\section{Quasi isometries are isometries} \label{Sdrum5}

In this section we prove that two connected Riemannian manifolds,
which are regular in capacity, are isomorphic if they
have isomorphic open subsets whose complements are polar.
Moreover, we give many useful tools to understand and to work with the $H^1_0$-spaces
defined on Riemannian manifolds.

Let $(M_1,g_1)$ and $(M_2,g_2)$ be Riemannian manifolds. 
We say that
\[ M_1 \simh M_2

\]
if there exist open sets $M_1'\subset M_1$ and $M_2'\subset M_2$  and
an isometry $\tau$ from $M_2'$ onto $M_1'$ such that 
$\capp_{M_1}(M_1\setminus M_1') = 0 = \capp_{M_2}(M_2\setminus M_2')$.

\smallskip

The following theorem is the main theorem in this section.
It shows that the relation $\simh$ defined on connected Riemannian manifolds
determines the manifold.


\begin{thm} \label{tdrum324}
Let $(M_1,g_1)$ and $(M_2,g_2)$ be two connected Riemannian manifolds which are regular
in capacity.
Then
\[
M_1\simh M_2 \Longleftrightarrow (M_1,g_1) \mbox{ and } (M_2,g_2) \mbox{ are isomorphic.}
\]
Explicitly, if $M_1'$ and $M_2'$ are open subsets
of $M_1$ and $M_2$ such that 
$\capp_{M_1}(M_1\setminus M_1') = 0 = \capp_{M_2}(M_2\setminus M_2')$
and $\tau \colon M_2' \to M_1'$ is an isometry,
then there exists an isometry 
$\hat \tau \colon M_2 \to M_1$ such that $\hat \tau|_{M_2'} = \tau$.
\end{thm}


We define the space $H^1_c(M)$ to be 
the set of all $\varphi \in H^1(M)$ such that there exists a compact subset 
$K$ of $M$ with $\varphi = 0$ a.e.\ on $M \setminus K$.

\begin{lemma} \label{ldrum310}
Let $N$ be an open subset of a Riemannian manifold $(M,g)$
such that $|M \setminus N| = 0$.
Then the following are equivalent.
\begin{tabel}
\item \label{ldrum310-1}
$\capp_M(M \setminus N) = 0$.
\item \label{ldrum310-2}
The restriction map $\psi \mapsto \psi|_N$ from $H^1(M)$ into $H^1(N)$
maps $H^1_0(M)$ onto $H^1_0(N)$.
\item \label{ldrum310-3}
$H^1_0(M) 
= \{ \psi|_N : \psi \in H^1_0(M) \} 
= H^1_0(N) 
$.
\end{tabel}
\end{lemma}
\proof\
Clearly \ref{ldrum310-3} is a reformulation of \ref{ldrum310-2}.
`\ref{ldrum310-3}$\Rightarrow$\ref{ldrum310-1}'.
Let $K \subset M$ be a compact set.
There exists a $\psi \in C_c^\infty(M)$ such that 
$\psi|_K \geq 1$.
Then $\psi|_N \in H^1_0(N)$ by assumption.
Let $\varepsilon > 0$.
There exists a $\varphi \in C_c^\infty(N)$ such that 
$\|\psi|_N - \varphi\|_{H^1(N)}^2 < \varepsilon$.
Then $\psi - \varphi \in C_c^\infty(M)$ and 
$\psi - \varphi \geq 1$ on $K \setminus N$.
So 
\[
\capp(K \setminus N) 
\leq \|\psi - \varphi\|_{H^1(M)}^2 
= \|\psi|_N - \varphi\|_{H^1(N)}^2 
< \varepsilon
\;\;\; ,  \]
where we used that $|M \setminus N| = 0$ in the 
equality.
Since $M$ is $\sigma$-compact one deduces that 
$\capp(M \setminus N) = 0$.

`\ref{ldrum310-1}$\Rightarrow$\ref{ldrum310-3}'.
Clearly $\{ \psi|_N : \psi \in H^1_0(M) \} \supset H^1_0(N)$.
Conversely, let $\psi \in C_c^\infty(M)$.
Let $\varepsilon > 0$.
There exists an open neighbourhood of 
$M \setminus N$ and a $\chi \in H^1(M)$ such that 
$0 \leq \chi \leq 1$, $\chi|_U = 1$ and 
$\|\chi\|_{H^1(M)} < \varepsilon$.
Then $(\psi (\one - \chi))|_N \in H^1_c(N) \subset H^1_0(N)$ and 
\[
\|\psi|_N - (\psi (\one - \chi))|_N\|_{H^1(N)}
= \|\psi \, \chi\|_{H^1(M)}
\leq 3 \|\psi\|_{W^{1,\infty}(M)} \, \|\chi\|_{H^1(M)}
\leq 3 \|\psi\|_{W^{1,\infty}(M)} \, \varepsilon
\;\;\; .  \]
So $\psi|_N \in H^1_0(N)$.
Since $C_c^\infty(M)$ is dense in $H^1_0(M)$ and 
$\varphi \mapsto \varphi|_N$ is isometric from 
$H^1(M)$ into $H^1(N)$ the lemma follows.\hfill$\Box$

