Rigidity and Non-Rigidity of $\mathbb{H}^n/\mathbb{Z}^{n-2}$ with Scalar Curvature Bounded from Below

We show that the hyperbolic manifold $\mathbb{H}^n/\mathbb{Z}^{n-2}$ is not rigid under all compactly supported deformations that preserve the scalar curvature lower bound $-n(n-1)$, and that it is rigid under deformations that are further constrained by certain topological conditions. In addition, we prove two related splitting results.

According to [21], a deformation can change not only the metric, but also the topology of a compact region in Z.If the deformation is topologically a connected sum with a closed nmanifold, Statement 1.1 is known to be true for (at least) Z = H n /Z n−1 , with idea of proof already outlined by [19, Section 5 5  6 ] (for a detailed treatment, see also [2,Theorem 1.1]).The situation turns out to be more subtle if broader types of deformations are considered, allowing, for example, surgeries along an embedded, non-contractible loop.In this latter case we construct a counterexample to Statement 1.1, which, more precisely, demonstrates the following.Theorem 1.2.For n ≥ 3, let H n /Z n−2 be equipped with the standard hyperbolic metric.There exists a complete Riemannian manifold (M n , g), not (globally) hyperbolic, and compact subsets K ⊂ M and K ′ ⊂ H n /Z n−2 , such that (1) R g ≥ −n(n − 1) and (2) Remark 1.3.
(1) While the theorem above concerns non-rigidity of H n /Z n−2 , it is also interesting to ask whether its statement still holds if H n /Z n−2 is replaced by H n /Z n−1 ; this will be answered in the affirmative in Section 2.4.1.Thus, we obtain counterexamples to the "weak rigidity of H n /Z n−1 " mentioned in [20, p. 678].
(2) En route to proving Theorem 1.2, we obtain counterexamples (see Proposition 2.6) to the following statement in [21, p. 12]: Represent H n /Z n−2 as a warped product H 2 ×T n−2 , g H (see formula (2.1)), and, for a geodesic 2-disk with the restricted metric g H | X ; then no Riemannian manifold M n , g with boundary isometric to ∂X can have scalar curvature R g ≥ −n(n − 1) and mean curvature 1 of ∂M greater than that of ∂X.
(3) Our proof of Theorem 1.2 is constructive, which, a little to our surprise, shows that M can be chosen to be homeomorphic to H n /Z n−2 (see Section 2.4.2);moreover, R g > −n(n − 1) for some points in K.
From the perspective of our construction, the non-rigidity of H n /Z n−2 seems closely related to the fact: A deformation supported in a compact subset K can 'break' the incompressibility 2 of some submanifold that is disjoint from K. On the other hand, rigidity does hold if one only considers deformations that preserve such incompressibility, as the next theorem shows (cf.[11,Theorem 1.8]).
Theorem 1.4.For 3 ≤ n ≤ 7, let (M n , g) be a complete Riemannian manifold 3 with scalar curvature R g ≥ −n(n−1).Suppose that there exist compact subsets K ⊂ M , K ′ ⊂ H n /Z n−2 , and an isometry f : M \K → H n /Z n−2 \K ′ .Representing H n /Z n−2 topologically as R 2 + ×T n−2 , let p ∈ R 2 + be such that T = {p}×T n−2 is disjoint from K ′ , and suppose that the map f −1 | T : T → M is incompressible.Then (M, g) is isometric to H n /Z n−2 .
Technically, we will derive Theorem 1.4 as a consequence of Theorem 1.5 below.The latter can be regarded as a kind of positive mass type theorem for manifolds with an ALH end; its statement relies on a gluing construction, which we now describe.
Gluing construction: Let N n be a smooth manifold, and suppose that ϕ : T k → N (1 ≤ k ≤ n − 2) is an embedding with trivial normal bundle.Moreover, write H n /Z n−1 (topologically) as the product R × T n−k−1 × T k , and define by ψ(p) = (t, q, p) for some fixed t ∈ R and q ∈ T n−k−1 .By removing tubular neighborhoods of ϕ T k ⊂ N and ψ T k ⊂ H n /Z n−1 and then identifying the respective boundaries in the obvious way, we obtain a manifold M .For brevity, M will be referred to as obtained by gluing N and H n /Z n−1 along T k via (ϕ, ψ).In particular, for c sufficiently large, (c, ∞) × T n−1 ⊂ H n /Z n−1 remains an 'end' of M , and this end is denoted by E.
Theorem 1.5.For 3 ≤ n ≤ 7, let N n be a smooth manifold that is either closed or non-compact without boundary, and let M n be obtained by gluing N with H n /Z n−1 along T k via (ϕ, ψ) (see description above).Suppose that (a) the map ϕ : T k → N is incompressible; 1 Unless specified otherwise, in this article the mean curvature along a boundary will always be computed with respect to the outward unit normal.
2 A continuous map f : X → Y between topological spaces is said to be incompressible if the induced map f * : π1(X) → π1(Y ) is injective; when f is an inclusion, we say 'X is incompressible in Y '. 3 In this article, all manifolds are assumed to be orientable, and all hypersurfaces 2-sided.
Then mE,g ≥ 0 (see Definition 3.2).In addition, suppose that (d) the curvature tensor of (M, g) and its first covariant derivatives are bounded; (e) there exists some α > 0 such that R g ≤ −α outside a compact set.
