Two rigidity results for stable minimal hypersurfaces

The aim of this paper is to prove two results concerning the rigidity of complete, immersed, orientable, stable minimal hypersurfaces: we show that they are hyperplane in $\mathbb{R}^4$, while they do not exist in positively curved closed Riemannian $(n+1)$-manifold when $n\leq 5$; in particular, there are no stable minimal hypersurfaces in $\mathbb{S}^{n+1}$ when $n\leq 5$. The first result was recently proved also by Chodosh and Li, and the second is a consequence of a more general result concerning minimal surfaces with finite index. Both theorems rely on a conformal method, inspired by a classical work of Fischer-Colbrie.


Introduction
It is well-known that a minimal surface M 2 ⊂ R 3 is a critical point of the area functional A t for all compactly supported variations, i. e. d dt t=0 A t = 0; equivalently, M 2 is minimal if and only if the mean curvature H, i.e. the (normalized) trace of the second fundamental form, is identically zero, or if and only if M 2 can be expressed, locally, as the graph Γ(u) of a solution u of the minimal surfaces equation (1 + u 2 x )u yy − 2u x u y u xy + (1 + u 2 y )u xx = 0.In 1914, S. Bernstein showed that an entire (i.e., defined on the whole plane R 2 ) minimal graph in R 3 is necessarily a plane; the so-called "Bernstein problem" in higher dimension can be then stated in the following way: if the graph Γ(u) of a function u : R n → R is a minimal hypersurface in R n+1 , does Γ(u) have to be necessarily a hyperplane?Many famous mathematicians worked on this problem in the Sixties, in particular Fleming [14] (who gave a new proof in the case n = 2), De Giorgi [10] (case n = 3), Almgren [1] (case n = 4), Simons [24] (the three remaining cases for n ≤ 7) and, eventually, Bombieri, De Giorgi and Giusti [2], who showed that, for n ≥ 8, there are minimal entire graphs that are not hyperplanes.We explicitly remark that a minimal graph is area-minimizing, i.e. it is not only a critical point of the area functional, but also a minimum, while this is not true for minimal hypersurfaces that are "non-graphical", and also that area-minimizing implies stability, that is the non-negativity of the second variation for the area functional d 2 dt 2 t=0 A t ≥ 0 for all compactly supported variations.A natural generalization of the classical Bernstein problem, thus, is the stable Bernstein problem, that is: if M n ֒→ R n+1 is a complete, orientable, isometrically immersed, stable minimal hypersurface, does M have to be necessarily a hyperplane?In the case n = 2 the (positive) answer was given in three different papers, which appeared between 1979 and 1981 (see do Carmo and Peng [11], Fischer-Colbrie and Schoen [16] and Pogorelov [19]).
In higher dimensions, the aforementioned result of Bombieri, De Giorgi and Giusti implies that there exist non-flat orientable, complete, stable minimal hypersurfaces in R n+1 for n ≥ 8, while for n ≤ 5 the stable Bernstein theorems is true with some additional assumptions (for instance, if one requires bounds on the volume growth of geodesic balls, see e.g.[21]; see also [12], [3], [4], [18] and references therein for other interesting results in the same spirit).Moreover, by [2] and [17], we also note that there are non-flat area-minimizing (and thus minimal and stable) complete orientable hypersurfaces M 7 ֒→ R 8 .
Up until recently, without additional hypothesis, the remaining cases (3 ≤ n ≤ 6) were still open, even if the study of minimal (in particular stable or in general with finite index) hypersurfaces immersed into a Riemannian manifold (not only the Euclidean space, then) is a very active field and has attracted a lot of interest.Then, in 2021, Chodosh and Li [6] (see also [7]) showed that a complete, orientable, isometrically immersed, stable minimal immersion M 3 → R 4 is a hyperplane.Their proof, clever and highly non-trivial, is based on the nonparabolicity of M : they perform careful estimates for the quantity (here u is a positive Green's function for the Laplacian and Σ t is the t-level set of u), relating it to Σt |A M | 2 (A M is the second fundamental form of M ).