\ruimte

If $N$ is an open subset of a Riemannian manifold $M$
and $A \subset N$, then $\capp_N(A) \leq \capp_M(A)$.
The next lemma is instrumental to deduce equality 
of the two capacities if $\capp_M(M \setminus N) = 0$.
It is a kind of $L_2$-version of Propositions~\ref{pdrum304} and \ref{pdrum305}.

\begin{lemma} \label{ldrum301}
Let $N$ be an open subset of a manifold $M$.
Suppose that $\capp_M(M \setminus N) = 0$.
Then $H^1(N) = H^1(M)$.
\end{lemma}
\proof\
Let $\varphi \in H^1(N) \cap L_\infty(N)$.
We shall prove that $\varphi \in H^1(M)$.
Let $n \in \Ni$.
Since $\capp_M(M \setminus N) = 0$ there exists a $\psi_n \in H^1(M)$ such that 
$\psi_n \geq 1$ in a neighbourhood of $M \setminus N$ and 
$\|\psi_n\|_{H_1(M)}^2 \leq n^{-1}$.
We may assume that $0 \leq \psi_n \leq 1$.
Then $\varphi - \varphi \, \psi_n \in H^1(N)$.
But $\varphi - \varphi \, \psi_n = \varphi (\one - \psi_n)$ vanishes in 
a neighbourhood of $M \setminus N$.
Therefore we can extend this function by zero
to a function $\varphi_n \in H^1(M)$.
Then 
\[
\|\varphi \, \psi_n\|_{H^1(N)}^2
\leq 2 \|\varphi\|_\infty^2 \|\psi_n\|_{H^1(M)}^2 + 2 \|\varphi\|_{H^1(N)}^2 \|\psi_n\|_\infty^2
\leq  2 \|\varphi\|_\infty^2 + 2 \|\varphi\|_{H^1(N)}^2
\]
for all $n \in \Ni$.
So the sequence $\varphi_1,\varphi_2,\ldots$ is uniformly bounded in $H^1(M)$.
Hence it has a weakly convergent subsequence.
Passing to a subsequence if necessary, there exists a $\tilde \varphi \in H^1(M)$
such that $\lim \varphi_n = \tilde \varphi$ weakly in $H^1(M)$.
Then $\lim \varphi_n = \tilde \varphi$ weakly in $L_2(M)$.
But 
\[
\|\varphi - \varphi_n\|_{L_2(M)}
= \|\varphi \, \psi_n\|_{L_2(M)} 
\leq \|\varphi\|_\infty \, \|\psi_n\|_{L_2(M)} 
\leq \|\varphi\|_\infty \, n^{-1/2}
\]
for all $n \in \Ni$.
So $\lim \varphi_n = \varphi$ strongly and therefore also weakly in $L_2(M)$.
Hence $\tilde \varphi = \varphi$ a.e.\ and $\varphi \in H^1(M)$.

Thus $H^1(N) \cap L_\infty(N) \subset H^1(M)$.
Since $H^1(N) \cap L_\infty(N)$ is dense in $H^1(N)$ 
the lemma follows.\hfill$\Box$


\begin{cor} \label{cdrum320}
If $M$ is a Riemannian manifold and $N \subset M$ is open with 
$\capp_M(M \setminus N)=0$ then $\capp_N(A) = \capp_M(A)$ 
for all $A \subset N$.
\end{cor}


\begin{cor} \label{cdrum321}
Let $M_1$ and $M_2$ be two Riemannian manifolds. 
If $M_1\simh M_2$ and $M_1'$, $M_2'$ and $\tau$ are as in the 
definition of $\simh$ then
$\capp_{M_1}(\tau(A)) = \capp_{M_2}(A)$ for every set $A\subset M_2'$.
\end{cor}
\proof\
One deduces from the previous corollary that 
$\capp_{M_1}(\tau(A)) = \capp_{M_1'}(\tau(A)) 
       = \capp_{M_2'}(A)=\capp_{M_2}(A)=0$.\hfill$\Box$

\ruimte

We emphasize that the next proposition does not require the manifolds
to be regular in capacity.


\begin{prop} \label{pdrum322}
The relation $\simh$ is an equivalence relation. 
\end{prop}
\proof\
The reflexivity and symmetry are trivial.

