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 dg.differential-geometry
Search options not deleted
user 82561
Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
30
votes
2
answers
1k
views
Polynomial $g:\mathbb R^n \rightarrow\mathbb R^n$ with no critical point may have no root
Version 1 (solved): If $g$ : $\mathbb R^n \rightarrow \mathbb R^n$ is a polynomial, $Dg(x)$ is non-degenerate for every $x$, then there exists $x$, such that $g(x)=0$.
Version 2: If $f$ : $\mathbb R^n …
- 423