For a Riemannian covering $M_1\to M_0$ of complete Riemannian manifolds with
boundary (possibly empty) and respective fundamental groups
$\Gamma_1\subseteq\Gamma_0$, we show that the bottoms of the spectra of $M_0$
and $M_1$ coincide if the right action of $\Gamma_0$ on
$\Gamma_1\backslash\Gamma_0$ is amenable.Comment: 8 pages, fixed a technical mistake concerning the volume of the
  boundary of fundamental domain

Ballmann, Werner

Matthiesen, Henrik

Polymerakis, Panagiotis

English

arXiv

Ballmann, W. 

Matthiesen, H.

Polymerakis, P. 

MPG.PuRe

Math. Z. (2018) 288:1029–1036https://doi.org/10.1007/s00209-017-1925-9 Mathematische ZeitschriftOn the bottom of spectra under coveringsWerner Ballmann1,2 · Henrik Matthiesen1,2 ·Panagiotis Polymerakis3Received: 9 January 2017 / Accepted: 16 July 2017 / Published online: 30 September 2017© The Author(s) 2017. This article is an open access publicationAbstract For a Riemannian covering M1 → M0 of connected Riemannian manifolds withrespective fundamental groups 1 ⊆ 0, we show that the bottoms of the spectra of M0 andM1 coincide if the right action of 0 on 1\0 is amenable.Keywords Bottom of spectrum · Amenable coveringMathematics Subject Classification 58J50 · 35P15 · 53C991 IntroductionIn this article, we study the behaviour under coverings of the bottom of the spectrum ofSchrödinger operators on Riemannian manifolds.Let M be a connected Riemannian manifold, not necessarily complete, and V : M → Rbe a smooth potential with associated Schrödinger operator  + V . We consider  + V asan unbounded symmetric operator in the space L2(M) of square integrable functions on Mwith domain C∞c (M), the space of smooth functions on M with compact support.We would like to thank the Max Planck Institute for Mathematics and the Hausdorff Center for Mathematicsin Bonn for their support.B Werner Ballmannhwbllmnn@mpim-bonn.mpg.deHenrik Matthiesenhematt@mpim-bonn.mpg.dePanagiotis Polymerakispolymerp@hu-berlin.de1 Max Planck Institute for Mathematics, Vivatsgasse 7, 53111 Bonn, Germany2 Hausdorff Center for Mathematics, Endenicher Allee 60, 53115 Bonn, Germany3 Institut für Mathematik, Humboldt-Universität zu Berlin, Unter den Linden 6, 10099 Berlin, Germany1231030 W. Ballmann et al.For a non-vanishing Lipschitz continuous function on M with compact support in M , wecallR( f ) =∫M (|∇ f |2 + V f 2)∫M f 2(1.1)the Rayleigh quotient of f . We letλ0(M, V ) = inf R( f ), (1.2)where f runs through all non-vanishing Lipschitz continuous functions on M with compactsupport in M . If λ0(M, V ) > −∞, then  + V is bounded from below on C∞c (M) andλ0(M, V ) is equal to the bottom of the spectrum of the Friedrichs extension of  + V . Ifλ0(M, V ) = −∞, then the spectrum of any self-adjoint extension of  + V is not boundedfrom below.Recall that +V is essentially self-adjoint on C∞c (M) if M is complete and inf V > −∞.Then the unique self-adjoint extension of  + V is its closure. In the case where M is theinterior of a complete Riemannian manifold N with smooth boundary and where V extendssmoothly to the boundary of N , λ0(M, V ) is equal to the bottom of the Dirichlet spectrumof  + V on N .In the case of the Laplacian, that is, V = 0, we also write λ0(M) and call it the bottomof the spectrum of M . It is well known that λ0(M) is the supremum over all λ ∈ R suchthat there is a positive smooth λ-eigenfunction f : M → R (see, e.g., [3, Theorem 7],[4, Theorem 1], or [5, Theorem 2.1]). It is