Let $M_1$, $M_2$ and $M_3$ be three Riemannian manifolds and assume that  
$M_1\simh M_2$ and $M_2\simh M_3$.
Then there exist open $M_1' \subset M_1$, $M_2',M_2'' \subset M_2$ and 
$M_3'' \subset M_3$ and isometries
$\tau \colon M_2' \to M_1$ and $\sigma \colon M_3'' \to M_2''$ such that 
$\capp_{M_1}(M_1 \setminus M_1') = \capp_{M_2}(M_2 \setminus M_2') = 
\capp_{M_2}(M_2 \setminus M_2'') = \capp_{M_3}(M_3 \setminus M_3'') = 0$.
Now let $M_2''' = M_2' \cap M_2''$.
Then $M_2'''$ is open in $M_2$ and
\[ 
\capp_{M_2}(M_2\setminus M_2''')
\leq \capp_{M_2}(M_2\setminus M_2')+\capp_{M_2}(M_2\setminus M_2'')=0
\;\;\; .  \]
Next set $M_1''' = \tau(M_2''') \subset M_1'$
and $M_3''' = \sigma^{-1}(M_2''') \subset M_3'$.
Then $M_1'''$ is open in $M_1$ and $M_3'''$ is open in $M_3$.
Moreover, $\tau|_{M_2'''} \circ \sigma|_{M_3'''}$ is an isometry from
$M_3'''$ onto $M_1'''$.
Since $M_1' \setminus M_1''' = \tau(M_2' \setminus M_2''')$ it follows from
Corollary~\ref{cdrum321} that 
$\capp_{M_1} (M_1' \setminus M_1''') = \capp_{M_2}(M_2' \setminus M_2''') = 0$.
Therefore 
$\capp_{M_1}(M_1 \setminus M_1''') 
   \leq \capp_{M_1}(M_1 \setminus M_1') + \capp_{M_1}(M_1' \setminus M_1''') = 0$.
So $\capp_{M_1}(M_1 \setminus M_1''') = 0$.
It similarly follows that $\capp_{M_3}(M_3 \setminus M_3''') = 0$.
Therefore $M_1 \simh M_3$.\hspace*{10pt}\hfill$\Box$

\ruimte

Also the next proposition does not assume regular in capacity. 
But it overshoots the conclusions in Theorem~\ref{tdrum324}
since the range of $\tilde\tau$ can be bigger than $M_1$.


\begin{prop} \label{pdrum323}
Let $M_1$ and $M_2$ be two connected Riemannian manifolds.
If $M_1\simh M_2$ and if $M_1'$, $M_2'$ and $\tau$ are
as in the definition of $\simh$,
then there exists a distance preserving isomorphism 
$\tilde\tau \colon \widetilde M_2 \to \widetilde M_1$
such that $\tilde\tau|_{M_2'}=\tau$.
\end{prop}
\proof\
The function $\tau|_{M_2'} \colon M_2' \to M_1'$ is distance preserving
with respect to the 
distances $d_{M_2'}$ and $d_{M_1'}$.
Then by Theorem~\ref{tdrum302} the map $\tau|_{M_2'}$ is also an 
distance preserving 
with respect to the induced distances from $M_2$ and $M_1$ on $M_2'$ and $M_1'$.
Since $M_2'$ is dense in $M_2$ and therefore also in $\widetilde M_1$
it follows that there exists a unique distance preserving map 
$\tilde \tau \colon \widetilde M_2 \to \widetilde M_1$
such that $\tilde \tau |_{M_2'} = \tau$.
Similarly there exists a unique distance preserving map
$\tilde \sigma \colon \widetilde M_1 \to \widetilde M_2$
such that $\tilde \sigma |_{M_1'} = \tau^{-1}$.
Then $\tilde \tau \circ \tilde \sigma|_{M_1'}$ is the identity function on $M_1'$.
So by density and continuity $\tilde \tau \circ \tilde \sigma = I_{\widetilde M_1}$
Similarly $\tilde \sigma \circ \tilde \tau = I_{\widetilde M_2}$ and the proposition follows.\hfill$\Box$

\ruimte

Now we are able to prove the main theorem of this section.

