2022

[1,2]Dr Stefano Spessato ORCID: 0000-0003-0069-5853

[1]Università degli Studi ”Niccolò Cusano”, via don Carlo Gnocchi 3, Rome,
00166, Italy

#  $L^{2}$-cohomology and quasi-isometries on the ends of unbounded geometry

<EMAIL_ADDRESS>*

###### Abstract

In this paper we study the minimal and maximal $L^{2}$-cohomology of oriented,
possibly not complete, Riemannian manifolds. Our focus will be on both the
reduced and the unreduced $L^{2}$-cohomology groups. In particular we will
prove that these groups are invariant under uniform homotopy equivalence
quasi-isometric on the unbounded ends. A uniform map is a uniformly continuous
map such that the diameter of the preimage of a subset is bounded in terms of
the diameter of the subset itself. A map $f$ between two Riemannian manifolds
$(X,g)$ and $(Y,h)$ is quasi-isometric on the unbounded ends if $X=M\cup
E_{X}$ where $M$ is the interior of a manifold of bounded geometry with
boundary, $E_{X}$ is an open of $X$ and the restriction of $f$ to $E_{X}$ is a
quasi-isometry. Finally some consequences are shown: the main ones are
definition of a mapping cone for $L^{2}$-cohomology and the invariance of the
$L^{2}$-signature.

###### keywords:

$L^{2}$-cohomology, signature, mapping cone, quasi-isometry

### Acknowledgments

I would like to thank Francesco Bei for his suggestions and for his very
helpful mails.

Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

## Introduction

In this paper a generalization of the work done by the author in Spes  is
proved. In that work, given a uniform map $f:(M,g)\longrightarrow(N,h)$
between manifolds of bounded geometry, an $\mathcal{L}^{2}$-bounded operator
$T_{f}:\mathcal{L}^{2}(N,h)\longrightarrow\mathcal{L}^{2}(M,g)$ between the
spaces of squared-integrable forms was introduced. A uniform map is a
uniformly continuous map such that for each compact subset $A$, the diameter
of the preimage of $A$ is bounded in terms of the diameter of $A$. The
operator $T_{f}$ induces an operator between the reduced and unreduced
$L^{2}$-cohomology groups and it replaces the pullback operator which is not a
well-defined operator between the $\mathcal{L}^{2}$-spaces. As a consequence
of this we obtain that the reduced and the unreduced $L^{2}$-cohomology are
invariant under uniform homotopy equivalences.
Our goal in this paper is to prove a similar result for the minimal and
maximal $L^{2}$-cohomologies, in both the reduced and unreduced version, of
possibly not complete Riemannian manifolds. In order to obtain such a result
we need some further assumptions on the homotopy equivalences. Let us briefly
introduce these assumptions.

Given a Riemannian manifold $(X,g)$, it is always possible to decompose it as
$X=M\cup E_{X}$ where $E_{X}$ is an open subset of $X$ and $M$ is the interior
of a manifold with boundary of bounded geometry (a definition can be found in
the paper of Schick Schick ). We will say that $M$ is an open of bounded
geometry and $E_{X}$ will be the unbounded ends of $X$. A manifold $(X,g)$
decomposed in this way is called a manifold of bounded geometry with unbounded
ends.
In this paper we will study uniform maps $f:(X,g)\longrightarrow(Y,h)$ between
oriented manifolds which are quasi-isometric on unbounded ends, i.e. given two
manifolds of bounded geometry with unbounded ends $(X=M\cup E_{X},g)$ and
$(Y=N\cup E_{Y},h)$, then $f(E_{X})\subseteq E_{Y}$ and $f_{|E_{X}}$ is a
quasi-isometry.
So we want to show that if $f:(X,g)\longrightarrow(Y,h)$ is a uniform homotopy
equivalence which is quasi-isometric on the unbounded ends, then

$\begin{cases}H^{k}_{2,max}(X,g)\cong H^{k}_{2,max}(Y,h)\\\
\overline{H}^{k}_{2,min}(X,g)\cong H^{k}_{2,min}(Y,h)\\\
H^{k}_{2,min}(X,g)\cong H^{k}_{2,min}(Y,h)\\\
\overline{H}^{k}_{2,min}(X,g)\cong\overline{H}^{k}_{2,min}(Y,h)\end{cases}$
(1)

where $H^{k}_{2,min\backslash max}(X,g)$ and $H^{k}_{2,min\backslash
max}(Y,h)$ are the minimal and maximal $k$-th group of $L^{2}$-cohomology and
$\overline{H}^{k}_{2,min\backslash max}(X,g)$ and
$\overline{H}^{k}_{2,min\backslash max}(Y,h)$ are their reduced versions. This
result is proved in Proposition 4.7. Finally we will show some consequences.

The structure of the paper is the following: in the first section we introduce
the notions of uniform maps quasi-isometric on the unbounded ends and some
examples are showed. In the second section we introduce the minimal and
maximal $L^{2}$-cohomology of a Riemannian manifold. In third section we
recall the notion of Radon-Nikodym-Lipschitz map. These are maps such that
their pullback induced a well-defined $\mathcal{L}^{2}$-bounded operator. In
particular we recall a technique from Spes  which allows to know if a
Lipschitz submersion is a Radon-Nikodym-Lipschitz map or not.
In the fourth section, given a uniform map isometric on the unbounded ends
$f:(X,g)\longrightarrow(Y,h)$ between oriented manifolds, we introduce a new
version of the operator $T_{f}$. In particular if $(X,g)$ and $(Y,h)$ are
manifolds of bounded geometry, then $T_{f}$ is exactly the same defined in
Spes . In the fifth section, Proposition 4.7, the main result stated above is
proved.
In the last section we will show some consequences of the existence of
$T_{f}$: first we define a mapping cone for minimal and maximal
$L^{2}$-cohomology. Then we will show that the $L^{2}$-signature of a complete
manifold of dimension $4k$ is invariant under uniform homotopy equivalences
isometric on the ends which preserve the orientations. We also prove a result
similar to point 1 of Proposition 5 of the work of Lott Lott : if two homotopy
equivalent manifolds are isometric outside two compact subsets, then their
maximal and minimal $L^{2}$-cohomology are equivalent. The same also holds for
their reduced versions.
Finally an example of Riemannian manifolds with the same minimal and maximal
$L^{2}$-cohomology (reduced and not) which are not quasi-isometric is shown.

##  1 Quasi-isometries on the unbounded ends and open subsets of bounded
geometry

In this section we introduce the geometric setting of our paper. In particular
we introduce the notions of open of bounded geometry and of uniform homotopy
quasi-isometric on the unbounded ends.

###  1.1 Manifold of bounded geometry with some unbounded ends

The following definition is the Definition 2.2 of the work of Schick Schick .

######  Definition 1.1.

Let $(\overline{M},g)$ an oriented Riemannian manifold with boundary
$\partial\overline{M}$ (possibily empty) and denote by $M$ the interior of
$\overline{M}$. Consider on $\partial\overline{M}$ the Riemannian metric
induced by $g$. Let us denote by $l$ the second fundamental form of $\partial
M$ and by $\overline{\nabla}$ the Levi-Civita connection on
$\partial\overline{M}$. Then we say that $(\overline{M},g)$ is a manifold with
boundary and of bounded geometry if the following holds.

  1. 1.

there exists a number $r_{C}>0$ such that the geodesic collar

$\begin{split}\partial\overline{M}\times[0,r_{c})&\longrightarrow M\\\
(x,t)&\longrightarrow exp_{x}(t\nu_{x})\end{split}$ (2)

is a diffeomorphism with its image ($\nu_{x}$ is the unit inward normal
vector).

  2. 2.

The injectivity radius $inj_{\partial\overline{M}}$ of $\partial\overline{M}$
is positive.

  3. 3.

Consider a number $r$ and define
$\mathcal{N}(r):=B_{r}(\partial\overline{M})\cap M$. There is a $r_{i}>0$ such
that if $r\leq r_{i}$ we have that for each $p$ in $M\setminus\mathcal{N}(r)$
the exponential map is a diffeomorphism on $B_{0_{p}}(r)\subset T_{p}M$.

  4. 4.

For every $k\in\mathbb{N}$ there is a constant $C_{k}$ so that
$|\nabla^{i}R|\leq C_{i}$ and $|\overline{\nabla}^{i}l|\leq C_{i}$ for each
$i=0,\ldots,k$.

Finally, if $M$ is the interior of a manifold of bounded geometry, then it is
an open of bounded geometry.

######  Remark 1.

If $M\subset(X,g)$ is an open of bounded geometry, then
$(\overline{M},g)\subset(X,g)$ is a submanifold with boundary of bounded
geometry.

######  Remark 2.

By Theorem 2.5 of Schick , we know that the condition 4. of Definition 1.1 can
be replaced by:

  * •

there exist $0<R_{1}\leq inj_{\partial M}$, $0<R_{2}\leq r_{C}$ and
$0<R_{3}<r_{i}$ and there are some constants $C_{k}$ for each $k\in\mathbb{N}$
such that whenever we have normal boundary coordinates of width
$(r_{1},r_{2})$ where $r_{1}\leq R_{1}$ and $r_{2}\leq R_{2}$ or Gaussian
coordinates of radius $r_{3}\leq R_{3}$ then, in this coordinates

$|\frac{\partial^{\alpha}g_{ij}}{\partial x_{1}^{\alpha_{1}}\ldots\partial
x_{m}^{\alpha_{m}}}|\leq C_{\alpha}\mbox{ and
}|\frac{\partial^{\alpha}g^{ij}}{\partial x_{1}^{\alpha_{1}}\ldots\partial
x_{m}^{\alpha_{m}}}|\leq C_{\alpha}$ (3)

where $\alpha:=\sum_{i=1}^{m}\alpha_{i}$.

######  Definition 1.2.

A manifold of bounded geometry with some (possibly) unbounded ends is an
oriented Riemannian manifold $(X,g)$ such that $X=M\cup E_{X}$ where $M$ is an
open of bounded geometry, $E_{X}$ is an open of $X$ such that $E_{X}\setminus
M\neq\emptyset$ and $\mathcal{N}_{X}(r_{X})\subset M\cap E_{X}$ for some
$r_{X}\leq r_{C,M}$. Moreover $E_{X}$ are the unbounded ends of $X$.

######  Remark 3.

Essentially each Riemannian manifold can be seen as a manifold of bounded
geometry with some possibly unbounded ends, indeed it is sufficient to choose
a point $p$ on $X$ and fix as $M=B_{\delta}(p)$ where $\delta$ is small
enough.
This definition will be useful for us when we will talk about uniform map
quasi-isometric on the unbounded ends.

######  Lemma 1.1.

Let $(X=M\cup E_{X},g)$ a manifold of bounded geometry with unbounded ends.
Let $p$ be a point in $\mathcal{N}_{X}(r_{X})$. Then since $r_{X}\leq r_{C,M}$
we can identify $p$ as a point $(x_{0},t_{0})$ of $\partial M\times(0,r_{X})$.
Then we have that

$d(p,\partial M)=t_{0}.$ (4)

Proof: Fix some normal coordinates $(U,x^{i})$ centered in $x_{0}$ in
$\partial M$. The normal collar provides some fibered coordinates
$(U\times[0,r_{c}),x^{i},t)$ on $\pi^{-1}(U)\subset M$. Notice that, thanks to
Proposition 2.8 of Schick , we know the Gram matrix of $g$ with respect to
$x^{i}$ and $t$ has the form

$g(x,t)=\begin{bmatrix}g_{ij}(x,t)&0\\\ 0&1\end{bmatrix}.$ (5)

It immediately follows that the curve
$\sigma:[0,t_{0}]\longrightarrow\mathcal{N}(r_{X})\subset M$ defined as
$\sigma(t)=(0,t)$ is length minimizing.

###  1.2 Uniform maps quasi-isometric on the unbounded ends

Let us recall the following definitions from Section 1 of Spes . Fix two
Riemannian manifolds $(X,g)$ and $(Y,h)$.

######  Definition 1.3.

A map $f:(X,g)\longrightarrow(Y,h)$ is uniformly continuous if for each
$\epsilon>0$ there is a $\delta(\epsilon)>0$ such that for each $x_{1}$,
$x_{2}$ in $X$

$d_{X}(x_{1},x_{2})\leq\delta(\epsilon)\implies
d_{Y}(f(x_{1}),f(x_{2}))\leq\epsilon.$ (6)

Moreover $f$ is uniformly (metrically) proper, if for each $R\geq 0$ there is
a number $S(R)>0$ such that for each subset $A$ of $(Y,d_{Y})$

$diam(A)\leq R\implies diam(f^{-1}(A))\leq S(R).$ (7)

We say that a map $f:(X,g)\longrightarrow(Y,h)$ is a uniform map if it is
uniformly continuous and uniformly proper.

######  Definition 1.4.

Two maps $f_{0}$ and $f_{1}:(X,d_{X})\longrightarrow(Y,d_{Y})$ are uniformly
homotopic if they are homotopic with a uniformly continuous homotopy
$H:(X\times[0,1],g+dt)\longrightarrow(Y,h)$. We will denote it by

$f_{1}\sim f_{2}.$ (8)

######  Definition 1.5.

A map $f:(X,g)\longrightarrow(Y,h)$ is a uniform homotopy equivalence if $f$
is uniformly continuous and there is map $s$ such that

  * •

$s$ is a homotopy inverse of $f$,

  * •

$s$ is uniformly continuous,

  * •

$f\circ s$ is uniformly homotopic to $id_{N}$ and $s\circ f$ is uniformly
homotopic to $id_{M}$.

In order to define the class of maps that we will study, we need to introduce
the notion of quasi-isometry.

######  Definition 1.6.

Let $(X,g)$ and $(Y,h)$ be two Riemannian manifolds. Then, a quasi-isometry is
a local diffeomorphism $f:(X,g)\longrightarrow(Y,h)$ such that $f^{*}h$ and
$g$ are quasi-isometric metric i.e. there is a constant $K\geq 1$ such that

$K^{-1}g\leq f^{*}h\leq Kg.$ (9)

######  Definition 1.7.

Let $(X=M\cup E_{X},g)$ and $(Y=N\cup E_{Y},h)$ be two manifolds of bounded
geometry with unbounded ends. A map $f:(X,g)\longrightarrow(Y,h)$ is quasi-
isometric on the unbounded ends if

  * •

$f(E_{X})\subseteq E_{Y}$, $f(M)\subset N$, $f(\partial M)\subseteq\partial
N$,

  * •

$f_{|_{E_{X}}}:E_{X}\longrightarrow E_{Y}$ is a quasi-isometry.

A uniform homotopy equivalence isometric on the ends is a uniform homotopy
equivalence which is isometric on the unbounded ends. Finally we say that $f$
is a uniform homotopy equivalence quasi-isometric on the ends if
$f_{|_{E_{X}}}$ is a quasi-isometry.

######  Remark 4.

Fix a uniform homotopy equivalence isometric on the ends
$f:(X,g)\longrightarrow(Y,h)$. Let $r_{0}:=\min\\{r_{X},r_{Y}\\}$. Then
$f(\mathcal{N}_{X}(r_{0}))\subseteq\mathcal{N}_{Y}(r_{0})$. Moreover, if we
choose some collar coordinates $\\{x^{i},t\\}$ on $\mathcal{N}_{X}(r_{0})$ and
$\\{y^{j},s\\}$ on $\mathcal{N}_{Y}(r_{0})$, then $f$ has the form

$f(x^{i},t)=(g(x^{i},t),t).$ (10)

This is a consequence of Lemma 1.1.

######  Lemma 1.2.

Let $f:X\rightarrow(Y,h)$ be a map between two manifolds. Let $g_{1}$ and
$g_{2}$ be two quasi-isometric Riemannian metrics on $X$. Then
$f:(X,g_{1})\longrightarrow(Y,h)$ is a uniform map if and only if
$f:(X,g_{2})\longrightarrow(Y,h)$ is a uniform map.

