November 19, 2009

A theorem of Mazur-Ulam on isometries of vector spaces

Posted by Akhil Mathew under analysis, functional analysis, MaBloWriMo | Tags: , |
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Today’s will be a relatively short post, and primarily algebraic. I had mentioned a couple of days back the following result:

Theorem 1 (Mazur-Ulam) An isometry {M: X \rightarrow X'} of a normed linear space {X} onto another normed linear space {X'} with {M(0)=0} is linear.

We needed the theorem to show that a distance-preserving map between Riemannian manifolds is an isometry.   Recall that M‘s being an isometry means that it preserves distances between points.

Apparently the theorem has no relation to Barry Mazur, but to another Mazur.

I will follow Lax’s Functional Analysis in the proof.

 

Reduction

It is enough to prove that

\displaystyle M\left( \frac{x+y}{2}\right) = \frac{1}{2}(M(x) + M(y))\ \ \ \ \ (1) 

for all {x,y \in X}.

Indeed, if this is the case, then by {M(0)=0}, we get {M\left( \frac{1}{2^n} x\right) = \frac{1}{2^n} M(x)} by induction. So

\displaystyle M\left( \frac{k}{2^n} x\right) = \frac{k}{2^n} M(x)

 for all {k}. By density of the dyadic fractions and continuity, we find {M(kx)=kM(x)} for all {k \in \mathbb{R}^+}. Also (1) and what’s already proved imply {M(x+y) = M(x)+M(y)}, so {M(-x)=-M(x)}, which proves linearity.

 

Strategy of the proof

The idea is to use a purely norm-theoretic way of describing the midpoint of {x,y}. This must be reflected by {M}, so it will prove (1). In particular, given {x,y \in X}, we will define sets {A_0, A_1, \dots} with {\bigcap A_i = \{ \frac{1}{2}(x+y)\}}, with the sets defined solely in terms of the norm structure on {X}. It will thus follow that {M} maps these sets onto their analogs on {X'}—which will prove (1) and the theorem. (more…)

 

November 18, 2009

Continuation of isometries

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , |
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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. (more…)

 

November 17, 2009

Isometries of Riemannian manifolds and a theorem of Myers-Steenrod

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: , , |
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An isometry between two Riemannian manifolds {(M,g), (N,h)} is a diffeomorphism {\phi: M \rightarrow N} such that {\phi^* h = g}. In other words, {\phi} respects the Riemannian structure as well as the differentiable structure.

It follows that if {d,d'} are the metrics on {M,N} induced by the Riemannian metrics {g,h}, then {d'(\phi(x),\phi(y)) = d(x,y)} for {x,y \in M}—that is, {\phi} is distance-preserving. Interestingly, a version of the converse is true:

Theorem 1 (Myers-Steenrod) If {\phi: M \rightarrow M} is distance-preserving and surjective, then it is an isometry (in particular, it is smooth).

I should observe that the proof applies to the case of a map of Riemannian manifolds {\phi: M \rightarrow N}! I don’t know why Helgason states it with the extra hypotheses. Initially I wrote this post with this question unresolved under the goal of figuring out what I had missed and why this hypothesis was necessary by blogging, but I was unable to do so. With that complete, however, a search reveals that it’s stated in the additional generality in Petersen’s book (the relevant parts of which can be viewed at Google Books). However, it’s easier to do the notations in the case of one manifold.

{\phi} is evidently a homeomorphism. Now pick {p \in M} and a neighborhood {D_r(p)} such that for any {q \in D_r(p)}, there is a unique geodesic in that neighborhood connecting {p,q}. Call it {\gamma}. I claim that {\phi \circ \gamma} is a geodesic in {N}. The geodesic {\gamma} can be assumed to be parametrized by unit length. We have for all {t},

\displaystyle d(p,q) = d(\gamma(t),p) + d(\gamma(t), q). 

Thus

\displaystyle d(\phi(p),\phi(q)) = d(\phi(\gamma(t)),\phi(p)) + d(\phi(\gamma(t)), \phi(q)) \ \ (*). 

In other words, we have strict equality in the triangle inequality. (more…)