\ruimte

\noindent
{\bf Proof of Theorem~\ref{tdrum324} \hspace{5pt}}\
The implication $\Leftarrow$ is trivial.
Suppose that $M_1 \simh M_2$ and let $M_1'$, $M_2'$ and $\tau$ be as in 
the definition of $\simh$.
Let $\tilde \tau$ be as in Proposition~\ref{pdrum323}.
If $\varphi \in H^1_0(M_1) \cap C_0(\widetilde M_1)$ then 
$\varphi \circ \tilde \tau \in C_0(\widetilde M_2)$.
Moreover, $\varphi|_{M_1'} \in H^1_0(M_1')$, so 
$(\varphi \circ \tilde \tau)|_{M_2'} \in H^1_0(M_2')$ and therefore
$(\varphi \circ \tilde \tau)|_{M_2} \in H^1_0(M_2)$ by Lemma~\ref{ldrum310}.
So we can define $V \colon H^1_0(M_1) \cap C_0(\widetilde M_1) \to H^1_0(M_2) \cap C_0(\widetilde M_2)$ by 
$V \varphi = \varphi \circ \tilde \tau$.

Next, let $p \in M_1$.
There exists a $\varphi \in C_c^\infty(M_1)$ such that $\varphi(p) = 1$.
Then $V \varphi \in H^1_0(M_2) \cap C_0(\widetilde M_2)$.
Moreover, $(V \varphi)(\tilde \tau^{-1}(p)) = \varphi(p) = 1$.
Hence $\tilde \tau^{-1}(p) \in M_2$ since $M_2$ is regular in capacity.
Similarly $\tilde \tau(q) \in M_1$ for all $q \in M_2$ since 
$M_1$ is regular in capacity.
Let $\hat \tau = \tilde \tau|_{M_2}$.
Then $\hat \tau$ is a topological 
homeomorphism from $M_2$ onto $M_1$.
It remains to show that $\hat \tau$ and its inverse are smooth and an isometry.

By (\ref{eSdrum1;1}) we can define $U \colon L_2(M_1) \to L_2(M_2)$ by
$U \varphi = \varphi \circ \hat \tau$.
If $\varphi \in H^1_0(M_1)$ then 
$\varphi \circ \tau \in H^1_0(M_2') = H^1_0(M_2)$ by isometry, (\ref{eSdrum1;1})
and Lemma~\ref{ldrum310}.
Therefore $U$ is a bijection from $H^1_0(M_1)$ onto $H^1_0(M_2)$.
Let $h_1$ and $h_2$ be the forms associated to the Dirichlet Laplace--Beltrami operators 
on $M_1$ and $M_2$, with form domains $H^1_0(M_1)$ and $H^1_0(M_2)$.
Then 
\begin{eqnarray*}
h_1(\varphi)
= \int_{M_1} |\nabla \varphi|^2
& = & \int_{M_1'} |\nabla \varphi|^2  \\
& = & \int_{M_2'} |\nabla (\varphi \circ \tau)|^2
= \int_{M_2} |\nabla (\varphi \circ \tau)|^2
= \int_{M_2} |\nabla (\varphi \circ \hat \tau)|^2
= h_2(U \varphi)
\end{eqnarray*}
for all $\varphi \in C_c^\infty(M_1)$.
Since $C_c^\infty(M_1)$ is a core for $h_1$ it follows that 
$h_1(\varphi) = h_2(U \varphi)$ for all $\varphi \in H^1_0(M_1)$.
Hence if $\varphi \in L_2(M_1)$, then $\varphi \in D(\Delta_1)$ if and only if 
$U \varphi \in D(\Delta_2)$, and 
$\Delta_2 U \varphi = U \Delta_1 \varphi$ if both conditions are valid.
Now let $\varphi \in C_c^\infty(M_1)$.
Then $U \varphi \in \bigcap_{n=1}^\infty D(\Delta_2^n) \subset C^\infty(M_2)$
by elliptic regularity.
(See \cite{GT} Theorem~9.11 if $p \neq 1$.
If $p = 1$ first apply a 
Sobolev embedding to embed $L_1$ into a Sobolev space $W^{s,p}$ with $s < 0$ and 
$p > 1$, and then apply \cite{GT} Theorem~9.11.)
So there exists a $\psi \in C^\infty(M_2)$ such that $U \varphi = \psi$ a.e.
But $\varphi = \varphi \circ \hat \tau$ is continuous.
Therefore $\varphi \circ \hat \tau = \psi$ pointwise.
Thus $\varphi \circ \hat \tau$ is smooth for all $\varphi \in C_c^\infty(M_1)$.
Therefore $\hat \tau$ is a $C^\infty$-map from $M_2$ onto $M_1$.
Similarly also $\hat \tau^{-1}$ is a $C^\infty$-map, 
so $\hat \tau$ is a $C^\infty$-diffeomorphism.
Finally, since $\tau$ is an isometry and $M_2'$ is dense in $M_2$ it
follows by continuity that also $\hat \tau$ is an isometry.
This proves Theorem~\ref{tdrum324}.\hfill$\Box$
{"config": "arxiv", "file": "0806.0437/drum3.tex"}