Torus stability under Kato bounds on the Ricci curvature

Abstract

We show two stability results for a closed Riemannian manifold whose Ricci curvature is small in the Kato sense and whose first Betti number is equal to the dimension. The first one is a geometric stability result stating that such a manifold is Gromov–Hausdorff close to a flat torus. The second one states that, under a stronger assumption, such a manifold is diffeomorphic to a torus: this extends a result by Colding and Cheeger–Colding obtained in the context of a lower bound on the Ricci curvature.

1 INTRODUCTION

The celebrated Bochner theorem states that if a closed Riemannian manifold $$(M^n,g)$$ has non-negative Ricci curvature, then its first Betti number satisfies

$$$$\begin{equation*} b_1(M)\leqslant n, \end{equation*}$$$$

with equality if and only if $$(M^n,g)$$ is isometric to a flat torus. This inequality was improved by Gromov [25] and Gallot [18] who found $$\varepsilon (n)>0$$ such that if

$$$$\begin{equation*} \mbox{Ric}\geqslant -\frac{\varepsilon (n)}{\mathrm{diam}(M,g)^2}\, g, \end{equation*}$$$$

then $$b_1(M)\leqslant n$$. Gromov also made a conjecture about the equality case, which was proven true by Colding [14] and Cheeger–Colding [5]: there exists $$\updelta(n)>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying

$$$$\begin{equation*} b_1(M)= n, \qquad \mbox{Ric}\geqslant -\frac{\updelta(n)}{D^2}\, g, \end{equation*}$$$$

then $$M$$ is diffeomorphic to a torus. The proof of this latter result consists in two steps. First, one shows that for any $$\varepsilon \in (0,1)$$ there exists $$\updelta(n,\varepsilon )>0$$ such that if $$(M^n,g)$$ with diameter $$D$$ satisfies

$$$$\begin{equation*} b_1(M)= n, \qquad \mbox{Ric}\geqslant -\frac{\updelta(n,\varepsilon )}{D^2}\, g, \end{equation*}$$$$

then there exists a flat torus $$\mathbb {T}^n$$ and an $$\varepsilon D$$-almost isometry† $$\Phi \colon M\rightarrow \mathbb {T}^n$$. Second, the intrinsic Reifenberg theorem of Cheeger–Colding allows to prove a topological stability result: if $$\varepsilon$$ is sufficiently small, then $$M$$ is diffeomorphic to $$\mathbb {T}^n$$ (see [6, Theorem A.1.1. and Theorem A.1.13]). For more details, we refer to the very instructive texts [9, 20] presenting the work of Cheeger and Colding.

The Bochner estimate has been generalized in several directions. For a Riemannian manifold $$(M^n,g)$$, let $$\mbox{Ric}_{\mbox{{-}}}: M \rightarrow \mathbb {R}_+$$ be the lowest non-negative function such that for any $$x \in M$$,

$$$$\begin{equation*} \mbox{Ric}_x \geqslant - \mbox{Ric}_{\mbox{{-}}}(x)g_x. \end{equation*}$$$$

Then Gallot obtained in [19] that for every $$p>n/2$$, there exists $$\varepsilon (n,p)>0$$ such that if $$(M^n,g)$$ with diameter $$D$$ satisfies

$$$$\begin{equation*} D^2{\left(\fint _M \mbox{Ric}_{\mbox{{-}}}^p \mathop {}\!\mathrm{d}\nu _g\right)}^{\frac{1}{p}}\leqslant \varepsilon (n,p), \end{equation*}$$$$

then $$b_1(M)\leqslant n$$; here and throughout, $$\nu _g$$ is the Riemannian volume measure induced by $$g$$ on $$M$$, and $$\fint _A f \mathop {}\!\mathrm{d}\nu _g \coloneqq \nu _g(A)^{-1}\int _A f \mathop {}\!\mathrm{d}\nu _g$$ for any Borel set $$A \subset M$$ and any measurable function $$f$$ defined on $$A$$. To our knowledge, no topological rigidity result has been obtained so far from this integral condition. It seems to us that the segment inequality proven in [10] and the results from [30, 31] may imply such a rigidity result. Another direction is the one of metric measure spaces satisfying a suitable synthetic Ricci curvature lower bound. In this context, a rigidity result à la Bochner and geometric stability results hold, see [24, 28, 29].

In this paper, we obtain geometric and topological results under a Kato bound. More precisely, let us introduce the following definition.

Definition 1.1.Let $$(M^n,g)$$ be a complete Riemannian manifold with heat kernel $$H(t,x,y)$$ and diameter $$D$$. For any $$T>0$$, we set

$$$$\begin{equation*} \mbox{k}_T(M^n,g):=\sup _{x\in M} \iint\nolimits _{[0,T]\times M} H(t,x,y)\mbox{Ric}_{\mbox{{-}}}(y)\mathop {}\!\mathrm{d}\nu _g(y) \mathop {}\!\mathrm{d}t. \end{equation*}$$$$

We say that the number $$\mbox{k}_{D^2}(M^n,g)$$ is the Kato constant of $$(M^n,g)$$.

The first occurrence of $$\mbox{k}_T(M^n,g)$$ in the study of Riemannian manifolds seems to be [23]. The geometric and analytic consequences of a bound on $$\mbox{k}_T(M^n,g)$$ have been extensively studied since then, see, for example, [1, 4, 12, 13, 15, 32, 34, 36]. It is useful to note that if $$\mbox{Ric}\geqslant -\kappa ^2 g$$, then $$\mbox{k}_T(M^n,g)\leqslant \kappa ^2 T$$; hence, a smallness assumption on $$\mbox{k}_T(M^n,g)$$ should be understood as a control on the part of the manifold where $$\mbox{Ric}_{\mbox{{-}}}\gtrsim T^{-1}$$.

The Bochner estimate extends to the case of Riemannian manifolds with small Kato constant, as proven in [32] and improved in [4]: there exists $$\updelta(n)>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ such that $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta(n)$$, then $$b_1(M)\leqslant n$$. Our first main result provides an answer to a question raised in [4] about the equality case.

Theorem A.For any $$\varepsilon \in (0,1)$$ there exists $$\updelta(n,\varepsilon )>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying

$$$$\begin{equation*} b_1(M)= n \qquad \text{and} \qquad \mbox{k}_{D^2}(M^n,g)\leqslant \updelta(n,\varepsilon ), \end{equation*}$$$$

then $$M$$ is $$\varepsilon D$$-almost isometric to a flat torus.

Remark 1.2.From [15, Theorem 4.3] we can replace the smallness assumption $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta(n,\varepsilon )$$ with an integral condition involving the Ricci curvature only, namely,

$$$$\begin{equation*} \sup _{x \in M}\int _0^Dr\fint _{B_r(x)}\mbox{Ric}_{\mbox{{-}}}(y)\mathop {}\!\mathrm{d}\nu _g(y)\,\mathop {}\!\mathrm{d}r\leqslant \updelta(n,\varepsilon ). \end{equation*}$$$$

Our second main result provides a topological stability theorem under a so-called strong Kato bound. This assumption appeared naturally in our previous work [12, 13] where we obtained, among other results, Reifenberg regularity.

Theorem B.Let $$f\colon [0,1]\rightarrow \mathbb {R}_+$$ be a non-decreasing function satisfying

$$$$\begin{equation} \int _0^1 \frac{\sqrt {f(t)}}{t} \mathop {}\!\mathrm{d}t<+\infty . \end{equation}$$$$(SK)

Then there exists $$\updelta(n,f)>0$$ such that if a closed Riemannian manifold $$(M^n,g)$$ of diameter $$D$$ satisfies

$$$$\begin{equation*} b_1(M)=n, \qquad \mbox{k}_{D^2}(M^n,g)\leqslant \updelta(n,f), \end{equation*}$$$$

and

$$$$\begin{equation*} \mbox{k}_{tD^2}(M^n,g)\leqslant f(t) \quad \text{for all $t \in (0,1]$}, \end{equation*}$$$$

then $$M$$ is diffeomorphic to a torus.

Remark 1.3.Our proof actually shows a stronger result: for any $$\upalpha\in (0,1)$$ there exists $$\updelta(n,f,\upalpha)>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying $$b_1(M)= n$$, $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta(n,f,\upalpha)$$, and $$\mbox{k}_{tD^2}(M^n,g)\leqslant f(t)$$ for any $$t \in (0,1]$$, with $$f$$ satisfying (SK), then there exist a flat torus $$(\mathbb {T}^n,\mathsf {d}_{\mathbb {T}^n})$$ and a diffeomorphism $$\mathcal {A}\colon M\rightarrow \mathbb {T}^n$$ such that for any $$x,y\in M$$,

$$$$\begin{equation*} \upalpha{\left(\frac{\mathsf {d}_g(x,y)}{D}\right)}^{\frac{1}{\upalpha}}\leqslant \frac{\mathsf {d}_{\mathbb {T}^n}{\left(\mathcal {A}(x),\mathcal {A}(y)\right)}}{D}\leqslant \upalpha^{-1}{\left(\frac{\mathsf {d}_g(x,y)}{D}\right)}^\upalpha. \end{equation*}$$$$

Here and throughout, $$\mathsf {d}_g$$ is the Riemannian distance induced by $$g$$.

According to [33], an $$L^p$$ smallness condition on $$\mbox{Ric}_{\mbox{{-}}}$$ yields the strong Kato bound, so that Theorem B has the following corollary.

Corollary 1.4.For any $$p>n/2$$, there exists $$\varepsilon (n,p)>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying

$$$$\begin{equation*} b_1(M)= n \quad \text{and} \quad D^2{\left(\fint _M \mbox{Ric}_{\mbox{{-}}}^p\mathop {}\!\mathrm{d}\nu _g\right)}^{\frac{1}{p}}\leqslant \varepsilon (n,p), \end{equation*}$$$$

then $$M$$ is diffeomorphic to a torus.

