HARMONICS FORMS ON NON COMPACT MANIFOLDS.

Transcription

1 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. GILLS CARRON The source of these notes is a series of lectures given at the CIPA s summer school Recent Topics in Geometric Analysis. I want to thank the organizers of this summer school : Ahmad l Soufi and ehrdad Shahsahani and I also want to thank ohsen Rahpeyma who solved many delicate problems. Theses notes aimed to give an insight into some links between L 2 cohomology, L 2 harmonics forms, the topology and the geometry of complete Riemannian manifolds. This is not a survey but a choice of few topics in a very large subject. The first part can be regard as an introduction ; we define the space of L 2 harmonics forms, of L 2 cohomology. We recall the theorems of Hodge and de Rham on compact Riemannian manifolds. However the reader is assumed to be familiar with the basic of Riemannian geometry and with Hodge theory. According to J. Roe ([55]) and following the classification of Von Neumann algebra, we can classify problems on L 2 harmonics forms in three types. The first one (type I) is the case where the space of harmonics L 2 forms has finite dimension, this situation is the nearest to the case of compact manifolds. The second (type II) is the case where the space of harmonics L 2 forms has infinite dimension but where we have a renormalized dimension for instance when a discrete group acts cocompactly by isometry on the manifold; a good reference is the book of W. Lueck ([47]) and the seminal paper of. Atiyah ([4]). The third type (type III) is the case where no renormalization procedure is available to define a kind of dimension of the space L 2 harmonics forms. Here we consider only the type I problems and at the end of the first part, we will prove a result of J. Lott which says that the finiteness of the dimension of the space of L 2 harmonics forms depends only on the geometry at infinity. any aspects 1 of L 2 harmonics forms will not be treated here : for instance we will not describe the important problem of the L 2 cohomology of locally symmetric spaces, and also we will not speak on the pseudo differential approach developped by R. elrose and his school. However the reader will find at the end of this first chapter a list of some interesting results on the topological interpretation of the space of L 2 harmonics forms. In the second chapter, we are interested in the space of harmonic L 2 1 forms. This space contains the differential of harmonic functions with L 2 gradient. We will not speak of the endpoint result of A. Grigory an ([32, 31]) but we have include a study of P.Li and L-F. Tam ([39]) and of A. Ancona ([1]) on non parabolic ends. In this chapter, we will also study the case of Riemannian surfaces where this space depends only on the complex structure. The last chapter focuses on the L 2 cohomology of conformally compact manifolds. The result is due to R. azzeo ([48]) and the proof present here is the one 1 almost all in fact! 1

2 2 GILLS CARRON of N. Yeganefar ([65]) who used an integration by parts formula due to H.Donnelly and F.Xavier ([24]). Contents 1. A short introduction to L 2 cohomology Hodge and de Rham s theorems Some general properties of reduced L 2 cohomology Lott s result Some bibliographical hints Harmonics L 2 1 forms nds H0 1 () versus H2 1 () The two dimensional case Bibliographical hints L 2 cohomology of conformally compact manifold The geometric setting The case where k = dim / A reduction to exact metric A Rellich type identity and its applications Application to conformally compact Riemannian manifold The spectrum of the Hodge-deRham Laplacian Applications to conformally compact manifolds azzeo s result Bibliographical hints 40 References A short introduction to L 2 cohomology In this first chapter, we introduce the main definitions and prove some preliminary results Hodge and de Rham s theorems de Rham s theorem. Let n be a smooth manifold of dimension n, we denote by C (Λ k T ) the space of smooth differential k forms on and by C0 (Λ k T ) the subspace of C (Λ k T ) formed by forms with compact support; in local coordinates (x 1, x 2,..., x n ), an element α C (Λ k T ) has the following expression α = α I dx I I={i 1<i 2<...<i k } α I dx i1 dx i2... dx ik = I where α I are smooth functions of (x 1, x 2,..., x n ). The exterior differentiation is a differential operator locally we have d : C (Λ k T ) C (Λ k+1 T ), ( ) d α I dx I = dα I dx I. I I

3 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 3 This operator satisfies d d = 0, hence the range of d is included in the kernel of d. Definition 1.1. The k th de Rham s cohomology group of is defined by { α C (Λ k T ), dα = 0 } H k dr() = dc (Λ k 1 T ) These spaces are clearly diffeomorphism invariants of, moreover the deep theorem of G. de Rham says that these spaces are isomorphic to the real cohomology group of, there are in fact homotopy invariant of : Theorem 1.2. H k dr() H k (, R). From now, we will suppress the subscript dr for the de Rham s cohomology. We can also define the de Rham s cohomology with compact support. Definition 1.3. The k th de Rham s cohomology group with compact support of is defined by { α C 0 (Λ k T ), dα = 0 } H k 0 () = dc 0 (Λk 1 T ) These spaces are also isomorphic to the real cohomology group of with compact support. When is the interior of a compact manifold with compact boundary = \, then H k 0 () is isomorphic to the relative cohomology group of : { α C (Λ k T ), dα = 0, ι α = 0 } H0 k () = H k (, ) := { dβ, β C (Λ k 1 T ) and ι β = 0 } where ι : is the inclusion map Poincaré duality. When we assume that is oriented 2 the bilinear map H k () H0 n k () R ([α], [β]) α β := I([α], [β]) is well defined, that is to say I([α], [β]) doesn t depend on the choice of representatives in the cohomology classes [α] or [β] (this is an easy application of the Stokes formula). oreover this bilinear form provides an isomorphism between H k () and ( H0 n k () ). In particular when α C (Λ k T ) is closed (dα = 0) and satisfies that [β] H0 n k (), α β = 0 then there exists γ C (Λ k 1 T ) such that α = dγ... 2 It is not a serious restriction we can used cohomology with coefficient in the orientation bundle.

4 4 GILLS CARRON L 2 cohomology a) The operator d. We assume now that is endowed with a Riemannian metric g, we can define the space L 2 (Λ k T ) whose elements have locally the following expression α = α I dx i1 dx i2... dx ik I={i 1<i 2<...<i k } where α I L 2 loc and globally we have α 2 L := α(x) 2 2 g(x) d vol g(x) <. The space L 2 (Λ k T ) is a Hilbert space with scalar product : α, β = (α(x), β(x)) g(x) d vol g (x). We define the formal adjoint of d : by the formula d : C (Λ k+1 T ) C (Λ k T ) α C 0 (Λ k+1 T ) and β C 0 (Λ k T ), d α, β = α, dβ. When is the Levi-Civita connexion of g, we can give local expressions for the operators d and d : let ( 1, 2,..., n ) be a local orthonormal frame and let (θ 1, θ 2,..., θ n ) be its dual frame : then (1.1) dα = and (1.2) d α = θ i (X) = g( i, X) n θ i i α, i=1 n int i ( i α), i=1 where we have denote by int i the interior product with the vector field i b) L 2 harmonic forms. We consider the space of L 2 closed forms : Z k 2 () = {α L 2 (Λ k T ), dα = 0} where it is understood that the equation dα = 0 holds weakly that is to say That is we have : β C 0 (Λ k+1 T ), α, d β = 0. Z k 2 () = ( d C (Λ k+1 T ) ), hence Z k 2 () is a closed subspace of L 2 (Λ k T ). We can also define H k () = ( d C 0 (Λ k+1 T ) ) ( dc 0 (Λ k 1 T ) ) = Z k 2 () {α L 2 (Λ k T ), d α = 0} = {α L 2 (Λ k T ), dα = 0 and d α = 0}.