Proof: Let $\gamma$ be a differentiable curve on $X$. Fix $i=1,2$ and denote
by $L_{i}(\gamma)$ the length of $\gamma$ with respect to the metric $g_{i}$.
Since $g_{1}$ and $g_{2}$ are quasi-isometric, we obtain that there is a
constant $K$, which does not depends on $\gamma$, such that

$K^{-1}\cdot L_{1}(\gamma)\leq L_{2}(\gamma)\leq K\cdot L_{1}(\gamma).$ (11)

So, this implies that for each couple of points $a$ and $b$ in $X$ we obtain

$K^{-1}\cdot d_{1}(a,b)\leq d_{2}(a,b)\leq K\cdot d_{1}(a,b)$ (12)

where $d_{i}$ is the distance induced by $g_{i}$. The claim immediately
follows.

######  Proposition 1.3.

Let $f:(X=M\cup E_{X},g)\longrightarrow(Y=N\cup E_{Y},h)$ be a smooth uniform
map which is quasi-isometric on the unbounded ends. Assume, moreover, that, in
normal coordinates, $f_{|M}$ has uniformly bounded derivatives of each order.
Then there is a metric $\tilde{g}$ on $X$ such that

  1. 1.

$g$ and $\tilde{g}$ are quasi-isometric,

  2. 2.

$(\overline{M},\tilde{g})$ is again a manifold with boundary of bounded
geometry,

  3. 3.

there is an open $E_{Y}$ of $Y$ such that $f:(X=M\cup
E_{X},\tilde{g})\longrightarrow(Y=N\cup\tilde{E}_{Y},h)$ is a uniform map
which is isometric on the unbounded ends.

Proof: Let $\phi:\mathbb{R}\longrightarrow\mathbb{R}$ a smooth function such
that $\phi\cong 1$ on $[1,+\infty)$, $\phi\cong 0$ on $(-\infty,0]$ and
$\phi(x)\in[0,1]$ otherwise. Let us define

$\chi(x):=\begin{cases}\phi(\frac{3}{r_{0}}d_{g}(\partial M,x)-1)\mbox{ if
}x\in M\\\ 0\mbox{ otherwise.}\end{cases}$ (13)

Notice that, with respect to the normal coordinates of $(X,g)$, the function
$\chi$ has bounded derivatives of each order. This fact is a consequence of
Theorem 2.5 of Schick  and Lemma 1.1. Then we define the metric $\tilde{g}$ on
$x\in X$ as

$\tilde{g}_{x}:=\chi(x)g_{x}+(1-\chi(x))f^{*}h_{x}.$ (14)

Observe that $\tilde{g}$ and $g$ are quasi-isometric. Indeed, since there is a
$K$ such that

$K^{-1}\cdot g\leq f^{*}h\leq K\cdot g,$ (15)

then

$\frac{1}{1+K}g_{x}\leq\tilde{g}\leq(1+K)g_{x}.$ (16)

Our next step is to prove that $(\overline{M},\tilde{g})$ is a manifold with
boundary of bounded geometry.
Let us check the conditions on Definition 1.1:

  * •

the conditions $1$ and $2$ are satisfied. Indeed the image of the
$\frac{r_{0}}{3}$-neighborhood of $\partial\overline{M}$ (with respect to $g$)
by $f$ contains on a $\delta_{0}$-neighborhood of $\partial\overline{N}$. This
implies that the $\delta_{0}$-neighborhood of $\partial\overline{M}$ (with
respect to $\tilde{g}$) is isometric to the $\delta_{0}$-neighborhood of
$\partial\overline{N}$. Then, since $\overline{N}$ is a manifold with boundary
of bounded geometry, the first two conditions of Definition 1.1 are satisfied.

  * •

Condition $3.$ is satisfied. Since the $\delta_{0}$-neighborhood of $\partial
M$ is isometric to the $\delta_{0}$-neighborhood of $\partial N$, we just have
to check that the injectivity radius on $M$ is bounded from below.
Let $p$ be a point on $M$, let $\\{e_{i}\\}$ be an orthonormal basis of
$T_{p}X$ and fix some normal coordinates $\\{U,x^{i}\\}$ (with respect to $g$)
referred to $p$ and $\\{e_{i}\\}$. We obtain the coordinates
$\\{x^{i},\mu^{j}\\}$ on $TX$, where $\\{\mu^{j}\\}$ are the components of
$\frac{\partial}{\partial x^{i}}$.
Observe that, with respect to these coordinates, the components of the metric
$\tilde{g}_{ij}$ are uniformly bounded and the same holds for its derivatives
of each order. This because $f$ and $\chi$ have uniformly bounded derivatives
of each order with respect to the coordinates $\\{x^{i}\\}$ and because $M$
and $N$ have bounded geometry.
Fix $s=\frac{inj_{M}}{K^{\prime}}$ where $K^{\prime}>1$ is a constant such
that $\tilde{g}\leq K^{\prime}\cdot g$. Notice that, on the ball111the radius
is given with respect to $g$ $B_{s}(0_{p})\subset T_{p}M$, the exponential map
$exp_{g.p}:B_{s}(0_{p})\longrightarrow M$ with respect to $g$ is injective. As
a consequence of Picard-Lindelöf Theorem we also know that the exponential map
$exp^{\prime}_{p}$ with respect to $\tilde{g}$ is well defined on the same
ball $B_{s}(0_{p})$.
We know that the Jacobian matrix $J_{exp^{\prime}_{p}}(0_{p})$ is the the
identity. Moreover, we also know that the derivatives of the entries of
$J_{exp^{\prime}_{p}}$ are uniformly bounded by a constant $C$: this is a
consequence of Lemma 3.4 of Schick . So this means that there is a constant
$D$ such that for each $v_{p}$ in $B_{s}(0_{p})$

$||J_{exp^{\prime}_{p}}(v_{p})-Id||\leq D\cdot||v_{p}||.$ (17)

Then, if $R=\min\\{\frac{1}{2D},s\\}$ on $B_{R}(0_{p})$ the exponential map
$exp^{\prime}_{p}$ is invertible. Then we can conclude by observing that
$B_{R}(0_{p})$ contains a ball of radius $\frac{R}{K}$ with respect to the
metric $\tilde{g}$. Indeed this fact implies that

$inj_{(M,\tilde{g})}(p)\geq\frac{\min\\{\frac{1}{2D},\frac{inj_{(M,g)}}{K^{\prime}}\\}}{K^{\prime}}.$
(18)

  * •

Condition $4.$ is satisfied. In particular the boundedness of the derivatives
of the second fundamental form of $\partial M$ follows because $M$ is
isometric to $N$ near $\partial M$. On the other hand, the boundedness of the
covariant derivatives of the Riemann tensor of $\tilde{g}$ is a consequence of
the fact that the norms induced by $g$ and $\tilde{g}$ on each fiber $F_{x}$
of a tensor multi-product of $TX$ and $T^{*}X$ are equivalent with some
constants which does not depends on $x$, and that $\chi$ and the components of
$f^{*}h$ have uniformly bounded derivatives in normal coordinates with respect
to $g$.

Finally, we can conclude the proof by observing that
$f:(X,\tilde{g})\longrightarrow(N,h)$ is a uniform map over $M$ (Lemma 1.2)
and it is an isometry on $(X\setminus M)\cup\mathcal{N}(\delta_{0})$ where
$\mathcal{N}(\delta_{0})$ is the $\delta_{0}$-neighborhood of $\partial M$ in
$M$ with respect to $\tilde{g}$.

##  2 Minimal and maximal domains

###  2.1 The space of squared-integrable differential forms

Let $(X,g)$ be an oriented Riemannian manifold and let $\Omega^{k}_{c}(X)$ be
the space of complex differential forms with compact support. The Riemannian
metric $g$ induces for every $k\in\mathbb{N}$ a scalar product on
$\Omega^{k}_{c}(X)$ as follows: for each $\alpha$ and $\beta$ in
$\Omega^{*}_{c}(X)$ then

$\langle\alpha,\beta\rangle_{\mathcal{L}^{2}(X)}:=\int\limits_{X}\alpha\wedge\star\bar{\beta},$
(19)

where $\star$ is the Hodge star operator induced by $g$.
This hermitian product gives a norm on $\Omega^{k}_{c}(X)$:

$\begin{split}|\alpha|^{2}_{\mathcal{L}^{2}(X)}:&=\langle\alpha,\alpha\rangle_{\mathcal{L}^{2}(X)}\\\
&=\int\limits_{M}\alpha\wedge\star\bar{\alpha}<+\infty.\end{split}$ (20)

######  Definition 2.1.

We will denote by $\mathcal{L}^{2}\Omega^{k}(X)$ the Hilbert space given by
the closure of $\Omega^{*}_{c}(X)$ with respect to the norm
$|\cdot|_{\mathcal{L}^{2}\Omega^{k}(X)}$. Moreover we can also define the
Hilbert space $\mathcal{L}^{2}(X)$ as

$\mathcal{L}^{2}(X):=\bigoplus\limits_{k\in\mathbb{N}}\mathcal{L}^{2}\Omega^{k}(X).$
(21)

The norm of $\mathcal{L}^{2}(X)$ it will be denoted by
$|\cdot|_{\mathcal{L}^{2}}$ or $|\cdot|_{\mathcal{L}^{2}(X)}$.

######  Remark 5.

Since $\Omega_{c}^{k}(X)$ is dense in $\mathcal{L}^{2}\Omega^{k}(X)$, then
$\Omega_{c}^{*}(X)$ is dense in $\mathcal{L}^{2}(X)$.

######  Definition 2.2.

Let $(X,g)$ be an oriented Riemannian manifold whose dimension is $m$. The
chirality operator is defined as the operator
$\tau_{X}:\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$ such that, for
each $\alpha$ in $\Omega_{c}^{*}(X)$,

$\tau_{X}(\alpha):=i^{\frac{m}{2}}\star\alpha$ (22)

if $m$ is even and

$\tau_{X}(\alpha):=i^{\frac{m+1}{2}}\star\alpha$ (23)

if $m$ is odd.

######  Remark 6.

The chirality operator is an $\mathcal{L}^{2}$-bounded operator. In
particular, with respect to $||\cdot||_{\mathcal{L}^{2}}$, it is an isometry
and an involution and so we obtain that $\tau_{X}^{2}=1$ and
$||\tau_{X}||_{\mathcal{L}^{2}}=1$. Finally, $\tau_{X}$ is also self-adjoint.

######  Definition 2.3.

Let $(X,g)$ and $(Y,h)$ be two Riemannian manifold. Fix
$A:dom(A)\subset\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(Y)$ an
operator and let
$A^{*}:dom(A^{*})\subset\mathcal{L}^{2}(Y)\longrightarrow\mathcal{L}^{2}(X)$
be its adjoint operator. Then we denote by $A^{\dagger}$ the operator defined
as

$A^{\dagger}:=\tau_{X}\circ A^{*}\circ\tau_{Y}.$ (24)

######  Remark 7.

Notice that if $\pi:(X,g)\longrightarrow(Y,h)$ is a submersion, then
$(\pi^{*})^{\dagger}=\pi_{star}$ which is the integration along the fibers of
$\pi$. If $\omega$ is smooth form on a Riemannian manifold $(X,g)$ and if
$e_{\omega}:\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$ is defined as
$e_{\omega}(\beta)=\beta\wedge\omega$, then $e_{\omega}^{\dagger}=r_{\omega}$
where $r_{\omega}(\beta):=\omega\wedge\beta$.

###  2.2 A regularizing operator

In this subsection we will talk about $C^{k}$-forms over a manifold $X$. This
are sections of class $C^{k}$ of the bundle $\Lambda^{*}(X)$. If $k>1$, then
the differential of a $C^{k}$-form is defined locally in the usual way and it
is a $C^{k-1}$-form.
Let $(X,g)$ be a Riemannian manifold and fix $\epsilon>0$. Gol’dshtein and
Troyanov in Gol  studied the de Rham regularizing operator $R_{\epsilon}^{X}$.
In their paper they proved that, given $k>0$ in $\mathbb{N}$, then for each
compactly supported $C^{k}$-form $\omega$

  * •

$R_{\epsilon}^{X}\omega$ is in $\Omega^{*}_{c}(X)$,

  * •

$\lim\limits_{\epsilon\rightarrow+\infty}||R_{\epsilon}^{X}\omega-\omega||_{\mathcal{L}^{2}}=0$.

  * •

$R_{\epsilon}^{X}d\omega=dR_{\epsilon}^{X}\omega$.

For the sake of completeness we would like briefly recall the definition and
the proofs of the main properties of $R_{\epsilon}^{X}$.
The operator $R_{\epsilon}^{X}$ is defined as follows: let $n$ be the
dimension of $X$ and fix a mollifier
$\rho:\mathbb{R}^{n}\longrightarrow\mathbb{R}$. Then let
$h:B_{1}(0)\subset\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}$ be a radial
diffeomorphism such that $h(x)=x$ if $||x||\leq\frac{1}{3}$ and

$h(x)=\frac{1}{||x||}exp(\frac{1}{1-||x||^{2}})\cdot x\mbox{ if
}||x||\geq\frac{2}{3}$ (25)

Then, we define the submersion
$s:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}$ as

$s(x,v):=\begin{cases}h^{-1}(h(x)+v)\mbox{if }||x||\leq 1\\\ x\mbox{
otherwise.}\end{cases}$ (26)

Let $U$ be a bounded convex domain of $\mathbb{R}^{n}$ which contains the ball
$B_{1}(0)$. Then, for each $\epsilon>0$, the local regularizing operator
$R_{\epsilon}:\mathcal{L}^{2}(U)\longrightarrow\mathcal{L}^{2}(U)$ is defined
as

$R_{\epsilon}\omega:=\int_{\mathbb{R}^{n}}s^{*}\omega\wedge\rho_{\epsilon}(v)dv^{1}\wedge\ldots\wedge
dv^{n},$ (27)

where $\rho_{\epsilon}(v):=\rho(\frac{v}{\epsilon})$. In order to obtain the
global regularizing operator we need to fix a constant $J$ and consider a
countable atlas $\\{(V_{i},\phi_{i})\\}$ of $M$ such that for each $i$ there
are at most $K$ charts $(V_{j},\phi_{j})$ such that $V_{i}\cap
V_{j}\neq\emptyset$. Assume that
$B_{1}(0)\subseteq\phi_{i}(V_{i})\subset\mathbb{R}^{n}$ for each $i$ and that
$\\{\phi_{i}^{-1}(B_{1}(0))\\}$ is again a cover of $M$. Then we define for
each $k$ in $\mathbb{N}$,

$R_{\epsilon,k}:=R_{\epsilon,V_{0}}\circ\ldots\circ R_{\epsilon,V_{k}}$ (28)

where $R_{\epsilon,V_{i}}:=\phi_{i}^{*}\circ
R_{\epsilon}\circ[\phi_{i}^{-1}]^{*}$ on $V_{i}$ and it is the identity
outside $V_{i}$. Then we can define the operator
$R^{X}_{\epsilon}:=\lim\limits_{k\rightarrow+\infty}R_{\epsilon,k}.$

It is a well known fact that if $\omega$ is a $C^{k}$-form, then
$R^{X}_{\epsilon}\omega$ is a smooth differential form (Gol ). Essentially
this follows because $R_{\epsilon}$ is a smoothing operator: the components of
its kernel $K(x,y)=k(x,y)_{J}^{I}dx^{J}\boxtimes\frac{\partial}{\partial
y^{I}}$ are given by the components of the form
$t^{*}(\rho_{\epsilon}(v)dv^{1}\wedge\ldots\wedge dv^{n})$ where $t$ is the
inverse map of
$S:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}\times\mathbb{R}^{n}$
which is defined for each $(x,y)$ as $S(x,y):=(x,s(x,y))$. 222Actually $S$ is
a diffeomorphism only on $\mathbb{R}^{n}\times B_{1}(0)$. However the
components of $K$ are given by the components of the form
$t^{*}(\rho(v)dv^{1}\wedge\ldots\wedge dv^{n})$ on $S(\mathbb{R}^{n}\times
B^{1}(0))$ and zero outside. The components of $K$ are smooth since the
support of $\rho(v)dv^{1}\wedge\ldots\wedge dv^{n}$ is contained in
$\mathbb{R}^{n}\times B^{1}(0)$.
For the same reason is quite easy to show that if $\omega$ is a compactly
supported form, then the same happens to $R^{X}_{\epsilon}\omega$. In
particular, if $\omega=f_{I}(x)dx^{I}$ is a differential form whose support is
contained on $B_{1}(0)\subset\mathbb{R}^{n}$, then
$R_{\epsilon}\omega=f_{I,\epsilon}(x)dx^{I}$ where
$|f_{I}(x)-f_{I,\epsilon}(x)|\leq C_{\epsilon,f_{I}}$ and $C_{\epsilon,f_{I}}$
goes to zero if $\epsilon$ goes to zero. In other words if
$\epsilon=\frac{1}{n}$ then the components $f_{I,\frac{1}{n}}$ uniformly
approximate $f_{I}$. This, in particular, implies that

$||R_{\epsilon}\omega-\omega||_{\mathcal{L}^{2}(B_{1}(0))}\rightarrow 0$ (29)

when $\epsilon\rightarrow 0$. Then for each compactly supported differential
form $\alpha$ on a Riemannian manifold $(X,g)$, immediately follows that

$\lim\limits_{\epsilon\rightarrow
0}||R^{X}_{\epsilon}\alpha-\alpha||_{\mathcal{L}^{2}(X)}=0.$ (30)

Finally, $R_{\epsilon}$ and the exterior derivative operator commute on
$C^{k}_{c}(\Lambda^{*}T^{*}\mathbb{R}^{n})$. Indeed $d$ and $s^{*}$ commute,
$d(\alpha)\wedge\phi(v)dv^{1}\wedge\ldots\wedge
dv^{n}=d(\alpha\wedge\phi(v)dv^{1}\wedge\ldots\wedge dv^{n})$ since
$\phi(v)dv^{1}\wedge\ldots\wedge dv^{n}$ is a closed form and finally the
integral along $\mathbb{R}^{n}$ and the exterior derivative operator commute
on the space of vertically compactly supported smooth forms on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ (Proposition 6.14.1. of the book of Bott
and Tu bottu ). As a consequence of this, also $R^{X}_{\epsilon}$ commute with
the exterior derivative operator on $\Omega^{*}(X)$.