Similarly, using [15, Theorem 4.3], we get the following corollary involving a suitable Morrey norm.

Corollary 1.5.For any $$\upalpha\in (0,2]$$, there exists $$\varepsilon (n,\upalpha)>0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying

$$$$\begin{equation*} b_1(M)= n \quad \text{and} \quad \sup _{\substack{x \in M\\ r \in (0,D)}} D^{\alpha }r^{2-\alpha }\fint _{B_r(x)}\mbox{Ric}_{\mbox{{-}}}(y)\mathop {}\!\mathrm{d}\nu _g(y)\leqslant \varepsilon (n,\alpha ), \end{equation*}$$$$

then $$M$$ is diffeomorphic to a torus.

Colding's original argument relied upon harmonic approximations of Busemann-like functions. Here we follow an alternative approach, closer to the one proposed by Gallot in [20]. We consider the Albanese map $$\mathcal {A}\colon M\rightarrow \mathbb {T}^n$$. We lift $$\mathcal {A}$$ to a harmonic map $$\widehat{\mathcal {A}}:=(\widehat{\mathcal {A}}_1,\dots ,\widehat{\mathcal {A}}_n)$$ defined on a suitable abelian cover $$\widehat{M}{\longrightarrow } M$$ which is equivariant under an action of $$\mathbb {Z}^n$$. Estimates from [4] imply that if $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta\leqslant 1/(16n)$$, then $$\widehat{\mathcal {A}}$$ is surjective, $$(1+C(n)\sqrt [3]{\updelta})$$-Lipschitz and for any $$r\in [D, \updelta^{-1/6}D]$$,

$$$$\begin{equation} {\left(r^2\fint _{B_r} |\nabla d\widehat{\mathcal {A}}|^2\mathop {}\!\mathrm{d}\nu _{\widehat{g}}\right)}^{\frac{1}{2}} + \fint _{B_r} {\left| \langle d\widehat{\mathcal {A}}_i,d\widehat{\mathcal {A}}_j\rangle -\updelta_{i,j}\right|} \leqslant C(n)\sqrt [3]{\updelta}. \end{equation}$$$$( <!--${ifMathjaxEnabled: 10.1112%2Fjlms.12704}-->$${{\star _r}}$$<!--${/ifMathjaxEnabled:}--><!--${ifMathjaxDisabled: 10.1112%2Fjlms.12704}-->$$<!--${/ifMathjaxDisabled:}-->)

One original point in our proof of Theorem A is the use of an almost rigidity result (Theorem 3.1) which implies that under such an estimate, the restriction of $$\widehat{\mathcal {A}}$$ to a ball of radius $$64n^2D$$ realizes an $$\varepsilon D$$-almost isometry with a Euclidean ball of the same radius. We prove this almost rigidity result by means of the analysis developed in [12]. Then we can follow the lines of Gallot's argument to get Theorem A.

We prove Theorem B by showing that $$\mathcal {A}$$ is a diffeomorphism, answering a question raised by Gallot [20, Section 6]. This differs from Colding's proof which used the intrinsic Reifenberg theorem to conclude. It is enough to show that the restriction of $$\widehat{\mathcal {A}}$$ to the ball $$B_D(\widehat{o})$$ is a diffeomorphism onto its image. In the context of almost non-negative Ricci curvature, according to the recent Reifenberg theorem established in [11, Theorem 7.10], this is the case if ($$\star _{2D}$$) holds and if the volume ratio

$$$$\begin{equation*} \frac{\nu _{\widehat{g}}(B_{2D}(\widehat{o}))}{\omega _n (2D)^n} \end{equation*}$$$$

is almost one. In our context, we have at our disposal an analogous Reifenberg type result: see Proposition 3.4. To apply this result, we must control a heat kernel ratio which plays the role, in our setting, of the volume ratio for Ricci curvature lower bounds. One main difference is that, unlike the volume, the heat kernel is a non-local quantity. Our key tools to get the desired control are a heat kernel comparison theorem á la Cheeger–Yau [17], and an almost Euclidean volume bound (Theorem A.1, after [7, Theorem 1.2]).

2 THE DYNKIN CONDITION AND CONSEQUENCES

In this section, we point out relevant properties of the so-called Dynkin condition. We say that a complete Riemannian manifold $$(M^n,g)$$ satisfies such a condition if there exists $$T&gt;0$$ such that

$$$$\begin{align} \mbox{k}_T(M^n,g)\leqslant \frac{1}{16n} \, \cdot \end{align}$$$$(Dy)

2.1 Closed manifolds

Let us first mention properties of closed Riemannian manifolds satisfying a Dynkin condition.

2.1.1 Volume doubling

See [4, Proposition 3.8] and [12, Proposition 3.3] for the next result.

Proposition 2.1.Let $$(M^n,g)$$ be a closed Riemannian manifold satisfying (Dy). Then there exists $$C(n)&gt;0$$ such that for any $$x\in M$$ and $$0&lt;s\leqslant r\leqslant \sqrt {T}$$,

$$$$\begin{equation*} \frac{\nu _g{\left(B_r(x)\right)}}{\nu _g{\left(B_s(x)\right)}}\leqslant C(n) {\left(\frac{r}{s}\right)}^{e^2 n} . \end{equation*}$$$$

2.1.2 Heat kernel bounds

See [13, Proposition 2.6] for the following.

Proposition 2.2.Let $$(M^n,g)$$ be a closed Riemannian manifold satisfying (Dy). Then there exists $$C(n)&gt;0$$ such that for any $$x,y\in M$$ and $$t\in (0,T)$$,

• (i) $$ \frac{C(n)^{-1}}{\nu _g(B_{\sqrt {t}}(x))}e^{-C(n)\frac{\mathsf {d}_g^2(x,y)}{t}}\leqslant H(t,x,y)\leqslant \frac{C(n)}{\nu _g(B_{\sqrt {t}}(x))}e^{-\frac{\mathsf {d}_g^2(x,y)}{5t}}$$,
• (ii) $$|d_x H(t,x,y)|\leqslant \frac{C(n)}{\sqrt {t}\nu _g(B_{\sqrt {t}}(x))}e^{-\frac{\mathsf {d}_g^2(x,y)}{5t}}$$.

When $$T\geqslant \mathrm{diam}^2(M^n,g)$$, we can get an estimate depending only on the volume of $$M$$.

Proposition 2.3.Let $$(M^n,g)$$ be a closed Riemannian manifold of diameter $$D$$ satisfying (Dy). If $$T\geqslant D^2$$, then for any $$x,y\in M$$ and $$t\in (D^2,T)$$,

$$$$\begin{equation*} H(t,x,y)\leqslant {\left(1+C(n)\frac{D}{\sqrt {t}}\right)}\frac{1}{\nu _g(M)} \, \cdot \end{equation*}$$$$

Proof.Let $$t\in (D^2,T)$$ and $$x,y\in M$$. By stochastic completeness, we have that $$\int _M H(t,z,y)\mathop {}\!\mathrm{d}\nu _g(z)=1$$; hence, there is some $$z_0\in M$$ such that $$H(t,z_0,y)=\frac{1}{\nu _g(M)}$$. By the previous proposition, we have that $$|d_z H(t,z,y)|\leqslant \frac{C(n)}{\sqrt {t}\nu _g(M)}$$. Then

$$$$\begin{eqnarray*} {\left| H(t,x,y)-H(t,z_0,y)\right|}\leqslant \mathsf {d}_g(x,z_0)\, \frac{C(n)}{\sqrt {t}\nu _g(M)}\leqslant \frac{C(n) D}{\sqrt {t}\nu _g(M)} \, \cdot\\[-45pt] \end{eqnarray*}$$$$

$$\Box$$

2.2 Non-compact manifolds

For complete non-compact manifolds, the results of [12, Subsection 3.1] yield the following.

Proposition 2.4.Let $$(M^n,g,o)$$ be a pointed complete Riemannian manifold. Assume that there exists a sequence $$\lbrace (M_\alpha ^n,g_\alpha ,o_\alpha )\rbrace _\alpha$$ of pointed closed Riemannian manifolds satisfying (Dy) for a same $$T&gt;0$$. Assume that for any $$R&gt;0$$ there exists $$\alpha _R$$ such that for any $$\alpha \geqslant \alpha _R$$ there exists a diffeomorphism onto its image

$$$$\begin{equation*} \Phi _\alpha \colon B_R(o)\rightarrow M_\alpha \end{equation*}$$$$

such that the following convergence holds:

$$$$\begin{equation*} \lim _{\alpha \rightarrow +\infty } {\left\Vert \Phi _\alpha ^*g_\alpha -g\right\Vert} _{\mathcal {C}^0}=0. \end{equation*}$$$$

Then $$(M^n,g)$$ satisfies (Dy), the volume doubling estimate from Proposition 2.1 and the heat kernel estimates from Proposition 2.2.

We recall that a group $$\Gamma$$ with neutral element 1 is residually finite if and only if it admits a sequence of normal subgroups $$\lbrace \Gamma _j\rbrace$$ with finite index such that

$$$$\begin{equation*} \bigcap _j \Gamma _j=\lbrace 1\rbrace . \end{equation*}$$$$

Then Proposition 2.4 has the following useful application.

Proposition 2.5.Let $$\pi :\widehat{M}\rightarrow M$$ be a normal covering of a closed Riemannian manifold $$(M^n,g)$$ with residually finite deck transformation group. If $$({M},g)$$ satisfies (Dy), then $$(\widehat{M},\pi ^*g)$$ satisfies (Dy), the volume doubling estimate from Proposition 2.1, and the heat kernel estimates from Proposition 2.2.