5 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 5 Because the operator d + d is elliptic, we have by elliptic regularity : H k () C (Λ k T ). We also remark that by definition we have Hence α C 0 (Λ k 1 T ), β C 0 (Λ k+1 T ) dα, d β = ddα, β = 0 dc 0 (Λ k 1 T ) d C 0 (Λ k+1 T ) and we get the Hodge-de Rham decomposition of L 2 (Λ k T ) (1.3) L 2 (Λ k T ) = H k () dc 0 (Λk 1 T ) d C 0 (Λk+1 T ), where the closures are taken for the L 2 topology. And also (1.4) H k () Z k 2 () dc 0 (Λk 1 T ) c) L 2 cohomology: We also define the (maximal) domain of d by D k (d) = {α L 2 (Λ k T ), dα L 2 } that is to say α D k (d) if and only if there is a constant C such that β C 0 (Λ k+1 T ), α, d β C β 2. In that case, the linear form β C 0 (Λ k+1 T ) α, d β extends continuously to L 2 (Λ k+1 T ) and there is γ =: dα such that β C 0 (Λ k+1 T ), α, d β = γ, β. We remark that we always have dd k 1 (d) Z k 2 (). Definition 1.4. We define the k th space of reduced L 2 cohomology by H k 2 () = Zk 2 () dd k 1 (d). The k th space of non reduced L 2 cohomology is defined by nr H k 2 () = Zk 2 () dd k 1 (d). These two spaces coincide when the range of d : D k 1 (d) L 2 is closed; the first space is always a Hilbert space and the second is not necessary Hausdorff. We also have C 0 (Λ k 1 T ) D k 1 (d) hence we always get a surjective map : H k () H k 2 () {0}. In particular any class of reduced L 2 cohomology contains a smooth representative d) Case of complete manifolds. The following result is due to Gaffney ([28], see also part 5 in [64]) for a related result) Lemma 1.5. Assume that g is a complete Riemannian metric then dd k 1 (d) = dc 0 (Λk 1 T ).

6 6 GILLS CARRON Proof. We already know that dc 0 (Λ k 1 T ) D k 1 (d), moreover using a partition of unity and local convolution it is not hard to check that if α D k 1 (d) has compact support then we can find a sequence (α l ) l N of smooth forms with compact support such that α l α L 2 + dα l dα L 2 1/l. So we must only prove that if α D k 1 (d) then we can build a sequence (α N ) N of elements of D k 1 (d) with compact support such that L 2 lim N dα N = dα. We fix now an origin o and denote by B(o, N) the closed geodesic ball of radius N and centered at o, because (, g) is assumed to be complete we know that B(o, N) is compact and = N N B(o, N). We consider ρ C 0 (R + ) with 0 ρ 1 with support in [0, 1] such that ρ = 1 on [0, 1/2] and we define ( ) d(o, x) (1.5) χ N (x) = ρ. N Then χ N is a Lipschitz function and is differentiable almost everywhere and ( ) d(o, x) dχ N (x) = ρ dr N where dr is the differential of the function x d(o, x). Let α D k 1 (d) and define α N = χ N α, the support of α N is included in the ball of radius N and centered at o hence is compact. oreover we have α N α L 2 α L 2 (\B(o,N/2)) hence L 2 lim α N = α N oreover when ϕ C0 (Λ k+1 T ) we have Hence α N D k 1 (d) and α N, d ϕ = α, χ N d ϕ = α, d (χ N ϕ) + α, int grad χ N ϕ = χ N dα + dχ N α, ϕ dα N = χ N dα + dχ N α. But for almost all x, we have dχ N (x) ρ L /N hence dα N dα L 2 ρ L N α L 2 + χ Ndα dα L 2 ρ L N α L 2 + dα L 2 (\B(o,N/2)).

7 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 7 Hence we have build a sequence α N of elements of D k 1 (d) with compact support such that L 2 lim N dα N = dα. A corollary of this lemma (1.5) and of (1.4) is the following : Corollary 1.6. When (, g) is a complete Riemannian manifold then the space of harmonic L 2 forms computes the reduced L 2 cohomology : H k 2 () H k (). With a similar proof, we have another result : Proposition 1.7. When (, g) is a complete Riemannian manifold then H k () = {α L 2 (Λ k T ), (dd + d d)α = 0}. Proof. Clearly we only need to check the inclusion : {α L 2 (Λ k T ), (dd + d d)α = 0} H k (). We consider again the sequence of cut-off functions χ N defined previously in (1.5). Let α L 2 (Λ k T ) satisfying (dd +d d)α = 0 by elliptic regularity we know that α is smooth. oreover we have : d(χ N α) 2 L = [ 2 dχn α dχ N α, χ N dα + χ 2 N dα 2] d vol g [ = dχn α 2 + dχ 2 N α, dα + χ 2 N dα 2] d vol g [ = dχn α 2 + d(χ 2 N α), dα ] d vol g [ = dχn α 2 + χ 2 N α, d dα ] d vol g Similarly we get : d (χ N α) 2 L = 2 [ ] int α 2 + χ 2 grad χ N N α, dd α d vol g Summing these two equalities we obtain : d (χ N α) 2 L + 2 d (χ N α) 2 L = dχ 2 N 2 α 2 d vol g ρ 2 L N 2 Hence when N tends to we obtain α 2 d vol g. d α 2 L 2 + d α 2 L 2 = 0. This proposition has the consequence that on a complete Riemannian manifold harmonic L 2 functions are closed hence locally constant. Another corollary is that the reduced L 2 cohomology of the uclidean space is trivial 3 : Corollary 1.8. H k 2 (R n ) = {0}. 3 This can also be proved with the Fourier transform.

