Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with closed-geodesic
Search options answers only
not deleted
user 104334
1
vote
Accepted
Is this $a(p)=\lim_{r\to \infty} \frac{VolS(p,r)}{e^{h r}}$ exists and applied for manifolds...
When $M$ has positive curvature, the limit should always be 0. Indeed if $M$ has non-negative Ricci curvature, Bishop-Gromov tells you that $\frac{Vol S(p, r)}{n\omega_nr^{n-1}}\leq 1$, so the volumes …
- 320