Proof.We start by noticing that if $$V\colon M\rightarrow \mathbb {R}$$ is a bounded function, then

$$$$\begin{equation*} e^{-t\Delta _{\pi ^*g}}(V\circ \pi )={\left(e^{-t\Delta _{g}}V\right)}\circ \pi . \end{equation*}$$$$

Indeed, since $$(M,g)$$ is closed, $$(\widehat{M}, \pi ^*g)$$ is stochastically complete and then we have uniqueness in $$L^\infty$$ of the solution of the heat equation with fixed initial condition

$$$$\begin{equation*} {\left\lbrace \def\eqcellsep{&}\begin{array}{l}{\left(\frac{\partial }{\partial t}+\Delta _{\pi ^*g}\right)}u=0,\\[3pt] u(0,\cdot )=V\circ \pi (\cdot ). \end{array} \right.} \end{equation*}$$$$

But $$e^{-t\Delta _{\pi ^*g}}(V\circ \pi )$$ and $$(e^{-t\Delta _{g}}V)\circ \pi$$ are both solutions of this Cauchy problem, hence we get the desired equality. Notice that $$\mbox{Ric}_{\mbox{{-}}}(\pi ^*g)=\mbox{Ric}_{\mbox{{-}}}(g)\circ \pi$$, hence for any $$x\in \widehat{M}$$ and $$t&gt;0$$

$$$$\begin{equation*} \int _{\widehat{M}} H_{\pi ^*g}(t,x,y) \mbox{Ric}_{\mbox{{-}}}(\pi ^*g)(y) \mathop {}\!\mathrm{d}\nu _{\pi ^*g}(y)=\int _M H_g(t,\pi (x),z) \mbox{Ric}_{\mbox{{-}}}(g)(z) \mathop {}\!\mathrm{d}\nu _{g}(z), \end{equation*}$$$$

where we have noted $$H_{\pi ^*g}$$ (respectively, $$H_g$$) the heat kernel of $$(\widehat{M}, \pi ^*g)$$ (resp. of $$({M}, g)$$). Hence we get that for any $$T&gt;0$$:

$$$$\begin{equation} \mbox{k}_T(M,g)=\mbox{k}_T(\widehat{M}, \pi ^*g). \end{equation}$$$$(1)

As $$\Gamma$$ is residually finite, there exists a sequence of normal subgroup $$\Gamma _j\triangleleft \Gamma$$ of finite index such that

$$$$\begin{equation*} \bigcap _j \Gamma _j=\lbrace 1\rbrace . \end{equation*}$$$$

For any $$j\in \mathbb {N}^*$$, we set $$\widehat{M}_j:=\widehat{M}/\Gamma _j$$. We get two covering maps

$$$$\begin{equation*} \widehat{M}\stackrel{p_j}{\longrightarrow }\widehat{M}_j\stackrel{\pi _j}{\longrightarrow }M. \end{equation*}$$$$

Note that $$\widehat{M}_j$$ is a closed manifold and the above argument implies that for any $$T&gt;0$$:

$$$$\begin{equation*} \mbox{k}_T(\widehat{M}, \pi ^*g)=\mbox{k}_T(\widehat{M}_j, \pi _j^*g). \end{equation*}$$$$

Hence each $$(\widehat{M}_j, \pi _j^*g)$$ satisfies (Dy). Moreover, if we consider $$\widehat{o}\in \widehat{M}$$, then for any $$R&gt;0$$ there is some $$j_R$$ such that for any $$j\geqslant j_R$$, the restriction of the covering map

$$$$\begin{equation*} p_j\colon B_R(\widehat{o})\rightarrow B_R(p_j(\widehat{o})) \end{equation*}$$$$

is an isometry. Then the conclusion follows from Proposition 2.4.$$\Box$$

3 ALMOST RIGIDITY RESULTS

In this section, we provide almost rigidity results which are consequences of our previous work [12, 13]. For any $$\rho &gt;0$$, we let $$\mathbb {B}^n_\rho$$ be the Euclidean ball in $$\mathbb {R}^n$$ centered at 0 with radius $$\rho$$. If $$B$$ is a ball in an $$n$$-dimensional Riemannian manifold and $$h = (h_1,\ldots ,h_n) :B \rightarrow \mathbb {R}^n$$ is a smooth map, we denote by

$$$$\begin{equation*} dh \, ^tdh = {\left[ \langle dh_i,dh_j \rangle \right]}_{1\leqslant i,j \leqslant n} \end{equation*}$$$$

its Gram matrix map. We write $$\mathrm{Id}_n$$ for the identity matrix of size $$n$$ and $$\omega _n$$ for the Lebesgue measure of the unit Euclidean ball in $$\mathbb {R}^n$$.

3.1 Harmonic almost splitting

Theorem 3.1.For any $$\varepsilon \in (0,1)$$, there exists $$\updelta=\updelta(n,\varepsilon )&gt;0$$ such that for any closed Riemannian manifold $$(M^n,g)$$ satisfying $$\mbox{k}_{\rho ^2}(M^n,g)\leqslant \updelta$$ for some $$\rho &gt;0$$, if for some $$x \in M$$, there exists a harmonic map

$$$$\begin{equation*} h\colon B_{\updelta^{-1}\rho }(x)\rightarrow \mathbb {R}^n \end{equation*}$$$$

such that for any $$r\in [\rho ,\updelta^{-1}\rho )$$,

$$$$\begin{equation*} {\left(r^2\fint _{B_r(x)} |\nabla dh|^2\mathop {}\!\mathrm{d}\nu _g\right)}^{\frac{1}{2}}+\fint _{B_{r}(x)}{\left| dh\,{}^tdh-\mathrm{Id}_n\right|} \mathop {}\!\mathrm{d}\nu _g\leqslant \updelta, \end{equation*}$$$$

then $$h$$ is an $$\varepsilon \rho$$-almost isometry between $$B_\rho (x)$$ and $$\mathbb {B}^n_\rho$$.

Remarks 3.2.

• (i) From the proof of Proposition 2.5, we easily see that the result also holds if we assume that $$(M^n,g)$$ is a normal covering of a closed Riemannian manifold whose deck transformation group is residually finite.
• (ii) A similar statement holds for $$\operatorname{RCD}(K,N)$$ spaces [3, Proposition 3.7].

Proof.By scaling, there is no loss of generality in assuming $$\rho =1$$, what we do from now on. We argue by contradiction. Assume that there exists $$\varepsilon \in (0,1)$$ and:

• a sequence of positive numbers $$\lbrace \updelta_\alpha \rbrace$$ such that $$\updelta_\alpha \downarrow 0$$,
• a sequence of pointed closed Riemannian manifolds $$\lbrace (M^n_\alpha ,g_\alpha ,x_\alpha )\rbrace$$ such that $$\mbox{k}_{1}(M_\alpha ,g_\alpha )\leqslant \updelta_\alpha$$ for any $$\alpha$$,
• a sequence of maps $$\lbrace h_\alpha \colon B_{\updelta_\alpha ^{-1}}(x_\alpha )\rightarrow \mathbb {R}^n\rbrace$$ such that for any $$\alpha$$,
  $$$$\begin{equation} \text{$h_\alpha $ is not an $\varepsilon $-almost isometry between $B_1(x_\alpha )$ and $\mathbb {B}_1^n$} \end{equation}$$$$(2)
  and for any $$r\in [1,\updelta_\alpha ^{-1}]$$,
  $$$$\begin{equation*} {\left(r^2\fint _{B_r(x_\alpha )} |\nabla dh_\alpha |^2\mathop {}\!\mathrm{d}\nu _{g_\alpha }\right)}^{\frac{1}{2}}+\fint _{B_{r}(x_\alpha )}{\left| dh_\alpha \,{}^tdh_\alpha -\mathrm{Id}_n\right|} \mathop {}\!\mathrm{d}\nu _{g_\alpha }\leqslant \updelta_\alpha . \end{equation*}$$$$

We can assume the following.

• (1) Thanks to [12, Corollary 2.5, Remark 4.9], the sequence $$\lbrace (M^n_\alpha ,g_\alpha ,x_\alpha )\rbrace$$ converges in the pointed measured Gromov–Hausdorff topology to a space $$(X,\mathsf {d},\mu ,x)$$ which is infinitesimally Hilbertian in the sense of [21]. This limit space is endowed with a carré du champ $$\Gamma$$ and a natural Laplace operator $$L$$, which is the Friedrichs extension of the quadratic form $$\varphi \in W^{1,2}\mapsto \int _X \mathop {}\!\mathrm{d}\Gamma (\varphi ,\varphi )$$. For the precise definitions of $$\Gamma$$ and $$L$$, see [12, Section 1.2] and references therein.
• (2) By [12, Proposition E.10], the sequence $$\lbrace h_\alpha \rbrace _\alpha$$ converges uniformly on compact sets to a harmonic function $$h=(h_1,\dots ,h_n)\colon X\rightarrow \mathbb {R}^n$$ and for any $$i,j\in \lbrace 1,\dots ,n\rbrace$$,
  $$$$\begin{equation*} \frac{\mathop {}\!\mathrm{d}\Gamma }{\mathop {}\!\mathrm{d}\mu }(h_i,h_j)=\updelta_{i,j} \qquad \text{$\mu $-a.e.\nobreakspace on $X$.} \end{equation*}$$$$

Moreover, we know that $$(X,\mathsf {d},\mu )$$ admits a locally Lipschitz heat kernel $$H:(0,+\infty )\times X\times X \rightarrow (0,+\infty )$$ which satisfies the following Li–Yau inequality [13, Proposition 2.9 and Remark 2.10]: for any $$x\in X$$, $$t&gt;0$$, and $$\mu$$-a.e. $$y\in X$$,

$$$$\begin{equation*} |d_y H(t,x,y)|^2- H(t,x,y)\frac{\partial H}{\partial t}(t,x,y)\leqslant \frac{n}{2t} H^2(t,x,y). \end{equation*}$$$$