8 8 GILLS CARRON Proof. On the uclidean space R n a smooth k form α can be expressed as α = α I dx i1 dx i2... dx ik I={i 1<i 2<...<i k } and α will be a L 2 solution of the equation (dd + d d)α = 0 if and only if all the functions α I are harmonic and L 2 hence zero because the volume of R n is infinite. Remark 1.9. When (, g) is not complete, we have not necessary equality between the space H k () (whose elements are sometimes called harmonics fields) and the space of the L 2 solutions of the equation (dd + d d)α = 0. For instance, on the interval = [0, 1] the space H 0 () is the space of constant functions, whereas L 2 solutions of the equation (dd + d d)α = 0 are affine. ore generally, on a smooth compact connected manifold with smooth boundary endowed with a smooth Riemannian metric, then again H 0 () is the space of constant functions, whereas the space {f L 2 (), d df = 0} is the space of harmonic L 2 function; this space is infinite dimensionnal when dim > e) Case of compact manifolds The Hodge s theorem says that for compact manifold cohomology is computed with harmonic forms : Theorem If is a compact Riemannian manifold without boundary then H k 2 () H k () H k (). When is the interior of a compact manifold with compact boundary and when g extends to (hence g is incomplete) a theorem of P. Conner ([19]) states that H k 2 () H k () H k abs() where H k abs() = {α L 2 (Λ k T ), dα = d α = 0 and int ν α = 0 along } and ν : T is the inward unit normal vector field. In fact when K is a compact subset of with smooth boundary and if g is a complete Riemannian metric on then for = \ K, we also have the equality H k 2 () H k abs() where if ν : T is the inward unit normal vector field, we have also denoted (1.6) H k abs() = {α L 2 (Λ k T ), dα = d α = 0 and int ν α = 0 along } Some general properties of reduced L 2 cohomology a general link with de Rham s cohomology. We assume that (, g) is a complete Riemannian manifold, the following result is due to de Rham (theorem 24 in [21]) Lemma Let α Z k 2 () C (Λ k T ) and suppose that α is zero in H k 2 () that is there is a sequence β l C 0 (Λ k 1 T ) such that α = L 2 lim l dβ l

9 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 9 then there is β C (Λ k 1 T ) such that α = dβ. In full generality, we know nothing about the behavior of β at infinity. Proof. We can always assume that is oriented, hence by the Poincaré duality (1.1.2), we only need to show that if ψ C0 (Λ n k T ) is closed then α ψ = 0. But by assumption, α ψ = lim l = lim l = 0. Hence the result. dβ l ψ d(β l ψ) because ψ is closed This lemma implies the following useful result which is due to. Anderson ([2]): Corollary There is a natural injective map Im ( H k 0 () H k () ) H k 2 (). Proof. As a matter of fact we need to show that if α C0 (Λ k T ) is closed and zero in the reduced L 2 cohomology then it is zero in usual cohomology: this is exactly the statement of the previous lemma (1.11) Consequence for surfaces. These results have some implications for a complete Riemannian surface (S, g) : i) If the genus of S is infinite then the dimension of the space of L 2 harmonic 1 forms is infinite. ii) If the space of L 2 harmonics 1 forms is trivial then the genus of S is zero and S is diffeomorphic to a open set of the sphere. As a matter of fact, a handle of S is a embedding f : S 1 [ 1, 1] S such that if we denote A = f(s 1 [ 1, 1]) then S \ A is connected. A S c

10 10 GILLS CARRON We consider now a function ρ on S 1 [ 1, 1] depending only on the second variable such that { 1 when t > 1/2 ρ(θ, t) = 0 when t < 1/2 then dρ is a 1 form with compact support in S 1 ] 1, 1[ and we can extend α = ( f 1) dρ to all S ; we obtain a closed 1-form also denoted by α which has compact support in A. oreover because S \ A is connected, we can find a continuous path c : [0, 1] S \ A joining f(1, 1) to f(1, 1); we can defined the loop γ given by { f(1, t) for t [ 1, 1] γ(t) = c(t 1) for t [1, 2]. It is easy to check that α = 1, γ hence α is not zero in H 1 (S). A little elaboration from this argument shows that genus(s) dim Im ( H 1 0 () H 1 () ) Lott s result. We will now prove the following result due to J.Lott ([44]): Theorem Assume that ( 1, g 1 ) and ( 2, g 2 ) are complete oriented manifold of dimension n which are isometric at infinity that is to say there are compact sets K 1 1 and K 2 2 such that ( 1 \ K 1, g 1 ) and ( 2 \ K 2, g 2 ) are isometric. Then for k [0, n] N dim H k ( 1, g 1 ) < dim H k ( 2, g 2 ) <. We will give below the proof of this result, this proof contains many arguments which will be used and refined in the next two lectures. In view of the Hodgede Rham theorem and of J. Lott s result, we can ask the following very general questions : (1) What are the geometry at infinity iensuring the finiteness of the dimension of the spaces H2 k ()? Within a class of Riemannian manifold having the same geometry at infinity : (2) What are the links of the spaces of reduced L 2 cohomology H2 k () with the topology of and with the geometry at infinity of (, g)? There is a lot of articles dealing with these questions, I mention some of them : (1) In the pioneering article of Atiyah-Patodi-Singer ([5]), the authors considered manifold with cylindrical end : that is to say there is a compact K of such that \ K is isometric to the Riemannian product K ]0, [. Then they show that the dimension of the space of L 2 -harmonic forms is finite ; and that these spaces are isomorphic to the image of the relative cohomology in the absolute cohomology. These results were used by Atiyah- Patodi-Singer in order to obtain a formula for the signature of compact anifolds with boundary. (2) In [48, 50], R. azzeo and R.Phillips give a cohomological interpretation of the space H k () for geometrically finite real hyperbolic manifolds. These manifolds can be compactified. They identify the reduced L 2 cohomology with the cohomology of smooth differential forms satisfying certain boundary conditions.

