There are some more useful facts about isometries that I want to gather together in this post.

Isometries are determined by one tangent space

Unlike smooth maps, isometries have to be much more rigid:

Proposition 1 Let ${\phi, \psi: M \rightarrow N}$ be isometries such that ${q=\phi(p)=\psi(p)}$ and ${\phi_*=\psi_*: T_p(M) \rightarrow T_q(N)}$. Then ${\phi = \psi}$.

(Recall that we’re assuming all manifolds to be connected.)

The idea is that we can take local “exponential coordinates” around ${p}$, ${q}$ respectively. Then as I showed yesterday, ${\phi}$ and ${\psi}$ are necessarily given by a linear map in these coordinates. Since ${\phi_*=\psi_*}$ on ${T_p(M)}$, this means the two linear maps coincide, so ${\phi}$ and ${\psi}$ are locally equal. So we can let ${S}$ be the set of ${r \in M}$ with ${\phi(r)=\psi(r), \phi_*=\psi_*}$ on ${T_r(M)}$. ${S}$ is evidently closed, and it is open by the argument just given, so by connectedness we get the result.

Continuation of isometries

Because of the previous result, it is possible to talk about “analytic continuation” of an isometry. So let ${c: I \rightarrow M}$ be a curve, and let ${\phi}$ be an isometry of a neighborhood ${U}$ of ${c(0)}$ into ${N}$ (i.e. onto a submanifold thereof). Suppose we have a family of isometries ${\phi_t}$, ${t \in I}$, of isometries of neighborhoods ${U_t}$ into ${N}$, with ${\phi_0 = \phi}$, satisfying the following condition: Whenever ${t,t'}$ are close enough, ${U_t \cap U_{t'} \neq \emptyset}$ and ${\phi_t |_{U_t\cap U_{t'}} = \phi_{t'} |_{U_t \cap U_{t'}}}$. This is exactly analogous to the treatment of analytic continuation of holomorphic functions, and I have to wonder if there is a general framework using fancy techniques in category theory or something like that.

The first basic result is that this is unique:

Proposition 2 If ${\phi_t, \psi_t}$ are continuations of ${\phi}$ along the curve ${c}$, then for all ${t}$, ${\phi_t = \psi_t}$ in some neighborhood of ${c(t)}$.

Indeed, the proof is now a direct corollary of the previous result: take the set of ${t}$ where the statement of the proposition is true, and it follows that said set is open, closed, and nonempty.

Existence of continuations

As in complex analysis, it is possible to perform a continuation, but one needs the assumptions of completeness and analyticity:

Proposition 3 If ${N}$ is a complete Riemannian manifold, then any (analytic) isometry ${\phi: U \subset M \rightarrow N}$ can be continued along the curve ${c}$.

The curve ${c}$ doesn’t have to be analytic.

The idea is that we only need to do this one small piece at a time. The small pieces rely on the following lemma, which allows us to extend on normal neighborhoods:

Lemma 4 Let ${D_r(p)}$ be a disk around ${p}$ with respect to the metric on ${M}$, and let ${\rho > r}$. Suppose ${U \supset D_r(p)}$ is the diffeomorphic image of an open subset of ${ T_p(M)}$ under the exponential map (which is true for ${U}$ small enough). Then an (analytic) isometry ${\phi}$ of ${D_r(p)}$ into ${N}$ can be extended to an isometry ${\tilde{\phi}}$ of ${U}$ into ${N}$.

Such a ${U}$ is called a normal neighborhood of ${U}$. The normal neighborhood theorem, which I have not yet proven, states that there is a basis consisting open sets which are normal neighborhoods of each of their points.

Since the map ${D_r(p) \rightarrow N}$ is an isometry, it sends geodesics to geodesics (cf. yesterday’s post). Now let ${A}$ be the linear transformation ${T_p(M) \rightarrow T_q(N)}$. The map ${\phi}$ can be described as follows: it maps ${\exp_p(v)}$ to ${\exp_q(Av)}$. This follows from the previous discussions. Moreover, by the description it offers a way of defining ${\tilde{\phi}}$ on ${U}$ since each element of ${U}$ can be written as ${\exp_p(v)}$ for a unique ${v \in T_p(M)}$. The fact that this is an isometry follows because everything here (especially ${\tilde{\phi}^*h - g}$ if ${g,h}$ are the metrics on ${M,N}$) is real-analytic, and ${\tilde{\phi}}$ is an isometry on an open subset of ${U}$.

Now let’s prove the existence proposition.

Consider the set ${S}$ of ${s \in I}$ such that an analytic continuation of ${\phi}$ on ${[0,s]}$ exists. It is open and nonempty. I claim it is closed.

Let ${s^* := \sup S}$, and suppose ${s<1}$. There is a curve ${\phi(c(s))}$ defined on ${[0,s^*)}$, well-defined by uniqueness. It tends to a limit ${c(s^*)}$ as ${s \rightarrow s^*}$ by completeness and since ${\phi}$ is an isometry.

Let ${U}$ be a neighborhood of ${c(s^*)}$ which is a normal neighborhood of each of its points. Then there is ${s \in [0,s^*)}$ very close to ${s^*}$ with ${c(s) \in U}$ and a normal neighborhood ${D_r(c(s))}$ of ${c(s)}$ with an isometry into ${N}$ via continuation. This extends to ${U}$ by the lemma, which gives the continuation to a neighborhood of ${[0,s^*]}$. This is a contradiction since ${s^*}$ was assumed to be the supremum.

Independence up to homotopy

Proposition 5 Let ${c,d: I \rightarrow M}$ be homotopic paths (i.e. with ${c(0)=d(0),c(1),d(1)}$, and the homotopy respecting endpoints). Then the continuations of ${\phi}$ along ${c,d}$ agree in a neighborhood of ${c(1)}$.

The proof is similar to the one for analytic continuation.

The Myers-Rinow Theorem

The following theorem—and its proof—are strikingly analogous to the monodromy theorem:

Theorem 6 (Myers-Rinow) Let ${M,N}$ be analytic, simply connected, complete manifolds and ${\phi}$ an (analytic) isometry between an open subset of ${M}$ and an open subset of ${N}$. Then ${\phi}$ can be extended to an isometry ${\tilde{\phi}}$ of ${M}$ onto ${N}$.

The idea is the same as the monodromy theorem: to define the extension ${\tilde{\phi}}$ on ${q \in M}$ by starting with a ${p \in M}$ with ${\phi(p)}$ defined, drawing a curve ${c}$ from ${p}$ to ${q}$, and continuing ${\phi}$ along ${c}$; this can be used to define ${\tilde{\phi}}$ in a neighborhood of ${q}$. Simple connectivity shows that the choice of ${c}$ is irrelevant. It is easy to check that ${\tilde{\phi}}$ is smooth locally by comparing paths ${c}$ with ${c}$ catenated with a small arc between ${q}$ and, say, ${q'}$ close to ${q}$.

We can do the same for ${\psi:=\phi^{-1}}$ and get an isometry ${\tilde{\psi}: N \rightarrow M}$. Now ${\tilde{\psi} \circ \tilde{\phi} = 1_M,\tilde{\phi} \circ \tilde{\psi}=1_N}$ on open subsets of ${M,N}$; thus this is true everywhere, and the extensions are diffeomorphisms. We already know they preserve the inner product.