Let $$U\colon (0,+\infty )\times X\times X\rightarrow \mathbb {R}$$ be such that for any $$x,y \in X$$ and $$t&gt;0$$,

$$$$\begin{equation*} H(t,x,y)=\frac{e^{-\frac{U(t,x,y)}{4t}}}{(4\pi t)^{\frac{n}{2}}} \, \cdot \end{equation*}$$$$

It is easy to check (see [12, Formula (83)]) that $$U$$ satisfies, for any $$(x,t)\in X \times (0,+\infty )$$,

$$$$\begin{equation*} L U(t,x,\cdot )\geqslant -2n \end{equation*}$$$$

in a weak sense, that is to say, $$v(\cdot )=U(t,x,\cdot )\in W^{1,2}_{loc}$$ and for any non-negative $$\varphi \in W^{1,2}_{c}$$:

$$$$\begin{equation*} \int _X \mathop {}\!\mathrm{d}\Gamma (\varphi ,v)\geqslant -2n\int _X\varphi \mathop {}\!\mathrm{d}\mu . \end{equation*}$$$$

According to Varadhan formula ([12, Proposition 1.6]), for any $$x,y\in X$$,

$$$$\begin{equation*} \lim _{t\rightarrow 0} U(t,x,y)=\mathsf {d}^2(x,y). \end{equation*}$$$$

For any $$x,y \in X$$, set

$$$$\begin{equation*} \rho (x,y)=|h(x)-h(y)|^2 \end{equation*}$$$$

and note that

$$$$\begin{equation*} L \rho (x,\cdot )=-2n, \end{equation*}$$$$

hence for any $$t&gt;0$$ the function $$\rho (x,\cdot )-U(t,x,\cdot )$$ is sub-harmonic; passing to the limit $$t \downarrow 0$$ we get that $$\rho (x,\cdot )-\mathsf {d}^2(x,\cdot )$$ is sub-harmonic. Moreover, for each $$\xi =(\xi _1,\dots ,\xi _n)$$, the function $$h_\xi :=\sum _i \xi _ih_i$$ satisfies $$|dh_\xi |^2=|\xi |^2$$, so that $$h_\xi$$ is $$|\xi |$$-Lipschitz. As a consequence, for any $$x,y \in X$$,

$$$$\begin{equation*} \rho (x,y) = \sup _{\substack{\xi \in \mathbb {R}^n\\ |\xi |=1}} |\langle \xi , h(x)-h(y) \rangle |^2 = \sup _{\substack{\xi \in \mathbb {R}^n\\ |\xi |=1}} |h_\xi (x)-h_\xi (y)|^2 \leqslant \mathsf {d}^2(x,y). \end{equation*}$$$$

Therefore, the function $$\rho (x,\cdot )-\mathsf {d}^2(x,\cdot )$$ is sub-harmonic, non-positive and it reaches its maximum value, zero, at $$y=x$$; hence, it is constantly equal to zero. Thus we have shown that $$h\colon X\rightarrow \mathbb {R}^n$$ is an isometry onto its image. But $$h(X)$$ is closed, convex and according to [16, Claim 4.1, p. 130], its convex hull is $$\mathbb {R}^n$$. Then $$h$$ is an isometry between $$(X,\mathsf {d})$$ and the Euclidean space $$\mathbb {R}^n$$. By uniform convergence, we get that for $$\alpha$$ large enough, $$h_\alpha \colon B_1(x_\alpha )\rightarrow \mathbb {R}^n$$ is an $$\varepsilon$$-almost isometry between $$B_1(x_\alpha )$$ and an Euclidean ball of radius 1; this contradicts (2).$$\Box$$

Remark 3.3.Our proof avoids the use of the $$\operatorname{RCD}$$ theory; however, a better result could be proven using a recent result by Brué–Naber–Semola [2]: There is some constant $$c(n)\in (0,1)$$ so that for any $$\varepsilon \in (0,1)$$, there exists $$\updelta=\updelta(n,\varepsilon )&gt;0$$ such that for any closed Riemannian manifold $$(M^n,g)$$ satisfying $$\mbox{k}_{\rho ^2}(M^n,g)\leqslant \updelta$$ for some $$\rho &gt;0$$, if for some $$p \in M$$ there exists a harmonic map

$$$$\begin{equation*} h\colon B_{c(n)\rho }(p)\rightarrow \mathbb {R}^n \end{equation*}$$$$

satisfying:

$$$$\begin{equation*} {\left(\rho ^2\fint _{B_{\rho }(p)} |\nabla dh|^2\mathop {}\!\mathrm{d}\nu _g\right)}^{\frac{1}{2}}+\fint _{B_{\rho }(p)}{\left| dh\,{}^tdh-\mathrm{Id}_n\right|} \mathop {}\!\mathrm{d}\nu _g\leqslant \updelta, \end{equation*}$$$$

then $$h$$ is an $$\varepsilon \rho$$-almost isometry between $$B_{c(n)\rho }(p)$$ and $$\mathbb {B}^n_{c(n)\rho }$$.

This improvement would be the consequence of [13, Theorem 3.8 and Remark 3.10], of [2, Theorem 3.8], and of the following corollary of [12, Theorem 4.11]: for every $$\upeta\in (0,1)$$ there is some $$\updelta_1 = \updelta_1(n,\eta )&gt;0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold $$(M^n,g)$$ satisfying $$\mbox{k}_{\rho ^2}(M^n,g)\leqslant \updelta_1$$ and $$p\in M$$, then there is a pointed $$\operatorname{RCD}(0,n)$$ space $$(X,\mathsf {d}_X,\mu _X,x)$$ (which depends on $$p$$) such that when $$M$$ is endowed with the geodesic distance and with the measure $$\mu _M=\nu _g/\nu _g(B_\rho (p))$$, then

$$$$\begin{equation*} \mathsf {d}_{mGH}{\left(B_\rho (p), B_\rho (x)\right)}\leqslant \upeta\rho . \end{equation*}$$$$

3.2 Reifenberg regularity result

Let $$(M^n,g)$$ be a complete Riemannian manifold. For any $$(t,x) \in \mathbb {R}_+ \times M$$, we set

$$$$\begin{equation*} \uptheta(t,x):=(4\pi t)^{\frac{n}{2}} H(t,x,x). \end{equation*}$$$$

The quantity $$\uptheta$$ is an on-diagonal heat kernel ratio, as $$(4\pi t)^{-\frac{n}{2}}$$ is the on-diagonal Euclidean heat kernel. In [12, 13], we showed that if $$(M^n,g)$$ is closed and satisfies a strong Kato bound, then the quantity $$\uptheta$$ is almost monotone and controls the geometry of $$M$$. In this regard, we have at our disposal the following Reifenberg regularity result, which is a consequence of [13, Theorem 5.19] and the proof of Proposition 2.5. The case of almost non-negative Ricci curvature was originally proven in [11, Theorem 7.10].

Proposition 3.4.Let $$f\colon (0,1]\rightarrow \mathbb {R}_+$$ be a non-decreasing function satisfying (SK). Then there exists $$\upbeta(n,f)&gt;0$$ such that for any complete Riemannian manifold $$(M^n,g)$$ which is a normal covering of a closed Riemannian manifold with residually finite deck transformation, if there exist $$x \in M$$, $$R&gt;0$$ and $$h:B_{R}(x)\rightarrow \mathbb {R}^n$$ harmonic such that:

• (1) $$\mbox{k}_{R^2}(M^n,g)\leqslant \upbeta(n,f)$$ and $$\mbox{k}_{tR^2}(M^n,g)\leqslant f(t)$$ for any $$t \in (0,1]$$,
• (2) $$\uptheta(R^2,x)\leqslant 1+\upbeta(n,f)$$,
• (3) $$\Vert dh\Vert _{L^\infty }\leqslant 2$$, $$h(x)=0$$ and
  $$$$\begin{equation*} {\left(R^2\fint _{B_R(x)} |\nabla dh|^2\mathop {}\!\mathrm{d}\nu _{g}\right)}^{\frac{1}{2}}+\fint _{B_R(x)}{\left| dh\,{}^tdh-\mathrm{Id}_n\right|} \mathop {}\!\mathrm{d}\nu _{g}\leqslant \upbeta(n,f), \end{equation*}$$$$

then the restriction of $$h$$ to $$B_{3R/4}(x)$$ is a diffeomorphism onto its image.

4 ALBANESE MAPS

In this section, we recall the construction of the Albanese maps and derive some relevant properties.

4.1 Construction of the Albanese maps

Let $$(M^n,g)$$ be a closed Riemannian manifold. We write $$H_1(M,\mathbb {Z})$$ for the first integer-valued homology group of $$M$$, $$H^1(M,\mathbb {R})$$ for its first real-valued cohomology group, $$H^1_{dR}(M)$$ for its first De Rham cohomology space, and

$$$$\begin{equation*} \mathcal {H}^1(M^n,g) \coloneqq {\left\lbrace \alpha \in \mathcal {C}^\infty (T^*M) \, : \, d\alpha =d^*\alpha =0\right\rbrace} \end{equation*}$$$$

for the space of harmonic 1-forms of $$(M^n,g)$$. Let $$b_1$$ be the first Betti number of $$M$$. Then the torsion free part $$\Gamma$$ of $$H_1(M,\mathbb {Z})$$ is isomorphic to $$\mathbb {Z}^{b_1}$$ (hence it is a residually finite group) and satisfies

$$$$\begin{equation*} \Gamma = \pi _1(M)/\Lambda , \end{equation*}$$$$

where

$$$$\begin{equation*} \Lambda \coloneqq {\left\lbrace \gamma \in \pi _1(M) \, : \, \int _\gamma \alpha =0 \quad \text{for all $[\alpha ]\in H^1(M,\mathbb {R})$}\right\rbrace} . \end{equation*}$$$$

Moreover, $$\Gamma$$ is the deck transformation group of the normal covering

$$$$\begin{equation*} \widetilde{M}/\Lambda \eqqcolon \widehat{M} \stackrel{\pi }{\longrightarrow } M, \end{equation*}$$$$

where $$\widetilde{M}$$ is the universal cover of $$M$$.