###  2.3 Maximal and minimal domains of the exterior derivative operator

Let $(X,g)$ be a Riemannian manifold. The exterior derivative operator
$d:\Omega^{*}_{c}(X)\longrightarrow\Omega^{*}_{c}(X)$ can be seen as an
unbounded operator with respect to the $\mathcal{L}^{2}$-norm on $(X,g)$. So
we can define some different closures
$\overline{d}:dom(\overline{d})\subset\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$
of $(d,\Omega^{*}_{c}(X))$. In this manuscript we will concentrate on the
maximal and the minimal closure of $d$.

######  Definition 2.4.

The minimal domain of the exterior derivative on $(X,g)$ is defined as the
subset $dom(d_{min})$ of $\mathcal{L}^{2}(X)$ given by the form $\alpha$ such
that there is a sequence of $\\{\omega_{k}\\}\subset\Omega^{*}_{c}(X)$ such
that
$\lim\limits_{k\rightarrow+\infty}||\omega_{k}-\alpha||_{\mathcal{L}^{2}(X)}=0$
and the sequence of $\\{d\omega_{k}\\}$ converges in $\mathcal{L}^{2}(X)$ with
respect to the $\mathcal{L}^{2}$-norm.
Then we can define
$d_{min}(\alpha):=\lim\limits_{k\rightarrow+\infty}d\omega_{k}$.

######  Definition 2.5.

The maximal domain of the exterior derivative on $(X,g)$ is defined as the
subset $dom(d_{max})$ of $\mathcal{L}^{2}(X)$ given by the $k$-forms $\alpha$
such that there a $\mathcal{L}^{2}$-form $\eta_{\alpha}$ such that for each
$\beta$ in $\Omega_{c}^{*}(X)$

$\int_{X}\alpha\wedge d\beta=(-1)^{k+1}\int_{X}\eta_{\alpha}\wedge\beta.$ (31)

Then we can define $d_{max}(\alpha):=\eta_{\alpha}$.

A well-known result about the minimal and the maximal closure of $d$, is the
following. Let $(\overline{d},dom(\overline{d}))$ be a closed extension of $d$
on $\mathcal{L}^{2}(X)$, then

$(dom(d_{min}),d_{min})\subseteq(\overline{d},dom(\overline{d}))\subseteq(dom(d_{max}),d_{max}).$
(32)

Moreover Gaffney in Gaf  proved that if $(X,g)$ is a complete Riemannian
manifold, then $dom(d_{min})=dom(d_{max})$ and so there is just one closed
extension of $d$.

######  Lemma 2.1.

Let $(X,g)$ be a Riemannian manifold and let $\alpha\in dom(d_{max})$. Then,
for each $k\geq 1$ and for each $\beta$ in $C^{k}_{c}(\Lambda^{*}T^{*}M)$, we
have

$\int_{M}\alpha\wedge d\beta=(-1)^{k}\int_{M}d_{max}\alpha\wedge\beta.$ (33)

Proof: Let us denote by $R_{\epsilon}^{X}$ the de Rham regularizing operator.
We can assume, without loss of generality, that there is a $j$ such that
$supp(\beta)$ is contained in $\phi_{j}^{-1}(B_{1}(0))\subseteq V_{j}$. Here
$\\{(V_{i},\phi_{i})\\}$ is the atlas used to define $R_{\epsilon}^{X}$.
Observe that $\alpha\wedge R^{X}_{\frac{1}{n}}\beta$ converges point-wise to
$\alpha\wedge\beta$.
Notice that on $V_{i}$, the form $\alpha$ has the form
$\alpha(x)=\alpha_{I}(x)dx^{I}$ such that

$\int_{V_{i}}g^{IJ}(x)\alpha_{I}(x)\alpha_{J}(x)dx^{1}\wedge\ldots\wedge
dx^{n}$ (34)

is bounded. Here the capital letter $I$ is a multi-index defined unless double
switching. Moreover, we also denote by $N_{I}$ the multi-index such that
$(I,N-I)=(1,\ldots,n)$. Again $N_{I}$ is defined unless to double switching.
So the integral of $\alpha\wedge d\beta$ has the form
$\int_{\phi_{i}^{-1}(B_{1}(0))}\alpha_{I}\cdot(d\beta)_{N-I}dx^{1}\wedge\ldots\wedge
dx^{n}$. Notice that $|\alpha_{I}\cdot(d\beta)_{N-I}|\leq
C_{\beta}\cdot|\alpha_{I}|$. This means that if $|\alpha_{I}|$ is in
$L^{1}(\phi_{j}^{-1}(B_{1}(0)))$ then we could apply the Dominated Convergence
Theorem to the sequence
$\alpha_{I}\cdot\lim\limits_{k\rightarrow+\infty}(dR^{X}_{\frac{1}{k}}\beta)_{N-I}$.
Observe that $|\alpha_{I}|$ is in $L^{2}(\phi_{j}^{-1}(B_{1}(0)))$. Indeed

$\begin{split}\int_{\phi_{j}^{-1}(B_{1}(0))}|\alpha_{I}|^{2}dx^{1}\wedge\ldots\wedge
dx^{n}&\leq\frac{1}{m_{j}}\int_{\phi_{j}^{-1}(B_{1}(0))}m_{j}|\alpha_{I}|^{2}dx^{1}\wedge\ldots\wedge
dx^{n}\\\
&\leq\frac{1}{m_{j}}\int_{\phi_{j}^{-1}(B_{1}(0))}g^{IJ}\alpha_{I}\alpha_{J}dx^{1}\wedge\ldots\wedge
dx^{n}\\\ &\leq C_{j}\end{split}$ (35)

where $m_{i}$ is the infimum of the eigenvalues of $g$ on the closure of
$V_{i}$. So, since $\phi_{j}^{-1}(B_{1}(0))$ is compact, then $|\alpha_{I}|$
is also an $L^{1}$ function. This means that, thanks to the Dominated
Convergence Theorem,

$\begin{split}\int_{V_{i}}\alpha\wedge
d\beta&=\int_{V_{i}}\alpha_{I}\cdot(d\beta)_{N-I}dx^{1}\wedge\ldots\wedge
dx^{n}\\\
&=\int_{V_{i}}\alpha_{I}\cdot\lim\limits_{k\rightarrow+\infty}(dR^{X}_{\frac{1}{k}}\beta)_{N-I}dx^{1}\wedge\ldots\wedge
dx^{n}\\\
&=\int_{V_{i}}\lim\limits_{k\rightarrow+\infty}\alpha_{I}\cdot(dR^{X}_{\frac{1}{k}}\beta)_{N-I}dx^{1}\wedge\ldots\wedge
dx^{n})\\\ &=\lim\limits_{k\rightarrow+\infty}\int_{V_{i}}\alpha\wedge
dR^{x}_{\frac{1}{k}}\beta.\end{split}$ (36)

Observe that $dR^{X}_{\frac{1}{k}}\beta$ is a compactly supported smooth form:
this means that

$\int_{V_{i}}\alpha\wedge
d\beta=\lim\limits_{k\rightarrow+\infty}\int_{V_{i}}d\alpha\wedge
R^{X}_{\frac{1}{k}}\beta.$ (37)

So, by applying again the Dominated Convergence Theorem, we conclude.

######  Corollary 2.2.

Let $(X,g)$ and $(Y,h)$ be two Riemannian manifolds. Fix an
$\mathcal{L}^{2}$-bounded operator $A:\Omega^{*}_{c}(Y)\longrightarrow
C^{k}_{c}(\Lambda^{*}(T^{*}X))$ for some $k>1$. Suppose that
$A^{\dagger}:\Omega^{*}_{c}(X)\longrightarrow C^{k}_{c}(\Lambda^{*}(T^{*}Y))$.
Finally assume that, for each smooth form $\omega$

$dA\omega\pm Ad\omega=B\omega$ (38)

where $B\omega$ is a $\mathcal{L}^{2}$-bounded operator. Then
$A(dom(d_{min}))\subseteq dom(d_{min})$ and $A(dom(d_{max}))\subseteq
dom(d_{max})$.

Proof: Let $\alpha$ be a $k$-form of $dom(d_{maxY})$: there is an
$\mathcal{L}^{2}$-form $d\alpha$ such that, for each $\beta$ in
$\Omega^{*}_{c}(Y)$,

$\int_{Y}d\alpha\wedge\beta=(-1)^{k}\int_{Y}\alpha\wedge d\beta.$ (39)

Let us concentrate on $A\alpha$. Let $\gamma$ be a form in
$\Omega^{*}_{c}(X)$. We obtain that

$\begin{split}\int_{X}A\alpha\wedge d\gamma&=\langle
A\alpha,\tau_{X}d\gamma\rangle_{X}\\\
&=\langle\alpha,A^{*}\tau_{X}d\gamma\rangle_{Y}\\\
&=\langle\alpha,\tau_{Y}A^{\dagger}\tau_{X}\tau_{X}d\gamma\rangle_{Y}\\\
&=\langle\alpha,\tau_{Y}A^{\dagger}d\gamma\rangle_{Y}\\\
&=(-1)^{k}\langle\alpha,\tau_{Y}A^{\dagger}d^{\dagger}\gamma\rangle_{Y}\\\
&=(-1)^{k}\langle\alpha,\tau_{Y}(d^{\dagger}\mp
A^{\dagger}+B^{\dagger})\gamma\rangle_{Y}\\\ &=\langle\alpha,\tau_{Y}(d\mp
A^{\dagger}+B^{\dagger})\gamma\rangle_{Y}\\\ &=\langle
d_{max}\alpha,\tau_{Y}\mp A^{\dagger}\gamma\rangle+\langle
B\alpha,\tau_{Y}\gamma\rangle\\\ &=\langle(\mp
Ad_{max}+B)\alpha,\gamma\rangle.\end{split}$ (40)

So, this means that $A(dom(d_{max}))\subseteq dom(d_{max})$ and $d_{max}A=\mp
Ad_{max}+B$.

Let $\beta$ be an element in $dom(d_{minY})$. Then there is a sequence
$\\{\beta_{n}\\}$ such that $\beta=\lim\limits_{n\rightarrow+\infty}\beta_{n}$
and $d\beta=\lim\limits_{n\rightarrow+\infty}d\beta_{n}$ with respect to the
$\mathcal{L}^{2}$-norm. Since $A$ is a $\mathcal{L}^{2}$-bounded operator, we
obtain that

$A\beta=\lim\limits_{n\rightarrow+\infty}A\beta_{n}.$ (41)

Fix $n\in\mathbb{N}$ and let $K(n)$ be a number such that

$||A\beta-A\beta_{K(n)}||_{\mathcal{L}^{2}}\leq\frac{1}{2n}$ (42)

and

$||Ad\beta-Ad\beta_{K(n)}||_{\mathcal{L}^{2}}\leq\frac{1}{2n}.$ (43)

Then let $\epsilon(n)$ be small enough to

$||A\beta_{K(n)}-R_{\epsilon(n)}^{X}A\beta_{K(n)}||_{\mathcal{L}^{2}}\leq\frac{1}{2n}$
(44)

and

$||Ad\beta_{K(n)}-R_{\epsilon(n)}^{X}Ad\beta_{K(n)}||_{\mathcal{L}^{2}}\leq\frac{1}{2n}.$
(45)

So let $\\{\gamma_{n}\\}$ be the sequence defined as
$R_{\epsilon(n)}^{X}A\beta_{K(n)}$ for each $n$. This is a sequence of
compactly supported smooth forms which converges, with respect to the
$\mathcal{L}^{2}$-norm, to $A\beta$. We just have to show that
$dR_{\epsilon(n)}^{X}A\beta_{K(n)}$ converges to an element of
$\mathcal{L}^{2}(X)$.
Observe that

$\begin{split}dR_{\epsilon(n)}^{X}A\beta_{K(n)}&=R_{\epsilon(n)}^{X}d_{max,X}A\beta_{K(n)}\\\
&=R_{\epsilon(n)}^{X}(\mp Ad_{max,X}+B)\beta_{K(n)}\\\
&=R_{\epsilon(n)}^{X}(\mp Ad+B)\beta_{K(n)}.\end{split}$ (46)

This means that

$\begin{split}||Ad\beta-R_{\epsilon(n)}^{X}(\mp
Ad+B)\beta_{K(n)}||_{\mathcal{L}^{2}}&\leq||Ad\beta-
Ad\beta_{K(n)}||_{\mathcal{L}^{2}}\\\
&+||Ad\beta_{K(n)}-R_{\epsilon(n)}^{X}(\mp
Ad+B)\beta_{K(n)}||_{\mathcal{L}^{2}}\\\
&\leq\frac{1}{2n}+\frac{1}{2n}=\frac{1}{n}.\end{split}$ (47)

So we just proved that $A(dom(d_{minY}))\subseteq dom(d_{minX})$ and
$d_{min}A=\mp Ad_{min}+B.$

###  2.4 Reduced and unreduced cohomologies

Let $(X,g)$ be a Riemannian manifold. It is a well-known fact that

$d_{min\backslash max}(dom(d_{min\backslash max}))\subseteq
dom(d_{min\backslash max})$ (48)

and that $d_{min\backslash max}^{2}:=d_{min\backslash max}\circ
d_{min\backslash max}=0$. So, we can define the cohomology groups of
$L^{2}$-cohomology as follows.

######  Definition 2.6.

The $k$-th group of minimal $L^{2}$-cohomology is the group defined as

$H^{k}_{2,min}(X,g):=\frac{ker(d_{min}^{k})}{im(d^{k-1}_{min})}.$ (49)

Moreover the $k$-th group of reduced minimal $L^{2}$-cohomology is given by

$\overline{H}^{k}_{2,min}(X,g):=\frac{ker(d_{min}^{k})}{\overline{im(d^{k-1}_{min})}}.$
(50)

where $d^{k}_{min}$ is the operator $d^{k}_{min}$ defined on
$dom(d_{min})\cap\mathcal{L}^{2}(\Omega^{k}(X))$.

######  Definition 2.7.

The $k$-th group of maximal $L^{2}$-cohomology is the group defined as

$H^{k}_{2,max}(X,g):=\frac{ker(d_{max}^{k})}{im(d^{k-1}_{max})}.$ (51)

Moreover the $k$-th group of reduced maximal $L^{2}$-cohomology is given by

$\overline{H}^{k}_{2,max}(X,g):=\frac{ker(d_{max}^{k})}{\overline{im(d^{k-1}_{max})}}.$
(52)

where $d^{k}_{max}$ is the operator $d^{k}_{max}$ defined on
$dom(d_{max})\cap\mathcal{L}^{2}(\Omega^{k}(X))$.

