Let $f: M \to M$ be a distance nonincreasing map between a closed Riemannian manifold $M$ and $f$ is homotopic to the idendity map. Is it then $f$ an isometry?
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2 Answers
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Combine the following:
If $X$ is a compact metric space, then any surjective distance non-increasing map $X\to X$ is an isometry (see e.g. Burago-Burago-Ivanov's "A course in metric geometry", theorem 1.6.15, or prove yourself-this is easy).
If a map of closed manifolds has nonzero $\mathbb Z_2$-degree, the map is surjective (because every point in the target has odd number of preimages).
answered Apr 18, 2014 at 19:10
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If $f$ is not an isometry then $Vol(f(M))< f(M)$, which then implies that $deg(f)=0$, which is a contradiction.
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$\begingroup$ Why if $f$ is not an isometry, then $Vol (f(M)< vol (M)$?For example, a covering map is distance nonincreasing and $Vol f(M)=vol (M)$, but a covering map is not an isometry. Of course a covering map is not homotopic to the idendity map. $\endgroup$ Commented Apr 18, 2014 at 16:48
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$\begingroup$ In your example with covering map, it is not a local isometry from the manifold to itself. To see why volume goes down, look for instance, at small balls: $f(B(x,R))\subset B(f(x),R)$ and one of this inclusions is proper. Alternatively, look at the map $f\times f: M^2\to M^2$ and show that this map strictly reduces volume by considering images of the subsets of the form $\{(x,y): d(x,y)\le R\}\subset M^2$. $\endgroup$– MishaCommented Apr 18, 2014 at 17:18
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$\begingroup$ @xiaoyang, Covering map is not distance nonincreasing. Look at 2 to 1 map from $S^1$ to itself. $\endgroup$– J. GECommented Apr 20, 2014 at 20:10
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$\begingroup$ Hi, why covering map is not distance nonincreasing? I think any submetry is distance nonincreasing? $\endgroup$ Commented Apr 20, 2014 at 22:42
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$\begingroup$ Just try to define carefully what you mean and you will see. $\endgroup$– MishaCommented Apr 20, 2014 at 23:15