By the Hodge–de Rham theorem, there exists a normalized $$L^2$$-orthonormal family of harmonic 1-forms $$\alpha _1,\dots ,\alpha _{b_1}$$, that is,

$$$$\begin{equation*} \fint _M \langle \alpha _i,\alpha _j \rangle = \delta _{i,j} \end{equation*}$$$$

for any $$i,j$$, such that $$[\alpha _1],\ldots ,[\alpha _{b_1}]$$ form a basis of $$H^1_{dR}(M)$$. We choose $$o\in M$$ and $$\widehat{o}\in \widehat{M}$$ such that $$\pi (\widehat{o})=o$$. Then for any $$i\in \lbrace 1,\ldots ,b_1\rbrace$$ there exists a unique harmonic function $${\widehat{\mathcal {A}}}_i : \widehat{M} \rightarrow \mathbb {R}$$ such that

$$$$\begin{equation*} d\widehat{\mathcal {A}}_i=\pi ^*\alpha _i \quad \text{ and } \quad \widehat{\mathcal {A}}_i{\left(\widehat{o}\right)}=0. \end{equation*}$$$$

This yields a harmonic map

$$$$\begin{equation} \widehat{\mathcal {A}} := (\widehat{\mathcal {A}}_1,\ldots ,\widehat{\mathcal {A}}_{b_1}) \colon \widehat{M}\rightarrow \mathbb {R}^{b_1}. \end{equation}$$$$(3)

Let us now consider the linear map

$$$$\begin{equation} \def\eqcellsep{&}\begin{array}{ccccl}\rho & : & \Gamma & \rightarrow & \mathbb {R}^{b_1}\\[3pt] & & \gamma & \mapsto & {\left(\int _\gamma \alpha _1,\dots , \int _\gamma \alpha _{b_1}\right)} \end{array} \end{equation}$$$$(4)

and set

$$$$\begin{equation*} {\bm \Gamma }\coloneqq \rho (\Gamma ). \end{equation*}$$$$

Then $$\rho \colon \Gamma \rightarrow {\bm \Gamma }$$ is an isomorphism and $${\bm \Gamma }$$ is a lattice of $$\mathbb {R}^{b_1}$$. We endow $$\mathbb {R}^{b_1} / {\bm \Gamma }$$ with the flat quotient Riemannian metric $$g_{\mathbb {R}^{b_1} / {\bm \Gamma }}$$. Note that $$\widehat{\mathcal {A}}$$ is $${\bm \Gamma }$$-equivariant, that is, for any $$\gamma \in \Gamma$$ and $$\widehat{x} \in \widehat{M}$$,

$$$$\begin{equation} \widehat{\mathcal {A}}(\gamma .\widehat{x})=\widehat{\mathcal {A}}(\widehat{x})+\rho (\gamma ). \end{equation}$$$$(5)

Then $$\widehat{\mathcal {A}}$$ induces a harmonic quotient map

$$$$\begin{equation} \def\eqcellsep{&}\begin{array}{ccccl}\mathcal {A}& : & M=\widehat{M}/\Gamma & \rightarrow & \mathbb {R}^{b_1} / {\bm \Gamma }\\[3pt] & & \Gamma .\widehat{x} & \mapsto & \widehat{\mathcal {A}}(\widehat{x})+{\bm \Gamma }. \end{array} \end{equation}$$$$(6)

We say that $$\mathcal {A}$$ is the Albanese map of $$M$$ and $$\widehat{\mathcal {A}}$$ is the lifted Albanese map of $$M$$. Note that by construction, the following diagram commutes:

4.2 Some estimates for harmonic 1-forms

In the next proposition, we derive some estimates for the elements of $$\mathcal {H}^1(M^n,g)$$ under a smallness assumption on the Kato constant.

Proposition 4.1.Let $$(M^n,g)$$ be a closed Riemannian manifold of diameter $$D$$ such that $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta$$ for some $$\updelta\in (0,1/(16n))$$. Then there exists $$C(n)&gt;0$$ such that for any $$\alpha \in \mathcal {H}^1(M^n,g)$$,

• (i) $$ \Vert \alpha \Vert ^2_{L^\infty }\leqslant (1+C(n)\sqrt [3]{\updelta\,}\,)\fint _M |\alpha |^2$$,
• (ii) $$ \fint _M | \, |\alpha |^2- \fint _M |\alpha |^2\, |\leqslant C(n)\sqrt [3]{\updelta\,}\,\fint _M |\alpha |^2$$,
• (iii) $$ D^2\fint _M |\nabla \alpha |^2\leqslant C(n)\updelta\,\fint _M |\alpha |^2$$.

Proof.Set $$N:=\lfloor \frac{1}{\updelta16n}\rfloor$$. Proceeding like in [4, Lemma 2.22], for instance, we get that for any $$\ell \in \lbrace 1,\dots , N\rbrace$$,

$$$$\begin{equation} \mbox{k}_{\ell D^2}(M^n,g)\leqslant \ell \, \mbox{k}_{D^2}(M^n,g) \leqslant \ell \updelta\leqslant \frac{1}{16n} \, \cdot \end{equation}$$$$(7)

We first prove (i). Let $$\alpha \in \mathcal {H}^1(M^n,g)$$. By the Bochner formula,

$$$$\begin{equation*} |\nabla \alpha |^2+\frac{1}{2} \Delta |\alpha |^2+\mbox{Ric}(\alpha ,\alpha )=0. \end{equation*}$$$$

We fix $$x \in M$$. We multiply the previous identity evaluated in $$y \in M$$ by the heat kernel $$H(t,x,y)$$ and integrate with respect to $$(t,y)\in [0,\ell D^2]\times M$$, where $$\ell \in \lbrace 1,\dots , N\rbrace$$ is suitably chosen later. This gives

$$$$\begin{align*} & - \iint\nolimits _{[0, \ell D^2]\times M} \! \mbox{Ric}(\alpha ,\alpha )(y) \Delta _y H(t,x,y)\mathop {}\!\mathrm{d}\nu _g(y) \mathop {}\!\mathrm{d}t\\ &\quad = \iint\nolimits _{[0, \ell D^2]\times M} \! |\nabla \alpha |^2(y)H(t,x,y) \mathop {}\!\mathrm{d}\nu _g(y) \mathop {}\!\mathrm{d}t + \frac{1}{2} \iint\nolimits _{[0, \ell D^2]\times M} \! |\alpha |^2(y) \Delta _y H(t,x,y)\mathop {}\!\mathrm{d}\nu _g(y) \mathop {}\!\mathrm{d}t. \end{align*}$$$$

Since $$\mbox{Ric}\geqslant -\mbox{Ric}_{\mbox{{-}}}g$$ and $$\Delta _y H (t,x,y)= \frac{\partial }{\partial t} H(t,x,y)$$, we get

$$$$\begin{align*} |\alpha |^2(x)&\leqslant \int _M H(\ell D^2,x,y) |\alpha |^2(y)\mathop {}\!\mathrm{d}\nu _g(y) +2\Vert \alpha \Vert ^2_{L^\infty }\iint\nolimits _{[0,\ell D^2]\times M} \! H(t,x,y)\mbox{Ric}_{\mbox{{-}}}(y)\mathop {}\!\mathrm{d}\nu _g(y)\mathop {}\!\mathrm{d}t\\ &\leqslant {\left(1+C(n)\frac{1}{\sqrt {\ell }}\right)}\fint _M |\alpha |^2+2\mbox{k}_{\ell D^2}(M^n,g)\,\Vert \alpha \Vert ^2_{L^\infty } \quad \text{by Proposition 2.3}\\ &\leqslant {\left(1+C(n)\frac{1}{\sqrt {\ell }}\right)}\fint _M |\alpha |^2+2\ell \updelta\,\Vert \alpha \Vert ^2_{L^\infty } \qquad \qquad \, \,\quad \text{by (7). } \end{align*}$$$$

Thus

$$$$\begin{equation*} {\left(1-2\ell \updelta\right)}\,\Vert \alpha \Vert ^2_{L^\infty }\leqslant {\left(1+C(n)\frac{1}{\sqrt {\ell }}\right)}\fint _M |\alpha |^2. \end{equation*}$$$$

Choosing $$\ell$$ of the same order as $$\updelta^{-\frac{2}{3}}$$ yields the desired estimate.

Let us now prove (ii). Consider $$\alpha \in \mathcal {H}^1(M^n,g)$$. Up to rescaling, we may assume that

$$$$\begin{equation*} \fint _M |\alpha |^2 = 1. \end{equation*}$$$$

Then

$$$$\begin{align*} \fint _M{\left| |\alpha |^2-1\right|}&=\frac{2}{\nu _g(M)}\int _{\lbrace |\alpha |^2\geqslant 1\rbrace } {\left(|\alpha |^2-1\right)}\\ & \leqslant \frac{2\nu _g{\left(\lbrace |\alpha |^2\geqslant 1\rbrace \right)}}{\nu _g(M)}C(n)\sqrt [3]{\updelta\,}\\ & \leqslant 2C(n)\sqrt [3]{\updelta\,}, \end{align*}$$$$

where we have used that (i) and then $$\nu _g(\lbrace |\alpha |^2\geqslant 1\rbrace )\leqslant \int _M |\alpha |^2$$.