Readers familiar with positive mass theorems may have noticed that the second half of Theorem 1.5 is not in an ideal form; in other words, one wants to know whether the vanishing of mE,g , and not just κ, implies that (M, g) is isometric to H n /Z n−1 , even without the assumptions (d) and (e).In our proof, these assumptions play a role in making sure that the normalized Ricci flow (NRF) starting at g has desired properties (see Lemma 3.4); on the other hand, it seems subtle to prove hyperbolicity from (M, g) being Einstein and the assumed ALH decay rate.Thus we decide to leave the stronger statement for future investigation.
Theorem 1.5 has the following corollary.
Corollary 1.6.For 3 ≤ n ≤ 7, let N n be a closed manifold, and suppose that M n is obtained by gluing N with H n /Z n−1 along T k via (ϕ, ψ).Suppose that g is a complete metric on M such that (M, g) is isometric to the hyperbolic manifold H n /Z n−1 outside a compact set, 4 and suppose that (a) the map ϕ : .
In fact, Corollary 1.6 remains true if N is allowed to be non-compact, which can be deduced as a corollary of Theorem 1.10 below (see Remark 4.5).
Besides rigidity problems modeled on complete manifolds, it is often natural to consider similar problems for manifolds with boundary and scalar/mean curvature bounds.In this regard, we present a splitting result of 'cuspidal-boundary' type (see [21,Section 4,last paragraph]).Our proof relies on an approximation scheme developed in [38] involving µ-bubbles.
Theorem 1.7.Let M 4 , g be a complete, non-compact Riemannian 4-manifold with compact, connected boundary ∂M .Suppose that π 2 (M ) = π 3 (M ) = 0 and that the scalar curvature where H is the mean curvature of ∂M .Moreover, if + e 2t g 0 , where t is the coordinate on (−∞, 0] and (Σ, g 0 ) is a closed 3-manifold with a flat metric.
Remark 1.8.Theorem 1.7 would fail if one allows M to be compact.Indeed, take where θ ∈ S 1 , ρ is the radial coordinate on B 3 , and g S 2 is the standard round metric on S 2 .In this example, M has contractible universal cover, so both its π 2 and π 3 vanish.Moreover, since g is hyperbolic, R g = −12, but the mean curvature H ∂M = 2 coth ρ 0 + tanh ρ 0 > 3.
Counterexamples also exist if one drops the assumption on π 2 (M ) and π 3 (M ).In fact, let us take the manifold (M ′ , g ′ ) in Section 2.4.1 and then, for sufficiently small z 0 > 0, remove the subset {0 < z < z 0 } from M ′ ; the result is a manifold M ′′ with Clearly, Finally, we present an analogue of Theorem 1.7 in more general dimensions.
Definition 1.9 (cf.[11]).We say that a closed, connected manifold Σ belongs to the class C deg , if Σ is aspherical,5 and any compact manifold Σ ′ that admits a map to Σ of nonzero degree cannot be endowed with a PSC metric (i.e., metric with positive scalar curvature).
It is well known that T n ∈ C deg for n ≤ 7; also note that the second item in Definition 1.9 is redundant when dim Σ ≤ 5, according to [13].
Theorem 1.10.For 3 ≤ n ≤ 7, let (M n , g) be a complete and non-compact Riemannian manifold with compact, connected boundary ∂M .Suppose that where H is the the mean curvature of ∂M .
Moreover, if where t is the coordinate on (−∞, 0] and (Σ, g 0 ) is a closed (n − 1)-manifold with a flat metric.[23,Section 11]).(d) To adapt to modern language, our Theorem 1.5 considers manifolds with a prescribed end and some 'arbitrary ends'; the study of positive-mass type theorems on such manifolds has generated considerable interest recently (see, for example, [10,12,29,40]).(e) While in this paper we focus on rigidity results for complete, non-compact manifolds with boundary and scalar curvature lower bounds, similar results in the compact case (with boundary) are obtained by Gromov in [21, Section 4].In both cases, the proofs rely on the µ-bubble technique.(f ) It would be interesting to compare Theorem 1.5 with some recent progress in proving positive mass and rigidity results for ALH manifolds (see [1,16,17,26]); in this latter development, manifolds are often assumed to have nonempty inner boundary with the mean curvature bound H ≤ n − 1 (now H is computed with respect to the inner unit normal); such mean curvature bounds serve as barrier conditions in the method of 'marginally outer trapped surfaces' (MOTS), which can be viewed as a generalization of the µ-bubble technique.
Organization of this article.The proof of Theorem 1.2 is technically independent from the rest of the work and is included in Section 2. Section 3 serves as a preliminary to proving Theorem 1.5, presenting results concerning NRF and conformal deformations.In Section 4, we prove Theorem 1.5, followed by proofs of Corollary 1.6 and Theorem 1.4.In Section 5, we prove Theorem 1.7 and Theorem 1.10.Several of the proofs rely on the so-called 'µ-bubble' technique, a brief discussion of which is included in Appendix A. Appendix B includes two topological lemmas.
2 Non-rigidity of H n /Z n−2 Let the hyperbolic n-space H n be represented by the upper half-space model R n + = {(x, y, z): x ∈ R, y ∈ R n−2 , z > 0}, and let Z n−2 act by translating along the orthogonal lattice 2πZ n−2 ⊂ R n−2 while keeping the x, z-coordinates fixed.The quotient space is denoted by H n /Z n−2 and has the hyperbolic metric where the subscript 'H' stands for 'hyperbolic', and g T n−2 is the associated flat metric on T n−2 .Henceforth, we will regard (x, z) as coordinates on the hyperbolic plane H 2 ; manifestly that H n /Z n−2 , g H is a warped product of H 2 and T n−2 , g T n−2 .
