ha a i = 0 or d i. Then for all j, a j = 0 or d j, and ω is a real number imes n j=1 b j, where b j = 1 if a j = 0, and b j = η j (he volume form on H j if a j = d j. We shall prove his proposiion nex, bu noe ha i immediael implies he heorem, since if a harmonic invarian form is of a pe where a i = 0 or d i for some i, hen b he proposiion above, i is in Huniv, m and if no, hen i is in Hcusp, m which proves he firs saemen of he heorem (as menioned earlier, he second saemen follows from he firs. The res of his secion is devoed o he proof of Proposiion 2.1. Wihou loss of generali, assume ha i = 1, where i is as in he saemen of he proposiion. For some finie se I, one can wrie ω uniquel as ω = k I f kω k such ha ω k = ω A1 ω An for some A i B i wih A i = a i for all i = 1,..., n and wih he requiremen ha he ω k s are disinc and f k s are nonzero smooh funcions. Le x i denoe (x i, i if H i is he upper half plane and x i = (x i, i, i if H i is he upper half space. Lemma 2.2. For each k I, f k is independen of x 1 (i.e., independen of he variables in x 1. Proof. Le X denoe he quoien of n i=1 H i under he acion of Γ; under our assumpions, i is a compac Riemannian manifold wihou boundar. If ω is a form on n i=1 H i ha is invarian under Γ, hen i induces a form on X ha we shall denoe b ω. We define an inner produc on p-forms on n i=1 H i ha are invarian under Γ b (ω 1, ω 2 = X ω 1 ω 2, where he laer inegral is well defined since X is compac. Wih his inner produc, he usual proof ha a harmonic form on a compac Riemannian manifold is closed generalizes o show ha a harmonic form on n i=1 H i ha is invarian under Γ is closed (as was poined ou o us b E. Klassen, his also follows because i is a local resul, which holds on he quoien since i is compac. Since ω is harmonic, we have dω = 0. We remark ha his is he onl place in his aricle where we use he hpohesis ha he quoien X is compac. Consider firs he case where a 1 = 0. Then noe ha for each k I, ω k = ω A2 ω An for some A 2 B 2,..., A n B n (he ω A1 is missing since a 1 = 0. For i = 1,..., n, if H i is he upper half plane, hen le B i = {dx i, d i }, and if H i is he upper half space, hen le B i = {dx i, d i, d i } (hus B i is a differen basis for he span of B i. Now, wih I as above, one can wrie ω uniquel as ω = k I g kω k such ha ω k = ω A ω A 2 for some n A i B i wih A i = a i for all i and wih he requiremen ha he ω k s are disinc, g k s are nonzero smooh funcions, and g k ω k = f kω k (we are jus doing a change of basis here. We have 0 = dω = g k k I b B 1 b db ω k + n g k k I i=2 b B i b db ω k. Now each erm in he firs sum conains db for some b B 1, while he erms in he second sum do no conain db for an b B 1 (considering he wa in which ω k could be wrien in he paragraph above, so he firs sum mus be zero. Secondl, in he firs sum, for each b B 1, he db ω k s are disinc for he various k s, so we conclude ha for all k, g k / b = 0 for all b B 1, i.e., g k is independen of x 1. Now for each k, f k differs from g k b n i=2 c i, where c i is a power of i if H i is he upper half planes and c i is a power of i if H i is he upper half space; his produc n i=2 c i is non-zero and is independen of x 1. Thus for each k, f k is also independen of x 1. Now consider he case where a 1 = d 1. Then ω is also invarian and harmonic, and for i, he corresponding a 1 is 0, while he se of f k s do no change up o signs. Thus he argumen above applied o ω shows ha he f k s are independen of x 1 in his case also. 3

