Timeline for Finite-dimensionality for de Rham cohomology
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2011 at 16:34 | vote | accept | M Turgeon | ||
| Feb 17, 2011 at 5:44 | comment | added | Pete L. Clark | Just to be sure: the isomorphism between de Rham and singular cohomology with $\mathbb{R}$-coefficients holds for all smooth manifolds, right? So this question isn't really about differential forms per se, but rather about the algebraic topology of smooth manifolds. | |
| Feb 17, 2011 at 1:53 | comment | added | Paul | @YBL. John K. knows much more about this than I do, but IIRC the "almost" in John's answer is that $M$ should be dominated by a finite complex, which is stronger than having finite dimensional homology. For example, the infinite connected sum of $RP^3$s has finite dimensional deRham cohomology, but isn't finitely dominated. My comment was meant to find out what was MT's context for the question. | |
| Feb 17, 2011 at 0:35 | comment | added | AFK | @Paul I don't understand your comment. Being finite or infinite dimensional is a well established notion for a vector space (like a Rham cohomology space). And you can't ask for more precise hypothesis when the question is "What hypothesis do I need?". | |
| Feb 17, 2011 at 0:15 | comment | added | Paul | The cohomology of a manifold can be infinite dimensional in many ways. You might get a better answer if you add a hypothesis like finitely many ends, or admits a complete riemannian metric. | |
| Feb 17, 2011 at 0:15 | answer | added | John Klein | timeline score: 17 | |
| Feb 16, 2011 at 22:53 | history | asked | M Turgeon | CC BY-SA 2.5 |