The following lemma is easily verified by standard computation, so we omit its proof.
Lemma 2.1.Let ∇, ∇ 2 denote the gradient and Hessian with respect to g H (same below).We have Next, we proceed to prove Theorem 1.2 by constructing an example that satisfies all its conditions.The idea is to remove a suitable compact subset, X p,r , from H n /Z n−2 and then 'glue' the result with a compact manifold, Xr , along their boundaries; X p,r and Xr will be defined in Sections 2.1 and 2.2 respectively, and then we handle the gluing step in Section 2.3.

First preliminary construction
Let p ∈ H 2 , and define where D r (p) ⊂ H 2 is the geodesic disc, centered at p, of radius r > 0; the inclusion in (2.2) makes sense since H n /Z n−2 is a warped product of H 2 and T n−2 , as we already noted.
Now we have two sets of coordinates for H 2 : (x, z) and the polar coordinates (ϱ, θ) centered at p.In terms of the polar coordinates, the metric on H 2 reads g H 2 = dϱ 2 + sinh 2 ϱ dθ 2 .Lemma 2.2.The boundary Y p,r ⊂ (X p,r , g H ) has the mean curvature (b) There exists a constant r 0 > 0 such that H p,r > 0 for all r ≤ r 0 .
Proof .The formula (2.3) is straightforward to check by using the representation Moreover, since both z −1 ∇z and ∇ϱ have unit norm with respect to g H , This implies (a), and (b) follows since coth r → ∞ as r → 0. ■ Lemma 2.3.There exists a constant C r > 0, depending only on r, such that on ∂D r (p) we have Proof .Since both z −1 ∇z and (sinh r) −1 (∂/∂θ) have unit norm with respect to g H , we have Moreover, a calculation shows that By Lemma 2.1 (b), (c), the left-hand side of (2.5) has its magnitude bounded by (sinh 2 ρ)z; thus, using (2.4) and evaluating (2.5) at ϱ = r, we get Taking C r = sinh r(sinh r + cosh r) finishes the proof.■

Second preliminary construction
Let D be a 2-disc with polar coordinates ρ, θ , where Equip D with the metric Thus, (D, g D ) is isometric to a 'cap' in the round sphere of radius 2.
Now let r > 0 and z(ϱ, θ) be as in Section 2.1 above.Consider and let Ȳr := ∂ Xr .By construction, the boundaries (Y p,r , g H | Yp,r ) and Ȳr , ḡ| Ȳr are isometric under the obvious identification.
Lemma 2.4.The boundary Ȳr ⊂ Xr , ḡ has the mean curvature Proof .Standard computation by using (2.6).■ Regarding the scalar curvature of a warped-product metric, the following is well-known.
Lemma 2.5 (cf.[23,Proposition 7.33]).Let N n−1 , h be an (n − 1)-dimensional Riemannian manifold with scalar curvature R h .Given any smooth function ϕ(θ) defined on an interval I and a constant a > 0, the warped product metric g = a 2 dθ 2 + ϕ(θ) 2 h defined on I × N has the scalar curvature In our case, to compute the scalar curvature of ḡ, it suffices to substitute h = g D + g T n−3 , ϕ(θ) = 1/z(r, θ) and a = sinh r into (2.8).Noting that R h = 1/2, we have where z, ∂ θ z and ∂ 2 θ z are evaluated at (r, θ).Now we are ready to observe the following.Proposition 2.6.For fixed r > 0, the manifold ( Xr , ḡ) satisfies: for a constant C n,r > 0 depending only on n and r.In particular, we have R ḡ > −n(n − 1) provided that p ∈ H 2 is chosen to have a large enough z-coordinate; (b) Under the obvious identification (isometry) between Y p,r and Ȳr , we have Hr > H p,r provided that the z-coordinate of p is large enough.

The gluing step
Lemma 2.7 ([8, Theorem 5]).Let Ω be a compact n-manifold with boundary ∂Ω, and let g and g be two smooth Riemannian metrics on Ω such that (a) g − g = 0 at each point on ∂Ω; (b) the mean curvatures satisfy H g − H g > 0 at each point on ∂Ω.
Then, given any ϵ > 0 and any neighborhood U of ∂Ω, there exists a smooth metric ĝ on Ω with the following properties: (2) ĝ = g in Ω \ U ; (3) ĝ = g in a neighborhood of ∂Ω.
Remark 2.8.By an arbitrary extension, in Lemma 2.7 it suffices to assume that g is defined only in a neighborhood of ∂Ω.
To prove Theorem 1.2, a basic idea is to apply Lemma 2.7 to obtain a metric ĝ on Xr which agrees with g H in a neighborhood of ∂ Xr ∼ = ∂X p,r , so ĝ extends smoothly into H n /Z n−2 \ X p,r by g H .A compromise is the ϵ-cost to the scalar curvature estimate.Thus, one would like to have a bit more scalar curvature to begin with, so that the cost can be absorbed, maintaining the desired lower bound R ĝ ≥ −n(n − 1).This can be achieved by a suitable deformation of g H in a neighborhood of Y p,r ⊂ H n /Z n−2 , as the following lemma shows.Lemma 2.9.Let and define As long as r 0 > 0 is small enough, we can find δ > 0 such that where We want to estimate the right-hand side of formula (2.11).To start with, by Lemma 2.5, Thus, by the proof of Lemma 2.3, there exists a constant C n,r 0 , depending on n, r 0 only, such that Next, by the definition of u, we have, for (2.13) Moreover, for sufficiently small r 0 , we have H p,ϱ ≥ 1 for any ϱ ≤ r 0 (Lemma 2.2 (b)), and so On combining (2.11), (2.12), (2.13) and (2.14), we obtain that Clearly, we can choose a small δ > 0 such that This completes the proof.■ Proof of Theorem 1.2.Let r 0 be small enough, and let u(ϱ), g ′ H and δ be as in Lemma 2.9.Define Take r := r 0 − δ, and note that we still have the freedom of choosing p ∈ H 2 .Suppose that the isometry between Y p,r and Ȳr maps q ∈ Y p,r to q ∈ Ȳr .Furthermore, by using Fermi coordinates, any point in a small neighborhood of Y p,r ⊂ X p,r is uniquely represented by a pair (q, d ′ ), where d ′ is the g ′ H -distance to Y p,r .Similarly, (q, d) coordinatizes a neighborhood of Ȳr ⊂ Xr .By identifying (q, d) with (q, d), we have arranged that g ′ H = ḡ along Ȳr .