In this paper we provide a completely different proof of Chodosh and Li result, based on a conformal deformation of the metric, a comparison result and integral estimates, and we also prove another rigidity result when the ambient space is a complete Riemannian manifold with non-negative sectional curvature and either uniformly positive bi-Ricci curvature or uniformly positive Ricci curvature.To be precise, and to fix the notation, we consider smooth, complete, connected, orientable, isometrically immersed hypersurfaces M n ֒→ (X n+1 , h), n ≥ 2, where (X n+1 , h) is a (complete) Riemannian manifold of dimension n + 1 endowed with metric h.We denote with g the induced metric on M and with H the mean curvature of M ; we have that M is minimal if H ≡ 0 on M .In this latter case we say that M is stable if where A = A M is the second fundamental form of M n , ν is a unit normal vector to M in X and dV g is the volume form of g .As we recalled before, stability is related to the non-negativity of the second variation or, equivalently, the non-positivity of the Jacobi operator The first result is thus the following: The second result concerns minimal hypersurfaces with finite index.We recall that a minimal immersion M n ֒→ (X n+1 , h) has finite index if the number of negative eigenvalues (counted with multiplicity) of the Jacobi operator L M on every compact domain in M with Dirichlet boundary conditions is finite; in particular stability implies finite (equal zero) index.Before presenting our next result, we need to recall the notion of bi-Ricci curvature tensor introduced in [23]: given two orthonormal tangent vectors u, v we define where Sect h (u, v) denotes the sectional curvature of the plane spanned by u and v. Our second result is the following: manifold with non-negative sectional curvature and either uniformly positive bi-Ricci curvature or uniformly positive Ricci curvature, then every complete, orientable, immersed, minimal hypersurface M n ֒→ (X n+1 , h) with finite index must be compact.
In particular, there is no complete, orientable, immersed, stable minimal hypersurface of the round spheres M n ֒→ (S n+1 , g std ), provided n ≤ 5.In dimension n = 2 this follows from a more general result proved in [22], while, in dimension n = 3, it was recently proved in [8,Corollary 1.5].We mention that Theorem 1.2 holds also for complete, orientable, immersed, stable minimal hypersurface of the cylinder M n ֒→ (R × S n , g std ) (observe that in this case Sect ≥ 0 and BRic ≥ 1), provided n ≤ 5.As far as we know, Corollary 1.3 is new in the cases n = 4, 5.We do not know if Theorem 1.2 and Corollary 1.3 hold also in dimension greater than five.We note that, in the same spirit, in [23] the authors obtained a compactness result for stable minimal hypersurfaces of dimension n ≤ 4 immersed in space with uniformly positive bi-Ricci curvature.

Proof of Theorem 1.1
In this section we give an alternative proof of [6, Theorem 1] (see Theorem 1.1).The main idea is to use a weighted volume comparison for a suitable conformal metric g together with a new weighted integral estimate inspired by [21].
Following the line in [15] (see also [13]), let k > 0 and consider the conformal metric g = u 2k g, where g = ı * h is the induced metric on M (and ı denotes the inclusion).First of all we prove the following lower bound for a modified Bakry-Emery-Ricci curvature of g.In particular, this implies the non-negativity of the 2-Bakry-Emery-Ricci curvature of g for a suitable k.
Lemma 2.1.Let f := k(n − 2) log u.Then the Ricci tensor of the metric g = u 2k g satisfies du u and On the other hand, from the standard formulas for a conformal change of the metric g = e 2ϕ g, ϕ ∈ C ∞ (M ), ϕ > 0 we get Note that, in our case, ϕ = k log u; now we exploit the facts that u is a solution of equation (2.1) to write From the Cauchy-Schwarz inequality we have from Gauss equations in the minimal case we get Ric g = −A 2 ; since A is traceless we have the inequality and substituting in the previous relation we conclude

2.2.
Completeness.In this subsection we are going to prove that the conformal metric g = u 2k g is complete, provided n = 3 and k = 2 3 .In order to do this we follow the strategy in [15] and we use some computations in [13].First, we recall that in the proof of [15, Theorem 1], given a reference point O ∈ M n , the author showed the existence of a g-minimizing geodesic, where s is the g-arclength and M n ֒→ R n+1 is the usual complete, connected, orientable, isometrically immersed, stable minimal hypersurface.For the sake of completeness, we report the argument here.First of all, for every R > 0, we consider the geodesic ball (of g) centered at O of radius R, B R (O).Then, we first claim that there exists a g-minimizing geodesic, γ R , joining O to any boundary point of B R (O).Indeed, consider u R := u + η, where η is a smooth function is complete, and these geodesics exist.Therefore, for every , then γ i is a g-minimizing geodesic.We parametrize γ i with respect to g-arclength.
In particular, since | γi (s)| g = 1 for every s, up to subsequences, the sequence γi (0) converges to a limit vector as R i → ∞.Thus, by ODE theory and Ascoli-Arzelà, γ i converges on compact sets of [0, ∞) to a limiting curve γ which is a g-minimizing geodesic ad is parametrized by g-arclength.