In general, for each $k\in\mathbb{N}$, the groups $H^{k}_{2,min}(X,g)$,
$\overline{H}^{k}_{2,min}(X,g)$, $H^{k}_{2,max}(X,g)$ and
$\overline{H}^{k}_{2,max}(X,g)$ can be different.
On the other hand, we know that if $(X,g)$ is a complete Riemannian manifold
then $d_{min}=d_{max}$. In this case, we obtain

$H^{k}_{2,min}(X,g)=H^{k}_{2,max}(X,g)\mbox{ and
}\overline{H}^{k}_{2,min}(X,g)=\overline{H}^{k}_{2,max}(X,g).$ (53)

######  Remark 8.

If two Riemannian manifolds $(X,g)$ and $(Y,h)$ are quasi-isometric manifolds,
then their $L^{2}$-cohomology groups (minimal or maximal, reduced or
unreduced) are isomorphic.

##  3 R.-N.-Lipschitz maps and vector bundles

###  3.1 Radon-Nikodym-Lipschitz maps

In this Subsection we will recall some notions introduced in subsections 2.3.
and 2.4 of Spes . This notions will be useful in order to understand when the
pullback along a map between Riemannian manifolds induces a
$\mathcal{L}^{2}$-bounded operator.
Let $(M,\nu)$ and $(N,\mu)$ be two measured spaces and let
$f:(M,\nu)\longrightarrow(N,\mu)$ be a function such that the pushforward
measure $f_{\star}(\nu)$ is absolutely continuous with respect to $\mu$.

######  Definition 3.1.

Let $(N,\mu)$ be $\sigma$-finite, then the Fiber Volume of  $f$ is the Radon-
Nikodym derivative

$Vol_{f,\nu,\mu}:=\frac{\partial f_{\star}\nu}{\partial\mu}.$ (54)

Let $(M,d_{M},\mu_{M})$ and $(N,d_{N},\mu_{N})$ be two measured and metric
spaces.

######  Definition 3.2.

A map $f:(M,d_{M},\mu_{M})\longrightarrow(N,d_{N},\mu_{N})$ is Radon-Nikodym-
Lipschitz or R.-N.-Lipschitz if

  * •

$f$ is Lipschitz

  * •

$f$ has a well-defined and bounded Fiber Volume.

######  Remark 9.

As showed in Remark 12 of Spes , an equivalent definition is the following. A
map $f:(M,d_{M},\nu_{M})\longrightarrow(N,d_{N},\mu_{N})$ is a R.-N.-Lipschitz
map if it is Lipschitz and there is a constant $C$ such that, for each
measurable set $A\subseteq N$,

$\mu_{M}(f^{-1}(A))\leq C\mu_{N}(A).$ (55)

So this implies that a composition of R.-N.-Lipschitz maps is a
R.-N.-Lipschitz map.

The main property of R.-N.-Lipschitz map is proved in Proposition 2.4 of Spes
.

######  Proposition 3.1.

Let $(M,g)$ and $(N,h)$ be Riemannian manifolds. Let
$f:(M,g)\longrightarrow(N,h)$ be a R.-N.-Lipschitz map. Then $f^{*}$, which is
the pullback along $f$, is an $\mathcal{L}^{2}$-bounded operator.

In general a Lipschitz map is not R.-N.-Lipschitz: for example if $(M,g)$ and
$(N,h)$ are two Riemannian manifolds such that $dim(M)<dim(N)$ then there is
no Lipschitz embedding $i:(M,g)\longrightarrow(N,h)$ which is R.-N.-Lipschitz.
However we are interested in the Fiber Volume of a submersion. Indeed if
$f:(M,g)\longrightarrow(N,h)$ is a Lipschitz submersion between oriented
Riemannian manifold, then we know how to compute its Fiber Volume. In order to
explain how compute it, we need the notion of quotient between two
differential forms.

######  Definition 3.3.

Let us consider a differentiable manifold $X$. Given two differential forms
$\alpha\in\Omega^{k}(X)$, $\beta\in\Omega^{n}(X)$ we define a quotient between
$\alpha$ and $\beta$, denoted by $\frac{\alpha}{\beta}$, as a (possibly not
continuous) section of $\Lambda^{k-n}(X)$ such that for all $p$ in $M$

$\alpha(p)=\beta(p)\wedge\frac{\alpha}{\beta}(p)..$ (56)

In general, given two smooth differential forms, we do not know if there is a
quotient between them. Moreover, if a quotient between $\alpha$ and $\beta$
exists, it may be not unique. However, given a submersion between Riemannian
manifolds $\pi:(X,g)\longrightarrow(Y,h)$, then it is possible to define a
locally smooth quotient between $Vol_{X}$ and $\pi^{*}Vol_{Y}$ and the
pullback of this quotient on a fiber of $\pi$ is a smooth form which does not
depends on the choice of the quotient (pag. 15 of Spes ). So this means that
if we denote by $i_{q}:F_{q}\longrightarrow X$ the embedding of the fiber of
$q$ in $X$, then

$\int_{F_{q}}i^{*}_{q}(\frac{Vol_{X}}{\pi^{*}Vol_{Y}})$ (57)

is a well-defined real number for each $q$ in $N$.

######  Proposition 3.2.

Let $(X,g)$ and $(Y,h)$ two oriented, Riemannian manifolds. Let
$\pi:(X,g)\longrightarrow(Y,h)$ be a submersion. Then, if $q$ is on $im(\pi)$,
then

$Vol_{\pi,\mu_{M},\mu_{N}}(q)=\int_{F}\frac{Vol_{X}}{\pi^{*}Vol_{Y}}(q)$ (58)

and $Vol_{\pi,\mu_{M},\mu_{N}}(q)=0$ otherwise.

######  Remark 10.

If the submersion $f:X\rightarrow Y$ is a diffeomorphism between oriented
manifold which preserves the orientations, then the integration along the
fibers of $f$ is the pullback $(f^{-1})^{*}$. This means that the Fiber Volume
of $f$ is given by $|(f^{-1})^{*}\frac{Vol_{X}}{f^{*}(Vol_{Y})}|$.

We also recall Proposition 2.6 of Spes : this Proposition allow to compute the
Fiber Volume of the composition of two submersions.

######  Proposition 3.3.

Let $f:(M,g)\longrightarrow(N,h)$ and $g:(N,h)\longrightarrow(W,l)$ be two
submersions between oriented Riemannian manifolds. Then

$Vol_{g\circ
f,\mu_{M},\mu_{W}}(q)=\int_{g^{-1}(q)}(\int_{f^{-1}g^{-1}(q)}\frac{Vol_{M}}{f^{*}(Vol_{N})})\frac{Vol_{N}}{g^{*}Vol_{N}}$
(59)

Finally we conclude this subsection by showing the relation between
R.-N.-Lipschitz maps and the quasi-isometries.

######  Proposition 3.4.

Let $f:(X,g)\longrightarrow(Y,h)$ be a quasi-isometry. Then $f$ is a
R.N.-Lipschitz map.

Proof: Notice that $f$ is Lipschitz: it directly follows from the definition
of quasi-isometry. So we just have to show that $f$ has a bounded Fiber
Volume. This can be proved by showing that $id:(X,g)\longrightarrow(X,f^{*}h)$
has bounded Fiber Volume. Let $x$ be a point of $X$. By a simultaneous
diagonalization argument we can choose a basis $\\{e_{1},\ldots,e_{n}\\}$ of
$T_{x}X$ such that the Gram matrices of $f^{*}h$ and $g$ are both diagonals.
Observe that for each $i=1,\ldots,n$ we obtain that $f^{*}h(e_{i},e_{i})\leq
Kg(e_{i},e_{i})$. So this means that

$det(f^{*}h_{jl}(x))=\prod_{i=1}^{n}f^{*}h(e_{i},e_{i})\geq
K^{-n}\prod_{i=1}^{n}g(e_{i},e_{i})=det(g_{jl}(x)).$ (60)

Notice that the Fiber Volume of $id$ on a point $y=f(x)$ is given by

$\sqrt{\frac{det(g_{jl}(x))}{det(f^{*}h_{jl}(x))}}\leq K^{\frac{n}{2}}.$ (61)

This conclude the proof.

###  3.2 Generalized Sasaki metrics

The definition of Sasaki metric on a vector bundle given in this paper is a
generalization of the metric defined by Sasaki in SSK  for the tangent bundle
of a Riemannian manifolds. The definition of this generalized Sasaki metric
can be found in page 2 of the paper of Boucetta and Essoufi Boucetta .
Let us consider a Riemannian manifold $(N,h)$ of dimension $n$,
$\pi_{E}:E\longrightarrow N$ a vector bundle of rank $m$ endowed with a bundle
metric $H_{E}\in\Gamma(E^{*}\otimes E^{*})$ and a linear connection
$\nabla_{E}$ which preserves $H_{E}$. Fix $\\{s_{\alpha}\\}$ a local frame of
$E$: if $\\{x^{i}\\}$ is a system of local coordinates over $U\subseteq N$,
then we can define the system of coordinates $\\{x^{i},\mu^{\alpha}\\}$ on
$\pi_{E}^{-1}(U)$, where the $\mu^{\alpha}$ are the components with respect to
$\\{s_{\alpha}\\}$.
Let us denote by $K$ the map $K:TE\longrightarrow E$ defined as

$K(b^{i}\frac{\partial}{\partial
x^{i}}|_{(x_{0},\mu_{0})}+z^{\alpha}\frac{\partial}{\partial\mu^{\alpha}}|_{(x_{0},\mu_{0})}):=(z^{\alpha}+b^{i}\mu^{j}\Gamma_{ij}^{\alpha}(x_{0}))s_{\alpha}(x_{0}),$
(62)

where the $\Gamma_{ij}^{l}$ are the Christoffel symbols of $\nabla_{E}$. The
Christoffel symbols $\Gamma^{\gamma}_{\eta j}(x)$ are defined by the formula

$\nabla^{E}_{\frac{\partial}{\partial
x^{j}}}s_{\eta}(x):=\Gamma^{\gamma}_{\eta j}(x)s_{\gamma}(x).$ (63)

######  Definition 3.4.

The Sasaki metric on $E$ is the Riemannian metric $h^{E}$ defined for all
$A,B$ in $T_{(p,v_{p})}E$ as

$h^{E}(A,B):=h(d\pi_{E,v_{p}}(A),d\pi_{E,v_{p}}(B))+H_{E}(K(A),K(B)).$ (64)

######  Remark 11.

Let us consider the system of coordinates $\\{x^{i}\\}$ on $N$ and
$\\{x^{i},\mu^{j}\\}$ on $E$. The components of $h^{E}$ are given by

$\begin{cases}h^{E}_{ij}(x,\mu)=h_{ij}(x)+H_{\alpha\gamma}(x)\Gamma^{\alpha}_{\beta
i}(x)\Gamma^{\gamma}_{\eta j}(x)\mu^{\beta}\mu^{\eta}\\\
h^{E}_{i\sigma}(x,\mu)=H_{\sigma\alpha}(x)\Gamma^{\alpha}_{\beta
i}(x)\mu^{\beta}\\\ h^{E}_{\sigma\tau}(x,\mu)=H_{\sigma,\tau}(x),\end{cases}$
(65)

where $i,j=1,\ldots,n$ and $\sigma,\tau=n+1,\ldots,n+m$. Consider a point
$x_{0}=(x^{1}_{0},\ldots,x^{n}_{0})$ in $N$. If all the Christoffel symbols of
$\nabla_{E}$ in $x_{0}$ are zero, then, in local coordinates, the matrix of
$h^{E}$ in a point $(x_{0},\mu)$ is

$\begin{bmatrix}h_{i,j}(x_{0})&&0\\\ 0&&H_{\sigma,\tau}(x_{0})\end{bmatrix}.$
(66)

Moreover, with respect to the coordinates $(x^{i},\mu^{\sigma})$, the matrix
$h_{E}:=(h^{E})^{-1}$ is given by

$\begin{cases}h^{ij}_{E}(x,\mu)=h^{ij}(x)\\\
h^{i\sigma}_{E}=-\Gamma^{\sigma}_{\beta j}(x)h^{ij}(x)\mu^{\beta}\\\
h_{E}^{\sigma\tau}=H^{\sigma\tau}(x)+h^{ij}(x)\Gamma^{\sigma}_{\beta
i}(x)\Gamma^{\tau}_{\eta j}(x)\mu^{\beta}\mu^{\eta}\end{cases}$ (67)

where $H^{\sigma\tau}$ and $h^{ij}(x)$ are the components of the inverse
matrices of $h_{ij}$ and $H_{\sigma,\tau}$.

######  Example 3.1.

Let $(M,g)$ a Riemannian manifold. Consider as $E$ the tangent bundle $TM$ and
as $h_{E}$ the metric $g$ itself. Choose the connection $\nabla_{E}$ as the
Levi-Civita connection $\nabla_{g}^{LC}$. We denote by $g_{S}$ the Sasaki
metric induced by $g$ and $\nabla^{LC}_{g}$.

######  Example 3.2.

Consider a smooth map $f:(M,g)\longrightarrow(N,h)$. Let
$\pi:f^{*}TN\longrightarrow M$ be the pullback bundle. Then the Riemannian
metric $h$ can be seen as a bundle metric on $TN$ and so we obtain a bundle
metric $f^{*}h$ on $f^{*}TN$. Fix the connection $f^{*}\nabla^{LC}_{h}$ on
$f^{*}TN$ which is the pullback of the Levi-Civita connection on $(N,g)$. Let
us denote by $g_{S,f}$ the Sasaki metric induced by $f^{*}\nabla^{LC}_{h}$,
$f^{*}h$ and $g$.

######  Remark 12.

Then the Christoffel symbols of the pullback connection $f^{*}\nabla^{E}$ with
respect to the pullback frame $\\{f^{*}e_{i}\\}$ and to the coordinates
$\\{x^{i}\\}$ are given by

$\tilde{\Gamma}^{\alpha}_{\beta,i}:=\frac{\partial f^{l}}{\partial
x^{i}}f^{*}(\Gamma^{\alpha}_{\beta,l}).$ (68)

A first consequence of this fact is that the map
$ID:(id^{*}TN,g_{S,id})\longrightarrow(TN,g_{S})$ which sends $(p,w_{p})$ to
$w_{p}$ is an isometry. So, from now on, we will identify
$(id^{*}TN,g_{S,id})$ and $(TN,g_{S})$.

######  Remark 13.

Let $f:(M,g)\longrightarrow(N,h)$ be a smooth map. Let us fix a chart
$\\{U,x^{i}\\}$ on $M$ and a chart $\\{V,y^{j}\\}$ on $N$ such that
$f(U)\subseteq V$. Fix a bundle $E$ on $N$ and let $\nabla^{E}$ be a
connection. Let us denote by $\Gamma^{\alpha}_{\beta l}$ the Christoffel
symbols of $\nabla^{E}$ with respect to a frame $\\{e_{i}\\}$ and the
coordinates $\\{y^{j}\\}$. Fix $\delta>0$ and let $T^{\delta}N$ be the subset
of $TN$ given by the vector whose norm is less or equal to $\delta$. If
$f:(M,g)\longrightarrow(N,h)$ is a smooth Lipschitz map, then also the induced
bundle map

$\begin{split}F:(f^{*}T^{\delta}N,g_{S,f})&\longrightarrow(T^{\delta}N,g_{S})\\\
(p,w_{f(p)})&\longrightarrow w_{f(p)}\end{split}$ (69)

is a smooth Lipschitz map.

######  Proposition 3.5.

Consider a vector bundle $(E,\pi,M)$ over a Riemannian manifold $(M,g)$. Fix
on $E$ a bundle metric $h$, a connection $\nabla^{E}$ and let us denote by
$h_{E}$ the Sasaki metric induced by $g$, $h$ and $\nabla^{E}$. Let us suppose
that for each point $p$ on $M$ there is a system of normal coordinates
$\\{x^{i}\\}$ around $p$ and a local frame $\\{s_{\sigma}\\}$ such that the
Christoffel symbols of $\nabla^{E}$ vanishes at $x=0$. Moreover let us suppose
that $\frac{\partial h_{\sigma\tau}}{\partial x^{k}}(0)=0$, where
$h_{\sigma\tau}$ are the components of the Gram matrix of $h$ with respect to
the coordinates $\\{x^{i}\\}$. Then the fibers of $\pi$ are totally geodesic
submanifolds (in our case this means that the straight lines on the fibers are
geodesics) and $\pi$ is a Riemannian submersion.

