Amod Agashe

Lydia Eldredge

Comptes Rendus Mathématique

C. R. Acad. Sci. Paris, Ser. I 355 (2017) 937–941Contents lists available at ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comNumber theory/Differential geometryOn the cohomology groups of certain quotients of products of upper half planes and upper half spacesSur les groupes de cohomologie de certains quotients de produits de demi-plans et demi-espaces supérieursAmod Agashe 1, Lydia EldredgeFlorida State University, Department of Mathematics, Tallahassee, FL, 32306, United Statesa r t i c l e i n f o a b s t r a c tArticle history:Received 12 February 2017Accepted 28 April 2017Available online 27 September 2017Presented by the Editorial BoardA theorem of Matsushima–Shimura shows that the space of harmonic differential forms on the quotient of products of upper half planes under the action of certain groups, when the quotient is compact, is the direct sum of two subspaces called the universal and cuspidal subspaces. We generalize this result to compact quotients of products of upper half planes and upper half spaces (hyperbolic three spaces) under the action of certain groups to obtain a similar decomposition.© 2017 Published by Elsevier Masson SAS on behalf of Académie des sciences.r é s u m éUn théorème de Matsushima et Shimura montre que l’espace des formes différentielles harmoniques sur un quotient compact d’un produit de demi-plans supérieurs sous l’action de certains groupes est la somme directe de deux sous-espaces appelés sous-espaces universel et cuspidal. Nous généralisons ce résultat aux quotients compacts d’un produit de demi-plans supérieurs et demi-espaces supérieurs (hyperboliques de dimension 3) sous l’action de certains groupes, obtenant une décomposition similaire.© 2017 Published by Elsevier Masson SAS on behalf of Académie des sciences.1. Introduction and main resultLet n be a positive integer. For i = 1, . . . , n, let Hi be either the upper half plane H2 = {z ∈ C : (z) > 0} or upper half space H3 = C × R>0 (a model for hyperbolic three space), and let Ci = R if Hi is the upper half plane and Ci = C if Hi is upper half space. The group S L2(R) acts on H2 as usual by fractional linear transformations, and the action of S L2(C) on H3is given byE-mail addresses: amod.agashe@gmail.com (A. Agashe), eldredge.lydia@gmail.com (L. Eldredge).1 The first author was partially supported by the Collaboration Grants for Mathematicians from the Simons Foundation (award number 245630).http://dx.doi.org/10.1016/j.crma.2017.04.0161631-073X/© 2017 Published by Elsevier Masson SAS on behalf of Académie des sciences.938 A. Agashe, L. Eldredge / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 937–941[a bc d](z, t) = 1δ((az + b)(cz + d) + act2, |ad − bc|t), where δ = |cz + d|2 + |ct|2 .Let  be a subgroup of∏ni=1 S L2(Ci), which thus acts on∏ni=1 Hi .For i = 1, . . . , n, if Hi is the upper half plane, then let xi and yi denote the x and y coordinates on Hi (i.e. for which the corresponding point is xi +√−1yi ), and if Hi is the upper half space, then let xi , yi and ti denote the x, y, and tcoordinates on Hi (i.e. for which the corresponding point is (xi +√−1 yi, ti)). The standard Riemannian metrics on the upper half plane and upper half space induce a metric (the product metric) on∏ni=1 Hi . For i = 1, . . . , n, let ηi denote the volume form (also called fundamental form) on Hi , i.e. if Hi is the upper half plane, then ηi = dxi/yi ∧ dyi/yi , and if Hiis the upper half space, then ηi = dxi/ti ∧ dyi/ti ∧ dti/ti . If C ⊆ {1, . . . , n}, then let ηC = ∧i∈Cηi . Note that ηC is invariant