11 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 11 (3) The solution of the Zucker s conjecture by L.Saper-.Stern and.looijenga ([46],[58]) shows that the spaces of L 2 harmonic forms on Hermitian locally symmetric space with finite volume are isomorphic to the middle intersection cohomology of the Baily-Borel-Satake compactification of the manifold. An extension of this result has been given by A.Nair and L.Saper ([52],[56]). oreover recently, L. Saper obtains the topological interpretation of the reduced L 2 cohomology of any locally symmetric space with finite volume ([57]). In that case the finiteness of the dimension of the space of L 2 harmonics forms is due to A. Borel and H.Garland ([8]). (4) According to Vesentini ([63]) if is flat outside a compact set, the spaces H k () are finite dimensional. J. Dodziuk asked about the topological interpretation of the space H k () ([22]). In this case, the answer has been given in [15]. (5) In a recent paper ([33]) Tamás Hausel, ugenie Hunsicker and Rafe azzeo obtain a topological interpretation of the L 2 cohomology of complete Riemannian manifold whose geometry at infinity is fibred boundary and fibred cusp (see [49, 62]). These results have important application concerning the Sen s conjecture ([34, 59]). (6) In [45], J. Lott has shown that on a complete Riemannian manifold with finite volume and pinched negative curvature, the space of harmonic L 2 forms has finite dimension. N. Yeganefar obtains the topological interpretation of these spaces in two cases, first when the curvature is enough pinched ([65]) and secondly when the metric is Kähler ([66]). Proof of J. Lott s result. We consider (, g) a complete oriented Riemannian manifold and K a compact subset with smooth boundary and we let = \ K be the exterior of K, we are going to prove that dim H k 2 () < dim H k 2 () < ; this result clearly implies Lott s result. The co boundary map b : H k 2 () H k+1 (K, K) is defined as follow: let c = [α] H k 2 () where α is a smooth representative of c, we choose ᾱ C (Λ k T ) a smooth extension of α, then dᾱ is a closed smooth form with support in K and if ι : K K is the inclusion we have ι (dᾱ) = 0. Some standard verifications show that b(c) = [dᾱ] H k+1 (K, K) is well defined, that is it doesn t depend of the choice of α c nor on the smooth extension of α. The inclusion map j : induced a linear map (the restriction map) Lemma We always have [j ] : H k 2 () H k 2 (). ker b = Im[j ]. Proof of lemma First by construction we have b [j ] = 0 hence we only need to prove that ker b Im[j]. Let c ker b and let α Z k 2 () a smooth representative of c, we know that α has a smooth extension ᾱ such that dᾱ is zero in H k+1 (K, K). That is to say

12 12 GILLS CARRON there is a smooth k form β C (Λ k T K) such that dᾱ = dβ on K and ι β = 0 We claim that the L 2 form α defined by { ᾱ β on K α = α on is weakly closed. As a matter of fact, we note ν : K T the unit normal vector field pointing into and ι : K the inclusion, then let ϕ C0 (Λ k+1 T ) with the Green s formula, we obtain ( α, d ϕ) = (ᾱ β, d ϕ) + (α, d ϕ) K = (ι (ᾱ β), int ν ϕ)dσ + (ι ᾱ, int ν ϕ)dσ = 0. K We clearly have j α = α, hence c = [α] Im[j ]. Now because H k+1 (K, K) has finite dimension, we know that Hence we get the implication : K dim Im[j ] < dim H k 2 () <. dim H k 2 () < dim H k 2 () <. To prove the reverse implication, we consider the reduced L 2 cohomology of relative to the boundary = K. We introduce Z2 k (, ) = { α L 2 (Λ k T ), such that ϕ C0 (Λ k T ), α, d ϕ = 0 } (. = d C0 (Λ k T )) We remark here that the elements of C0 (Λ k T ) have compact support in in particular their support can touch the boundary; in fact a smooth L 2 closed k form belongs to Z2 k (, ) if and only its pull-back by ι is zero. This is a consequence of the integration by part formula (α, d ϕ)d vol g = (dα, ϕ)d vol g + (ι α, int ν ϕ)dσ where dσ is the Riemannian volume on induced by the metric g and ν : T is the unit inward normal vector field. We certainly have dc0 (Λ k T ) Z2 k (, ) and we define H k 2 (, ) = Zk 2 (, ) dc 0 (Λk T ). These relative (reduced) L 2 cohomology space can be defined for every Riemannian manifold with boundary. In fact, these relative (reduced) L 2 cohomology spaces also have an interpretation in terms of harmonics forms : (1.7) H k 2 (, ) H k rel() where H k rel() = { } α L 2 (Λ k T ), dα = d α = 0 and ι α = 0.

13 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 13 There is a natural map : the extension by zero map : e : H k 2 (, ) H k 2 (). When α L 2 (Λ k T ) we define e(α) to be α on and zero on K, e : L 2 (Λ k T ) L 2 (Λ k T ) is clearly a bounded map moreover ( ) e dc0 (Λ k T ) dc0 (Λ k T ), ( ) hence by continuity of e : e dc0 (Λk T ) dc0 (Λk T ). oreover when α Z2 k (, ) then e(α) Z2 k () : because if ϕ C0 (Λ k+1 T ) then e(α), d α = (α, d (jϕ))d vol g but (j ϕ) C 0 (Λ k T ) hence this integral is zero by definition of Z2 k (, ). Let j K : K the inclusion map, it induces as before a linear map [j K] : H k 2 () H k 2 (K) H k (K). We always have [j K ] e = 0 hence Im e ker[j K]. In fact as before Lemma We have the equality Im e = ker[j K]. Proof of the lemma If c ker[jk ] and α L2 (Λ k T ) is a smooth representative of c (for instance α is L 2 and harmonic). By definition we know that there is β C (Λ k 1 T K) such that j Kα = dβ Now consider β any smooth extension of β with compact support. We clearly have oreover by construction [α d β] = [α] = c in H k 2 (). α d β = e(j (α d β)) If we verify that j (α d β) Z2 k (, ) we have finish the proof of the equality the lemma In fact, this verification is straightforward. Let ϕ C0 (Λ k T ) and consider ϕ any smooth extension of ϕ : (j(α d β), d ϕ)d vol g = (α d β, d ϕ)d vol g = 0. Again since H k (K) have finite dimension, the kernel of [jk ] have finite codimension in H2 k () and we have obtain the implication dim H k (, ) < dim H k 2 () <. In order to conclude, we use the Hodge star operator; because our manifold is oriented, the Hodge star operator is an isometry which exchanges k forms and (n k) forms : : Λ k T x Λ n k T x.

