Timeline for Is the singular homology of a real algebraic set always finitely generated?
Current License: CC BY-SA 3.0
8 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| May 30, 2012 at 9:14 | history | edited | Mark Grant | CC BY-SA 3.0 |
fixed gap in argument
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| May 30, 2012 at 0:54 | comment | added | Vitali Kapovitch | @Dan Ramras the singular case follows from above because any algebraic set has finitely many smooth strata and you can choose $f_v$ that works for all of them. this easily yields a smooth gradient like vector field for $f_v$ tangent to $M$ on $M$ and without critical points outside a large ball. | |
| May 29, 2012 at 22:07 | comment | added | Dan Ramras | Vitali, that's a nice observation. I am specifically interested in the case of singular algebraic sets, however. | |
| May 28, 2012 at 18:28 | comment | added | Vitali Kapovitch | @Dan Ramras If you were only concerned with possible noncompactness of $M=Z(p)$ that is easy to deal with directly. For any $v\in \mathbb R^n$ look at $f_v(x)=|x-v|^2$. If $M$ is smooth then $f_v$ is a Morse function on $M$ for most $v\in \mathbb R^n$ (so long as $v$ is not a focal point of $M$). For any such $v$ the set of critical points of $f_v$ on $M$ is real algebraic and discrete and hence finite (e.g. by the Theorem of Whitney you mention). Therefore $M$ deformation retracts onto its intersection with the closed ball $\bar B(v,R)$ for some finite $R$. | |
| May 28, 2012 at 14:54 | history | edited | Mark Grant | CC BY-SA 3.0 |
ANR to ENR, accent
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| May 28, 2012 at 14:34 | comment | added | Dan Ramras | Ah, the trick I was missing is that any semialgebraic set is equivalent to a bounded one, by the trick explained just prior to Proposition 1.13. Thanks! | |
| May 28, 2012 at 12:58 | history | answered | Mark Grant | CC BY-SA 3.0 |