under∏ni=1 S L2(Ci). For i = 1, . . . , n, let di = 2 if Hi is the upper half plane, then let di = 3 if Hi is the upper half space. DefineHmuniv ={ ∑C⊆{1,...,n}:∑i∈C di=m RηC , if m = 2i + 3 j for some integers i and j0 otherwise.Note that if Hi is the upper half plane for all i, then Hmuniv ⊗ C is isomorphic to the usual universal cohomology group associated with , of degree m (e.g., see [4, §III.1]). Let Hm() denote the group of harmonic differential forms of degree mon∏ni=1 Hi that are invariant with respect to . Since the volume form on any Hi is harmonic, Hmuniv ⊆Hm().Now assume that  is discrete in∏ni=1 S L2(Ci), has no element of finite order other than the identity, and is such that the quotient by its action on∏ni=1 Hi is compact. For i = 1, . . . , n, if Hi is the upper half plane, then let Bi = {dxi/yi, dyi/yi}, and if Hi is the upper half space, then let Bi = {dxi/ti, dyi/ti, dti/ti}. For any i, if Ai ⊆ Bi , then let ωAi = ∧ω∈Ai ω. If a1, . . . , an are integers, then we say that a differential form is of type a1, . . . , an if it can be written as a linear combination (with coefficients being smooth functions) of some ωA1 ∧ · · · ∧ ωAn , such that for each i, Ai ⊆ Biand |Ai | = ai . Let Hmcusp denote the subgroup of Hm() generated by elements that are of type a1, . . . , an , such that none of the ai are equal to 0 or di (note that ∑i ai = m necessarily).Assume now that the image of  under the projection to any partial product of∏ni=1 S L2(Ci) is dense; this is called an “irreducibility condition” in [4, I.2.13] (cf. pp. 420–421 of [5]), and is often true for arithmetic subgroups if n > 1.Theorem 1.1. We have Hm() = Hmuniv + Hmcusp and Hmuniv ∩ Hmcusp = {0}. Hence Hm(, R) ∼= Hm(\(∏ni=1 Hi), R) ∼=HmdR(\(∏ni=1 Hi), R) ∼= Hm() ∼= Hmuniv ⊕ Hmcusp , where the first cohomology is group cohomology, the second one is Betti co-homology, and the third one is deRham cohomology.Note that the first three isomorphisms in the second statement are standard, and for this reason, this statement follows from the first one. The proof of the first statement is given in the next section. If Hi is the upper half plane for all i, then the statement of the theorem, when tensored with C, follows from [5] (cf. Theorem 1.6 in Chapter III of [4]). The proof given in [5] (see also [4, §III.1] for an exposition) uses crucially the fact that there is a basis of complex differential forms on the upper half plane for which the action of S L2(R) is diagonal (e.g., one such basis is {dz, dz}, where z denotes the coordinate on the upper half plane); this does not seem to be true for the upper half space (as far as we know), and so the proof of [5]does not generalize in an obvious way to our situation. While our proof does borrow ideas from [5], even in the case where Hi is the upper half plane for all i, it differs from the one in [5] and is more elementary, in the sense that the proof in [5]uses the maximum principle for harmonic functions, while we just use the fact that harmonic forms on compact manifolds are closed. Moreover, we get a result with coefficients in R, as opposed to in C, so our result is more general as well.The motivation for the authors to generalize the result of Matsushima–Shimura (to include upper half spaces) is for a potential application to the construction of Stark–Heegner points (also called Darmon points) on elliptic curves over arbitrary number fields (the case of totally real fields is considered in [1]); for an example of the application of the original theorem of Matsushima–Shimura in this context, see the proof of Theorem 8.9 on p. 92 of [3].2. Proof of Theorem 1.1We continue to use the notation introduced in the previous section. If ω is a differential form, then it can be written uniquely as ω = ∑α∈S ωα , where S is some finite indexing set and the ωα ’s are non-trivial forms of distinct types. The Laplacian operator and any element of  take a form of a given type to a form of the same type. Thus, if ω as above is an invariant harmonic form, then each ωα is also an invariant harmonic form (here and henceforth, when we say that a form is invariant, we shall mean with respect to the full group ). Thus it suffices to show that a nonzero harmonic invariant differential form of a given type is either in Hmuniv or in Hmcusp, but not in both.Proposition 2.1. Let ω be a nonzero invariant harmonic form of type a1, . . . , an for some a1, . . . , an. Suppose there is an i such that ai = 0 or di . Then for all j, a j = 0 or d j , and ω is a real number times ∧nj=1b j , where b j = 1 if a j = 0, and b j = η j (the volume form on H j ) if a j = d j .A. Agashe, L. Eldredge / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 937–941 939We shall prove this proposition next, but note that it immediately implies the theorem, since if a harmonic invariant form is of a type where ai = 0 or di for some i, then by the proposition above, it is in Hmuniv, and if not, then it is in Hmcusp, which proves the first statement of the theorem (as mentioned earlier, the second statement follows from the first one).The rest of this section is devoted to the proof of Proposition 2.1. Without loss of generality, assume that i = 1, where i is as in the statement of the proposition. For some finite set I , one can write ω uniquely as ω = ∑k∈I fkωk such that ωk = ωA1 ∧ · · · ∧ ωAn for some Ai ⊆ Bi with |Ai | = ai for all i = 1, . . . , n and with the requirement that the ωk ’s are distinct and fk ’s are nonzero smooth functions. Let xi denote (xi, yi) if Hi is the upper half plane and xi = (xi, yi, ti) if Hi is the upper half space.Lemma 2.2. For each k ∈ I , fk is independent of x1 (i.e. independent of the variables in x1).Proof. Let X denote the quotient of∏ni=1 Hi under the action of ; under our assumptions, it is a compact Riemannian manifold without boundary. If ω′ is a form on∏ni=1 Hi i.e. invariant under , then it induces a form on X that we shall denote by ω′ . We define an inner product on p-forms on∏ni=1 Hi that are invariant under  by (ω1, ω2) =∫X ω1 ∧ ∗ω2, where the latter integral is well defined since X is compact. With this inner product, the usual proof that a harmonic form on a compact Riemannian manifold is closed generalizes to show that a harmonic form on∏ni=1 Hi that is invariant under is closed (as was pointed out to us by E. Klassen, this also follows because it is a local result, which holds on the quotient since it is compact). Since ω is harmonic, we have dω = 0. We remark that this is the only place in this article where we use the hypothesis that the quotient X is compact.Consider first the case where a1 = 0. Then note that for each k ∈ I , ωk = ωA2 ∧ · · · ∧ ωAn for some A2 ⊆ B2, . . . , An ⊆ Bn(the ωA1 is missing since a1 = 0). For i = 1, . . . , n, if Hi is the upper half plane, then let B ′i = {dxi, dyi}, and if Hi is the upper half space, then let B ′i = {dxi, dyi, dti} (thus B ′i is a different basis for the span of Bi ). Now, with I as above, one can write ω uniquely as ω = ∑k∈I gkω′k such that ω′k = ωA′2 ∧ · · · ∧ ωA′n for some A′i ⊆ B ′i with |A′i | = ai for all i and with the requirement that the ω′k ’s are distinct, gk ’s are nonzero smooth functions, and gkω′k = fkωk (we are