crucial that these eigenfunctions are not requiredto be square-integrable. In fact, λ0(M) is exactly the border between the positive and the L2spectrum of  (see, e.g., [5, Theorem 2.2]).Suppose now that M is simply connected and let π0 : M → M0 and π1 : M → M1 beRiemannian subcovers of M . Let 0 and 1 be the groups of covering transformations ofπ0 and π1, respectively, and assume that 1 ⊆ 0. Then the resulting Riemannian coveringπ : M1 → M0 satisfies π ◦ π1 = π0. Let V0 : M0 → R be a smooth potential and setV1 = V0 ◦ π .Since the lift of a positive λ-eigenfunction of  on M0 to M1 is a positive λ-eigenfunctionof , we always have λ0(M0) ≤ λ0(M1) by the above characterization of the bottom of thespectrum of  by positive eigenfunctions. In Sect. 4, we present a short and elementary proofof the inequality which does not rely on the characterization of λ0 by positive eigenfunctions:Theorem 1.1 For any Riemannian covering π : M1 → M0 as above,λ0(M0, V0) ≤ λ0(M1, V1).Brooks showed in [2, Theorem 1] thatλ0(M0) = λ0(M1) in the case where M0 is complete,has finite topological type, and π is normal with amenable group 1\0 of covering transfor-mations. Bérard and Castillon extended this in [1, Theorem 1.1] to λ0(M0, V0) = λ0(M1, V1)in the case where M0 is complete, π1(M0) is finitely generated [this assumption occurs inpoint (1) of their Section 3.1], and the right action of 0 on 1\0 is amenable. We generalizethese results as follows:Theorem 1.2 If the right action of 0 on 1\0 is amenable, thenλ0(M0, V0) = λ0(M1, V1).Here a right action of a countable group  on a countable set X is said to be amenable ifthere exists a -invariant mean on L∞(X). This holds if and only if the action satisfies the123On the bottom of spectra under coverings 1031Følner condition: For any finite subset G ⊆  and ε > 0, there exists a non-empty, finitesubset F ⊆ X , a Følner set, such that|F\Fg| ≤ ε|F | (1.3)for all g ∈ G. By definition,  is amenable if the right action of  on itself is amenable, andthen any action of  is amenable.In comparison with the results of Brooks, Bérard, and Castillon, the main point ofTheorem 1.2 is that we do not need any assumptions on metric and topology of M0. Amain new point of our arguments is that we adopt our constructions more carefully to thedifferent competitors for λ0 separately.2 Fundamental domains and partitions of unityChoose a complete Riemannian metric h on M0. In what follows, geodesics, distances, andmetric balls in M0, M1, and M are taken with respect to h and its lifts to M1 and M ,respectively.Fix a point x in M0. For any y ∈ π−1(x), letDy = {z ∈ M1 | d(z, y) ≤ d(z, y′) for all y′ ∈ π−1(x)} (2.1)be the fundamental domain of π centered at y. Then Dy is closed in M1, the boundary ∂ Dyof Dy has measure zero in M1, and π : Dy\∂ Dy → M0\C is an isometry, where C is asubset of the cut locus Cut(x) of x in M0. Recall that Cut(x) is of measure zero. Moreover,M1 = ∪y∈π−1(x) Dy , y ∈ π−1(x).Lemma 2.1 For any ρ > 0, there is an integer N (ρ) such that any z in M1 is contained inat most N (ρ) metric balls B(y, ρ), y ∈ π−1(x).Proof Let z ∈ B(y1, ρ) ∩ B(y2, ρ) with y1 = y2 in π−1(x) and γ1, γ2 : [0, 1] → M1be minimal geodesics from y1 to z and y2 to z, respectively. Then σ1 = π ◦ γ1 andσ2 = π ◦ γ2 are geodesic segments from x to π(z). Since y1 = y2, σ1 and σ2 are nothomotopic relative to {0, 1}. Hence, if z lies in in the intersection of n pairwise different ballsB(yi , ρ) with y1, . . . , yn ∈ π−1(x), then the concatenations σ−11 ∗ σi represent n pairwisedifferent homotopy classes of loops at x of length at most 2ρ. Hence n is at most equal to thenumber N (ρ) of homotopy classes of loops at x with representatives of length at most 2ρ. unionsqLemma 2.2 If K ⊆ M0 is compact, then π−1(K ) ∩ Dy is compact. More precisely, ifK ⊆ B(x, r), then π−1(K ) ∩ Dy ⊆ B(y, r).Proof Choose r > 0 such that K ⊆ B(x, r). Let z ∈ π−1(K ) ∩ Dy and γ0 be a minimalgeodesic from π(z) ∈ K to x . Let γ be the lift of γ0 to M1 starting in z. Then γ is a minimalgeodesic from z to some point y′ ∈ π−1(x). Since z ∈ Dy , this impliesd(z, y) ≤ d(z, y′) ≤ L(γ ) = L(γ0) < r.Hence π−1(K ) ∩ Dy ⊆ B(y, r). unionsqLet K ⊆ M0 be a compact subset and choose r > 0 such that K ⊆ B(x, r). Letψ : R → Rbe the function which is equal to 1 on (−∞, r ], to t + 1 − r for r ≤ t ≤ r + 1, and to 0on [r + 1,∞]. For y ∈ π−1(x), let ψy = ψy(z) = ψ(d(z, y)). Note that ψy = 1 onπ−1(K ) ∩ Dy and that supp ψy = B¯(y, r + 1).1231032 W. Ballmann et al.Lemma 2.3 Any z in M1 is contained in the support of at most N (r + 1) of the functionsψy , y ∈ π−1(x).Proof This is clear from Lemma 2.1 since supp ψy is contained in the ball B(y, r + 1). unionsqIn particular, each point of M1 lies in the support of only finitely many of the functionsψy . Therefore the function ψ1 = max{1−∑ψy, 0} is well defined. By Lemma 2.2, we havesupp ψ1 ∩ π−1(K ) = ∅. Together with ψ1, the functions ψy lead to a partition of unity onM1 with functions ϕ1 and ϕy , y ∈ π−1(x), given byϕ1 = ψ1ψ1 + ∑z∈π−1(x) ψzand ϕy = ψyψ1 + ∑z∈π−1(x) ψz. (2.2)Note that supp ϕ1 = supp ψ1 and supp ϕy = supp ψy for all y ∈ π−1(x).Lemma 2.4 The functions ϕy , y ∈ π−1(x), are Lipschitz continuous with Lipschitz constant3N (r + 1).Proof The functions ψy , y ∈ π−1(x), are Lipschitz continuous with Lipschitz constant 1 andtake values in [0, 1]. Hence ψ1 is Lipschitz continuous with Lipschitz constant N = N (r+1),by Lemma 2.3, and takes values in [0, 1]. Therefore the denominator χ = ψ1+∑z∈π−1(x) ψzin the fraction defining the ϕy is Lipschitz continuous and takes values in [1, N ]. Hence|ϕy(z1) − ϕy(z2)| ≤ |(χ(z2) − χ(z1))ψy(z1) + χ(z1)(ψy(z1) − ψy(z2))|χ(z1)χ(z2)≤ (2N + N )d(z1, z2)χ(z1)χ(z2)≤ 3Nd(z1, z2).unionsqAs a consequence of Lemma 2.4, we get that ϕ1 = 1 −∑ϕy is also Lipschitz continuouswith Lipschitz constant 6N (r + 1)2.3 Pulling upLet f be a non-vanishing Lipschitz continuous function on M0 with compact support and letf1 = f ◦ π . We will construct a cutoff function χ on M1 such that R(χ f1) is close to R( f ).Let g be the given Riemannian metric on M0 and h be a complete background Riemannianmetric on M0 as in Sect. 2. Then there is a constant A ≥ 1 such thatA−1g ≤ h ≤ Ag (3.1)on the support of f . We continue to take distances and metric balls in M0, M1, and M withrespect to h and its respective lifts to M1 and M .Fix a point x in M0. With K = supp f and r > 0 such that K ⊆ B(x, r), we get a partitionof unity with functions ϕ1 and ϕy , y ∈ π−1(x), as above.Fix preimages u ∈ M and y = π1(u) ∈ M1 of x under π0 and π , respectively. Writeπ−10 (x) = 0u as the union of 1-orbits 1gu, where g runs through a set R of representativesof the right cosets of 1 in 0, that is, of the elements of 1\0. Then π−1(x) = {π1(gu) |g ∈ R}. Let123On the bottom of spectra under coverings 1033S = {s ∈ R | d(y, π1(su)) ≤ 2r + 2}= {s ∈ R | d(u, tsu) ≤ 2r + 2 for some t ∈ 1},T = {t ∈ 1 | d(u, tsu) ≤ 2r + 2 for some s ∈ S},G = T S ⊆ 0.Since the fibres of π and π0 are discrete, S and T are finite subsets of 0, hence also G.Let ε > 0 and F ⊆ 1\0 be a Følner set for G and ε satisfying (1.3). LetP = {g ∈ R | 1g ∈ F} ⊆ Rand setχ =∑g∈Pϕπ1(gu).Since |P| = |F | < ∞, supp χ is compact. Hence, by Lemma 2.4, χ f1 is compactly supportedand Lipschitz continuous on M1. LetQ = {y ∈ π−1(x) | (χ f1)(z) = 0 for some z ∈ Dy}.To estimate the Rayleigh quotient of χ f1, it suffices to consider χ f1 on the union of the Dy ,y ∈ Q. We first observe thatP1 = {π1(gu) | g ∈ P} ⊆ Q.To show this, let y = π1(gu) and observe that f1 does not vanish identically on π−1(K )∩ Dyand that ϕy is positive on π−1(K )∩ Dy . Since R is a set of representatives of the right cosetsof 1 in 0, there exists a one-to-one correspondence between P and P1, and hence|P| = |P1| ≤ |Q|.The problematic subset of Q isQ− = {y ∈ Q | 0 < χ(z) < 1 for some z ∈ π−1(K ) ∩ Dy}.Let now y ∈ Q− and z ∈ π−1(K ) ∩ Dy with 0 < χ(z) < 1. Since π1(gu), g ∈ R,runs through all points of π−1(x), we have∑g∈R ϕπ1(gu)(z) = 1. Hence there areg1, . . . , gk ∈ R\P such that ϕπ1(gi u)(z) = 0 andχ(z) +∑ϕπ1(gi u)(z) = 1.Furthermore, there has to be a g ∈ P with ϕπ1(gu)(z) = 0. Then the supports of the functionsϕπ1(gu) and ϕπ1(gi u) intersect and we get d(π1(gu), π1(gi u)) ≤ 2r + 2. That is, we haved(gu, hi gi u) ≤ 2r + 2 for some hi ∈ 1. We conclude thatd(u, g−1hi gi u) = d(gu, hi gi u) ≤ 2r + 2.Since π1 is distance non-increasing, we get that there are si ∈ S and ti ∈ T such thatg−1hi gi = ti si , and then hi gi = gti si . Since gi /∈ P , we conclude that 1gti si /∈ F , i.e.,1g ∈ F\F(ti si )−1. Since (ti si )−1 ∈ G, there are at most ε|F ||G| such elements g ∈ P .Since d(y, z) ≤ r and d(z, π1(gu)) ≤ r + 1, we conclude with Lemma 2.1 that for fixedg ∈ P there are at most N (2r + 1) such y ∈ Q. We conclude that|Q−| ≤ ε|F ||G|N (2r + 1)= ε|P||G|N (2r + 1) ≤ ε|Q||G|N (2r + 1). (3.2)1231034 W. Ballmann et al.We now estimate the Rayleigh quotient of χ f1. For any y ∈ Q+ = Q\Q−, we have χ = 1on π−1(K ) ∩ Dy and therefore∫Dy{|∇(χ f1)|2 + V1(χ f1)2} =∫Dy{|∇ f1|2 + V1 f 21 }=∫M0{|∇ f |2 + V0 f 2}and ∫Dyχ2 f 21 =∫Dyf 21 =∫M0f 2,where, here and below, integrals, gradients, and norms are taken with respect to the originalRiemannian metric g on M .For any y ∈ Q−, we have∫Dyχ2 f 21 ≤∫M0f 2 and∫Dy|V1|χ2 f 21 ≤ C0∫M0f 2,where C0 is the maximum of |V0| on supp f = K . By Lemma 2.3, Lemma 2.4, and (3.1),we have |∇χ |2 ≤ 9N (r + 1)4 A on the support of f . Therefore∫Dy|∇(χ f1)|2 ≤ 2∫Dy{|∇χ |2 f 2 + χ2|∇ f ◦ π |2|}≤ 18N (r + 1)4 A∫M0f 2 + 2∫M0|∇ f |2.In conclusion, ∫Dy{|∇(χ f1)|2 + |V1|χ2 f 21 } ≤ Cfor any y ∈ Q−, where C > 0 is an appropriate constant, which depends on f , but not on yor the choice of ε and F . With D = |G|N (2r + 1), we obtain from (3.2) that|Q−| ≤ εD1 − εD |Q+|,and conclude thatR(χ f1) =∫ {|∇(χ f1)|2 + V1χ2 f 21 }∫(χ f1)2=∑y∈Q∫Dy {|∇ f1|2 + V1 f 21 }∑y∈Q∫Dy f 21≤∑y∈Q+∫Dy {|∇ f1|2 + V1 f 21 } + εC D|Q+|/(1 − εD)∑y∈Q+∫Dy f 21=∫M0{|∇ f |2 + V0 f 2} + εC D/(1 − εD)∫M0 f 2= R( f ) + εC D(1 − εD) ∫M0 f 2.For ε → 0, the right hand side converges to R( f ).123On the bottom of spectra under coverings 1035Proof of Theorem 1.2 By Theorem 1.1, we have λ0(M0, V0) ≤ λ0(M1, V1). By (1.2), thebottom of the spectrum of Schrödinger operators is given by the infimum of correspond-ing Rayleigh quotients R( f ) of Lipschitz continuous functions with compact support. Thearguments above show that, for any such function