14 14 GILLS CARRON This operator satisfies the following properties = ± Id where the sign depends on the degree. oreover we have : d = ± d. Hence the Hodge star operator maps the space H k () to H n k (). oreover it is also a good exercise to check that in our setting : ι ( α) = ± int ν α. Hence the Hodge star operator maps the space Habs k () to Hn k rel (). But our last result says that dim H k rel() < dim H k 2() <. Hence using the Hodge star operator we get dim H n k abs () < dim Hn k 2 () <. That is dim H2 n k () < dim H2 n k () <. It is now clear that we have prove Lott s results. Remarks i) The reader can verify that using forms with coefficients in the orientation bundle, we can removed the orientability condition. ii) These two properties are in fact a heritage of the following two exact sequences for the de Rham cohomology:... H k (K, K) H k () H k () H k+1 (K, K) H k (, ) H k () H k (K) H k+1 (, ) Some bibliographical hints. We recommend the reading of the classical book of G. de Rham [21] or W. Hodge [35]. For a modern treatment of the de Rham s theorem a very good reference is the book of Bott-Tu [9]. The Hodge theorem for compact manifold with boundary have been proved by P.. Conner [19]. Other proofs of the Hodge-de Rham theorem can be found in other classical book (for instance in the book of Griffith-Harris (chapter 0 section 6 in [27]) or in the book of. Taylor (chap. 5 in [60]). We should also mentioned a sheaf theoretical proof of the Hodge-de Rham theorem by N. Telemann [61]. About the general feature on non compact manifold, you can read the paper of J. Dodziuk [22], the now classical paper of J. Cheeger [18], the first section of the paper by J. Lott [44] or look at [13] and also read the beautiful paper of. Anderson [2]. The paper of J. Brüning and. Lesch deals with an abstract approach about the identification between the space of L 2 harmonic form and L 2 cohomology [10]. For a first approach on the L 2 cohomology of symmetric space, the paper of S. Zucker [67] is very nice, the first parts of the survey of W. Casselman is also instructive [17]. 2. Harmonics L 2 1 forms In this second lecture, we survey some results between the space of harmonics L 2 1 forms, cohomology with compact support and the geometry of ends nds.

15 L 2 HARONICS FORS ON NON COPACT ANIFOLDS Definitions. When U is a subset of an open manifold we say that U is bounded if U is a compact subset of, when U is not bounded we say that U is unbounded. When g is complete Riemannian metric on, then U is bounded if and only if there is some R > 0 and o such that U B(o, R). Let be a smooth manifold, we say that has only one end if for any compact subset K, \ K has only one unbounded connected component. For instance, when n 2, the uclidean space R n has only one end. ore generaly, we say that has k ends (where k N) if there is a compact set K 0 such that for every compact set K containing K 0, \ K has exactly k unbounded connected components. A compact manifold is a manifold with zero end. For instance, R or R S 1 have two ends. The topology at infinity of manifold with only one end can be very complicated Number of ends and cohomology. Here is a smooth non compact connected manifold. Lemma 2.1. If has at least two ends, then In fact, if has at least k ends then H 1 0 () {0}. dim H 1 0 () k 1. Proof. If has more than two ends, then we can find a compact set K such that \ K = U U + where U, U + are unbounded and U U + =. U K U + Let u C () such that u = ±1 on U ±, then clearly α = du is a closed 1 form with compact support. If the cohomology class of α in H0 1 () is trivial, then we find some f C0 () such that du = α = df. But is connected hence we have a constant c such that u = f + c. We look at this equation outside the support of f and on U ± we find that c = ±1. In fact there is a weak reciprocal to this result (see proposition 5.2 in [16]) :

16 16 GILLS CARRON Proposition 2.2. If n is an open manifold having one end, and if every twofold normal covering of has also one end, then H 1 0 (, Z) = {0}. In particular, H 1 0 () = {0} and if furthermore is orientable, then H n 1 (, Z) = {0} H 1 0 () versus H 1 2 (). From now we assume that (, g) is a complete Riemannian manifold An easy case. Lemma 2.3. Assume that has only one end, then {0} H 1 0 () H 1 2 (). Proof. Let α C 0 (T ) be a closed 1 form which is zero in H 1 2 (). By the result (1.11), we know that there is a smooth function f C () such that α = df. But \ supp α has only one unbounded connected component U, hence on U df = 0 hence there is a constant c such that f = c on U. Now by construction the function f c has compact support and α = d(f c). Hence α is zero in H0 1 () In fact the main purpose of this lecture is to go from geometry to topology : we want to find geometrical conditions insuring that this map is injective Condition involving the spectrum of the Laplacian. Proposition 2.4. Assume that all ends of have infinite volume 4 and assume that there is a λ > 0 such that (2.1) f C0 (), λ f 2 d vol g df 2 d vol g. Then {0} H 1 0 () H 1 2 (). Proof. Let [α] H 1 0 () is mapped to zero in H 1 2 (). Hence there is a sequence (f k ) of smooth functions with compact support on such that α = L 2 lim df k. Since we have the inequality df k df l 2 L 2 λ f k f l 2 L 2, and since λ > 0, we conclude that this sequence (f k ) converges to some f L 2, so that α = df. But α has compact support, hence f is locally constant outside the compact set supp(α). Since all unbounded connected components of supp(α) have infinite volume and since f L 2, we see that f has compact support, hence [α] = [df] = 0 in H0 1 (). Before giving some comments on the hypothesis on this proposition, let s give a consequence of this injectivity : 4 that is ouside every compact subset of, all unbounded connected components have infinite volume.

17 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 17 Proposition 2.5. Assume that and that has at least k ends then {0} H 1 0 () H 1 2 (), (2.2) dim { h C (), h = 0 and dh L 2} k. A general formula for the dimension of the space of bounded harmonic function with L 2 gradient has been obtained by A. Grigor yan ([31]). Proof. Assume that k = 2 (the other cases are similar), there is a compact set K such that \ K = U U + with U, U + unbounded and U U + =. Let u C () such that u = ±1 on U ± we know that α = du is not zero in cohomology with compact support, hence we know that there is a non zero harmonic L 2 1 form η such that [α] = [η] in H 1 2 (); in particular with (1.11), we can find v C () such that du = α = η + dv that is if h = u v then dh = η L 2 and h = d d(u v) = d η = 0. And because η 0 we know that h is not the constant function. The linear span of h and of the constant function 1 is of dimension 2 and is include in dim { h C (), h = 0 and dh L 2}. Remark 2.6. In the setting of the proposition (2.4), we can give a direct and more classical proof of this inequality : we assume that (, g) satisfies that the assumption made in proposition (2.4), we are going to show the inequality (2.2). Again we assume that k = 2 : there is a compact set K such that \ K = U U + with U, U + unbounded and U U + =. Let o be a fixed point and for large k we have K B(o, k). We consider the solution of the Dirichlet problem : { uk = 0 on B(o, k) u k = ±1 on B(o, k) U ± Then u k is the minimizer of the functional u B(o,k) du 2 amongst all functions in W 1,2 (B(o, k)) (functions in L 2 (B(o, k)) whose derivatives are also in L 2 ) such that u = ±1 on U ± B(o, k) ). We extend u k to all by setting u k = ±1 on U ± \ B(o, k). Then this extension (also denote by u k ) is the minimizer of the functional u du 2 amongst all functions in W 1,2 loc (functions in L2 loc whose derivatives are also in L2 loc ) such that u = ±1 on U ± \ B(o, k). Hence we always have when k < l : du l 2 du k 2.