4 Now suppose here is a j such ha a j is neiher 0 nor d i. We will show ha his leads o a conradicion. Wihou loss of generali, suppose j = 2. Le M SL 2 (C 2 (recall ha C 2 = R if H 2 is he upper half plane and C 2 = C if H 2 is he upper half space. Then b he irreducibili condiion, for some marix A SL 2 (C 1 ha depends on M, here are elemens of Γ ha are arbiraril close o (A, M, I,..., I, where I is he ideni marix, as usual. Now ω is invarian wih respec o he firs componen (i.e., wih respec o all possible A s as above; here, if a 1 = d 1, we use he fac ha he volume form is invarian under SL 2 (C 1. Since ω is smooh, b pulling back ω o H 2, we ge a harmonic differenial form ω on H 2 or on H 3 ha is invarian under M and is no of degree zero or he op degree (which is 2 for H 2 and 3 for H 3. I is perhaps well known ha his canno happen for all such M (and in fac, his ma follow from a more general fac, bu we prove his anwa for he sake of compleeness: Lemma 2.3. There is no nonzero harmonic differenial form on H 2 or on H 3 ha is invarian under SL 2 (R or SL 2 (C respecivel and is of degree oher han zero or he op degree (which is 2 for H 2 and 3 for H 3. Proof. Assume ha a form ω of he pe above acuall exiss. Firs consider he case of H 2. Then ω is of degree one and so ω = f dz d z + g( for some f(z and g(z. A direc calculaion shows ha if P 1 = (x 1, 1 and P 2 = (x 2, 2 are poins in H 2, hen wih e = x 2 x and d = 1 2, he marix [ 1 ] M = d ed SL 0 d 2 (R (1 [ ] a b saisfies M(P 1 = P 2. The acion of he marix SL c d 2 (R on complex differenials in he basis ( dz, d z is given b he marix [ ] ɛ ɛ 2, where ɛ = (cz + d/ cz + d. (2 Using his, one sees ha he marix in (1 leaves dz d z and invarian (keeping in mind ha d > 0. Since ω is invarian under SL 2 (R, we conclude ha f(z and g(z are consans. Now consider he following marix γ = [ ] 0 1 SL (R Since f(z and g(z are consans, we shall wrie hem as jus f and g respecivel. Then ( dz ( f + g d z = γ ( f dz ( + g d z = fγ ( dz + gγ ( d z = f z 2 dz z 2 + g z2 ( z 2 d z, where we used formula (2 above in he las sep. Puing z = i (for example in he equaion above, we see ha f = g = 0 and so ω = 0, which is a conradicion. Now consider he case of H 3. Firs consider he case where he degree of ω is one. Le β 0 = dz/, β 1 = d/, and β 2 = dz/. Then ω = f 0 β 0 + f 1 β 1 + f 2 β 2 for some funcions f 0, f 1, and f 2. A direc calculaion shows ha if P 1 = (x 1, 1, 1 and P 2 = (x 2, 2, 2 are poins in H 3, hen wih z 1 = x 1 + i 1, z 2 = x 2 + i 2, e = z 2 z 2 1 1, and d = 1 2, he marix [ 1 ] M = d ed SL 0 d 2 (C (3 4