######  Remark 14.

The vector bundles of the Examples 3.1 and 3.2 satisfy the assumptions of
Proposition 3.5.

######  Corollary 3.6.

Consider $0_{E}$ the null section of a vector bundle $\pi:E\longrightarrow M$.
Under the same assumptions of the Proposition 3.5, the disk bundle
$E^{\delta}:=\\{v_{p}\in E|\sqrt{h(v_{p},v_{p})}\leq\delta\\}$ coincides with
$B_{\delta}(0_{E}):=\\{v_{p}\in E|d_{h^{E}}(v_{p},0_{E})\leq\delta\\}$ for
each $\delta>0$.

######  Proposition 3.7.

Consider a Riemannian manifold $(M,g)$ and let $\pi:E\longrightarrow M$ be a
vector bundle. Fix on $E$ a metric bundle $h_{E}$ and a connection
$\nabla_{E}$. Let $h^{E}$ be the Sasaki metric on $E$ defined by using $g$,
$h_{E}$ and $\nabla_{E}$. Fix a $\delta>0$. Then, under the assumptions of
Proposition 3.5 $\pi:E^{\delta}\longrightarrow M$ is a R.-N.-Lipschitz map. In
particular the Fiber Volume on a point $q$ is the Volume of an Euclidean ball
of radius $\delta$. Finally the integration along the fibers
$\pi_{\star}:\Omega_{c}^{*}(E^{\delta})\longrightarrow\Omega^{*}(M)$ is an
$\mathcal{L}^{2}$-bounded operator.

###  3.3 Mathai-Quillen-Thom forms

Let us start by introducing the notion of Thom form.

######  Definition 3.5.

Let $\pi:E\longrightarrow M$ be a vector bundle. A smooth form $\omega$ in
$\Omega_{cv}^{*}(E)$ is a Thom form if it is closed and its integral along the
fibers of $\pi$ is equal to the constant function $1$.

Given a Thom form $\omega$ of $f^{*}TN$ such that $supp(\omega)$ is contained
in a $\delta_{0}<\delta$ neighborhood of the null section, let us define an
operator
$e_{\omega}:\Omega^{*}(f^{*}(T^{\delta}N))\longrightarrow\Omega^{*}(f^{*}(T^{\delta}N))$,
for every smooth form $\alpha$ as

$e_{\omega}(\alpha):=\alpha\wedge\omega.$ (70)

Here $f^{*}(T^{\delta}N)$ is the $\delta$-neighborhood of the $0$-section of
$f^{*}TN$.
In this paper, we use the Thom forms introduced by Mathai and Quillen in
mathai . In their work, indeed, they defined a Thom form for a vector bundle
$\pi:E\longrightarrow(M,g)$ induced by the Riemannian metric $g$, a metric
bundle $h_{E}$ and a connection $\nabla_{E}$. In particular, for each
$\delta>0$ they found a Thom form whose support is contained on

$B_{\delta}(0_{E}):=\\{w\in E|h_{E}(w,w)\leq\delta^{2}\\}.$ (71)

So, we will consider, on the tangent bundle of a Riemannian manifold $(Y,h)$
the Mathai-Quillen-Thom $\omega_{Y}$ form induced by $h$ and
$\nabla^{LC}_{h}$. On the other hand, given a map $f:X\longrightarrow Y$, we
fix on $f^{*}TY$ the pullback $F^{*}\omega_{Y}$ where
$F:f^{*}(TY)\longrightarrow TY$ is the bundle map induced by $f$.
As shown in subsection 4.2. of Spes , if $h$ has the same bounds of a
manifolds of bounded geometry and if $f$ is a smooth Lipschitz map, then there
is a constant $C$ such that for each $w$ in $f^{*}TY$ the norm
$|f^{*}\omega_{Y}|_{q}\leq C$. In this case the operator
$\Omega^{*}_{c}(TY)\longrightarrow\Omega^{*}_{c}(TY)$ is an
$\mathcal{L}^{2}$-bounded operator (Propositions 4.3 and 4.4 of Spes ).

##  4 Stability of cohomologies

###  4.1 A submersion related to a uniform map

Let $(X,g)$ and $(Y,h)$ two manifolds of bounded geometry with unbounded ends.
Fix $f:(X,g)\longrightarrow(Y,h)$ which is a uniform map isometric on the
ends.

######  Proposition 4.1.

Assume the existence of a close subset $C$ on $X$ and a number $R$ such that
$f_{X\setminus C}$, in local normal coordinates defined on a ball of radius at
most $r_{i}$, the map $f$ has uniformly bounded derivatives of each order.
Then, for each $\epsilon>0$ there is a smooth map
$f_{\epsilon}:(X,g)\longrightarrow(Y,h)$ such that $f_{\epsilon}\sim f$ and
$f_{\epsilon}=f$ on $X\setminus B_{\epsilon}(C)$.

Proof: This is essentially the same proof given in Proposition 1.7 and in
Corollary 1.8 of Spes . Observe, indeed, that since $f$ is a quasi-isometry on
the ends, then $C\subset M$. Then, since the proof of Proposition 1.7 works
locally, we can assume that $C$ is a subset of a manifold of bounded geometry
and we can consider $f$ as a $C^{0}_{b}$-map. Fix $f:(X=M\cup
E_{X},g)\longrightarrow(Y=N\cup E_{Y},h)$ be a map which is isometric on
unbounded ends. Let $r_{0}$ be a number such that the restriction of $f$ to a
$r_{0}$-collar of $\partial M$ is an isometry. Fix $\delta=\frac{1}{4}r_{0}$
and denote by $\pi:f^{*}T^{\delta}Y\longrightarrow X$ be the fiber bundle
given by the vectors on $f^{*}T^{\delta}Y$ whose norm is less or equal to
$\delta$. Consider the map

$\begin{split}p_{f}:dom(p_{f})\subset&f^{*}T^{\delta}Y\longrightarrow Y\\\
v_{f(p)}&\longrightarrow exp_{f(p)}(v_{f(p)}).\end{split}$ (72)

If $f$ is not a smooth map, then we will use an $\epsilon$-approximation
$f_{\epsilon}$ of it.
Notice that outside $\pi^{-1}(M)\subseteq f^{*}T^{\delta}Y$ the map $p_{f}$
could be not defined on all $f^{*}T^{\delta}Y$ since $Y$ is not complete.
However it is defined on a neighborhood of the $0$-section.
Since Lemma 3.3 of Spes  we know that, on its domain, $p_{f}$ is a smooth
submersion, such that $p_{f}(0_{f(p)})=f(p)$ and that if $f$ has on $M$
uniformly bounded derivatives with respect to some coordinates $\\{x^{i}\\}$
on $X$ and $\\{y^{j}\\}$ on $Y$, the same happens to $p_{f}$ with respect to
the fibered coordinates $\\{x^{i},\mu^{j}\\}$ referred to the frame
$\\{\frac{\partial}{\partial x^{i}}\\}$. This implies that on

$A_{f}:=\\{(p,w_{f(p})\in f^{*}T^{\delta}Y\mbox{ such that }f(p)\subset
N\mbox{ and }|w_{f(p)}|_{h}\leq d(f(p),\partial N)\\}$ (73)

the map $p_{f}$ is Lipschitz. This can be proved exactly as point 5 of Lemma
3.5 of Spes  is proved: it is a consequence of Lemma 3.4 of the work of Schick
Schick .
Let $\psi:\mathbb{R}\longrightarrow\mathbb{R}$ be the map defined as follows:

$\psi(t):=\begin{cases}0\mbox{ if }t\leq 0\\\ 1\mbox{ if }t\geq\delta\\\
P(\frac{t}{\delta})\mbox{ if }t\in(0,\delta)\end{cases}$ (74)

where $P(s):=-20s^{7}+70s^{6}-84s^{5}+35s^{4}$. Thanks to Lemma 2.34 of the
work of Eldering bound , we can fix $\epsilon<\frac{\delta}{3}$ and
$\nu\leq\frac{r_{0}}{4}$ and to find a map
$\psi_{\epsilon}:\mathbb{R}\longrightarrow\mathbb{R}$ such that

  * •

$\psi=\psi_{\epsilon}$ on $\mathbb{R}\setminus(\delta-\nu,\delta+\nu)$,

  * •

$d(\psi_{\epsilon}(t),\psi(t))\leq\epsilon$ for each $t$,

  * •

$\psi$ is a smooth function on $\mathbb{R}\setminus\\{0\\}$,

  * •

$\psi_{\epsilon}$ is a map of class $C^{3}$ on $\mathbb{R}$.

Then we can define the map $\phi:N\longrightarrow\mathbb{R}$ as

$\phi(q):=\begin{cases}0\mbox{ if }q\notin N\\\ 1\mbox{ if }q\in
N\setminus\mathcal{N}_{X}(r_{0})\\\ \psi_{\epsilon}(d(q,\partial N))\mbox{ if
}q\in\mathcal{N}_{X}(r_{0})\end{cases}$ (75)

By some easy computations, it is possible to notice that $\phi$ is a Lipschitz
function in $C^{3}(N)$ and it is smooth outside $\partial N$. Indeed, because
of Remark 4, if we fix some collar coordinates $(x,t)$ on
$\mathcal{N}_{Y}(r_{0})$, then $\phi$ only depends on $t$.
Let $\tilde{p}_{f}:f^{*}T^{\delta}Y\longrightarrow Y$ be the map defined as

$\tilde{p}_{f}(v_{f(p)}):=p_{f}(\phi(f(p))\cdot v_{f(p)}).$ (76)

Observe that $\tilde{p}_{f}$ is defined on all $f^{*}T^{\delta}Y$ and it is a
submersion. Indeed, on $M$, $\tilde{p}_{f}$ is a submersion because $\phi\neq
0$ and $p_{f}$ is a submersion; on $X\setminus M$, $\tilde{p}_{f}$ is a
submersion since $f$ is an isometry.

######  Proposition 4.2.

Let $f:(X,g)\longrightarrow(N,h)$ be a uniform map isometric on the unbounded
ends. Let $g_{S,f}$ be the Sasaki metric induced by $g$, $f^{*}h$ and
$f^{*}\nabla^{LC}_{h}$ on $f^{*}TN$. The map
$\tilde{p}_{f}:(f^{*}T^{\delta},g_{S,f})\longrightarrow(Y,h)$ is a
R.-N.-Lipschitz map.

Proof: Observe that $\tilde{p}_{f}$ is a Lipschitz map since $p_{f}$ is
Lipschitz on the subset $A_{f}$ defined in 73, since $f$ is an isometry on the
unbounded ends and since $\phi$ is Lipschitz. We have to show that the Fiber
Volume is uniformly bounded. Observe that if $q\in Y\setminus N$, then
$p_{f}^{-1}(q)$ is just the fiber over $p_{0}=f^{-1}(q)$ of
$f^{*}T^{\delta}N$. Then, since $f$ is an isometry on the ends, the Fiber
Volume is equal to the volume of a ball of radius $\delta$ in
$\mathbb{R}^{n}$.
Fix $q$ in $N$ such that $B_{\delta}(q)\cap\mathcal{N}_{Y}(\delta)=\emptyset$.
Observe that for each $p_{0}$ in $X$, since $\phi\leq 1$,

$d(f(p_{0}),\tilde{p}_{f}(w_{f(p_{0})}))\leq\delta\cdot
f^{*}\phi(p_{0})\leq\delta.$ (77)

This means that,

$\tilde{p}_{f}^{-1}(q)\subset\pi^{-1}f^{-1}(B_{\delta}(q))$ (78)

Then, since $B_{\delta}(q)\cap\mathcal{N}_{Y}(r_{0})=\emptyset$, we obtain
that $f^{*}\phi$ on $f^{-1}(B_{\delta}(q))$ is equal to $1$. Moreover, since
$\pi(\tilde{p}_{f}^{-1}(q))$ is contained in $f^{-1}(B_{\delta}(q))$, and

$\emptyset=f^{-1}(B_{\delta}(q))\cap
f^{-1}(\mathcal{N}_{Y}(r_{0}))=f^{-1}(B_{\delta}(q))\cap\mathcal{N}_{X}(r_{0})$
(79)

this means that $\tilde{p}_{f}$, on the fiber of $q$, is equal to the
submersion $p_{f}$ defined in Spes . So the Fiber Volume of $q$ can be
computed exactly as in Corollary 4.2 of Spes  and it is bounded by a constant
which only depends on the curvatures of $M$ and $N$.

Finally we have to calculate the Fiber Volume of a point $q$ in $N$ such that
$B_{\delta}(q)\cap\mathcal{N}_{Y}(\delta)\neq\emptyset$. Since
$\delta=\frac{r_{0}}{4}$ we obtain that $B_{\delta}(q)\cap
N\subset\mathcal{N}_{Y}(r_{0})$,
Since $\tilde{p}_{f}=f\circ\pi$ outside $M$ and since $f_{|E_{X}}$ is an
isometry, then $\tilde{p}_{f}^{-1}(q)$ is contained in
$\pi^{-1}(\mathcal{N}_{X}(r_{0}))$. Let us define the map

$\begin{split}\tilde{t}_{f}:\pi^{-1}(\mathcal{N}_{X}(r_{0}))&\longrightarrow
M\times N\\\
w_{f(x,t)}&\longrightarrow(x,t,\tilde{p}_{f}(w_{f(x,t)}))\end{split}$ (80)

Observe that $\tilde{p}_{f}=pr_{N}\circ\tilde{t}_{f}$ where $pr_{N}:M\times
N\longrightarrow N$ is the projection on the second component.
Notice that, since Proposition 3.3, we obtain that

$Vol_{\tilde{p}_{f}}(q)=\int_{X}Vol_{\tilde{t}_{f}}(p,q)d\mu_{X}.$ (81)

Let us focus on $Vol_{\tilde{t}_{f}}$. Fix some normal coordinates
$\\{V,y^{i}\\}$ around $q$ and some collar coordinates $\\{U,x^{j},t\\}$
around $p_{0}=(x_{0},t_{0})=f^{-1}(q)$. Assume that $f(U)\subseteq V$. Then on
$\pi^{-1}(U)\subset f^{*}T^{\delta}Y$ we have the fibered coordinates
$\\{x^{i},t,\mu^{j}\\}$ where $\\{\mu^{j}\\}$ are the coordinates relative to
the frame $\\{\frac{\partial}{\partial y^{i}}\\}$. Observe that

$\tilde{t}_{f}(x_{0},t_{0},\mu^{j})=(x_{0},t_{0},\psi_{\epsilon}(t_{0})\cdot\mu^{j}),$
(82)

where $\psi_{\epsilon}:\mathbb{R}\longrightarrow\mathbb{R}$ is the pa we used
to define $\phi$.
Moreover, since the Christoffell symbols of the pullback connection
$f^{*}\nabla^{LC}_{h}$ vanish in $(x_{0},t_{0})$, then the volume form of
$f^{*}T^{\delta}Y$ in $(x_{0},t_{0},\mu^{j})$ is given by

$Vol_{f^{*}T^{\delta}Y}(x_{0},t_{0},\mu^{j})=det(g_{ij}(x_{0},t_{0}))dx^{I}\wedge
dt\wedge d\mu^{J}.$ (83)

Observe that $det(g_{ij}(x_{0},t_{0}))$ is uniformly bounded since $M$ is an
open of bounded geometry. Let us study the Volume form on $M\times N$: this is
given by

$Vol_{M\times N}(x,t,y)=det(g_{ij}(x,t))\cdot det(h_{rs}(y))dx^{I}\wedge
dt\wedge dy^{J}.$ (84)

Observe that $\tilde{t}_{f}$ is a diffeomorphism with its image. This means
that its Fiber Volume is null outside the image of $\tilde{t}_{f}$. Moreover,
on $im(\tilde{t}_{f})$, the Fiber Volume is given by