18 18 GILLS CARRON oreover by the maximum principle, we always have 1 u k 1. After extraction of a subsequence we can assume that uniformly on compact set lim u k = u. k The function u is then a harmonic function whose value are in [ 1, 1] and it satisfies du 2 du k 2 <. We must show that this u is not a constant function. We apply our estimate to the function u k u l where l k, then we get λ u k u l 2 du k du l 2 4 du k 2 In particular if we let l, we find that that is u k u L 2 U ± u (±1) 2 <. But by hypothesis the volume of U ± are infinite hence u cannot be the constant function. We know make some comments on the assumption of the proposition (2.4) : On the first condition. We have the following useful criterion : Lemma 2.7. Assume that there is v > 0 and ε > 0 such that for all x then all ends of have infinite volume. vol B(x, ε) v Proof. We fix a based point o and let K be a compact set and U an unbounded connected component of \ K. Let R > 0 large enough so that K B(o, R) (recall that we have assumed here that the Riemannian manifold (, g) is complete). For k N we choose x B(o, R + (2k + 1)ε) U and we consider γ : [0, R+(2k +1)ε] a minimizing geodesic from o to x (parametrize by arc- length). Necessary for all t ]R, R + (2k + 1)ε] we have γ(t) U ; moreover for l = 0, 1,..., k, the open geodesic ball B(γ(R + (2l + 1)ε), ε) are in U and disjoint hence k vol U vol B ( γ(r + (2l + 1)ε), ε ) (k + 1)v. l=0

19 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 19 γ(r) γ(r + 11ε) o R K For instance, when the injectivity radius of (, g) is positive, C. Croke has shown in ([20]) that for all x and all r inj() then vol B(x, r) C n r n. Hence a complete Riemannian manifold with positive injectivity radius has all its ends with infinite volume. About the second condition The condition (2.1) is clearly equivalent to { } λ 0 (, g) := inf f C0 () df 2 d vol g f 2 d vol g > 0. This condition is linked with the spectrum of the Laplacian on functions. In fact λ 0 (, g) is the bottom of the spectrum of the operator d d = acting on functions. Because the Riemannian manifold (, g) is complete, the operator d d = : C 0 () L 2 (, d vol g ) has a unique self adjoint extension with domain : D( ) = {f L 2 (), f L 2 ()} = {f L 2 (), df L 2 () and d df L 2.} Hence f D( ) if and only if f D 0 (d) and there is C R such that ϕ C 0 (), df, dϕ C ϕ L 2. The spectrum of g is a closed subspace of [0, [ and the condition λ 0 (, g) > 0 is equivalent to zero not being in the spectrum of. Also this condition depends only of the geometry at infinity, in fact we have that λ 0 (, g) > 0 if and only if there is K a compact subset such that λ 0 ( \ K, g) > 0 and vol() =.. From the proof of the proposition (2.4) it is clear that we have the following result : Proposition 2.8. Assume that (, g) is a complete Riemannian manifold such that λ 0 (, g) > 0, then if we introduce H 1 v () = {α C 0 (T ), such that dα = 0} {df, f C (), such that df C 0 (T ) and vol(supp f) < }

20 20 GILLS CARRON then {0} H 1 v () H 1 2 (). Optimality of the results : The real line (R, (dt) 2 ) has clearly all his ends with infinite volume but λ 0 (R, dt 2 ) = 0 ; as a matter of fact if u C0 (R) is not the zero function, then for u n (t) = u(t/n) we have u n(t) 2 dt = 1 u (t) 2 dt n R and u n (t) 2 dt = n u (t) 2 dt. R R But the first group of cohomology with compact support of R has dimension 1 where as H 1 (R, dt 2 ) = {fdt, f L 2 and f = 0} = {0}. We consider now the manifold Σ = R S 1 endowed with the warped product metric : g = (dt) 2 + e 2t (dθ) 2. have two ends one with finite volume and the other of infinite volume. Again the first group of cohomology with compact support of Σ has dimension 1. oreover as a matter of fact let f C0 (Σ) df 2 d vol g Σ R λ 0 (Σ) 1 4 ; Σ 2 f (t, θ) t e t dtdθ Let f = e t/2 v then 2 f (t, θ) Σ t e t dtdθ = v Σ t v dtdθ [ 2 = v R S t 1 [ 2 = v R S t v [ 2 = v R S t + 1 ] 4 v 2 dtdθ 1 1 v 2 dtdθ 4 R S 1 = 1 f 2 d vol g. 4 Σ v 2 + v v ] dtdθ t v 2 t ] dtdθ In fact we can show (using an appropriate test function u n (t, θ) = χ n (t)e t/2 ) that λ 0 (Σ) = 1 4. In fact we can show that H 1 (Σ) has infinite dimension (see the first part of theorem 2.16), however if the conclusion of the proposition 2.4) were true for (Σ, g) then by the alternative proof of (2.2) in (2.6), we would find a non constant harmonic

21 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 21 function h with L 2 gradient. If we look at the construction of this function, it is not hard to check that h will only depends on the first variable that is where f solves the O.D.. h(t, θ) = f(t) (e t f ) = 0 ; hence there is constant A and B such that h(t, θ) = Ae t + B. The fact that dh L 2 implies that (f ) 2 (t)e t dt < R that is A = 0 and h is the constant function. This simple proposition 2.4 or variant of it has been used frequently, I will mention only two results, the first one is the following very beautifully result of P. Li and J. Wang ([40]): (see also the other articles of P. Li and J. Wang for other related results [41, 43]) Theorem 2.9. Assume that (, g) is a complete Riemannian manifold of dimension n > 2 with λ 0 () (n 2) assume moreover that its Ricci curvature satisfies the lower bound Ric g (n 1)g then either (, g) has only one end of infinite volume either (, g) has two ends with infinite volume and is isometric to the warped product R N endowed with the metric (dt) 2 + cosh 2 (t)h where (N, h) is a compact Riemannian manifold with Ric g (n 2)g The second concerns the locally symmetric space ([16]) : Theorem Let G/K be a symmetric space without any compact factor and without any factor isometric to a real or complex hyperbolic space. Assume that Γ G is a torsion-free, discrete subgroup of G such that Γ\G/K is non compact and that all ends of Γ\G/K have infinite volume. Then Γ\G/K has only one end, and H m 1 (Γ\G/K, Z) = {0}, where m = dim(g/k) Condition involving a Sobolev inequality. Proposition Assume that (, g) is a complete manifold that satisfies for a ν > 2 and µ > 0 the Sobolev inequality : ( ) 1 2 (2.3) f C0 (), µ f 2ν ν ν 2 d volg df 2 d vol g. Then {0} H 1 0 () H 1 2 ().