Let us prove (iii). We integrate the Bochner formula over $$M$$, divide by $$\nu _g(M)$$, and use (i) to get

$$$$\begin{equation*} \fint _M |\nabla \alpha |^2=-\fint _M \mbox{Ric}(\alpha ,\alpha )\leqslant C(n)\fint _M \mbox{Ric}_{\mbox{{-}}}\ \fint _M |\alpha |^2. \end{equation*}$$$$

But

$$$$\begin{align*} D^2\fint _M \mbox{Ric}_{\mbox{{-}}}&=\frac{1}{\nu _g(M)}\int _{[0,D^2]\times M\times M} H(t,x,y)\mbox{Ric}_{\mbox{{-}}}(y)\mathop {}\!\mathrm{d}\nu _g(y)\mathop {}\!\mathrm{d}\nu _g(x)\mathop {}\!\mathrm{d}t\\ & \leqslant \mbox{k}_{D^2}(M^n,g) \leqslant \updelta.\\[-45pt] \end{align*}$$$$

$$\Box$$

Remark 4.2.By the Grothendieck lemma [35, Theorem 5.1] (see also [27] and [22, Théoréme 4]), the previous proposition implies that for any closed Riemannian manifold $$(M^n,g)$$ of diameter $$D$$ such that $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta$$ for some $$\updelta\in (0,1/(16n))$$,

$$$$\begin{equation*} b_1(M)\leqslant {\left(1+C(n)\sqrt [3]{\updelta\,}\,\right)}\, n, \end{equation*}$$$$

so that:

$$$$\begin{equation*} \updelta&lt;\updelta_n:={\left(n C(n)\right)}^{-3} \qquad \Rightarrow \qquad b_1(M)\leqslant n. \end{equation*}$$$$

In particular, this provides another proof of [4, Proposition 3.12].

4.3 Consequences for the Albanese maps

The previous estimate implies the following.

Proposition 4.3.There exists $$\updelta(n)&gt;0$$ such that if $$(M^n,g)$$ is a closed Riemannian manifold of diameter $$D$$ satisfying $$b_1(M)=n$$ and $$\mbox{k}_{D^2}(M^n,g)\leqslant \updelta$$ for some $$\updelta\in (0,\updelta(n)]$$, then the Albanese maps $$\mathcal {A}$$ and $$\widehat{\mathcal {A}}$$ satisfy the following properties.

• (i) They are $$(1+C(n)\sqrt [3]{\updelta\,})$$-Lipschitz.
• (ii) They are surjective.
• (iii) For any $$r\in [D,\updelta^{-1/6}D]$$,
  $$$$\begin{equation} {\left(r^2\fint _{B_r(\widehat{o})} |\nabla d\widehat{\mathcal {A}}|^2\mathop {}\!\mathrm{d}\nu _{\widehat{g}}\right)}^{\frac{1}{2}}+\fint _{B_{r}(\widehat{o})}{\left| d\widehat{\mathcal {A}}\,{}^td\widehat{\mathcal {A}}-\mathrm{Id}_n\right|} \mathop {}\!\mathrm{d}\nu _{\widehat{g}}\leqslant C(n)\sqrt [3]{\updelta} \end{equation}$$$$(8)
  and
  $$$$\begin{equation} \nu _{\widehat{g}}(B_r(\widehat{o}))\geqslant {\left(1-C(n)\sqrt [3]{\updelta}\right)} \omega _n r^n. \end{equation}$$$$(9)

Proof.Let $$\pi$$ be the projection map from $$\widehat{M}$$ to $$M$$, and let $$(\alpha _1,\ldots ,\alpha _n)$$ be the orthonormal basis of harmonic 1-forms used to build $$\widehat{\mathcal {A}}$$. We let $$C(n)&gt;0$$ be a generic constant depending only on $$n$$ whose value may change from line to line.

Let us prove (i). For $$\xi =(\xi _1,\dots ,\xi _n)\in \mathbb {R}^n$$, set $$ \alpha :=\sum _i\xi _i\alpha _i \in \mathcal {H}^1(M^n,g)$$ and observe that

$$$$\begin{equation*} |\xi |^2 = \fint _M |\alpha |^2. \end{equation*}$$$$

Then for any $$ x\in \widehat{M}$$ and $$ v\in T_{x} \widehat{M}$$, since $$d \widehat{\mathcal {A}}= (\pi ^*\alpha _1,\ldots ,\pi ^*\alpha _n)$$,

$$$$\begin{equation*} {\left|\langle \xi ,d_{x}\widehat{\mathcal {A}}({v})\rangle \right|}={\left|\alpha (\pi (x))(d_{x}\pi (v))\right|}\leqslant |\xi | {\left(1+C(n)\sqrt [3]{\updelta\,}\,\right)}^{\frac{1}{2}}\sqrt {\widehat{g}_{x}(v,v)}, \end{equation*}$$$$

where we have applied (i) in Proposition 4.1. This yields (i).

Let us now prove (ii). Proposition 4.1 implies that $$\Omega :=\alpha _1\wedge \dots \wedge \alpha _n$$ satisfies

$$$$\begin{equation*} \fint _M {\left| |\Omega |-1\right|}\mathop {}\!\mathrm{d}\nu _g\leqslant C(n)\sqrt [3]{\updelta\,}. \end{equation*}$$$$

Hence if $$M$$ is oriented and up to permutation of $$\alpha _1$$ and $$\alpha _2$$, then

$$$$\begin{equation*} {\left|\int _M \Omega -\nu _g(M)\right|}\leqslant C(n)\sqrt [3]{\updelta\,} \nu _g(M), \end{equation*}$$$$

but by construction, setting $${\bf \Omega }:=dx_1\wedge \dots \wedge dx_n$$, we have $$\Omega =\mathcal {A}^{*}{\bf \Omega }$$, hence

$$$$\begin{equation*} {\left| \deg \mathcal {A}-\frac{\nu _g(M)}{\text{vol}\, \mathbb {R}^n/{\bm \Gamma }} \right|} = \frac{1}{\text{vol}\, \mathbb {R}^n/{\bm \Gamma }} {\left|\int _M\mathcal {A}^{*}{\bf \Omega }-\nu _g(M)\right|}\leqslant C(n)\sqrt [3]{\updelta\,}\frac{\nu _g(M)}{\text{vol}\, \mathbb {R}^n/{\bm \Gamma }} \, \cdot \end{equation*}$$$$

Hence if $$C(n)\sqrt [3]{\updelta\,}&lt;1$$, then $$\deg \mathcal {A}\not=0$$ and $$\mathcal {A}$$ is surjective. If $$M$$ is not oriented, then using the twofold oriented cover $$M_o\stackrel{\pi _o}{\longrightarrow } M$$, the same argumentation can be applied to $$\pi _o^*\Omega$$ in order to get that $$\mathcal {A}\circ \pi _o\colon M_o\rightarrow \mathbb {R}^n/{\bm \Gamma }$$ is surjective.

Let us now prove (iii). Observe that (9) is a direct consequence of (8) and [7, Theorem 1.2] (see also Theorem A.1). Thus we are left with proving (8). To this aim, we use the following result: for any $$f \in L^1(M)$$ and $$r\in [D,\sqrt {N}D]$$,

$$$$\begin{equation} \fint _{B_r(\widehat{o})}{\left|f\circ \pi \right|}\leqslant C(n) \fint _M |f|. \end{equation}$$$$(10)

Together with (ii) in Proposition 4.1, this implies that for any $$r\in [D,\sqrt {N}D]$$,

$$$$\begin{equation*} \fint _{B_r(\widehat{o})} {\left| d\widehat{\mathcal {A}}\, ^td\widehat{\mathcal {A}}-\mathrm{Id}_n\right|} \leqslant C(n)\sqrt [3]{\updelta}, \end{equation*}$$$$

and similarly (iii) in Proposition 4.1 yields that if $$r\in [D,\updelta^{-\frac{1}{6}}D]$$, then

$$$$\begin{equation*} r^2\fint _{B_r(\widehat{o})} {\left| \nabla d\widehat{\mathcal {A}}\right|}^2\leqslant C(n) \updelta{\left(\frac{r}{D}\right)}^2\leqslant C(n)\updelta^{\frac{2}{3}}. \end{equation*}$$$$

Thus we are left with proving (10). Let $$\mathcal {D}\subset B_D(\widehat{o})$$ be a fundamental domain for $$\Gamma \longrightarrow \widehat{M}\stackrel{\pi }{\longrightarrow } M$$. Set

$$$$\begin{equation*} G(r)={\left\lbrace \gamma \in \Gamma : \gamma \mathcal {D}\cap B_r(\widehat{o})\not=\emptyset \right\rbrace} . \end{equation*}$$$$

Then

$$$$\begin{equation*} B_r(\widehat{o})\subset \bigcup _{\gamma \in G(r)} \gamma \mathcal {D}\subset B_{r+D}(\widehat{o}), \end{equation*}$$$$

so that

$$$$\begin{equation*} \nu _{\hat{g}}(B_r(\widehat{o}))\leqslant \#G(r)\nu _{\hat{g}}(\mathcal {D})=\#G(r)\nu _g(M)\leqslant \nu _{\hat{g}}(B_{r+D}(\widehat{o})). \end{equation*}$$$$

The group $$\Gamma \simeq \mathbb {Z}^n$$ is residually finite; hence, Proposition 2.5 and Proposition 2.1 imply the volume doubling estimate: for any $$0&lt;r\leqslant R\leqslant \sqrt {N}D$$,

$$$$\begin{equation*} \nu _{\hat{g}}(B_R(\widehat{o}))\leqslant C(n) {\left(\frac{R}{r}\right)}^{e^2n} \nu _{\hat{g}}(B_r(\widehat{o})). \end{equation*}$$$$