$[\tilde{t}_{f}^{-1}]^{*}(\frac{Vol_{f^{*}T^{\delta}Y}}{\tilde{t}_{f}^{*}(Vol_{M\times
N})})$ (85)

Notice that, in $(x_{0},t_{0},q)$, we have

$\frac{Vol_{f^{*}T^{\delta}Y}}{\tilde{t}_{f}^{*}(Vol_{M\times
N})}(x_{0},t_{0},\mu^{j})=[det(h_{rs}(\psi_{\epsilon}(t_{0})\cdot
y))]^{-1}\frac{1}{\psi_{\epsilon}(t_{0})^{n}}.$ (86)

This means that, for each $(x,t,q)$ in $im(\tilde{t}_{f})$,

$Vol_{\tilde{t}_{f}}(x,t,q)\leq C\cdot\frac{1}{\psi_{\epsilon}(t)^{n}}.$ (87)

Let us decompose $M$ as the union of $\mathcal{N}_{X}(\frac{3}{10}\delta)$ and
$M_{0}:=M\setminus\mathcal{N}_{X}(\frac{3}{10}\delta)$. We are choosing
$\frac{3}{10}\delta$ because on $\mathcal{N}_{X}(\frac{3}{10}\delta)$ the
inequality $\phi(x,t)=\psi_{\epsilon}(t)\leq\frac{1}{2}$ holds. This
inequality will be useful later.
However we can see the Fiber Volume of $\tilde{p}_{f}$ as the integral

$Vol_{\tilde{p}_{f}}(q)=\int_{\mathcal{N}_{X}(\frac{3}{10}\delta)}Vol_{\tilde{t}_{f}}(p,q)d\mu_{M}+\int_{M_{0}}Vol_{\tilde{t}_{f}}(p,q)d\mu_{M}.$
(88)

Notice that, thanks to (87), on $M_{0}$ the Fiber Volume of $\tilde{t}_{f}$ is
uniformly bounded by $C_{\psi}$ since $\psi_{\epsilon}(t)$ is bounded from
below by $\phi(x,\frac{3}{10}\delta)=\psi_{\epsilon}(\frac{3}{10}\delta)$.
Moreover, since $f$ is uniformly proper, then also $\pi\circ f$ and
$\tilde{p}_{f}$ are uniformly proper and so this means that there is a number
$C$ (which does not depends on $q$) such that

$diam(\pi(\tilde{p}_{f}^{-1}(q)))\leq C.$ (89)

As a consequence of this and of the bounded geometry of $M$ we obtain a
constant $K$ such that

$\mu_{M}(\pi(\tilde{p}_{f}^{-1}(q))\cap M_{0})\leq K.$ (90)

Finally, observe that
$im(\tilde{t}_{f})\cap[M\times\\{q\\}]=\pi(\tilde{p}_{f}^{-1}(q))\times\\{q\\}$
and we obtain that

$\int_{M_{0}}Vol_{\tilde{t}_{f}}(p,q)d\mu_{M}\leq\int_{\pi(\tilde{p}_{f}^{-1}(q))}Vol_{\tilde{t}_{f}}(p,q)d\mu_{M}\leq
K\cdot C_{\psi}.$ (91)

We only have to prove the boundedness of the integral over
$\mathcal{N}_{X}(\frac{3}{10}\delta).$
Recall that if $(x,t,q)$ is not in $im(\tilde{t}_{f})$, then Fiber Volume of
$Vol_{\tilde{t}_{f}}(x,t,q)=0.$ Observe that $im(\tilde{t}_{f})$ is given by
the elements $(x,t,y)$ of $M\times N$ such that there is a tangent vector
$w_{f(x,t)}$ on $f(x,t)$ whose norm is less or equal to $\delta\cdot\phi(t)$
and which satisfies

$\tilde{p}_{f}(w_{f(x,t)})=exp_{f(x,t)}(w_{f(x,t)})=q=f(x_{0},t_{0}).$ (92)

This means that, if $(x,t,y)$ is in $im(\tilde{t}_{f})$, then

$d_{M}((x_{0},t_{0}),(x,t))=d_{N}(f(x_{0},t_{0}),f(x,t))=d_{N}(q,f(x,t))\leq||w_{f(x,t)}||\leq\delta\cdot\psi_{\epsilon}(t).$
(93)

Observe that for each $(x,t)$ in $\mathcal{N}_{X}(r_{0})$

$|t-t_{0}|=d_{M}((x_{0},t_{0}),(x_{0},t))\leq d_{M}((x_{0},t_{0}),(x,t))$ (94)

indeed, if $(x_{1},t)$ is a point such that
$d_{M}((x_{0},t_{0}),(x_{1},t))<|t-t_{0}|$ then $d((x_{1},t),\partial M)<t$
which is a contradiction with Lemma 1.1. Moreover, on
$\mathcal{N}_{X}(\frac{3}{10}\delta)$, we also have that
$\psi_{\epsilon}(t)\leq\frac{1}{2}t$ for each $(x,t)$. Then, because of the
inequality (93), if $(x,t,y)$ is in
$im(\tilde{t}_{f})\cap\mathcal{N}_{X}(\frac{3}{10}\delta)\times N$, then

$|t-t_{0}|\leq\delta\cdot\psi_{\epsilon}(t)\leq\frac{\delta}{2}t\implies\frac{2}{2+\delta}t_{0}\leq
t\leq\frac{2}{2-\delta}t_{0}$ (95)

and so, since $\psi_{\epsilon}$ is a monotone polynomial of degree $7$, we
obtain that

$d_{M}((x_{0},t_{0}),(x,t))\leq\delta\cdot\psi_{\epsilon}(t)\leq\delta\cdot\psi_{\epsilon}(\frac{2}{2-\delta}t_{0})\leq\delta\cdot\frac{2^{7}}{(2-\delta)^{7}}\cdot\psi_{\epsilon}(t_{0}).$
(96)

This means that $\pi(\tilde{p}_{f}^{-1}(q))$ is contained on a ball on $M$ of
radius $\delta\cdot\frac{2^{7}}{(2-\delta)^{7}}\cdot\psi_{\epsilon}(t_{0})$
where $t_{0}=d(q,\partial N)$ and so

$\mu_{M}(\pi(\tilde{p}_{f}^{-1}(q)))\leq
J_{1}\cdot\psi_{\epsilon}(t_{0})^{n}.$ (97)

Moreover, if $(x,t,y)$ is an element of
$im(\tilde{t}_{f})\cap\mathcal{N}_{X}(\frac{3}{10}\delta)\times N$, thanks to
formula (95) we obtain

$\frac{1}{\psi_{\epsilon}(t)^{n}}\leq\frac{1}{\psi_{\epsilon}(\frac{2}{2+\delta}t_{0})^{n}}=\frac{(2+\delta)^{7n}}{2^{7n}}\cdot\frac{1}{\psi_{\epsilon}(t_{0})^{n}}\leq
J_{2}\cdot\frac{1}{\psi_{\epsilon}(t_{0})^{n}}.$ (98)

So we conclude as follows: given $q=(y_{0},t_{0})$ in $N$ we obtain

$\int_{\mathcal{N}_{X}(\frac{3}{10}\delta)}Vol_{\tilde{t}_{f}}(p,q)d\mu_{M}\leq
J_{1}\cdot\phi(t_{0})^{n}\cdot J_{2}\cdot\frac{1}{\phi(t_{0})^{n}}\leq B.$
(99)

###  4.2 The pullback operator

Let $\omega$ be the Mathai-Quillen-Thom form of $f^{*}TY$ with support
contained in $f^{*}T^{\delta}Y$ defined in subsection 3.3. Then, for each
uniform map isometric on the unbounded ends, we can define the pullback
operator

$T_{f}(\alpha):=\int_{B^{\delta}}\tilde{p}_{f}^{*}\alpha\wedge\omega.$ (100)

######  Proposition 4.3.

The operator $T_{f}$ satisfies the following properties:

  1. 1.

$T_{f}$ is $\mathcal{L}^{2}$-bounded,

  2. 2.

$T_{f}(dom(d_{min}))\subset dom(d_{min})$ and $T_{f}(dom(d_{max}))\subset
dom(d_{max})$,

  3. 3.

$T_{f}$ and $d$ commute.

Proof: Let $\chi_{N}$ and $\chi_{Y\setminus N}$ be the characteristic
functions relative to $M$ and to $X\setminus M$. Observe that if $\alpha$ is a
$\mathcal{L}^{2}$-form, then $\chi_{N}\alpha$ and $\chi_{Y\setminus N}\alpha$
are $\mathcal{L}^{2}$-forms and

$||\alpha||^{2}=||\chi_{N}\alpha||^{2}+||\chi_{Y\setminus N}\alpha||.$ (101)

Observe that

$T_{f}(\chi_{Y\setminus N}\alpha)=\chi_{X\setminus M}f^{*}\alpha$ (102)

and so, since $f$ on $X\setminus M$ is an isometry, we have that

$||T_{f}(\chi_{Y\setminus N}\alpha)||^{2}=||\chi_{Y\setminus N}\alpha||^{2}.$
(103)

Let us focus on $\chi_{N}\alpha$. Observe that

$T_{f}(\chi_{N}\alpha)=pr_{M\star}\circ
e_{\omega}\circ\tilde{p}_{f}^{*}(\chi_{N}\alpha)$ (104)

where $e_{\omega}(\beta):=\beta\wedge\omega$ and $pr_{M\star}$ is the
integration along the fiber of $f^{*}(T^{\delta}N)$.
We already know that $\tilde{p}_{f}^{*}$ is a R.-N.-Lipschitz map and so this
means that $\tilde{p}_{f}^{*}$ is a $\mathcal{L}^{2}$-bounded operator.
Moreover, since Proposition 3.4. of Spes , we also know that $pr_{M\star}$ is
$\mathcal{L}^{2}$-bounded. The main problem is $e_{\omega}$: indeed, since the
norm $|\omega|_{p}$ could be not uniformly bounded in $p$ in $X$, the operator
$e_{\omega}$ can be not $\mathcal{L}^{2}$-bounded. However, observe that on
$\pi^{-1}(M)$ the norm $|\omega|_{p}$ actually is uniformly bounded since $M$
and $N$ have bounded geometry. Then, as a consequence of Proposition 4.4, if
we restrict
$e_{\omega}:\mathcal{L}^{2}(\overline{M})\longrightarrow\mathcal{L}^{2}(\overline{M})$,
we obtain a $\mathcal{L}^{2}$-bounded operator. Moreover we also have that
$\tilde{p}_{f}^{*}(\chi_{N}\alpha)$ has support contained in
$\pi^{-1}(\overline{M})$ and so this means that

$||T_{f}(\chi_{N}\alpha)||^{2}\leq C\cdot||\chi_{N}\alpha||^{2}.$ (105)

Observe that $T_{f}(\chi_{N}\alpha)=0$ on $X\setminus M$ and
$T_{f}(\chi_{Y\setminus N}\alpha)=0$ on $M$. Then we obtain

$\begin{split}||T_{f}\alpha||^{2}&=||T_{f}\chi_{N}\alpha||^{2}+||T_{f}\chi_{Y\setminus
N}\alpha||^{2}\\\ &\leq K\cdot||\chi_{N}\alpha||^{2}+||\chi_{Y\setminus
N}\alpha||^{2}\\\
&\leq\mbox{max}(K,1)(||\chi_{N}\alpha||^{2}+||\chi_{Y\setminus
N}\alpha||^{2})\\\ &=\mbox{max}(K,1)||\alpha||^{2}.\end{split}$ (106)

So we proved 1.
Then we will focus on proving that
$T_{f}(dom(d_{maxY}))\subseteq(dom(d_{maxX}))$.
Notice that, since $\tilde{p}_{f}$ is a map of class $C^{3}$, the pullback
along $\tilde{p}_{f}:f^{*}T^{\delta}N\longrightarrow N$ sends $\Omega^{*}(N)$
on $C^{2}(\Lambda^{*}(f^{*}T^{\delta}N))$, which is the space of differential
forms of class $C^{2}$ over $f^{*}T^{\delta}N$. Observe that on this space the
exterior derivative of a form $\alpha$ is defined exactly as it is defined for
smooth forms (pag. 549 of lang ) and all the properties of $d$ are the same.
In particular $\tilde{p}^{*}_{f}$ and the exterior derivative operator
commute.
The operator $e_{\omega}(\alpha):=\alpha\wedge\omega$ also commute with the
exterior differential and sends $C^{2}(\Lambda^{*}(f^{*}T^{\delta}N))$ in
itself. Finally, if the support of $\alpha$ is vertically compact, the
integration along the fibers $\pi_{\star}$ of $f^{*}T^{\delta}N$ also
satisfies $\pi_{\star}d\alpha=d\pi_{\star}\alpha$ (the proof is the same of
the classical one: see for example Proposition 6.14.1. of bottu ) and,
moreover, if $\alpha$ is a $C^{k}$-form, the same holds for
$\pi_{\star}\alpha$.
So we obtain that if $\alpha$ is a smooth form on $Y$, then $T_{f}\alpha$ is a
$C^{2}$-form on $X$ and $T_{f}d\alpha=dT_{f}\alpha$. Moreover, since
$\tilde{p}_{f}$ is uniformly proper, if $\alpha$ is an element of
$\Omega^{*}_{c}(N)$, then $T_{f}\alpha$ is compactly supported. Finally also
observe that outside $\partial M$, which has null measure on $X$,
$T_{f}\alpha$ is a smooth differential form.
Observe that $T_{f}^{\dagger}=\tilde{p}_{f\star}\circ r_{\omega}\circ\pi^{*}$,
where $\tilde{p}_{f\star}$ is the integration along the fibers of
$\tilde{p}_{f}$ and $r_{\omega}(\beta)=\omega\wedge\beta$. Notice that since
$\tilde{p}_{f}$ is a $C^{3}$ map, then for each $q$ in $N$ and for each $p$ in
$f^{*}T^{\delta}N$ there are a couple of $C^{3}$-charts
$(U,x^{1},\ldots.,x^{n},y^{1},\ldots,y^{n})$ and $(V,y^{1},\ldots,y^{n})$ such
that
$\tilde{p}_{f}(x^{1},\ldots.,x^{n},y^{1},\ldots,y^{n})=(y^{1},\ldots,y^{n})$.
This implies, by a partition of unity argument, that the integral along the
fibers of $\tilde{p}_{f}$ of a smooth form is $C^{3}$-form. So
$dT_{f}^{\dagger}\gamma$ is a compactly supported $C^{3}$-form. Then we can
conclude by applying Corollary 2.2.

###  4.3 The isomorphisms induced by the pullback

Let $(X,g)$ and $(Y,h)$ be two manifolds of bounded geometry with unbounded
ends and let $f:(M,g)\longrightarrow(N,h)$ be a uniform homotopy equivalence
isometric on the ends. In this section we introduce a couple of operators
$y:\mathcal{L}^{2}(Y)\longrightarrow\mathcal{L}^{2}(Y)$ and
$z:\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$ such that
$y(dom(d_{max})\subseteq dom(d_{max})$, $y(dom(d_{min})\subseteq
dom(d_{min})$, $z(dom(d_{max})\subseteq dom(d_{max})$,
$z(dom(d_{min})\subseteq dom(d_{min})$ and

$1\pm T_{f}^{\dagger}T_{f}=dy+yd\mbox{ and }1\pm T_{f}T_{f}^{\dagger}=dz+zd$
(107)

on the minimal or also in the maximal domain of the exterior derivative
operator. In this formula we have a $+$ if $f$ reverses the orientations on
$-$ otherwise.