Remark 2.2.(i) We observe that the completeness of the metric g = u 2k g will follow if we can show that the g-length of γ is infinite, i.e.
Indeed, by construction, the g-length of every other divergent geodesic starting from O (i.e. its image does not lie in any ball B R (O)) must be greater or equal than the one of γ.
(ii) Note that γ has unit speed with respect to g and to g, when it is parametrized by the arclength s and s, respectively.
From now on n = 3 and k = 2 3 .
Proof.We do part of the computations for every n.We consider the g-minimizing geodesic γ just constructed and as observed in Remark 2.2 the completeness of g is equivalent to prove that γ has infinite g length, i.e.Since γ is minimizing, by the second variation formula, following the computations in the proof of Theorem 1 (with H = 0) in [13], we obtain u 2 ds for every smooth function ϕ with compact support in (0, +∞) and for every k > 0. Since A is trace-free, we have In particular, if Using this estimate, the fact that |∇u| 2 ≥ (u s ) 2 and integrating by parts, we obtain Let now ϕ = u k ψ, with ψ smooth with compact support in (0, +∞).We have and substituting in the previous relation we get thus we have Moreover, for every t > 1 and using Young's inequality for every ε > 0, we have ψψ ss u k ds for every t > 1 and every k satisfying (2.2) and (2.4).Let for every p ∈ [4, 4 + 8/n] and for some C = C(n, p) > 0. For the sake of completeness we report here the proof of (2.5).We take ϕ = |A| 1+q ψ, q ≥ 0, with ψ ∈ C ∞ 0 (M ), in the stability inequality (1.1) obtaining for every ε > 0, where we used Young's inequality.On the other hand, multiplying Simons' inequality (see [9, Lemma 2.1] for a proof) by |A| 2q ψ 2 and integrating by parts, we get for every ε > 0, where we used again Young's inequality.Since q ≥ 0, for ε > 0 sufficiently small, we obtain Let q := p−4 2 .For ε > 0 small enough, we have /n] and we finally obtain Taking ψ = ϕ p/2 , by Holder's inequality we get (2.5).Take ϕ = u α ψ, with ψ smooth with compact support, u the solution of (2.1) and α < 0. Since, from Cauchy-Schwarz and Young's inequalities, Now we tackle the first integral on the right-hand side of (2.8) firstly integrating by parts secondly we use the fact that together with Cauchy-Schwarz and Young's inequalities to get for all ε > 0. From (2.1) we find 2 ∇|A| , then, from Cauchy-Schwarz and Young's inequalities we obtain for every t 1 > 0. Now, multiplying by |A| p−4 f 2 the Simons' inequality (2.6), integrating by parts and using Young's inequality we obtain for every t 2 > 0. Choosing f = u α ψ we get for every ε > 0, since Now let δ > 0. Using (2.10) in (2.9) with Proof of Theorem 1.2.Let (X n+1 , h) be a complete n-dimensional, n ≤ 5, manifold with nonnegative sectional curvature and either uniformly positive bi-Ricci curvature or uniformly positive Ricci curvature and consider an orientable, immersed, minimal hypersurface M n → (X n+1 , h) with finite index.Suppose, by contradiction, that M is non-compact.It is well known (see [15, Proposition 1]) that there exist 0 < u ∈ C ∞ (M ) and a compact subset Let k > 0 and consider the conformal metric where g is the induced metric on M .Let s be the arc length with respect to the metric g.Following the construction in [15, Theorem 1], we can construct a minimizing geodesic γ(s) : [0, +∞) → M \ K in the metric g which has infinite length in the metric g.Now we can argue exactly as in the proof of estimate ( 6) in [13], using H ≡ 0, obtaining for every smooth function ϕ such that ϕ(0) = ϕ(a) = 0 and for every k > 0. Arguing as in [23, Section 2] we have the following identity Since (N n+1 , h) is with non-negative sectional curvature and either uniformly positive bi-Ricci curvature or uniformly positive Ricci curvature, we have R h 1j1j ≥ 0 for every j = 2, . . ., n and either BRic h (e 1 , ν) ≥ R 0 or Ric h (ν, ν) ≥ R 0 for some R 0 > 0. Therefore, if k ≤ 1, we get We conclude that the length (in the metric g) of the geodesic γ(s) is finite and this gives a contradiction.Therefore (M n , g) must be compact and this concludes the proof of Theorem 1.2.
Integrating over M we get a contradiction, since Ric h > 0 on M .Equivalently, one can use f ≡ 1 in the stability inequality (1.1) to get a contradiction.