just doing a change of basis here).We have 0 = dω = ∑k∈I ∑b∈B ′1 ∂ gk∂b db ∧ ωk + ∑k∈I ∑ni=2 ∑b∈B ′i ∂ gk∂b db ∧ ωk . Now each term in the first sum contains dbfor some b ∈ B ′1, while the terms in the second sum do not contain db for any b ∈ B ′1 (considering the way in which ωkcould be written in the paragraph above), so the first sum must be zero. Secondly, in the first sum, for each b ∈ B ′1, the db ∧ωk ’s are distinct for the various k’s, so we conclude that for all k, ∂ gk/∂b = 0 for all b ∈ B ′1, i.e. gk is independent of x1. Now for each k, fk differs from gk by∏ni=2 ci , where ci is a power of yi if Hi is the upper half planes and ci is a power of ti if Hi is the upper half space; this product∏ni=2 ci is non-zero and is independent of x1. Thus for each k, fk is also independent of x1.Now consider the case where a1 = d1. Then ∗ω is also invariant and harmonic, and for it, the corresponding a1 is 0, while the set of fk ’s do not change up to signs. Thus the argument above applied to ∗ω shows that the fk ’s are independent of x1in this case also. Now suppose there is a j such that a j is neither 0 nor di . We will show that this leads to a contradiction. Without loss of generality, suppose j = 2. Let M ∈ S L2(C2) (recall that C2 = R if H2 is the upper half plane and C2 = C if H2 is the upper half space). Then by the “irreducibility condition”, for some matrix A ∈ S L2(C1) that depends on M , there are elements of that are arbitrarily close to (A, M, I, . . . , I), where I is the identity matrix, as usual. Now ω is invariant with respect to the first component (i.e. with respect to all possible A’s as above; here, if a1 = d1, we use the fact that the volume form is invariant under S L2(C1)). Since ω is smooth, by pulling back ω to H2, we get a harmonic differential form ω′ on H2 or on H3 that is invariant under M and is not of degree zero or the top degree (which is 2 for H2 and 3 for H3). It is perhaps well known that this cannot happen for all such M (and in fact, this may follow from a more general fact), but we prove this anyway for the sake of completeness as follows.Lemma 2.3. There is no nonzero harmonic differential form on H2 or on H3 that is invariant under S L2(R) or S L2(C) respectively and is of degree other than zero or the top degree (which is 2 for H2 and 3 for H3).Proof. Assume that a form ω′ of the type above actually exists. First consider the case of H2. Then ω′ is of degree one and so ω′ = f dzy + g(− dz̄y ) for some f (z) and g(z). A direct calculation shows that if P1 = (x1, y1) and P2 = (x2, y2) are points in H2, then with e = x2 − x1 y2y1 and d =√y1y2, the matrixM =[1d ed0 d]∈ S L2(R) (1)satisfies M(P1) = P2. The action of the matrix [a bc d]∈ S L2(R) on complex differentials in the basis ( dzy , − dz̄y ) is given by the matrix940 A. Agashe, L. Eldredge / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 937–941[ε−2 00 ε2], where ε = (cz + d)/|cz + d| . (2)Using this, one sees that the matrix in (1) leaves dzy and − dz̄y invariant (keeping in mind that d > 0). Since ω′ is invariant under S L2(R), we conclude that f (z) and g(z) are constants.Now consider the following matrixγ =[0 −11 0]∈ S L2(R).Since f (z) and g(z) are constants, we shall write them as just f and g , respectively. Thenf(dzy)+ g(− dz̄y)= γ ∗(fdzy+ g(− dz̄y))= f γ ∗(dzy)+ gγ ∗(− dz̄y)= f |z|2z2dzy+ g z2|z|2(− dz̄y),where we used formula (2) above in the last step. Putting z = i (for example) in the equation above, we see that f = g = 0and so ω′ = 0, which is a contradiction.Now consider the case of H3. First consider the case where the degree of ω′ is