In order to define these operators, we need a metric on the bundle
$f^{*}(TY)\oplus f^{*}(TY)$ over $X$. We consider the generalized Sasaki
metric $g_{S}$ induced by $g$, $h$ and $\nabla^{LC}_{h}$. In general, even if
$f_{1}=f_{2}=f$, we will denote the bundle $f^{*}(TY)\oplus f^{*}(TY)$ by
$f^{*}_{1}(TY)\oplus f^{*}_{2}(TY)$. Moreover, given $i=1,2$, we denote by
$pr_{i}:f^{*}_{1}(TY)\oplus f^{*}_{2}(TY)\longrightarrow f^{*}(TY)$ the
projection on the $i$-th component, i.e. $pr_{i}(w_{f_{1}(p)}\oplus
w_{f_{2}(p)}):=w_{f_{i}(p)}$ and so we obtain the maps
$\tilde{p}_{f,i}:=\tilde{p}_{f}\circ pr_{i}$ and $\pi_{i}:=\pi\circ pr_{i}.$
Finally we denote by

$\mathcal{B}:=\\{w_{f_{1}(p)}\oplus w_{f_{2}(p)}\in f^{*}_{1}(TY)\oplus
f^{*}_{2}(TY)s.t.|w_{f_{1}(p)}|_{h}\leq\delta,|w_{f_{2}(p)}|_{h}\leq\delta\\}.$
(108)

######  Lemma 4.4.

Assume that $(X,g)$ and $(Y,h)$ are two Riemannian manifolds and let
$f:(X,g)\longrightarrow(Y,h)$ be a smooth uniform map isometric on the
unbounded ends of $X$. Then there are two $\mathcal{L}^{2}$-bounded operators
$y_{0}:\mathcal{L}^{2}(Y)\longrightarrow\mathcal{L}^{2}(Y)$ and
$z_{0}:\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$ such that they
preserve the minimal and the maximal domains of the exterior derivative
operators

$\tilde{p}_{f,2}^{*}-\tilde{p}_{f,1}^{*}=dy_{0}+y_{0}d$ (109)

and finally

$\pi_{2}^{*}-\pi_{1}^{*}=dz_{0}+z_{0}d$ (110)

on the maximal domains of exterior derivative operators.

Proof: Let
$\int_{0,\mathcal{L}}^{1}:\Omega^{*}(\mathcal{B}\times[0,1])\longrightarrow\Omega^{*}(\mathcal{B})$
be the operator defined in Lemma 4.13 of Spes . It is defined as follows: if
$\alpha$ is a $0$-form with respect to $[0,1]$, then

$\int_{0,\mathcal{L}}^{1}\alpha:=\int_{0,\mathcal{L}}^{1}g(x,t)p^{*}\omega=0$
(111)

and, if $\alpha$ is a $1$-form with respect to $[0,1]$,

$\int_{0,\mathcal{L}}^{1}\alpha:=(\int_{0}^{1}f(x,t)dt)\omega.$ (112)

This is an $\mathcal{L}^{2}$-bounded operator and it sends compactly supported
$C^{1}$-forms in $C^{1}_{c}(\Lambda^{*}(T^{*}\mathcal{B})).$ Moreover, if
$\alpha$ is a $C^{1}$-form on $\mathcal{B}\times[0,1]$, then

$j_{1}^{*}\alpha-
j_{0}^{*}\alpha=d\int_{0,\mathcal{L}}^{1}\alpha+\int_{0,\mathcal{L}}^{1}d\alpha,$
(113)

where $j_{i}:\mathcal{B}\longrightarrow\mathcal{B}\times[0,1]$ is given by
$j_{i}(x):=(x,i).$
Let $\phi_{1}$ and $\phi_{0}$ be two R.-N.-Lipschitz maps which are uniformly
homotopic with a uniformly proper, R.-N.-Lipschitz homotopy $H$. Then, as a
consequence of Corollary 2.2, the operator $\int_{0,\mathcal{L}}^{1}\circ H$
is an $\mathcal{L}^{2}$-bounded operator which preserves the minimal and the
maximal domains of the exterior derivative operators and, on the maximal
domain, we have

$\phi_{1}^{*}-\phi_{0}^{*}=d(\int_{0,\mathcal{L}}^{1}\circ
H)+(\int_{0,\mathcal{L}}^{1}\circ H)d$ (114)

So, in order to conclude the proof it is sufficient to find a couple of
uniformly proper, R.-N.-Lipschitz homotopies.
Let
$A:(\mathcal{B}\times[0,1],g_{S}+dt)\longrightarrow(f^{*}T^{\delta}Y,g_{f})$
be defined as

$A(w_{f(p)}\oplus v_{f(p)},s):=s\cdot w_{f(p)}+(1-s)v_{f(p)}.$ (115)

It is an easy exercise to prove that $A$ is a uniformly proper R.-N.-Lipschitz
map: a proof of this fact can be also found in the work of the author Spes2 .
Then we can conclude by observing that $\tilde{p}_{f}\circ A$ is a homotopy
between $\tilde{p}_{f_{1}}$ and $\tilde{p}_{f_{2}}$ and $\pi\circ A$ is a
homotopy between $\pi_{1}$ and $\pi_{2}$. We also need the following Lemma:
the proof is very similar to the same given in Spes2 , however, for the sake
of completeness, here it is proved.

######  Lemma 4.5.

Let $(Y,g)$ be a manifold of bounded geometry with some possibly unbounded
ends. Fix $\tilde{p}_{id}:T^{\delta}Y\longrightarrow Y$ and let $\omega$ be a
Thom form of the bundle $\pi:TY\longrightarrow Y$, where $\pi(v_{p})=p$, such
that $supp(\omega)\subset T^{\delta}Y$. Then for all $q$ in $N$

$\int_{F_{q}}\omega=1$ (116)

where $F_{q}$ is the fiber of $\tilde{p}_{id}$.

Proof: For all $q$ in $N$ the fiber $F_{q}$ is an oriented compact submanifold
with boundary. The same also holds for $B^{\delta}_{q}$ which is the fiber of
the projection $\pi:T^{\delta}N\longrightarrow N$ defined as $\pi(v_{q}):=q$.
Define $H:T^{\delta}N\times[0,1]\longrightarrow N$ as

$H(v_{p},s)=\tilde{p}_{id}(s\cdot v_{p}).$ (117)

Since $H$ is a proper submersion, the fiber along $H$ given by $F_{H,q}$ is
submanifold of $T^{\delta}N\times[0,1]$. Its boundary, in particular is

$\partial F_{H,q}=B^{\delta}_{q}\times\\{0\\}\sqcup F_{q}\times\\{1\\}\cup A$
(118)

where $A$ is contained in

$S^{\delta}N:=\\{v_{p}\in TN||v_{p}|=\delta\\}.$ (119)

Then, if $\omega$ is a Thom form of $TN$ whose support is contained in
$T^{\delta}N$, then

$0=\int_{F_{H_{q}}}d\omega=d\int_{F_{H_{q}}}\omega+\int_{\partial
F_{H_{q}}}\omega.$ (120)

Observe that $\omega$ is a $k$-form and $dim(F_{H_{q}})=k+1$. Then the first
integral on the right side of 120 is $0$. Moreover, we obtain that $\omega$ is
null on $A$, and so

$\int_{A}\omega=0.$ (121)

Then, by the equality (120),

$0=\mp\int_{B^{\delta}}\omega\pm\int_{F_{q}}\omega.$ (122)

and we conclude.

######  Lemma 4.6.

Let $(X,g)$ and $(Y,h)$ be two manifolds of bounded geometry with unbounded
ends and let $f:(M,g)\longrightarrow(N,h)$ be a uniform homotopy equivalence
isometric on the ends. Then there are a couple of operators
$y:\mathcal{L}^{2}(Y)\longrightarrow\mathcal{L}^{2}(Y)$ and
$z:\mathcal{L}^{2}(X)\longrightarrow\mathcal{L}^{2}(X)$ such that they
preserve the minimal and maximal domain of the exterior derivative operators
and

$1\pm T_{f}^{\dagger}T_{f}=dy+yd\mbox{ and }1\pm T_{f}T_{f}^{\dagger}=dz+zd$
(123)

on the maximal domain of the exterior derivative operator. In this formula
there is a $+$ if $f$ reverses the orientations and $-$ otherwise.

Proof: We will start by $z$. We have the following diagrams.

---
$\textstyle{\mathcal{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{P}_{f,2}}$$\scriptstyle{\tilde{P}_{f,1}}$$\textstyle{f^{*}_{2}(T^{\delta}Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{p}_{f}}$$\textstyle{f^{*}_{1}(T^{\delta}Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{p}_{f}}$$\textstyle{Y}$
(124)

where $\tilde{P}_{f_{1}}$ and $\tilde{P}_{f_{2}}$ are the bundle maps induced
by $p_{f}$ and

---
$\textstyle{\mathcal{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Pi_{2}}$$\scriptstyle{\Pi_{1}}$$\textstyle{f^{*}_{2}(T^{\delta}Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{f^{*}_{1}(T^{\delta}Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{X}$
(125)

where $\Pi_{1}$ and $\Pi_{2}$ are the bundle maps induced by $\pi$.
Fix $\alpha$ and $\beta$ in $\Omega^{k}_{c}(X)$ for some $k$ in $\mathbb{N}$.
Then

$\begin{split}\langle T_{f}T_{f}^{\dagger}\alpha\mathchar
24635\relax\;\beta\rangle_{X}&=\langle T_{f}^{\dagger}\alpha\mathchar
24635\relax\;\tau T_{f}^{\dagger}\tau\beta\rangle_{X}\\\
&=\int_{N}(\int_{F_{2}}\omega_{2}\wedge\pi^{*}_{2}\alpha)\wedge(\int_{F_{1}}\omega_{1}\wedge\pi^{*}_{1}\tau\beta)\\\
&=deg(f)\int_{f^{*}_{1}TN}(\int_{F_{2}}\omega_{2}\wedge\pi^{*}_{2}\alpha)\wedge\omega_{1}\wedge\pi^{*}_{1}\tau\beta\\\
&=deg(f)(-1)^{j(n+j)}\int_{f^{*}_{1}TN}\omega_{1}\wedge\pi^{*}_{1}\tau\beta\wedge(\int_{F_{2}}\omega_{2}\wedge\pi^{*}_{2}\alpha)\\\
&=deg(f)(-1)^{j(n+j)}\int_{\mathcal{B}}\omega_{1}\wedge\pi^{*}_{1}\tau\beta\wedge\omega_{2}\wedge\pi^{*}_{2}\alpha\\\
&=deg(f)\int_{\mathcal{B}}\pi^{*}_{2}\alpha\wedge\pi^{*}_{1}\tau\beta\wedge\omega_{1}\wedge\omega_{2}.\end{split}$
(126)

Then, as a consequence of $\pi_{2}^{*}=\pi_{1}^{*}+dz_{0}+z_{0}d$, we obtain

$\begin{split}\langle T_{f}T_{f}^{\dagger}\alpha\mathchar
24635\relax\;\beta\rangle_{X}&=deg(f)\int_{\mathcal{B}}\pi^{*}_{1}\alpha\wedge\pi^{*}_{1}\tau\beta\wedge\omega_{1}\wedge\omega_{2}\\\
&+deg(f)\int_{\mathcal{B}}(dz_{0}+z_{0}d)\alpha\wedge\pi^{*}_{1}\tau\beta\wedge\omega_{1}\wedge\omega_{2}\\\
&=\langle deg(f)\cdot 1\alpha\mathchar
24635\relax\;\beta\rangle_{X}+\langle[d(-z)+(-z)d]\alpha\mathchar
24635\relax\;\beta\rangle_{X}\end{split}$ (127)

where $z:=-deg(f)\cdot\pi_{\star}\circ e_{\omega}\circ\Pi_{2\star}\circ
e_{\Pi_{2}^{*}\omega}\circ z_{0}.$
Observe that $z$ is a composition of $\mathcal{L}^{2}$-bounded operators and
$z(\Omega^{*}_{c}(X))\subseteq C^{2}_{c}(\Lambda^{*}(T^{*}X))$. Moreover we
also have that

$z^{\dagger}=-deg(f)z_{0}^{\dagger}\circ
r_{\Pi_{2}^{*}\omega}\circ\Pi_{2}^{*}\circ r_{\omega}\circ\pi^{*}$ (128)

is an $\mathcal{L}^{2}$-bounded operators and
$-z^{\dagger}(\Omega^{*}_{c}(Y))\subseteq C^{2}_{c}(\Lambda^{*}(T^{*}Y))$. So
by Corollary 2.2, we have that $z$ preserves the minimal and the maximal
domain of $d$ and

$1\pm T_{f}T_{f}^{\dagger}=dz+zd$ (129)

where we have a $-$ if $f$ preserves the orientations and a $+$ otherwise.

Let us focus on the operator $y$. Given $\alpha$ and $\beta$ in
$\Omega^{*}_{c}(Y)$, we obtain that

$\begin{split}\langle T^{\dagger}_{f}T_{f}\alpha,\beta\rangle&=\langle
T_{f}\alpha,\tau T_{f}\tau\beta\rangle\\\
&=\int_{M}(\int_{B^{\delta}_{2}}\tilde{p}_{f,2}^{*}\alpha\wedge\omega_{2})\wedge(\int_{B^{\delta}_{1}}\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1})\\\
&=\int_{f^{*}(T^{\delta}Y)_{1}}(\int_{B^{\delta}_{2}}\tilde{p}_{f,2}^{*}\alpha\wedge\omega_{2})\wedge\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\\\
&=(-1)^{(n+j)j}\int_{f^{*}(T^{\delta}Y)_{1}}\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\wedge(\int_{B^{\delta}_{2}}\tilde{p}_{f,2}^{*}\alpha\wedge\omega_{2})\\\
&=(-1)^{(n+j)j}\int_{\mathcal{B}}\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\wedge\tilde{p}_{f,2}^{*}\alpha\wedge\omega_{2}\\\
&=(-1)^{(n+j)j}(-1)^{(n+j)j}\int_{\mathcal{B}}\tilde{p}_{f,2}^{*}\alpha\wedge\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\wedge\omega_{2}\\\
&=\int_{\mathcal{B}}\tilde{p}_{f,2}^{*}\alpha\wedge\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\wedge\omega_{2}\end{split}$
(130)

Then, if we consider the operator $y_{0}$ defined before, we obtain that

$\begin{split}\langle
T^{\dagger}_{f}T_{f}\alpha,\beta\rangle&=\int_{\mathcal{B}}\tilde{p}_{f,1}^{*}(\alpha\wedge\tau\beta)\wedge\omega_{1}\wedge\omega_{2}\\\
&=\int_{\mathcal{B}}(dy_{0}+y_{0}d)\alpha\wedge\tilde{p}_{f,1}^{*}\tau\beta\wedge\omega_{1}\wedge\omega_{2}\\\
&=deg(f)\int_{TN}\tilde{p_{id}}^{*}(\alpha\wedge\tau\beta)\wedge\omega\\\
&+\langle(dy+yd)\alpha,\beta\rangle_{Y}\\\ &=\langle deg(f)\cdot
1(\alpha)\mathchar
24635\relax\;\beta\rangle_{Y}+\langle(dy+yd)\alpha,\beta\rangle_{Y}\end{split}$
(131)

where $y:=p_{f,\star}\circ e_{\omega}\circ\Pi_{1,\star}\circ
e_{\Pi^{*}_{1}\omega}\circ y_{0}$. Notice that $y$ is a composition of
$\mathcal{L}^{2}$-bounded operators and $y(\Omega^{*}_{c}(Y))\subseteq
C^{2}_{c}(\Lambda^{*}(T^{*}Y))$. Moreover

$y^{\dagger}=y_{0}^{\dagger}\circ r_{\Pi^{*}_{1}\omega}\circ\Pi_{1}^{*}\circ
r_{\omega}\circ p_{f}^{*}$ (132)

and so $y^{\dagger}(\Omega^{*}_{c}(X))\subseteq
C^{2}_{c}(\Lambda^{*}(T^{*}X))$. So, as a consequence of Corollary 2.2, we
obtain that $y$ preserves the maximal and the minimal domain of $d$ and

$1\pm T^{\dagger}_{f}T_{f}=dy+yd.$ (133)

######  Proposition 4.7.

Let $f:(M,g)\longrightarrow(N,h)$ be a uniform homotopy equivalence quasi-
isometric on the unbounded ends. Then, for each $k$ in $\mathbb{N}$, the
operators $T_{f}$ and $T^{\dagger}_{f}$ induce the following isomorphisms

$\begin{split}H^{k}_{2,max}(M,g)&\cong H^{k}_{2,max}(N,h)\\\
\overline{H}^{k}_{2,max}(M,g)&\cong\overline{H}^{k}_{2,max}(N,h)\\\
H^{k}_{2,min}(M,g)&\cong H^{k}_{2,min}(N,h)\\\
\overline{H}^{k}_{2,min}(M,g)&\cong\overline{H}^{k}_{2,min}(N,h).\end{split}$
(134)

Proof: Thanks to Lemma 1.2, we can assume, without loss of generality, that
$f$ is an isometry on th unbounded ends. Then the proof immediately follows by
Lemma 4.6.