To apply Lemma 2.7, assign Ω = Xr , g = g ′ H (defined in a neighborhood U of Ȳr ⊂ Xr , via the identification above) and g = ḡ (defined on Xr ).As noted above, Lemma 2.7 (a) is satisfied.Furthermore, the mean curvature of Y p,r ⊂ X p,r with respect to g ′ H is H ′ p,r := H p,r /u(r) ≥ H p,r , but by choosing p to have large z-coordinate, we can still arrange that Hr > H ′ p,r (see the proof of Proposition 2.6 (b)).Next, by shrinking U if needed, we can assume that R g ′ H ≥ c−n(n−1) on U , and we can assume the same lower bound for R ḡ by choosing p suitably (Proposition 2.6 (a)).Finally, take ϵ = c/2.
With the above setting, Lemma 2.7 applies and yields a metric ĝ defined on Xr , satisfying Thus, ĝ and g ′ H piece together to become a smooth metric g defined on where ∼ indicates boundary identification, with (non-constant) scalar curvature R g ≥ −n(n−1).
(For the reader's convenience, Figure 1 includes a schematic, 1-dimensional illustration of the construction.) In the statement of Theorem 1.2, take (M, g) = (M , g), K = Xr ∪ (X p,r 0 \X p,r ) ⊂ M and K ′ = X p,r 0 , and the proof is complete.■ 2.4 Further remarks The construction above only modifies a portion of H n /Z n−2 that is contained in between x 0 < x < x 1 for some x 0 , x 1 ∈ R. By a translation, we can always arrange that x 0 = 0. Now, T : (x, y, z) → (x + x 1 , y, z) maps a neighborhood of {x = 0} isometrically to a neighborhood of {x = x 1 }.Thus, by removing the subsets {x < 0} and {x > x 1 } from M and then identifying {x = 0} and {x = x 1 } via T , we obtain a smooth Riemannian manifold (M ′ , g ′ ) that satisfies R g ′ ≥ −n(n − 1).In fact, (M ′ , g ′ ) can be viewed as a compactly supported 'deformation' of a hyperbolic cusp H n /Z n−1 , where Z n−1 acts on (x, y) ∈ R n−1 by translating along the lattice x 1 Z × 2πZ n−2 .This serves as yet another counterexample to Gromov's Statement 1.1.

A note on topology
It is interesting to determine the topology of both M and M ′ above.Topologically, M is obtained by a surgery along {p} × S 1 ⊂ R 2 × S 1 and then taking product with T n−3 .The result of that surgery is homeomorphic to S 1 × R 2 .To see this, view S 3 as the union of D 2 × S 1 and S 1 × D 2 with the boundaries identified.Then R 2 × S 1 is simply S 3 with the core circle via a map that switches the first two factors.
Regarding M ′ , note that by identifying x = 0 and x = x 1 in {0 ≤ x ≤ x 1 } ⊂ H 2 , one obtains an open annulus, or equivalently R 2 \{0}.Thus, M ′ is obtained by a surgery along {p} × S 1 ⊂ R 2 \{0} × S 1 and then taking product with T n−3 .In this case, a similar argument as the above applies, and the result of the surgery is homeomorphic to S 1 × R 2 \ D 2 ϵ × S 1 , i.e., the result of removing a solid torus that is contained in a 3-ball B 3 ⊂ S 1 × R 2 (see Figure 2).Thus In particular, the two ends of M ′ are separated by a hypersurface with the topology S 2 × T n−3 ; the same is not true for H n /Z n−1 .

ALH manifolds, mass and deformations
This section includes basic notions and results concerning ALH manifolds, possibly with arbitrary ends, and their NRF and conformal deformations.These results will be used in proving Theorem 1.5.

ALH manifolds and mass
Definition 3.1.Let (M n , g) be a complete Riemannian manifold without boundary.Suppose that (1) for some (sufficiently large) compact set K ⊂ M , M \ K has a connected component E that is diffeomorphic to (0, 1) × T n−1 , and (2) restricted to E, the metric g admits an asymptotic expansion of the form where τ is the coordinate on the interval (0, 1); h denotes a flat metric on T n−1 , which represents metric at the conformal infinity E 0 (i.e., when τ = 0); κ = κ AB dy A dy B is a symmetric tensor defined on T n−1 , where y A are flat coordinates on T n−1 ; finally, O τ n+1 stands for a remainder Q = Q AB dy A dy B with the asymptotics for some constant C, where α = (α 1 , . . ., α n−1 ) are multi-indices.