22 22 GILLS CARRON The proof follows essentially the same path : if [α] H0 1 () is map to zero in H2 1 (), then we find f L 2ν ν 2 such that α = df. Hence f is locally constant outside the support of α. But according to (proposition 2.4 in [12]), we know that the Sobolev inequality (2.3) implies a uniform lower bound on the volume of geodesic balls : x, r 0 : vol B(x, r) C(ν) ( µr 2) ν/2 for some explicit constant C(ν) > 0 depending only on ν. Hence by (2.7), we know that such an estimate implies that all the unbounded connected components of \ supp α have infinite volume. Hence f has compact support and [α] = 0 in H0 1 (). xamples of manifolds satisfying the Sobolev inequality : i) The uclidean space R n (n 3) satisfies the Sobolev inequality for ν = n. ii) If n R N is a minimal submanifold then n with the induced metric satisfies the Sobolev inequality for ν = n [51, 36]. iii) A Cartan-Hadamard manifold (a simply connected non positively curved complete Riemannian manifold) of dimension n 3 satisfies the Sobolev inequality for ν = n. iv) The hyperbolic space of dimension n, satisfies the Sobolev inequality for any ν [n, [ ]2, [. In fact the validity of the Sobolev inequality depends only of the geometry at infinity : according to [14], if K is a compact subset of then the Sobolev inequality (2.3) holds for if and only if it holds for \ K: f C 0 ( \ K), µ ( \K ) 1 2 ν f 2ν ν 2 d volg \K df 2 d vol g. This proposition (2.11) has the following beautiful application due to Cao-Shen- Zhu [11] (see also [42]) however these authors haven t prove the vanishing of the first group of cohomology with compact support : Corollary If n R n+1 is a stable complete minimal hypersurface then has only one end, moreover H 1 0 () = {0}. Proof. We only need to show that H 1 () = {0}. Let α H 1 () then it satisfies the Bochner identity [ α 2 + Ric(α, α) ] d vol g = 0. However the Gauss equations imply that Ric(α, α) A 2 α 2, where A is the second fundamental form of the hypersurface R n+1. The stability condition says that the second variation of the area is non negative that is f C0 [ (), J(f) := df 2 A 2 f 2] d vol g 0.

23 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 23 But the refined Kato inequality (due in this case to S.T. Yau) shows that α 2 n d α 2. n 1 Hence we get [ 0 = α 2 + Ric(α, α) ] d vol g [ ] n n 1 d α 2 + Ric(α, α) d vol g [ ] 1 n 1 d α 2 + d α 2 A 2 α 2 d vol g J( α ) + 1 d α 2 d vol g. n 1 1 d α 2 d vol g. n 1 Hence α has constant length, but because of the Sobolev inequality, the volume of (, g) is infinite and α = What is behind the injectivity of the map H 1 0 () H 1 2 (). In fact there is a general notion from potential theory which is related to the injectivity of this map (see the survey of A. Ancona [1] or [39]). Definition Let be an open connected set with smooth compact boundary, the following properties are equivalent: i) there is a positive super-harmonic function : s : R + with and lim inf x s(x) = inf s(x) = 0 x inf s(x) 1. x ii) There is a positive harmonic function : h : R + with lim inf x h(x) = inf h(x) = 0 x and h = 1 on. oreover dh L 2. iii) The capacity of is positive : { } cap() = inf dv 2, v C0 () and v 1 on > 0. iv) For any U bounded open subset of there is a constant C U > 0 such that f C0 (), C U f 2 df 2. U v) For some U bounded open subset of there is a constant C U > 0 such that f C0 (), C U f 2 df 2. When one of these properties holds we say that is non parabolic and if one of these properties fails we say that is parabolic. U

24 24 GILLS CARRON Proof. We clearly have ii) i) and iv) v). We first prove that i) iv). Let ν : T be the unit inward normal vector field along. When v C0 () we set ϕ = v/ s so that dv 2 = s dϕ 2 ds 2 + ϕ ds, dϕ + 4s ϕ2. But 2 ϕ ds, dϕ = = ds, dϕ 2 ( s)ϕ 2 ϕ 2 ds( ν). But s is assumed to be super-harmonic hence s 0 and ds( ν) 0 along, hence we finally obtain the lower bound dv 2 ds 2 4s ϕ2 = 1 d log s 2 v 2. 4 By assumption, s is not constant hence we can find an open bounded set U and ε > 0 such that d log s > ε on U and we obtain that for all v C0 () dv 2 ε2 v 2. 4 We prove now that v) iv). Let U such as in iv). For o a fixed point and R such that U B(o, R), we will prove that there is a constant C R > 0 such that f C0 (), C R f 2 df 2. U B(o,R) Let V U be an non empty open set with V U and let ρ C () such that supp ρ U and 0 ρ 1 and ρ = 1 in V. Then for f C0 () we have : f 2 2 (ρf) ((1 ρ)f) 2 B(o,R) 2 B(o,R) U 2 C U f df λ B(o,R) B(o,R) ((1 ρ)f) 2 B(o,R) d((1 ρ)f) 2 Where λ > 0 is the first eigenvalue of the Laplacian on functions on (B(o, R) )\V for the Dirichlet boundary condition on V. But d((1 ρ)f) 2 2 df dρ 2 L f 2 B(o,R) Hence the result for C R = B(o,R) ( dρ 2 L C U ) df 2. ( ( )) dρ 2 L. C U λ C U U

25 L 2 HARONICS FORS ON NON COPACT ANIFOLDS. 25 Now we consider the implication iii) ii). We introduce C(R) = inf dv 2 B(o,R) where the infimum runs over all functions v C ( B(o, R)) such that v 1 on and v = 0 on B(o, R) ; where o is a fixed point and where R > 0 is chosen large enough so that B(o, R). We have C(R) > 0 and we have assumed that C( ) = inf R C(R) > 0. ach C(R) is realized by the harmonic function h R such that h R = 1 on and h R = 0 on B(o, R). We extend h R by zero on \ B(o, R), an application of the maximum principle implies that when R R then h R h R. We let h(x) = sup R h R (x). On compact subset of, h R converge to h in the smooth topology moreover h is a harmonic function with 0 h 1. The Green formula shows that C(R) = dh R ( ν), hence when R we obtain C( ) = dh( ν) > 0. In particular h is not the constant function and h > 0 on by the maximum principle. We also have dh 2 lim inf dh R 2 = lim inf C(R) = C( ) <. R R We must show that inf h = 0. Let v C 0 () such that v 1 on, we let v = hϕ then using the Green formula we get : dv 2 = h 2 dϕ hϕ dh, dϕ + dh 2 ϕ 2 = h 2 dϕ dh 2, dϕ 2 + dh 2 ϕ 2 2 = h 2 dϕ (h 2 )ϕ 2 ϕ 2 dh( ν) + 2 h 2 dϕ 2 dh( ν) (inf h)2 C( ) + C( ) Taking the infimum over such v we obtain that 0 (inf h)2 C( ) dh 2 ϕ 2 hence inf h = 0 It remains to show that iv) iii). We re consider the preceding notation. Under the hypothesis iv), we must show that C( ) > 0. Or by contraposition that if C( ) = 0 then iv) can t be true. So we assume that C( ) = 0, in that case we know that h is the constant function 1 and we get for the functions h R : h 2 R = vol U lim R U