######  Remark 15.

The morphisms induced by $T_{f}$ in maximal and minimal
$\mathcal{L}^{2}$-cohomology, do not depend on the choice of the smooth
$\varepsilon$-approximation $f_{\varepsilon}.$ The proof of this fact is
essentially the same given in points 2 and 3 of Proposition 4.17 of Spes .

##  5 Consequences

###  5.1 Mapping cone

In this subsection, given a Riemannian manifold $(M,g)$, we will denote by
$\Omega^{*}_{m\backslash\mathcal{M}}(M,g):=dom(d_{M,min\backslash max})$ and
by $d_{M,m\backslash\mathcal{M}}:=d_{M,min\backslash max}$.
Let $f:(M,g)\longrightarrow(N,h)$ a uniform map quasi-isometric on the
unbounded ends. It is not required that $f$ is a uniform homotopy equivalence.
Thanks to Proposition 4.3, we know that there is a $\mathcal{L}^{2}$-bounded
operator $T_{f}:\mathcal{L}^{2}(N,h)\longrightarrow\mathcal{L}^{2}(M,g)$ such
that
$T_{f}(\Omega^{*}_{m\backslash\mathcal{M}}(M,g))\subseteq\Omega^{*}_{m\backslash\mathcal{M}}(N,h)$.
In this subsection we will define the $L^{2}$-mapping cone of a map $f$ and we
will see some properties of this cone.

######  Definition 5.1.

Let $f:(M,g)\longrightarrow(N,h)$ be a uniform map quasi-isometric on the
unbounded ends between two Riemannian manifold and let us denote by
$(\Omega^{*}_{2,min\backslash max}(f),d_{f,min\backslash max})$ the cochain
complexes

$0\rightarrow\Omega^{0}_{2,m\backslash\mathcal{M}}(f)\xrightarrow{d_{f,m\backslash\mathcal{M},0}}\Omega^{1}_{2,m\backslash\mathcal{M}}(f)\xrightarrow{d_{f,m\backslash\mathcal{M},1}}\Omega^{3}_{2,m\backslash\mathcal{M}}(f)\xrightarrow{d_{f,m\backslash\mathcal{M},2}}\ldots$
(135)

where
$\Omega^{*}_{2,m\backslash\mathcal{M}}(f):=\Omega^{*}_{m\backslash\mathcal{M}}(N,h)\oplus\Omega^{*-1}_{m\backslash\mathcal{M}}(M,g)$
and
$d_{f,m\backslash\mathcal{M}}(\alpha,\beta):=(-d_{N,m\backslash\mathcal{M}}\alpha\mathchar
24635\relax\;T_{f}\alpha-d_{M,m\backslash\mathcal{M}}\beta).$

######  Definition 5.2.

The $k$-th group of the $L^{2}$-mapping cone of $f$ is the cohomology group of
the $L^{2}$-mapping cone, i.e.

$H^{k}_{2,m\backslash\mathcal{M}}(f):=\frac{ker(d_{f,m\backslash\mathcal{M},k})}{im(d_{f,m\backslash\mathcal{M},k})}.$
(136)

The reduced $k$-th group of the $L^{2}$-mapping cone of $f$ is the group
defined as

$\overline{H}^{k}_{2,m\backslash\mathcal{M}}(f):=\frac{ker(d_{f,m\backslash\mathcal{M},k})}{\overline{im(d_{f,m\backslash\mathcal{M},k})}}.$
(137)

Exactly as the mapping cone in the de Rham case, we have a short exact
sequence

$0\rightarrow\Omega^{*-1}_{2,m\backslash\mathcal{M}}(M,g)\xrightarrow{A}\Omega^{*}_{2,m\backslash\mathcal{M}}(f)\xrightarrow{B}\Omega^{*}_{2,m\backslash\mathcal{M}}(N,h)\rightarrow
0,$ (138)

where $A(\omega):=(0,\omega)$ and $B(\alpha,\omega):=\alpha$. This sequence
induces a long exact sequence on cohomology, and so

$0\rightarrow H^{0}_{2,m\backslash\mathcal{M}}(f)\rightarrow
H^{0}_{2,m\backslash\mathcal{M}}(M,g)\xrightarrow{\delta}H^{0}_{2,m\backslash\mathcal{M}}(N,h)\rightarrow
H^{1}_{2,m\backslash\mathcal{M}}(f)\rightarrow\ldots$ (139)

where $\delta$ is the connecting homomorphism.
Following the same proof given at pag. 78 of the book of Bott and Tu bottu ,
we obtain that $\delta[\omega]:=[T_{f}\omega]$. So we obtain the following
Proposition.

######  Proposition 5.1.

Let $f:(M,g)\longrightarrow(N,h)$ be a uniform map quasi-isometric on the
unbounded ends of $M$ and $N$. Then, the two following statements are
equivalent:

  1. 1.

the morphism induced by $T_{f}$ on $L^{2}$-cohomology is an isomorphism,

  2. 2.

all the cohomology groups of $\Omega^{*}_{2,m\backslash\mathcal{M}}(f)$ are
null.

Proof: This is a classical proof which holds for each cochain morphism
$T_{f}:A\longrightarrow B$ between cochain complexes on an additive category.

######  Remark 16.

The author used a bounded geometry version of Proposition 5.1 in Spes2  in
order to prove the invariance of the Roe Index of signature operator of a
manifold of bounded geometry under uniform homotopy equivalences which
preserve the orientations. As a consequence of Proposition 5.1, it could be
interesting to prove to generalize this result in a larger setting, for
example in the case of complete Riemannian manifolds with uniform homotopy
equivalence which are quasi-isometric on the unbounded ends.

In the reduced case we also obtain a long sequence

$0\rightarrow\overline{H}^{0}_{2,m\backslash\mathcal{M}}(f)\xrightarrow{B^{*}}\overline{H}^{0}_{2,m\backslash\mathcal{M}}(M,g)\xrightarrow{T_{f}}\overline{H}^{0}_{2,m\backslash\mathcal{M}}(N,h)\xrightarrow{A^{*}}\overline{H}^{1}_{2,m\backslash\mathcal{M}}(f)\rightarrow\ldots$
(140)

but it is not exact this time. Indeed it is exact only on
$\overline{H}^{k}_{2,m\backslash\mathcal{M}}(f)$ and on
$\overline{H}^{k}_{2,m\backslash\mathcal{M}}(M,g)$ while on
$\overline{H}^{k}_{2,m\backslash\mathcal{M}}(N,h)$ is just weakly exact which
means that $ker(T_{f})=\overline{im(B^{*})}.$

###  5.2 Uniform homotopy invariance of signature

Let $(M,g)$ be a complete Riemannian manifold. Recall that in this case there
is only one closure for the operator $d$. This means that the maximal and
minimal $L^{2}$-cohomology groups coincide, both the reduced and the unreduced
ones. In next pages we will denote by $d$ the unique closed extension of the
exterior derivative operator.
In this subsection we introduce the $L^{2}$-signature $\sigma_{M}$ of a
manifold $(M,g)$ with $dim(M)=4k$ such that $\overline{H}^{2k}_{2}(M,g)$ is
finite dimensional. This $L^{2}$-signature is the signature of a pairing
defined on $\overline{H}^{2k}_{2}(M,g)$. Consider the operator $d+d^{*}$. This
operator switches the eigenspaces of the chiral operator $\tau$, so it is
possible to define the $L^{2}$-signature operator $(d+d^{*})^{+}$ as the
restriction of $d+d^{*}$ to the $+1$-eigenspace of $\tau$.
We obtain that $(d+d^{*})^{+}$ is a Fredholm operator and the its index equals
the $L^{2}$-signature. The definitions and the proofs of all these facts can
be found in the work of Bei Bei3 , in particular at pag. 19-21.

######  Proposition 5.2.

Let $f:(M,g)\longrightarrow(N,h)$ be a uniform homotopy equivalence quasi-
isometric on the unbounded ends. In this subsection we will prove that

  1. 1.

$T_{f}$ well-behaves with respect to the pairings on $M$ and $N$, i.e.
$\langle
T_{f}[\alpha],T_{f}\beta]\rangle_{M}=deg(f)\cdot\langle[\alpha],[\beta]\rangle_{N}$,

  2. 2.

from the previous point we obtain that $\sigma_{M}=deg(f)\cdot\sigma_{N}$ and
the index of the $L^{2}$-signature operator is an invariant under uniform
homotopy equivalences quasi-isometric on the unbounded ends which preserve the
orientations.

Proof: In order to prove the first point, we need to introduce the pairing
$\langle\cdot\mathchar 24635\relax\;\cdot\rangle_{M}$. This is defined in pag.
19 of Bei3  as

$\begin{split}\langle\cdot,\cdot\rangle_{M}:&\overline{H}^{i}_{2}(M,g)\times\overline{H}^{i}_{2}(M,g)\longrightarrow\mathbb{R}\\\
&([\eta],[\omega])\longrightarrow\int_{M}\eta\wedge\omega\end{split}$ (141)

Let $id$ be the identity map on $N$ and let
$\tilde{p}_{id}:T^{\delta}N\longrightarrow N$ be the submersion related to
$id$. Because of Lemma 4.6, we know that this is a R.-N.-Lipschitz map and so
$T_{id}$ is a $L^{2}$-bounded operator. Let us denote by
$K:=\int_{0\mathcal{L}}^{1}\circ\tilde{p}_{h}^{*}$, where
$\tilde{p}_{h}:T^{\delta}N\times[0,1]\longrightarrow N$ is defined as
$\tilde{p}_{h}(w_{p},s):=\tilde{p}_{id}(s\cdot w_{p})$. By applying the same
proof of Lemma 4.11 of Spes , we obtain that $\tilde{p}_{h}$ is a
R.-N.-Lipschitz map and so $K$ is an $L^{2}$-bounded operator such that

$T_{id}-Id=dK+Kd$ (142)

for each smooth form $\alpha$. Then, as a consequence of Lemma 4.6, we obtain
that $T_{id}$ and $K$ both preserve the minimal and the maximal domain of $d$.
Observe that the operator $T_{f}=f^{*}\circ T_{id}$: this directly follows by
the definition of $T_{f}$ and by
$\int_{F_{0}^{\prime}}F^{*}\alpha=f^{*}\int_{F_{0}}\alpha$. Here $F^{*}$ is
the bundle map induced by $f$, and $F_{0}$ and $F_{0}^{\prime}$ are the fibers
of $TN$ and of $f^{*}TN$. Observe that $T_{id}\alpha$ is a smooth form outside
a subset of null measure (that is $\partial N$) and the same happens to
$f^{*}T_{id}\alpha$.
Then we can easily conclude as follows

$\begin{split}\langle[T_{f}\alpha],[T_{f}\beta]\rangle_{M}&=\int_{M}T_{f}\alpha\wedge
T_{f}\beta\\\ &=\int_{M}f^{*}(T_{id}\alpha)\wedge
f^{*}(T_{id}\beta)=\int_{M}f^{*}(T_{id}\alpha\wedge T_{id}\beta)\\\
&=deg(f)\cdot\int_{N}T_{id}\alpha\wedge
T_{id}\beta=deg(f)\int_{N}(\alpha+d\eta)\wedge(\beta+d\nu)\\\
&=deg(f)\int_{N}\alpha\wedge\beta+deg(f)\int_{N}d(\alpha\wedge\nu)+\\\
&+deg(f)\int_{N}d(\eta\wedge\beta)+deg(f)\int_{N}d(\eta\wedge d\nu)\\\
&=deg(f)\int_{N}\alpha\wedge\beta+0=deg(f)\langle[\alpha],[\beta]\rangle_{N}.\end{split}$
(143)

We proved point 1. The proof of point 2. is a direct consequence of point 1.
and Theorem 4.2 of Bei3 .

###  5.3 Compact manifolds with unbounded ends

Notice that Proposition 4.7 is coherent with Proposition 5 of Lott in Lott .
In the work of Lott, it is proved that if two complete Riemannian manifolds
$(M,g)$ and $(N,h)$ are isometric outside a compact set, then, for each $k$ in
$\mathbb{N}$,

$dim(\overline{H}^{k}_{2}(M,g))=+\infty\iff
dim(\overline{H}^{k}_{2}2(N,h))=+\infty.$ (144)

Then, as a consequence of Proposition 4.7, we can say something more if we add
some assumptions on $f$ on the compact subsets. On the other hand we are not
assuming the completeness of $(M,g)$ and $(N,h)$ and we also relax the
assumptions on $f$ on the ends.

######  Corollary 5.3.

Let $(M,g)$ and $(N,h)$ be two (possibly not complete) oriented Riemannian
manifolds. Let $K$ be a compact subset of $M$. We denote by
$K^{\prime}:=f(K)$. Assume the existence of a homotopy equivalence
$f:(M,g)\longrightarrow(N,h)$ such that $f_{|M\setminus K}:(M\setminus
K,g)\longrightarrow(N\setminus K^{\prime},h)$ is a quasi-isometry. Then both
the minimal and maximal $L^{2}$-cohomology groups are isomorphic. The same
also happens for the reduced $L^{2}$-cohomology groups.

Proof: Thanks to Proposition 1.3 we can assume that $f_{|K}$ is an isometry.
Fix a number $r>0$ and denote by $B_{r}(K)$ (resp. $B_{r}(K^{\prime})$) the
subset of the points whose distance from $K$ (resp. $K^{\prime}$) is less to
$r$. Let $\delta$ be a number small enough such that $B_{\delta}(K)$ and
$B_{\delta}(K^{\prime})$ are two tubular neighborhoods of $K$ and
$K^{\prime}$. Then $B_{\delta}(K)$ and $B_{\delta}(K^{\prime})$ are two open
of bounded geometry and $f$ is uniform homotopy equivalence isometric on the
unbounded ends. Then we conclude by applying Proposition 4.7.

######  Remark 17.

Probably, this result for unreduced $L^{2}$-cohomology could be also achieved
by using the results of Spes  and a Mayer-Vietoris argument. However, the
difficult part of this approach would be the reduced case. Indeed it is a
well-known fact that, in general, there is no Mayer-Vietoris sequence for the
reduced $L^{2}$-cohomology.

###  5.4 Manifolds not quasi-isometric with same cohomologies

We conclude this paper by showing an easy example of two metrics on
$\mathbb{R}$ which are not quasi-isometric but they have the same minimal and
maximal $L^{2}$-cohomology groups, both in their reduced and unreduced
versions.
Let $\phi:\mathbb{R}\longrightarrow\mathbb{R}_{>0}$ be a smooth map such that

  * •

$\phi(0)=1$ and, for each $k$ in $\mathbb{N}$, the derivative of order $k$ in
$0$ of $\phi$ is $0$,

  * •

outside $[-4,4]$ we have $\phi(t)=e^{t^{2}}$.

So denote by $g_{1}$ the Riemannian metric on $\mathbb{R}$ as
$g_{1}(t):=\phi(t)dt^{2}$. Moreover let
$\psi:\mathbb{R}\longrightarrow\mathbb{R}$ be the function defined as follows:

$\psi(t):=\begin{cases}\phi(t-sign(t)1)\mbox{ if }|t|>1\\\ 1\mbox{
otherwise.}\end{cases}$ (145)

Denote by $g_{2}$ the Riemannian metric defined as $g_{2}(t):=\psi(t)dt^{2}$.
Notice that
$\lim\limits_{t\rightarrow\pm\infty}\frac{\phi(t)}{\psi(t)}=+\infty.$ Then
$g_{1}$ and $g_{2}$ are not quasi-isometric. However, if we define the map
$f:(\mathbb{R},g_{2})\longrightarrow(\mathbb{R},g_{1})$ as

$f(t):=\begin{cases}t-sign(t)1\mbox{ if }|t|>1\\\ 0\mbox{
otherwise.}\end{cases}$ (146)

this is a uniform map isometric on the unbounded ends
$(-\infty,-5]\cup[5,+\infty)$ and it is also a uniform homotopy equivalence.
So, as a consequence of Subsection 5.3, we conclude that minimal and maximal
$L^{2}$-cohomology groups, both in their reduced and unreduced versions, are
isomorphic.

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