Such an (M, g) is called asymptotically locally hyperbolic (ALH ), and E an ALH end.Moreover, if M \ E is non-compact, we say that (M, g) is ALH with arbitrary ends.the mass aspect function associated to the pair (E, g).Furthermore, define mE,g := sup Throughout, let each τ -level set in E be denoted by E τ .The following lemma shows how mE,g is related to the mean curvature of E τ ⊂ E. Lemma 3.3.Let M n , g be a Riemannian manifold with an ALH end E.If mE,g < 0, then there exist constants τ 0 , C > 0 such that where H Eτ is the mean curvature of E τ computed with respect to the 'outward normal' −∂/∂τ .
Thus, up to constant factors, the curvature tensor Rm(t) of g(t) satisfies the same estimates as Rm(Φ(t)) of g(Φ(t)).In particular, it follows from [33] that, for all t ∈ (0, T ], g(t) is complete, and |Rm(t)| is uniformly bounded.Now we turn to proving the properties.(i) follows from [5, Proposition 3.1].(ii) can be verified by applying the maximum principle (see [15,Theorem 7.42]) to the evolution equation 6satisfied by e 2(n−1)t (R g(t) + n(n − 1)); to prove (iii), invoke the strong maximum principle on the domain Ω × [0, t], where Ω ⊂ M is compact on which g 0 is not Einstein, and then let Ω exhaust M .(iv) would follow once we show that the integral is uniformly bounded; to see this, note that the first covariant derivatives of Rm(0) are assumed to be bounded (assumption (d) in Theorem 1.5), by [14, Theorem 14.16], we have for some constant C > 0; in addition, the evolution equation of Rm reads7 of course, 1/ √ t is integrable; combining these, it is easy to see that (3.5) is uniformly bounded for small enough T ; since, by assumption (e) in Theorem 1.5, R g ≤ −α outside a compact set, (iv) follows.(v) follows from [5,Proposition 4.3].Finally, (vi) follows from (v) and [5, formulas (3.19)-(3.21)](note that g ij (τ ) provides an extra factor of τ 2 ).■
Lemma 3.5.Let (M, g) be complete with an ALH end E, and let f ∈ C ∞ (M ) be a non-negative function that satisfies where τ is the function occurring in the expansion (3.1).
Then there exist a function v ∈ C ∞ (M ) and a constant δ 0 such that 0 < δ 0 ≤ v ≤ 1 and Proof .Let {Ω i } ∞ i=0 be a sequence of smooth, bounded domains satisfying Ω i ⋐ Ω i+1 and i Ω i = M .For each i, the Dirichlet problem has a positive solution v i .By the maximum principle, 0 < v i ≤ 1.Thus, v := lim i→∞ v i is well defined on M , satisfying 0 ≤ v ≤ 1 and (3.6).It remains to show that v has a positive lower bound.
Without loss of generality, assume that Σ i ⊂ ∂Ω i is the only component of ∂Ω i that is contained in E; in fact, let us assume that each Σ i is a τ -level set.Denote τ 0 := τ | Σ 0 .
To refine the estimate of v i , we construct an auxiliary function ξ and compare it with v i via the maximum principle.Indeed, let α ∈ (0, n − 1) be any constant, and define Using the fact that − ln τ is, up to adding a constant, the distance function to Σ 0 , one easily computes that Thus, by (3.3), for sufficiently small τ 0 , there exists a constant C n,α,τ 0 > 0 such that ∆ g ξ ≥ C n,α,τ 0 τ α for any τ ≤ τ 0 .Now, (3.7), the fact that v i ≤ 1, and the assumption that f ∈ O(τ n ) together imply where C ′ f,n is a constant depending only on f and n.In comparison, Thus, for sufficiently small τ 0 , the maximum principle implies that v i ≥ ξ in (Ω i \Ω 0 ) ∩ E. Upon taking limit, v ≥ ξ > 0 on E\Ω 1 .Since v ≥ 0, the strong maximum principle, applied to (3.6), implies that v > 0 on M .
When M \ E is compact (i.e., M having no arbitrary end), the above already implies that v has a positive lower bound.When M \ E is non-compact, since f is supported in K ∪ E, by choosing Ω 0 to include K, we have that each v i (i ≥ 1) is harmonic on Ω i \(Ω 0 ∪ E); using the maximum principle again, we get min To finish the proof, it suffices to take δ 0 = min{δ arb , inf Ω 1 v, inf E\Ω 1 ξ}.■ Proposition 3.6.Let (M n , g) be complete, with an ALH end E and with R g ≥ −n(n−1) on M .Let R ∈ C ∞ (M ) be a function that satisfies Then the Yamabe equation has a solution u with 0 < δ 0 ≤ u ≤ 1 for some constant δ 0 .In particular, the metric u 4/(n−2) g is complete and has the scalar curvature R.
Proof .The proof follows a super/sub-solution argument.To start with, define L g by Note that L g 1 = R g − R ≥ 0 by assumption.Thus, 1 is a super-solution of (3.9).To find a sub-solution to (3.9), take f := R g − R ≥ 0. Note that R g = −n(n − 1) + O(τ n ) in E. Thus, Lemma 3.5 applies and yields a solution v to (3.6), satisfying 0 < δ 0 ≤ v ≤ 1 for some constant δ 0 .Now we compute where the inequality follows from the assumption that R ≤ 0 and the bounds for v. Thus, v is a sub-solution of (3.9).
Then one finishes the proof by following the argument of [4, Proposition 2.1].■ Next, we will focus on the behavior of u towards the ALH infinity E 0 .