5 [ ] a b saisfies M(P 1 = P 2. The acion of he marix SL c d 2 (C on complex differenials in he basis (β 0, β 1, β 2 is given b he marix r 1 2 2rs s 2 r 2 + s 2 rs r 2 s 2 rs, where = ad bc = 1, r = cz + d, and s = c. (4 s 2 2rs r 2 Using his, one sees ha he marix in (3 leaves each of β 0, β 1, and β 2 invarian (considering ha d is real. Since ω is invarian under SL 2 (C, we conclude ha f 0, f 1, and f 2 are consans. I is eas o check ha β 0 = iβ 1 β 2, β 1 = i 2 β 2 β 0, and β 2 = iβ 0 β 1, and ha dβ 0 = β 0 β 1, dβ 1 = 0, and dβ 2 = β 2 β 1 (see, e.g., he proof of Lemma 61 in of [Bg99]. Using hese formulas, and he fac ha ω is harmonic, we have 0 = ω = (dδ + δdω = dδ(f 0 β 0 + f 1 β 1 + f 2 β 2 + δd(f 0 β 0 + f 1 β 1 + f 2 β 2 = d(( 1 3(1+1+1 d (f 0 β 0 + f 1 β 1 + f 2 β 2 + δ(f 0 β 0 β f 2 β 2 β 1 = d d(if 0 β 1 β 2 + i 2 f 1β 2 β 0 + if 2 β 0 β 1 + ( 1 3(2+1+1 d (f 0 β 0 β 1 + f 2 β 2 β 1 = d (0 + i 2 f 1( 2β 0 β 1 β d(f 0 (iβ 2 + f 2 ( iβ 0 = d (if 1 β 0 β 1 β 2 + d(f 0 iβ 2 f 2 iβ 0 = d( if 1 (2i + (f 0 iβ 2 β 1 f 2 iβ 0 β 1 = d(2f 1 f 0 i( iβ 0 + f 2 i(iβ 2 = 0 f 0 iβ 0 f 2 β 2 I follows ha f 0 = f 2 = 0. Recall ha f 1 is a consan. Le [ ] 0 1 γ = SL (C. Then f 1 β 1 = ω = γ (ω = γ (f 1 β 1 = f 1 γ (β 1 = f 1 1 z ( 2zβ 0 + ( z 2 2 β 1 + 2zβ 2, where we used formula (4 above in he las sep. Puing z = 0, = 1 in he equaion above, we see ha f 1 = 0 and so ω = 0, which is a conradicion. If he degree of ω is 2, hen ω is a non-zero harmonic form of degree one ha is invarian under SL 2 (C. Thus from he argumen above, we again ge a conradicion. Thus, for all j, a j = 0 or d j. Then one sees ha I consiss of onl one elemen k (since he dimension of he space of forms of degree 0 or d j is one for each j, and appling Lemma 2.2 above for an j insead of for i = 1, we see ha f k is indepeden of x j for all j. Thus f k is a real number. This finishes he proof of Proposiion 2.1. Remark 2.4. We ake he opporuni o poin ou a correcion o he lieraure (his remark and he res of his aricle are independen of each oher. Recall ha H 3 = C R >0 is he upper half space (a model for hperbolic hree space. Le f 0, f 1, andf 2 be smooh funcions on H 3. I has been wrongl saed in he lieraure ha ω = f 0 ( dz/ + f 1 d/ + f 2 dz/ is harmonic if and 5

6 onl if f 0 z + f 2 z = 0, f 1 z + f 0 1 f 0 = 0, f 1 z f f 2 = 0, f 1 2 f 1 2 f 0 z = 0 (5 The ssem of equaions above is equivalen o saing ha ω is closed and coclosed (e.g., see he proof of Lemma 61 in of [Bg99]. On a compac manifold, a differenial form is harmonic if and onl if i is closed and coclosed. However H 3 is no compac. Now ( d δ ( d = ( 1 3(1+1+1 d d ( dx = d d ( 1 = d 2 dx d ( = 2 1 ( d 3 d dx d = 2 dx d = 2 1 = 2, and ( d ( d ( d ( 1 = (δd + dδ = δd + dδ = δ 2 d d + d(2 = = 0. Thus he differenial d/ is harmonic, bu no coclosed. I does no saisf equaion (5 above (in our case, f 1 = 1 and f 0 = 0. I also follows ha ( d is harmonic, bu no closed. References [AT] Amod Agashe and Mak Trifkovic, Darmon poins for oall real fields, submied, available a arxiv: [Bg99] J. Bgo, Modular forms and modular smbols over imaginar quadraic fields, Ph.D. hesis, Exeer (1999, available a hp://hdl.handle.ne/10871/8322. [Dar04] Henri Darmon, Raional poins on modular ellipic curves, CBMS Regional Conference Series in Mahemaics, vol. 101, Published for he Conference Board of he Mahemaical Sciences, Washingon, DC, MR (2004k:11103 [Fre90] Eberhard Freiag, Hilber modular forms, Springer-Verlag, Berlin, MR (91c:11025 [MS63] Yozô Masushima and Goro Shimura, On he cohomolog groups aached o cerain vecor valued differenial forms on he produc of he upper half planes, Ann. of Mah. (2 78 (1963, MR (27 #5274 6