Timeline for LS category of 4-manifolds with free fundamental group
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8 at 18:14 | answer | added | Achim Krause | timeline score: 3 | |
| Jul 8 at 16:55 | comment | added | Achim Krause | Yes, if $\pi_1(M)$ is free, then $H_1(M;\mathbb{Z})$ is free abelian (of the same rank), and then UCT determines $H^1(M;R)$. | |
| Jul 8 at 13:05 | comment | added | Tyrone | Suppose $X$ is a connected complex satisfying Poincaré duality over $\mathbb{Z}/2$. If $Cat(X)=1$, then $X$ is simply connected integral homology sphere. Thus the only closed 4-manifold with $Cat(X)=1$ is $S^4$ (no orientability assumptions are necessary). The proof is mostly spelled out above. | |
| Jul 6 at 21:58 | comment | added | Achim Krause | One version of Poincare duality for unoriented Manifolds is that the cup-product pairing $H^i(M;\mathbb{F}_2)\otimes H^{n-i}(M;\mathbb{F}_2) \to H^n(M;\mathbb{F}_2)\cong \mathbb{F}_2$ is nondegenerate. I think this is for example discussed in Hatcher. | |
| Jul 6 at 21:45 | comment | added | Jeremy | @AchimKrause can you please explain the existence of such a non-zero class in $H^3(M;\mathbb{F}_2)$ so that its cup product with the above-mentioned non-zero class in $H^1(M;\mathbb{F}_2)$ is non-trivial? Why is the cup product non-trivial? | |
| Jul 6 at 20:51 | comment | added | LSpice | The time stamp of a comment is a link, which you can use to link to that particular comment. (There are some subtleties—see Links to comments when the title has been edited and a comment by @rene—but there's no need to worry about that unless you want to do so.) I have edited to refer to the comments you mentioned. | |
| Jul 6 at 20:45 | history | edited | LSpice | CC BY-SA 4.0 |
Links to comments
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| Jul 6 at 20:38 | comment | added | Achim Krause | Ah, I see. Then let me try the following: $H^1(M;\mathbb{F}_2)$ is nontrivial since $\pi_1$ is free and nontrivial, and by Poincare duality there exists a class in $H^3(M;\mathbb{F}_2)$ such that their cup product is nontrivial, so the cup-length is $2$, which should imply LS category $\geq 2$ | |
| Jul 6 at 20:26 | history | edited | Jeremy | CC BY-SA 4.0 |
added 50 characters in body
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| Jul 6 at 20:25 | comment | added | Jeremy | @AchimKrause in the above context (and in the linked post), the LS category is normalized: it is $0$ if and only if space is contractible. So, for non-contractible spaces, LS category is at least $1$ (this does not explain why it should be $2$ in my above setting). | |
| S Jul 6 at 20:23 | history | suggested | Katrina | CC BY-SA 4.0 |
Fixed the spelling mistake in the name of the author Hellen Colman.
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| Jul 6 at 20:05 | comment | added | Achim Krause | Well, compact manifolds (without boundary) are not contractible, for example since they have nontrivial mod 2 homology | |
| Jul 6 at 19:44 | review | Suggested edits | |||
| S Jul 6 at 20:23 | |||||
| Jul 6 at 19:33 | history | asked | Jeremy | CC BY-SA 4.0 |