f on M0 and any δ > 0, there is aLipschitz continuous function χ f1 on M1 with compact support and Rayleigh quotient atmost R( f ) + δ. Therefore we also have λ0(M0, V0) ≥ λ0(M1, V1). unionsq4 Pushing downLet f be a Lipschitz continuous function on M1 with compact support. Define the push downf0 : M0 → R of f byf0(x) =⎛⎝∑y∈π−1(x)f (y)2⎞⎠1/2.Since supp f is compact, the sum on the right hand side is finite for all x ∈ M0, and hence f0is well defined. We have supp f0 = π(supp f ), and hence supp f0 is compact. Furthermore,f0 is differentiable at each point x , where f is differentiable at all y ∈ π−1(x) and f (y) = 0for some y ∈ π−1(x), and then∇ f0(x) = 1f0(x)∑y∈π−1(x)f (y)π∗(∇ f (y)).For the norm of the differential of f0 at x , we get|∇ f0(x)|2 ≤ 1f0(x)2∣∣∣∣∣∣∑y∈π−1(x)f (y)π∗(∇ f (y))∣∣∣∣∣∣2≤ 1f0(x)2∑y∈π−1(x)f (y)2∑y∈π−1(x)|∇ f (y)|2=∑y∈π−1(x)|∇ f (y)|2.Furthermore, f0 is differentiable with vanishing differential at almost any point of { f0 = 0}.Therefore f0 is Lipschitz continuous and∫M0f 20 =∫M1f 2,∫M0V0 f 20 =∫M1V1 f 2,∫M0|∇ f0|2 ≤∫M1|∇ f |2.In particular, we have R( f0) ≤ R( f ).Proof of Theorem 1.1 For any non-vanishing Lipschitz continuous function f on M1 withcompact support, the push down f0 as above is a Lipschitz continuous function on M0 withcompact support and Rayleigh quotient R( f0) ≤ R( f ). The asserted inequality follows nowfrom the characterization of the bottom of the spectrum by Rayleigh quotients as in (1.2). unionsq1231036 W. Ballmann et al.5 Final remarksIt is well-known that any countable group is the fundamental group of a smooth four-manifold.(A variant of the usual argument for finitely presented groups, taking connected sums ofS1 × S3 and performing surgeries, can be used to produce five-manifolds with fundamentalgroup any countable group.) In particular, for a non-finitely generated, amenable group G,e.g., G = ⊕n∈N Z or G = Q, there is a smooth manifold M with π1(M) ∼= G. In contrastto the results in [1,2], our main result also applies to such examples.Moreover, we do not assume λ0(M0, V0) > −∞. Given any non-compact manifold M0,it is indeed easy to construct a smooth potential V0 such that λ0(M0, V0) = −∞. In fact, itsuffices that V0(x) tends to −∞ sufficiently fast as x → ∞.Acknowledgements Open access funding provided by Max Planck Society.Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 Interna-tional License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, andreproduction in any medium, provided you give appropriate credit to the original author(s) and the source,provide a link to the Creative Commons license, and indicate if changes were made.References1. Bérard, P., Castillon, P.: Spectral positivity and Riemannian coverings. Bull. Lond. Math. Soc. 45(5),1041–1048 (2013)2. Brooks, R.: The bottom of the spectrum of a Riemannian covering. J. Reine Angew. Math. 357, 101–114(1985)3. Cheng, S.Y., Yau, S.T.: Differential equations on Riemannian manifolds and their geometric applications.Commun. Pure Appl. Math. 28(3), 333–354 (1975)4. Fischer-Colbrie, D., Schoen, R.: The structure of complete stable minimal surfaces in 3-manifolds ofnonnegative scalar curvature. Commun. Pure Appl. Math. 33(2), 199–211 (1980)5. Sullivan, D.: Related aspects of positivity in Riemannian geometry. J. Differ. Geom. 25(3), 327–351 (1987)123

On the bottom of spectra under coverings

On the bottom of spectra under coverings

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