On the Strong Liouville Property of Covering Spaces

Lyons and Sullivan conjectured in Lyons and Sullivan (J. Differential Geom. 19(2), 299–323, 1984) that if p : M → N is a normal Riemannian covering, with N closed, and M has exponential volume growth, then there are non-constant, positive harmonic functions on M. This was proved recently in Polymerakis (Adv. Math. 379, 107552–107558, 2021) exploiting the Lyons-Sullivan discretization and some sophisticated estimates on the green metric on groups. In this note, we provide a self-contained proof relying only on elementary properties of the Brownian motion.


Introduction
It is well known that some properties of the fundamental group of a closed Riemannian manifold are reflected in the geometry of its universal covering space.A classic result [6,10,13] illustrating this fact asserts that the growth rate of the fundamental group and the volume growth rate of the universal covering space coincide.In this direction, Lyons and Sullivan [9] focused on the strong Liouville property of the covering space; that is, the non-existence of non-constant positive harmonic functions.It is noteworthy that it is still unknown whether the validity of the strong Liouville property on the universal covering space depends only on the fundamental group or if the Riemannian metric plays a role.
To set the stage, let p : M → N be a normal Riemannian covering of a closed manifold with deck transformation group .Lyons and Sullivan proved in [9] that if M has polynomial volume growth, then M has the strong Liouville property.They conjectured in [9, p. 305] that if M is the universal covering space of N , then M has the strong Liouville property if and only if M has subexponential volume growth.If is polycyclic, this was established in [2,5].More generally, this was confirmed recently in [11], in the case where M has exponential volume growth.Theorem 1.1 Let p : M → N be a normal Riemannian covering, with N closed.If the deck transformation group of the covering has exponential growth, then there are non-constant, positive harmonic functions on M.
The main result of [11] is the existence of non-constant positive harmonic functions on groups of exponential growth, which relies on the highly non-trivial estimates of [4] involving the green metric on finitely generated groups.Then Theorem 1.1 follows from the Lyons-Sullivan discretization, which yields that the cone of positive harmonic functions on M is isomorphic to the cone of positive harmonic functions on with respect to an induced probability measure (cf.[3, Theorem A]).In this note, we give an elementary proof of Theorem 1.1 relying only on basic properties of the Brownian motion.Our method is inspired by [1], which is devoted to harmonic functions on groups.
In view of Theorem 1.1 and [9, Theorem 1], it remains to examine the existence of non-constant positive harmonic functions on M, in the case where is of intermediate growth; that is, has superpolynomial and subexponential growth.Since there exist finitely generated groups of intermediate growth, it is evident that there exist normal coverings of closed manifolds with such deck transformation groups.It should be emphasized that these may not be universal coverings, because it remains open whether there exists a finitely presentable group of intermediate growth.
The known results on the Liouville and the strong Liouville property of normal covering spaces can be established either in a direct way or by proving a group-theoretic analogue and using the Lyons-Sullivan discretization.However, the study of positive harmonic functions on groups of intermediate growth seems to be quite complicated.For instance, with respect to a probability measure that does not have finite first moment, there may exist non-constant bounded harmonic functions on such a group (cf.[7]).But such measures do not occur from the Lyons-Sullivan discretization.So, the aforementioned problem may be more accessible with a straightforward approach relying on the Brownian motion on M, as pursued in this note.

Preliminaries
We begin by recalling some basic facts about the Brownian motion, which may be found for instance in [8].The Brownian motion on a Riemannian manifold M is the diffusion process corresponding to the Laplace operator.Let M be a stochastically complete Riemannian manifold.We view the Brownian motion on M starting at a point x ∈ M as a probability measure P x on the path space It is worth to point out that S(ω) is finite for P x -almost any ω ∈ , and is a stopping time.The exit measure ε for any Borel subset A of ∂K.A subset of ∂K is called ε K x -measurable if it belongs to the completion of the sigma-algebra of Borel subsets of ∂K with respect to ε K x .Since S(ω) is finite P x -almost surely, we readily see that ε K x is a probability measure on ∂K.It is important that the Brownian motion on M enjoys the strong Markov property.Thus, for compact domains K 1 , K 2 with K 1 ⊂ K 2 , and x ∈ K • 1 , the exit measures satisfy for any ε K 2 x -measurable subset A of ∂K 2 .There is a remarkable relation between the Brownian motion on M and harmonic functions on M (that is, smooth functions whose Laplacian is identically zero).More precisely, given an open domain U of M, a locally bounded f : U → R is harmonic if and only if holds for any compact domain K ⊂ U and any x ∈ K • .Moreover, for a smoothly bounded, compact domain K, the unique harmonic extension Hf of a function f ∈ C(∂K) is given by , and is continuous up to the boundary of K (cf.for example [8, p. 149]).Therefore, for for any x, y ∈ K 1 (cf.for instance [12, p. 336]).For a compact domain K of M, and x, y ∈ K • , consider the quantity where the supremum is taken over all ε K y -measurable subsets A of ∂K with K y (A) > 0. It is apparent that We are interested in this quantity due to its relation with the strong Liouville property on M, which is established in the following proposition.Taking into account (2), it is not hard to prove the converse of this proposition, but since we do not use it in the sequel, we omit it.
Proposition 2.1 If any positive harmonic function on M is constant, then for any exhausting sequence (K n ) n∈N of M, and x, y ∈ K • 1 , we have that ε(K n ; x, y) → 0 as n → +∞.
Proof Assume to the contrary that the conclusion does not hold.Then there exists c > 0, an exhausting sequence for any n ∈ N. We define the bounded function f n in the interior of K n by This shows that f n is a positive harmonic function in the interior of K n , with f n (y) = 1, for any n ∈ N. From Eq. 3, it follows that after passing to a subsequence, if necessary, we have that f n → f locally uniformly for some positive, harmonic function f ∈ C ∞ (M).This can be established by arguing as in the proof of [12,Theorem 2.1].It is immediate to verify that f (y) = 1 and which implies that f is non-constant.This is a contradiction.
y -measurable subset of ∂K 2 .From Eqs. 1 and 4, we derive that The asserted inequality is an immediate consequence of this estimate.
Let p : M → N be a normal Riemannian covering, with N closed.Consider a finite, smooth triangulation of N , and the triangulation of M obtained by lifting the simplices of N .For each full-dimensional simplex of N choose a lift on M, so that the union of their images is connected.Such a union F is called finite sided fundamental domain for the covering.The boundary of F consists of images of lower dimensional simplices, which are called faces of F .A face that corresponds to a codimension one simplex is called a side of the fundamental domain.Faces and sides are defined in a similar way for translates and unions of translates of F .Proof Let C be a side of F .Notice that there exists a smoothly bounded, compact domain K ⊂ F , with x ∈ K • , such that K ∩C has non-empty interior in ∂K.Then the exit measures satisfy Taking into account that K is smoothly bounded and K ∩ C has non-empty interior in ∂K, we obtain that ε K x (K ∩ C) > 0, which yields that ε F x (C) > 0. Since F has finitely many sides, this completes the proof.

