Timeline for Is it impossible for the dimension of a topological space to increase under a smooth map?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2013 at 14:22 | vote | accept | Ritwik | ||
| Jun 30, 2013 at 18:50 | comment | added | Neil Strickland | I have heard that one can deduce this kind of thing from the results in Section 4.3.20 of "Geometric measure theory" by Federer. However, I have never tried to work out the details myself. | |
| Jun 30, 2013 at 11:48 | comment | added | Ritwik | My reason is the following fact: Let $V \rightarrow M$ be a rank $k+1$ vector bundle over $M$ and $X$ a smooth submanifold of $M$ of dimension $k$. Then the zero set of a generic smooth section $s: M \rightarrow V$ does not intersect $X$. Would such a statement be true if $X$ was a space with hausdorf dimension $k$? | |
| Jun 30, 2013 at 11:38 | answer | added | Anton Petrunin | timeline score: 3 | |
| Jun 30, 2013 at 10:42 | comment | added | Neil Strickland | Is there some reason why you want to use this particular definition of dimension? If you use Hausdorff dimension instead, for example, then there is a well-developed theory for this sort of thing. | |
| Jun 30, 2013 at 9:27 | answer | added | Peter Michor | timeline score: 1 | |
| Jun 30, 2013 at 9:23 | history | asked | Ritwik | CC BY-SA 3.0 |