Timeline for When distance nonincreasing map is an isometry
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10 events
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| Aug 30, 2016 at 16:49 | comment | added | Asaf Shachar | @Misha (Please see my 2 previous comments above, I accidentally directed them to another user). Actually I am not sure that even the "local argument" above holds. It is possible that balls of the same radius around different points will have different volumes (this depends on the curvature in the respective neighbourhoods, which can be different, since we do not assume the manifold has a constant curvature) | |
| Aug 28, 2016 at 12:09 | comment | added | Asaf Shachar | (A-priori the "missing" parts in $(B(f(x),R)$ could be covered by the images of other points. For example if $f$ is surjective. In this case, it is true that $f$ must be an isometry (see Igor's answer) but this is not trivial). Would you mind elaborating? | |
| Aug 28, 2016 at 12:09 | comment | added | Asaf Shachar | @XiaoyangChen I am still not sure about why $Vol(f(M)) < Vol(M)$? I agree that in some sense $f$ "locally" reduces volume, i.e there exists closed balls of radius $R$ such that $Vol(f(B(x,R)) < Vol((B(f(x),R))$ (because of the strict containment like you mentioned). However, the inequality $Vol(f(M)) < Vol(M)$ is a global property, and I do not see how the local property implies it. | |
| Apr 21, 2014 at 0:19 | comment | added | Xiaoyang Chen | Hi, J. Ge, the $2$ to $1$ map from $S^1$ to itself is not a "Riemannian" covering map if the metric on $S^1$ is fixed. I agree with you, it is neither distance non-increasing, nor distance non-decreasing. But if we use two different metrics on $S^1$ such that this map is a Riemannian covering map (just pull back the metric), then it is distance non-increasing. Then it is not a local isometry from $S^1$ to itself, of course. That is what Misha means? | |
| Apr 20, 2014 at 23:15 | comment | added | Misha | Just try to define carefully what you mean and you will see. | |
| Apr 20, 2014 at 22:42 | comment | added | Xiaoyang Chen | Hi, why covering map is not distance nonincreasing? I think any submetry is distance nonincreasing? | |
| Apr 20, 2014 at 20:10 | comment | added | J. GE | @xiaoyang, Covering map is not distance nonincreasing. Look at 2 to 1 map from $S^1$ to itself. | |
| Apr 18, 2014 at 17:18 | comment | added | Misha | 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 $\{(x,y): d(x,y)\le R\}\subset M^2$. | |
| Apr 18, 2014 at 16:48 | comment | added | Xiaoyang Chen | 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. | |
| Apr 18, 2014 at 16:44 | history | answered | Misha | CC BY-SA 3.0 |