Verification of (C2).By (C1), we have Further information about u nj is obtainable by computing (∆ g − n)w using this expansion and then comparing the result with (3.11).In fact, direct computation and (3.3) yield: with 1 ≤ j ≤ N i .Now, since u nj are all defined on E 0 , we have ∆ g u nj ∈ O τ 2 ; and since the remainder O(τ n+1−ϵ ) does not contribute to the coefficients s j of τ n (log τ ) j in (∆ g − n)w, we have that s j equals to (n + 1)u n1 + 2u n2 for j = 0, 2(n + 1)u n2 + 6u n3 for j = 1, . . .
By (3.11), all s j must vanish, which implies that
Regarding boundary data, first note that the assumption about R g implies that w cannot be identically zero for τ ∈ (0, τ 0 ]; thus, the strong maximum principle implies, in particular, that w < 0 along τ = τ 0 .This allows us to choose δ such that w ≤ ζ along τ = τ 0 .Moreover, both w, ζ → 0 as τ → 0. Now, by the maximum principle, we get This proves that u n0 < 0, verifying (C3).■ Lemma 3.9.Let (M n , g) be a Riemannian manifold with an ALH end E, on which the asymptotic expansion (3.1) applies.Suppose that u = 1 + φτ n + O(τ n+1 ) is a function defined on E, where φ ∈ C ∞ (E 0 ).Then, up to a diffeomorphism that restricts to be the identity on E 0 , the deformed metric ḡ = u

Two rigidity results
The goal of this section is to prove Theorem 1.5, Corollary 1.6 and Theorem 1.4.The reader may consult Appendix A before proceeding.
Proposition 4.1 (cf.[11,Theorem 1.1]).For 3 ≤ n ≤ 7, let M n be a (connected) non-compact manifold with connected, compact boundary Σ.Let ι : Σ → M be the inclusion map.Suppose that Σ ∈ C deg (see Definition 1.9) and that the map ι is incompressible.Then M admits no complete metric g with R g ≥ −n(n − 1) and Proof .To begin with, by the classification of covering spaces, there exists a covering of M , say p : M → M , that satisfies where base points for the fundamental groups are omitted.Moreover, by the homotopy lifting property, there exists an embedding ι : Σ → M such that ι = p • ι.By (4.1) and the incompressibility of ι, the composition is a well-defined group homomorphism.Since Σ is aspherical, by [24, Proposition 1B.9], there exists a map j : M → Σ such that j * : π 1 ( M ) → π 1 (Σ) is equal to J; in particular, j * •ι * = id π 1 (Σ) ; then, by applying the uniqueness part of [24, Proposition 1B.9] to Σ, it is easy to see that j • ι is in fact homotopic to id Σ .
Since ι is an embedding, each boundary component of M , which is a lifting of Σ, must be diffeomorphic to Σ.In particular, denote Σ = ι(Σ).Since j • ι is homotopic to id Σ , we have Now, for the sake of deriving a contradiction, suppose that g is a complete metric on M with R g ≥ −n(n − 1) and H Σ ≥ (n − 1)(1 + δ) for some constant δ > 0. Let ĝ := p * g be the pull-back metric on M , and define ρ(x) := dist ĝ(x, Σ) for x ∈ M .
For an arbitrarily large T > 0, let and let Σi (0 where Σ0 = Σ.Define (see Figure 3 below) Since M is complete, connected and non-compact, so is M , and we have ŪT ⋐ U T,ϵ .Moreover, for small enough ϵ, Thus, by the smooth Urysohn lemma, there exists a function η ∈ C ∞ ( M ) with Let a ∈ (0, 1) be a regular value of η.Automatically, η −1 (a) is a smooth, closed hypersurface of M , and By the above arrangement, B a := η −1 [0, a] , equipped with the restriction of the metric ĝ, is a Riemannian band with By letting f = j| Ba and using Lemma A.8, one easily sees that B a satisfies the NSep + property (see Definition A.6). Then take With these choices, all assumptions of Lemma A.9 are satisfied for (B a , ĝ| Ba ; ∂ − , ∂ + ) and ∂ ⋆ .Since M , ĝ is complete and non-compact, the distance dist ĝ(∂ ⋆ , ∂ + ) can get arbitrarily large as one chooses large T .This contradicts Lemma A.9. ■ Remark 4.2.Proposition 4.1 still holds if M is allowed to be compact.In fact, proceeding along the same proof, we still have Σ ̸ = 0 ∈ H n−1 M ; Z , so M cannot be compact with a single boundary component.Hence, either (1) M is non-compact, and the previous proof applies verbatim; or (2) M is itself a Riemannian band with ∂ + = Σ that satisfies the NSep + property and the curvature bounds R ĝ ≥ −n(n − 1), H ∂ M ≥ (n − 1)(1 + δ); however, by Remark A.10 (A), such a band cannot exist, reaching a contradiction.
Proof .Suppose, on the contrary, that mE,g < 0. Let τ be a defining function compatible with the ALH structure of E (see (3.1)).Then by Lemma 3.3, there exists a small τ 0 > 0 such that the mean curvature of the level set E τ 0 satisfies for some δ 0 > 0. Now, remove {0 < τ < τ 0 }, a subset of E, from M and denote the resulting manifold by M ′ .By using the assumptions, it is easy to see that Clearly, ∂M ′ ∈ C deg .By Proposition 4.1, we get a contradiction.This proves the inequality mE,g ≥ 0.