If $$r \geqslant D$$ and $$r + D \leqslant \sqrt {N} D$$, this yields

$$$$\begin{equation*} \#G(r)\nu _g(M)\leqslant C(n){\left(\frac{r+D}{r}\right)}^{e^2n} \nu _{\hat{g}}(B_r(\widehat{o}))\leqslant C(n)\nu _{\hat{g}}(B_r(\widehat{o})), \end{equation*}$$$$

so that

$$$$\begin{align*} \fint _{B_r(\widehat{o})}{\left|f\circ \pi \right|}=\frac{1}{\nu _{\hat{g}}(B_r(\widehat{o}))} \sum _{\gamma \in G(r)}\int _{\mathcal {D}\cap B_r(\widehat{o})} {\left|f\circ \pi \right|} &\leqslant \frac{\#G(r)\nu _g(M)}{\nu _{\hat{g}}(B_r(\widehat{o}))} \fint _{M} {\left|f\right|}\\ &\leqslant C(n) \fint _{M} {\left|f\right|}. \end{align*}$$$$

If $$r + D &gt; \sqrt {N} D$$, then the volume doubling estimate yields (see, for example, [26, Subsection 2.3])

$$$$\begin{equation*} \nu _{\hat{g}}(B_{r+D}(\widehat{o})) \leqslant C(n)^{\frac{r+D}{r}}\nu _{\hat{g}}(B_{r}(\widehat{o}))\leqslant C(n)\nu _{\hat{g}}(B_{r}(\widehat{o})), \end{equation*}$$$$

and we can conclude in the same way as above.$$\Box$$

5 PROOF OF THEOREM A

In this section, we prove Theorem A. All the way through we consider $$\varepsilon \in (0,1)$$ and

$$$$\begin{equation*} \upeta:=\frac{\varepsilon }{640 n^2}\quad \text{ and }\quad R:=64n^{2} D. \end{equation*}$$$$

Let $$(M^n,g)$$ be a closed Riemannian manifold of diameter $$D$$ such that

$$$$\begin{equation*} b_1(M)= n \quad \text{ and } \quad \mbox{k}_{D^2}(M^n,g)\leqslant \updelta \end{equation*}$$$$

for some $$\updelta\in (0,1/(16n))$$. We consider the covering $$\Gamma \longrightarrow \widehat{M} \stackrel{\pi }{\longrightarrow } M$$ built in the previous section, and associated Albanese maps $$\widehat{\mathcal {A}}$$ and $$\mathcal {A}$$. Then the following holds.

Claim 5.1.There exists $$\updelta_0(\varepsilon ,n)&gt;0$$ such that if $$\updelta\leqslant \updelta_0(\varepsilon ,n)$$, then $$\widehat{\mathcal {A}}$$ satisfies the following.

• (a) $$\widehat{\mathcal {A}}$$ is $$(1+\upeta)$$-Lipschitz.
• (b) $$\widehat{\mathcal {A}}\colon B_{R}(\widehat{o})\rightarrow \mathbb {B}_R^n$$ is an $$\upeta R$$-almost isometry.
• (c) $$ \mathbb {B}_{(1-\upeta)R}^n\subset \widehat{\mathcal {A}}(B_{R}(\widehat{o}))$$.

This is a consequence of Proposition 4.3, Theorem 3.1 and (i) in Remark 3.2. The last assertion may be proven with degree theory, see [9], [20, Proof of 3.2 and 3.3], and [12, Proof of theorem 7.2].

From now on, we assume that

$$$$\begin{equation*} \updelta\leqslant \updelta_0(\varepsilon ,n). \end{equation*}$$$$

Step 1. Proceeding like in [14, 20], we construct a normal subgroup $$\Gamma _0$$ of $$\Gamma$$ with finite index, such that $$\widehat{\mathcal {A}}$$ induces a map

$$$$\begin{equation*} \mathcal {A}_0\colon M_0 := \widehat{M}/\Gamma _0 \rightarrow \mathbb {R}^n/\rho (\Gamma _0). \end{equation*}$$$$

Let $$(e_1,\dots , e_n)$$ be the canonical basis of $$\mathbb {R}^n$$. Since $$4 \sqrt {n}D \leqslant (1-\upeta)R$$, it follows from $$(c)$$ that for any $$i \in \lbrace 1,\dots ,n\rbrace$$, there exists $$x_i \in B_R(\widehat{o})$$ such that

$$$$\begin{equation*} 4\sqrt {n} De_i=\widehat{\mathcal {A}}(x_i). \end{equation*}$$$$

Moreover, for any $$i \in \lbrace 1,\dots ,n\rbrace$$, there exists $$\gamma _i \in \Gamma$$ such that

$$$$\begin{equation*} \mathsf {d}_{\widehat{g}}{\left(\gamma _i.\widehat{o},x_i\right)}\leqslant D. \end{equation*}$$$$

Then we set

$$$$\begin{equation*} \Gamma _0:=\langle \gamma _1,\dots ,\gamma _n\rangle \subset \Gamma \end{equation*}$$$$

and

$$$$\begin{equation*} {\bm \Gamma }_0:= \rho (\Gamma _0) = \bigoplus _{i=1}^n \rho (\gamma _i)\mathbb {Z}. \end{equation*}$$$$

Let us show that

$$$$\begin{equation} \text{$\rho (\gamma _1),\dots ,\rho (\gamma _n)$ form a basis of $\mathbb {R}^n$.} \end{equation}$$$$(11)

We know by $$(a)$$ that the map $$\widehat{\mathcal {A}}$$ is $$(1+\upeta)$$-Lipschitz. Then for any $$i \in \lbrace 1,\ldots ,n\rbrace$$, since the equivariance (5) of $$\widehat{\mathcal {A}}$$ yields $$\rho (\gamma _i)=\widehat{\mathcal {A}}(\gamma _i.\widehat{o})$$, we have

$$$$\begin{equation} | 4\sqrt {n} De_i -\rho (\gamma _i) | ={\left|\widehat{\mathcal {A}}(x_i)- \widehat{\mathcal {A}}(\gamma _i.\widehat{o})\right|}\leqslant (1+\upeta)\mathsf {d}_{\widehat{g}}(\gamma _i.\widehat{o},x_i)\leqslant (1+\upeta) D.\end{equation}$$$$(12)

Then for any $$\xi =\sum _{i=1}^n \xi _i e_i \in \mathbb {R}^n$$,

$$$$\begin{align*} {\left| 4\sqrt {n} D \xi - \sum _{i=1}^n \xi _i\rho (\gamma _i) \right|} & = {\left|\sum _{i=1}^n \xi _i(4\sqrt {n} De_i-\rho (\gamma _i))\right|} \\ & \leqslant (1+\upeta)D\sum _{i=1}^n |\xi _i| \leqslant 2 D \sqrt {n}|\xi | , \end{align*}$$$$

so that

$$$$\begin{equation} | \xi | \leqslant \frac{1}{2D\sqrt {n}}{\left|\sum _{i=1}^n \xi _i\rho (\gamma _i) \right|}. \end{equation}$$$$(13)

Hence we get (11). This implies that the quotient $$\mathbb {R}^n/{\bm \Gamma }_0$$ is a torus $$\mathbb {T}^n$$ which we equip with the natural flat quotient Riemannian metric $$g_{\mathbb {R}^n/{\bm \Gamma }_0}$$. We also equip $$M_0$$ with the quotient Riemannian metric $$g_0$$ induced by $$\widehat{g}$$.

Step 2. We establish the following diameter bound on $$(M_0,g_0)$$:

$$$$\begin{equation} \mathrm{diam}(M_0)\leqslant 4(n+\sqrt {n})D. \end{equation}$$$$(14)

To this aim, let us prove an intermediary result: if $$\gamma \in \Gamma _0$$ is such that

$$$$\begin{equation} |\rho (\gamma )|\leqslant \frac{R}{2\sqrt {n}+1}, \end{equation}$$$$(15)

then

$$$$\begin{equation} {\left| \mathsf {d}_{\widehat{g}}(\widehat{o},\gamma .\widehat{o})-|\rho (\gamma )|\, \right|}\leqslant \upeta R. \end{equation}$$$$(16)

Write $$\gamma \in \Gamma _0$$ as

$$$$\begin{equation*} \gamma =\gamma _1^{k_1}\dots \gamma _n^{k_n} \end{equation*}$$$$

for some $$k_1,\ldots ,k_n \in \mathbb {Z}$$. Consider $$i \in \lbrace 1,\ldots ,n\rbrace$$. From $$(b)$$, we know that $$|\mathsf {d}_{\widehat{g}}(x_i,\widehat{o})- |\widehat{\mathcal {A}}(x_i) - \widehat{\mathcal {A}}(\widehat{o}) | \, | \leqslant \upeta R$$. Since $$\widehat{\mathcal {A}}(\widehat{o})=0$$ and $$\widehat{\mathcal {A}}(x_i) =4 \sqrt {n}De_i$$, this implies

$$$$\begin{equation*} \mathsf {d}_{\widehat{g}}(x_i,\widehat{o}) \leqslant \upeta R + 4 \sqrt {n}D \leqslant (1+4\sqrt {n})D. \end{equation*}$$$$

Then

$$$$\begin{equation*} \mathsf {d}_{\widehat{g}}(\widehat{o},\gamma .\widehat{o})\leqslant \sum _i|k_i|\, \max _i \mathsf {d}_{\widehat{g}}(\widehat{o},\gamma _i.\widehat{o}) \leqslant (2+4\sqrt {n})D\sum _i|k_i|. \end{equation*}$$$$

Since $$\rho (\gamma )=\sum _{i}k_i \rho (\gamma _i)$$, it follows from (13) that

$$$$\begin{equation*} \sum _i|k_i|\leqslant \sqrt {n}{\left(\sum _i k_i^2\right)}^{1/2}\leqslant \frac{| \rho (\gamma )|}{2D} \, \cdot \end{equation*}$$$$

Then we get $$\mathsf {d}_{\widehat{g}}(\widehat{o},\gamma .\widehat{o})\leqslant (2\sqrt {n} +1)| \rho (\gamma )|$$, so that (15) implies

$$$$\begin{equation} \gamma .\widehat{o} \in B_R(\widehat{o}) \end{equation}$$$$(17)

and the conclusion (16) follows from $$(b)$$.