one. Let β0 = −dz/t , β1 = dt/t , and β2 = dz/t . Then ω′ = f0β0 + f1β1 + f2β2 for some functions f0, f1, and f2. A direct calculation shows that if P1 = (x1, y1, t1)and P2 = (x2, y2, t2) are points in H3, then with z1 = x1 + iy1, z2 = x2 + iy2, e = z2 − z1 t2t1 , and d =√t1t2, the matrixM =[1d ed0 d]∈ S L2(C) (3)satisfies M(P1) = P2. The action of the matrix [a bc d]∈ S L2(C) on complex differentials in the basis (β0, β1, β2) is given by the matrix1|r|2 + |s|2⎡⎢⎣r2 −2rs s2rs |r|2 − |s|2 −rss2 2rs r2⎤⎥⎦ , where  = ad − bc = 1, r = cz + d, and s = ct . (4)Using this, one sees that the matrix in (3) leaves each of β0, β1, and β2 invariant (considering that d is real). Since ω′ is invariant under S L2(C), we conclude that f0, f1, and f2 are constants.It is easy to check that ∗β0 = iβ1 ∧ β2, ∗β1 = i2 β2 ∧ β0, and ∗β2 = iβ0 ∧ β1, and that dβ0 = β0 ∧ β1, dβ1 = 0, and dβ2 = β2 ∧ β1 (see, e.g., the proof of Lemma 61 in §4.2.5 of [2]). Using these formulas, and the fact that ω′ is harmonic, we have:0 = ω = (dδ + δd)ω = dδ( f0β0 + f1β1 + f2β2) + δd( f0β0 + f1β1 + f2β2)= d((−1)3(1+1)+1 ∗ d ∗ ( f0β0 + f1β1 + f2β2)) + δ( f0β0 ∧ β1 + 0 + f2β2 ∧ β1)= −d ∗ d(i f0β1 ∧ β2 + i2f1β2 ∧ β0 + i f2β0 ∧ β1) + (−1)3(2+1)+1 ∗ d ∗ ( f0β0 ∧ β1 + f2β2 ∧ β1)= −d ∗ (0 + i2f1(−2β0 ∧ β1 ∧ β2) + 0) + ∗d( f0(iβ2) + f2(−iβ0))= d ∗ (i f1β0 ∧ β1 ∧ β2) + ∗d( f0iβ2 − f2iβ0) = d(−i f1(2i)) + ∗( f0iβ2 ∧ β1 − f2iβ0 ∧ β1)= d(2 f1) − f0i(−iβ0) + f2i(iβ2) = 0 − f0iβ0 − f2β2.It follows that f0 = f2 = 0. Recall that f1 is a constant. Letγ =[0 −11 0]∈ S L2(C) .Thenf1β1 = ω′ = γ ∗(ω′) = γ ∗( f1β1) = f1γ ∗(β1) = f1 1|z|2 + t2 (−2ztβ0 + (|z|2 − t2)β1 + 2ztβ2) ,where we used formula (4) above in the last step. Putting z = 0, t = 1 in the equation above, we see that f1 = 0 and so ω′ = 0, which is a contradiction.If the degree of ω′ is 2, then ∗ω′ is a non-zero harmonic form of degree one, which is invariant under S L2(C). Thus from the argument above, we again get a contradiction. A. Agashe, L. Eldredge / C. R. Acad. Sci. Paris, Ser. I 355 (2017) 937–941 941Thus, for all j, a j = 0 or d j . Then one sees that I consists of only one element k (since the dimension of the space of forms of degree 0 or d j is one for each j), and applying Lemma 2.2 above for any j instead of for i = 1, we see that fk is independent of x j for all j. Thus fk is a real number. This finishes the proof of Proposition 2.1.AcknowledgementWe are very grateful to E. Aldrovandi for several discussions and advice on the work that led to this article.References[1] A. Agashe, M. Trifkovic, Darmon points for totally real fields, submitted for publication, available at arXiv:1310.4237.[2] J. Bygott, Modular Forms and Modular Symbols Over Imaginary Quadratic Fields, Ph.D. thesis, Exeter, UK, 1999, available at http://hdl.handle.net/10871/8322.[3] H. Darmon, Rational Points on Modular Elliptic Curves, CBMS Reg. Conf. Ser. Math., vol. 101, 2004, published for the Conference Board of the Mathe-matical Sciences, Washington, DC. MR 2020572 (2004k:11103).[4] E. Freitag, Hilbert Modular Forms, Springer-Verlag, Berlin, 1990. MR 1050763 (91c:11025).[5] Y. Matsushima, G. Shimura, On the cohomology groups attached to certain vector valued differential forms on the product of the upper half planes, Ann. of Math. (2) 78 (1963) 417–449. MR 0155340 (27 #5274).

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

Agashe, Amod

Eldredge, Lydia

Numérisation de Documents Anciens Mathématiques

http://www.numdam.org/item/10.1016/j.crma.2017.04.016.pdf

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

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