Next we turn to the second part of the proposition.Again we argue by contradiction.Assume that κ = 0 without (M, g) being Einstein.Let g(t) be the NRF initiated at g. Then by Lemma 3.4, for some small t 0 , we have It is easy to check that a function R as described in Proposition 3.6 exists; thus, there is a positive function u ∈ C ∞ (M ) such that ḡ := u 4/(n−2) g(t 0 ) is complete with R ḡ = R ≥ −n(n − 1).Furthermore, thanks to (i) and (iii) above, both Proposition 3.8 and Lemma 3.9 apply.As a consequence, E, ḡ| E remains ALH and satisfies where u n0 < 0, and h is a flat metric on T n−1 .Clearly, mE,ḡ = tr h κ < 0. This contradicts the first part of the proposition.■ Proof of Theorem 1.5.For convenience, let N o (resp., H o ) denote the result of removing a tubular neighborhood of ϕ T k from N (resp., ψ By Proposition 4.3, to prove the theorem it suffices to show that the boundary Σ of To show this, it in turn suffices to show that the T k -factor of ∂N o is incompressible in M , according to Lemma B.2.
If this was not the case, let L be a non-contractible loop in {x}×T k ⊂ ∂N o that is contractible in M .Now consider where B is an (n − k)-ball embedded in S 1 × T n−k−1 .Topologically, M can be viewed as a subset of M ′ := H ′ ⊔ Φ N o , so L is also contractible in M ′ .By [11,Lemma A.3], H ′ satisfies the 'lifting property' (see [11,Definition A.2]).Thus, [11,Lemma A.4]  shows that ∂N o is incompressible in M , and it follows that the T k -factor of ∂N o is also incompressible in M .
Proof of Corollary 1.6.In this setting, the assumptions (a − e) in Theorem 1.5 are satisfied.Since κ automatically vanishes, we conclude that g is Einstein.Write the metric on H n /Z n−1 as dt 2 + e 2t g 0 where g 0 is a flat metric on T n−1 .Since H n /Z n−1 is isometric to (M, g) outside a compact set, one can remove the corresponding cusp (i.e., {t < −a} for some a ≫ 0) from M and obtain a complete, non-compact manifold (M ′ , g ′ ) with boundary ∂M ′ ∼ = T n−1 , satisfying define τ k by 3 coth(2τ k ) = 3 + k −1 , and then define μk : Thus, {τ k } ∞ k=1 is increasing and tends to infinity, and Now, choose a k , regular values of ρ, such that τ k ≤ a k < min{τ k+1 , τ k + 1}, and then define Proof .Suppose on the contrary that Σ k ∩K = ∅.This implies that η •ρ k = 0 on Σ k .Moreover, by assumption, R g ≥ −12, and by construction, |dρ k | g < 1.Thus, we have (see (A.1)) This, along with Fact A.5, implies that Σ k admits a PSC metric; since Σ k is separating, we get a contradiction, by Lemma A.11. ■

Convergence of Σ k
By using [37,Theorem 3.6], one can show that the second fundamental form II Σ k is uniformly bounded within any compact subset of M .Thus, by Lemma 5.1, Σ k subconverges to a smooth hypersurface Σ in M (for convenience, denote the subsequence by the same symbol Σ k ).Within compact subsets of M , the convergence is uniform and has multiplicity one; moreover, Σ bounds a 'minimizing 3-bubble' for which minimality is interpreted with respect to compactly supported perturbations (cf.[25,Lemma 4.10]).Depending on whether Σ is compact, we consider the two cases below.
Thus, a neighborhood of Σ is foliated by the t-level sets.Note that moving along the foliation leaves the energy functional invariant; thus, each t-slice also bounds a minimizing 3-bubble, to which the same analysis above applies.
This implies that a maximal neighborhood U of Σ on which the metric splits as By construction, R µ k + ≥ 0 outside K, and there exist δ k > 0, satisfying lim δ k = 0, such that R µ k + ≥ −δ k on M .Since R g k cannot be positive and λ k ≥ 0, by (5.4), we must have lim λ k = 0. Next, choose q k ∈ Σ k ∩ K so that lim q k = q ∈ Σ, and let p k = (q k , t 0 ) ∈ Σ k × S 1 and p = (q, t 0 ) ∈ M = Σ × S 1 .Normalize u k such that u k (q k ) = 1.By the Harnack inequality, u k converges smoothly to a positive function u on Σ with u(q) = 1.Thus, (M k , g k ) converges in the pointed smooth topology to M , g , where g = g Σ + u 2 dt 2 .Now one can follow the proof 8 of [38, Proposition 3.2] to show that Ric g = 0, and then follow the proof 9 of [38, Theorem 1.1] to show that u is constant, which implies Ric g Σ = 0.
In summary, (Σ, g Σ ) is complete, non-compact, Ricci-flat, and with finite area; this contradicts [32,p. 25,Theorem 4.1].■ 8 The proof of [38,Proposition 3.2] only relies on M admitting no PSC metric and the properties of R µ k + mentioned above. 9In particular, the boundedness of area(Σ) follows from A Remark To prove the second part of the theorem, first obtain a covering M , ĝ of (M, g) as in the proof of Proposition 4.1, and then apply (essentially) the same proof of Theorem 1.7 to M , ĝ ; to assist the reader, we list a few points that may need attention.
∂ M may not be connected, but Riemannian bands can still be constructed in a similar manner as in the proof of Proposition 4.1.To avoid clash of symbols, denote S := ∂M and let Ŝ be a fixed lifting of S in M .Thus we have H ≥ n − 1; one can check that the barrier condition is satisfied, and the Σ k s exist; restricting j : M → S to Σ k yields a map Σ k → Ŝ of nonzero degree.