We are now in a position to prove the diameter bound (14). Introduce the Dirichlet domain

$$$$\begin{equation*} \mathcal {D}_0 \coloneqq {\left\lbrace x\in \widehat{M} : \mathsf {d}_{\widehat{g}}(x,\widehat{o})\leqslant \mathsf {d}_{\widehat{g}}(\gamma .x,\widehat{o}) \,\,\, \text{for all $\gamma \in \Gamma _0\setminus \lbrace 1\rbrace $}\right\rbrace} . \end{equation*}$$$$

We are going to show that

$$$$\begin{equation} \mathcal {D}_0 \cap {\left\lbrace x\in \widehat{M} : \mathsf {d}_{\widehat{g}}(x,\widehat{o})= 2(n+\sqrt {n})D \right\rbrace} = \emptyset ; \end{equation}$$$$(18)

then the connectedness of $$\mathcal {D}_0$$ will imply $$\mathcal {D}_0 \subset B_{ 2 (n+\sqrt {n})D}(\widehat{o})$$ and (14) will be established.

The set $$\mathcal {F}_0 \coloneqq \sum _{i=1}^n [-\frac{1}{2},\frac{1}{2})\rho (\gamma _i)$$ is a fundamental domain for the action of $${\bm \Gamma }_0$$ on $$\mathbb {R}^n$$; it is included in the Euclidean ball centered at the origin with radius

$$$$\begin{equation*} \frac{\sqrt {n}}{2}\max _i |\rho (\gamma _i)|. \end{equation*}$$$$

By (12), for any $$i$$,

$$$$\begin{equation*} |\rho (\gamma _i)| \leqslant |4 \sqrt {n} D e_i|+ (1+\upeta)D \leqslant {\left(4 \sqrt {n}+2\right)}D, \end{equation*}$$$$

so that

$$$$\begin{equation*} \frac{\sqrt {n}}{2}\max _i |\rho (\gamma _i)| \leqslant (2n+\sqrt {n})D. \end{equation*}$$$$

For any $$x\in \widehat{M}$$ there exists $$\gamma _0 \in \Gamma _0$$ such that $$\widehat{\mathcal {A}}(\gamma _0.x) \in \mathcal {F}_0$$. By the equivariance (5) of $$\widehat{\mathcal {A}}$$ and the previous inequality, we get that

$$$$\begin{equation} {\left| \widehat{\mathcal {A}}(x)+\rho (\gamma _0)\right|} = {\left| \widehat{\mathcal {A}}(\gamma _0.x)\right|}\leqslant (2n+\sqrt {n})D. \end{equation}$$$$(19)

Now assume that $$\mathsf {d}_{\widehat{g}}(x,\widehat{o})= 2(n+\sqrt {n})D$$. We are going to show that $$x \notin \mathcal {D}_0$$. Since $$2(n+\sqrt {n})D\leqslant R$$, from $$(b)$$ we get

$$$$\begin{equation*} {\left| \widehat{\mathcal {A}}(x)\right|} = {\left| \widehat{\mathcal {A}}(x)- \widehat{\mathcal {A}}(\widehat{o})\right|} \leqslant 2(n+\sqrt {n})D +\upeta R. \end{equation*}$$$$

Consequently,

$$$$\begin{equation*} |\rho (\gamma _0)| \leqslant {\left|\widehat{\mathcal {A}}(x) + \rho (\gamma _0)\right|} + {\left|\widehat{\mathcal {A}}(x)\right|} \leqslant (4n+3\sqrt {n})D +\upeta R. \end{equation*}$$$$

By our choices of $$\upeta$$ and $$R$$ we have

$$$$\begin{equation*} (4n+3\sqrt {n})D +\upeta R \leqslant \frac{R}{2 \sqrt {n}+1} \, \cdot \end{equation*}$$$$

Then we are in a position to apply (16). We get

$$$$\begin{equation*} \mathsf {d}_{\widehat{g}}(\gamma _0.\widehat{o},\widehat{o})\leqslant (4n+3\sqrt {n})D + 2\upeta R \end{equation*}$$$$

and then

$$$$\begin{align*} \mathsf {d}_{\widehat{g}}(\gamma _0.x,\widehat{o}) & \leqslant \mathsf {d}_{\widehat{g}}(\gamma _0.x,\gamma _0.\widehat{o})+\mathsf {d}_{\widehat{g}}(\gamma _0.\widehat{o},\widehat{o}) \leqslant 2(n+\sqrt {n})D+(4n+3\sqrt {n})D + 2\upeta R. \end{align*}$$$$

Since

$$$$\begin{equation*} 2(n+\sqrt {n})D+(4n+3\sqrt {n})D + 2\upeta R \leqslant R, \end{equation*}$$$$

we can use $$(b)$$ and (19) to deduce that

$$$$\begin{equation*} \mathsf {d}_{\widehat{g}}(\gamma _0.x,\widehat{o})\leqslant {\left| \widehat{\mathcal {A}}(\gamma _0.x)\right|}+\upeta R \leqslant (2n+\sqrt {n})D+\upeta R &lt;2(\sqrt {n}+2n)D=\mathsf {d}_{\widehat{g}}(x,\widehat{o}). \end{equation*}$$$$

Thus $$x \notin \mathcal {D}_0$$ and (18) is proven.

Step 3. We prove that $$\mathcal {A}_0:M_0 \rightarrow \mathbb {R}^n/{\bm \Gamma }_0$$ is a $$3 \upeta R$$-almost isometry. From Proposition 4.3, we know that $$\widehat{\mathcal {A}}$$ is surjective; hence, $$\mathcal {A}_0$$ is surjective too. Thus we are left with proving the distance estimate. Let us introduce the following intermediate projection maps $$\pi _0$$ and $$p_0$$:

Let $$x,y\in M_0$$. Since $$\mathcal {D}_0 \subset B_{ 2(n+\sqrt {n})D}(\widehat{o})$$, we can choose $$\widehat{x}\in B_{ 2(n+\sqrt {n})D}(\widehat{o})$$ such that $$\pi _0(\widehat{x})=x$$. Let $$c\colon [0,1]\rightarrow M_0$$ be a minimizing geodesic joining $$x$$ and $$y$$. Let $$\widehat{c}\colon [0,1]\rightarrow \widehat{M}$$ be such that $$\pi _0\circ \widehat{c}=c$$ and $$\widehat{x}=\widehat{c}(0)$$. By the diameter bound (14), we know that $$\widehat{y}:=\widehat{c}(1)$$ belongs to $$B_{6(n+\sqrt {n})D}(\widehat{o})\subset B_{R}(\widehat{o})$$. Moreover, $$\pi _0(\widehat{y})=y$$. Thus

$$$$\begin{align} \mathsf {d}_{g_{\mathbb {R}^n/{\bm \Gamma }_0}}(\mathcal {A}_0(x),\mathcal {A}_0(y))=\mathsf {d}_{g_{\mathbb {R}^n/{\bm \Gamma }_0}}{\left(p_0(\widehat{\mathcal {A}}(\widehat{x})),p_0(\widehat{\mathcal {A}}(\widehat{y}))\right)} & \leqslant {\left|\widehat{\mathcal {A}}(\widehat{x})-\widehat{\mathcal {A}}(\widehat{y})\right|} \nonumber \\ & \leqslant \mathsf {d}_{g_0}(x,y)+\upeta R\end{align}$$$$(20)

thanks to $$(b)$$.

It remains to prove that

$$$$\begin{equation*} \mathsf {d}_{g_0}(x,y) - \mathsf {d}_{g_{\mathbb {R}^n/{\bm \Gamma }_0}}(\mathcal {A}_0(x),\mathcal {A}_0(y)) \leqslant 3 \upeta R. \end{equation*}$$$$

We start by showing that if $$\mathcal {A}_0(x)=\mathcal {A}_0(y)$$, then $$\mathsf {d}_{g_0}(x,y) \leqslant \upeta R$$. Since $$\mathcal {A}_0 \circ c = \mathcal {A}_0 \circ \pi _0 \circ \widehat{c} = p_0 \circ \widehat{\mathcal {A}} \circ \widehat{c}$$, the curve $$\widehat{\mathcal {A}} \circ \widehat{c}$$ is a lift of the curve $$\mathcal {A}_0 \circ c$$ joining $$\vec{v}:=\widehat{\mathcal {A}}(\widehat{x})\in \mathbb {R}^n$$ to $$\vec{w}:=\widehat{\mathcal {A}}(\widehat{y})$$. Moreover, the length of $$\widehat{c}$$ is less than $$4(n+\sqrt {n})D$$; hence, $$(b)$$ implies that the length of $$\widehat{\mathcal {A}} \circ \widehat{c}$$ is less than $$4(1+\upeta)(n+\sqrt {n})D$$. Since $$p_0(\vec{v})=p_0(\vec{w})$$, there exists $$\gamma \in \Gamma _0$$ such that

$$\widehat{\mathcal{A}}(\widehat{y})=\vec{w}=\vec{v}+\rho (\gamma )=\widehat{\mathcal{A}}(\widehat{x})+\rho ($$