26 26 GILLS CARRON where as That is iv) is not true lim dh R 2 = 0. R Proposition If all ends of are non parabolic then {0} H 1 0 () H 1 2 (). In fact if has more than k non parabolic ends then dim { h C (), h = 0 and dh L 2} k. The above inequality is due to P. Li and L-F Tam ([39]). Proof. We remark that because contains at most one non parabolic end, the above proof of (2.13 v) iv) ) shows that for any U bounded open subset there is C U > 0 such that f C0 (), C U f 2 df 2. We consider a closed form α C 0 (T ) which is map to zero in H 1 2 (). Hence there is a sequence v k C 0 () such that U α = L 2 lim k dv k. By the above remark, we know that there is v C () such that dv = α and v = L 2 loc lim k v k. Let be a unbounded connected component of \ supp α, v is then constant on v = v() on. We want to show that this constant is zero. We can enlarge \ and assume that has smooth boundary. On we have dv k 2 = α 2 = 0. lim k Choose U a non empty bounded open subset of, we know that for a certain constant C U > 0 we have the estimate dv k 2 C U vk, 2 When k tends to we obtain U 0 C U vol U v() 2. Hence v is zero on and v has necessary compact support The two dimensional case. In dimension 2, a remarkable property of the space of L 2 harmonic 1 forms is that it is an invariant of conformal structure (or of the complex structure).

27 L 2 HARONICS FORS ON NON COPACT ANIFOLDS Conformal invariance. Proposition Let ḡ = e 2u g be two conformally equivalent Riemannian metric 5 on a smooth manifold 2m, then then we have the equality : Proof. When α Λ k T x, we have H m (, ḡ) = H m (, g). α 2 ḡ = e 2ku(x) α 2 g. and d volḡ = e 2mu d vol g. As a consequence, the two Hilbert spaces L 2 (Λ m T, ḡ) and L 2 (Λ m T, g) are isometric, hence by definition the two space Z2 m (, ḡ) = {α L 2 (Λ m T, ḡ), dα = 0} and Z2 m (, ḡ) = {α L 2 (Λ m T, g), dα = 0} are the same. oreover the orthogonal of dc0 (Λ m 1 T ) is the same in L 2 (Λ m T, g) or L 2 (Λ m T, ḡ). But by definition H m (, ḡ) = Z m 2 (, ḡ) ( dc 0 (Λ m 1 T ) ). is also equal to H m (, g). Another proof is to compute the codifferential d ḡ : for α C (Λ k T ) we obtain ( ) d ḡα = e 2u d gα 2(m k) int α. grad u Application. Theorem Let (S, g) be a complete connected Riemannian surface with finite topology (finite genus and finite number of ends) then either dim H 1 (S, g) = or dim H 1 (S, g) < and is conformally equivalent to a compact Riemannian surface ( S, ḡ) with a finite number of points removed : and (S, g) ( S \ {p 1,..., p k }, ḡ) H 1 (S, g) Im ( H 1 0 (S) H 1 (S) ) H 1 ( S). Proof. We know that a Riemannian surface with finite topology is necessary conformally equivalent to a compact Riemannian surface ( S, ḡ) with a finite number of points and disks removed : ( ( ) ) (S, g) S \ b l=1 D l {p 1,..., p k }, ḡ Hence from our previous result, ( we have ( ) ) H 1 (S, g) = H 1 S \ b l=1 D l {p 1,..., p k }, ḡ. We first show that if b 1 then dim H 1 (S) =. Let f C ( D 1 ), then on S \D 1 we can solve the Dirichlet problem : { ḡu = 0 on S \ D 1 u = f on D 1 5 The metric ḡ and g are not necessary complete.

28 28 GILLS CARRON Then u C ( S \ D 1 ) hence du L 2 (T ( ) S \ D1, ḡ) is closed and coclosed hence du H 1 ( S \ D 1, ḡ) its restriction to S = S ( ) \ b l=1 D l {p 1,..., p k } is also L 2 closed and co closed, hence we have build a linear map f C ( D 1 ) du S H 1 (S, g) = H 1 (S, ḡ), the kernel of this map is the set of constant functions hence we have proved that Now we assume that b = 0 that is we have dim H 1 (S, g) =. (S, g) ( S \ {p 1,..., p k }, ḡ), H 1 (S, g) = H 1 ( S \ {p 1,..., p k }, ḡ). The main point is that when α H 1 ( S \ {p 1,..., p k }, ḡ) then α extends across the point {p 1,..., p k }. This is a direct consequence of the following lemma Lemma Let D be the unit disk, and let α L 2( T (D \ {0}) ) satisfying the equation (d + d )α = 0, then α extends smoothly across 0 that is H 1 (D \ {0}) = H 1 (D). In this lemma the metric on the disk can be considered as the uclidean one, as a matter of fact any other Riemannian metric on D is conformally equivalent to the flat metric. We offer two proof of this result. First proof of the lemma. Let α H 1 (D \ {0}), we are going to prove that (d + d )α = 0 holds weakly on D, then because the operator (d + d ) is elliptic it will hold strongly by elliptic regularity. Consider ϕ C0 (D) C0 (Λ 2 T D) we must show that α, (d + d )ϕ = 0. We consider the following sequence of cutoff functions 0 when r 1/n 2 χ n (r, θ) = log(r/n2 ) log n when 1/n 2 r 1/n 1 when r 1/n By hypothesis α, (d + d )(χ n ϕ) = 0. But (d + d )(χ n ϕ) = χ n (d + d )ϕ + dχ n ϕ + int grad χ n ϕ. We have then 0 = α, (d + d )(χ n ϕ) = α, χ n (d + d )ϕ) + α, dχ n ϕ + int grad χ n ϕ But when n, the first term in the right hand side goes to α, (d + d )ϕ : lim α, χ n(d + d )ϕ) = α, (d + d )ϕ) n and the second term is estimate as follow : α, dχ n ϕ + int ϕ grad χ n α L 2 ϕ L dχ n L 2 But a direct computation shows that dχ n 2 L 2 = 2π log n.