An adapted version of Lemma 5.1 holds; in the proof, invoke Lemma A.8 instead of Lemma A.11.It follows that Σ k converges to some Σ.
When Σ is compact, the corresponding part in Section 5.3 applies, apart from dimensional adjustments and the fact that Ricci-flatness may no longer imply flatness.
When Σ is non-compact, we need to argue, without relying on [9, Corollary 1.10], that M k = Σ k × S 1 admits no PSC metric, and this is already addressed by Remark 5.2.
The consequence is that M , ĝ is of the form (−∞, 0] × Σ, dt 2 + e 2t g Σ , where g Σ is Ricci-flat.In particular, the covering M → M is 1-fold and hence an isometry.Since Σ = ∂M is assumed to be aspherical, g Σ must be flat, which can be seen by applying the Cheeger-Gromoll splitting theorem to the universal cover; for details, see the beginning paragraph of [12,Section 6].■

A µ-bubbles
This section collects some 'definitions' and 'facts' concerning the µ-bubble technique, about which we make no claim to originality.For detailed expositions and proofs, the reader may consult [9,12,37,39] and [22,Section 5].This section also includes three supplementary 'lemmas'.
A common setting for µ-bubbles is a Riemannian band, namely a compact, connected Riemannian manifold M n , g whose (nonempty) boundary is expressed as a disjoint union ∂M = ∂ − ⊔ ∂ + , where each of ∂ ± is a smooth, closed and possibly disconnected (n − 1)-manifold.
Given a Riemannian band M n , g; ∂ − , ∂ + and a function µ ∈ C ∞ M , consider the following variational problem: Let Ω 0 be a smooth open neighborhood of ∂ − ; among all Caccioppoli sets Ω ⊂ M that satisfy ∂ − ⊂ Ω and Ω∆Ω 0 ⋐ M , seek a minimizer of the functional where H k is the induced k-dimensional Hausdorff measure, and χ Ω , χ Ω 0 are characteristic functions.Such a minimizer is called a µ-bubble.
Existence and regularity of µ-bubbles are well-established when µ satisfies the following 'barrier condition'.
The concept of separating hypersurfaces can also be defined for complete, non-compact Riemannian manifolds (M, g) with compact boundary-just require that S intersects with all paths connecting ∂M and infinity.The NSep + property can be extended to such manifolds.
Proof .Suppose that S ⊂ M is a (closed) separating hypersurface that admits a PSC metric, and let S ′ ⊂ S be as indicated in Remark A.7.In particular, S ′ admits a PSC metric, and [S ′ ] ̸ = 0 ∈ H 3 (M, Z).Since π 2 (M ) is trivial, the topological classification of closed 3-manifolds admitting a PSC metric implies that S ′ is homologous to a spherical class in H 3 (M, Z) (see [35, p. 112]).Since Suppose that Φ : ∂M 1 → ∂M 2 is a diffeomorphism, and let M := M 1 ⊔ Φ M 2 be the manifold obtained by identifying ∂M 1 , ∂M 2 via Φ.Let t ∈ R be such that {t} × T n−k−1 is disjoint from B.
Then the hypersurface Σ = {t} × T n−1 is incompressible in M if and only if the T k -factor10 of ∂M 2 is incompressible in M .
Proof .In M , the T k -factor of Σ is homotopic to that of ∂M 1 and hence to that of ∂M 2 .Thus, (⇒) is clear.For (⇐), we prove its contrapositive.Suppose that L ⊂ Σ is a non-contractible loop that is contractible in M .Write [L] = (m i α i , n j β j ) ∈ π 1 T n−k−1 × π 1 T k ∼ = π 1 (Σ), where α i generates the fundamental group of the i-th S 1 -factor in T n−k−1 and m i ∈ Z, similarly for β j and n j .Let us write αi , βj for the corresponding elements in the homology class H 1 (Σ; Z).
It will be convenient to view the R-factor in R × T n−k−1 as S 1 minus a point, and to view M as a subset of M := M1 ⊔ Φ M 2 , where M1 := S 1 × T n−k−1 − B × T k .
Let ι : Σ → M be the inclusion map.For 1 ≤ i ≤ n − k − 1, let θ i be the coordinate on the i-th S 1 -factor of T n−k−1 .By construction, there exists t i ∈ S 1 such that θ i = t i defines a hypersurface S i in M that is 'dual' to ι * αi , in the sense that the intersection products Since L is contractible in M ⊂ M , we have i m i ι * αi + j n j ι * βj = 0 ∈ H 1 ( M ; Z); by taking intersection products with [S i ], we see that m i = 0 for all i = 1, . . ., n − k − 1, so L is homotopic to a loop in the T k -factor of Σ.Thus, the T k -factor of Σ is not incompressible in M .By homotopy, the same is true for the T k -factor of ∂M 2 .This completes the proof.■

Figure 3 .
Figure 3.A schematic picture showing U T , U T,ϵ (left figure) and B a (right figure).The complement of U T,ϵ , which may include more boundary components of M , is not displayed.
[9,.The PSC obstruction, provided by[9, Corollary 1.10], for manifolds of the form Σ×S 1 only works when dim Σ ̸ = 4. On the other hand, if Σ (2 ≤ dim Σ ≤ 6) is closed, orientable, and if it admits a map of nonzero degree to some Σ ′ ∈ C deg , then by a similar argument as [11, Theorem 1.1], one can show that Σ × S 1 admits no PSC metric.Proof of Theorem 1.10.The inequality inf ∂M H ≤ n−1 follows directly from Proposition 4.1.