Proof of Theorem 1.1
We choose a finite sided fundamental domain F for the covering, and x ∈ F • .Let be the deck transformation group of the covering, denote by d the dimension of M, and set where Area(•) stands for the (d − 1)-dimensional Hausdorff measure.Then G is a symmetric, finite set of generators of , and consists of all g ∈ such that gF contains a side of F .For n ∈ N, denote by S n and W n the set of words of length n and at most n, respectively, with respect to G, and let K n be the union of the translates hF , with h ∈ W n .It is not difficult to verify that ∂K n is written as the union of C h := ∂K n ∩ hF , with h ∈ S n .It should be noticed that the set consists of all those g ∈ S n such that gF contributes a side to K n .Bearing in mind that ∂K n is written as the union of the sides of K n , we derive that From Lemma 2.3, there exists c > 0 such that ε F x (C) ≥ c for any side C of F .From the fact that x, gx ∈ K • 1 for any g ∈ G, and Eq. 3, we get that there exists c 0 > 0 such that Assume to the contrary that any positive harmonic function on M is constant, and let 0 < δ < 1.From Proposition 2.1, there exists n 0 ∈ N such that for any n ≥ n 0 and any g ∈ G.
Consider n > n 0 and h ∈ S n .Then the set C h contains a side of hF .Write h = g 1 . . .g n , with g i ∈ G for any 1 ≤ i ≤ n, and set x 0 := x, x i := g 1 . . .g i x for 1 ≤ i ≤ n.It is easy to see that where we used the invariance of the Brownian motion under isometries, and that h −1 C h contains a side of F .Since the measures ε K n x i are equivalent, this shows that ε K n x i (C h ) > 0 for any 0 ≤ i ≤ n.Hence, ε K n x (C h ) is given by For 0 ≤ i < n − n 0 , it is immediate to verify that For j ∈ N, set K j (x i ) := g 1 . . .g i K j .Since K n−i (x i ) ⊂ K n and ε(•; •, •) is invariant under isometries, Lemma 2.2 and Eq. 6 yield that Thus, we deduce that for 0 ≤ i < n − n 0 .
For n − n 0 ≤ i ≤ n − 1, from Eq. 1 and the invariance of the Brownian motion under isometries, we derive that where we used that g i+1 ∈ G. Finally, from Eqs. 7, 8, 9 and 10, we conclude that Taking into account that the action of on M is properly discontinuous, we observe that there exists k ∈ N such that any point z ∈ M belongs to at most k different translates of F .From the fact that ε K n x is a probability measure on ∂K n , and Eq. 5, we obtain that the cardinality of S n satisfies To estimate the cardinality of S n , consider g ∈ S n .Then there exists h ∈ S n−1 such that g −1 h ∈ G. From the definition of G, this yields that a side of hF is contained in gF .Since the latter one is not contained in K n−1 , we deduce that this side is contained in C h and therefore, h ∈ S n−1 .In other words, for any g ∈ S n there exist h ∈ S n−1 and ḡ ∈ G such that g = h ḡ.This gives the estimate for any n > n 0 + 1.In particular, it follows that lim sup Since 0 < δ < 1 is arbitrary, this implies that has subexponential growth, which is a contradiction.Therefore, there are non-constant positive harmonic functions on M.
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Lemma 2 . 3
Let p : M → N be a normal Riemannian covering, with N closed.Fix a finite sided fundamental domain F for the covering, and x ∈ F • .Then there exists c > 0 such that ε F x (C) ≥ c for any side C of F .