Timeline for Which 3-manifolds have positive rank gradient?
Current License: CC BY-SA 3.0
13 events
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| Nov 16, 2015 at 16:42 | comment | added | Pablo | Dear Ian, thank you for this illuminating discussion. I am asking a more general question now: mathoverflow.net/questions/223747/… | |
| Nov 16, 2015 at 16:16 | comment | added | Ian Agol | @Pablo: correct | |
| Nov 16, 2015 at 16:12 | vote | accept | Pablo | ||
| Nov 16, 2015 at 16:11 | comment | added | Pablo | Is it true that in these cases the fundamental group is a free product amalgamating $\mathbb{Z}/2\mathbb{Z} \cong \pi_1(\mathbb{RP}^2$)? | |
| Nov 16, 2015 at 16:01 | comment | added | Ian Agol | For non-orientable manifolds, one has to consider the $\mathbb{RP}^2$ decomposition. So, for example, if a hyperbolic 3-manifold $M$ admits an orientation-reversing isometry with two fixed points, the quotient will be an oribfold with two point which are cones over $\mathbb{RP}^2$. Remove neighborhoods of these and glue the $\mathbb{RP}^2$'s together. This manifold has orientable double cover $M\# (S^2\times S^1)$, so has positive rank gradient. You can imagine generalizations of this. See: plms.oxfordjournals.org/content/s3-11/1/469 | |
| Nov 16, 2015 at 15:34 | comment | added | Pablo | And what nonorientable examples exist? | |
| Nov 16, 2015 at 15:33 | comment | added | Ian Agol | Correct, in the orientable case. | |
| Nov 16, 2015 at 15:23 | comment | added | Pablo | So, the only way to get a $3$-manifold whose fundamental group has positive rank gradient is to take a connected sum of $3$-manifolds not both $\mathbb{R}\mathbb{P}^3$? | |
| Nov 16, 2015 at 14:57 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Nov 16, 2015 at 14:51 | comment | added | Ian Agol | @Pablo: see my revision. | |
| Nov 16, 2015 at 14:51 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Nov 16, 2015 at 14:41 | comment | added | Pablo | You assume that $M$ is closed and orientable. What about the general case? | |
| Nov 16, 2015 at 14:32 | history | answered | Ian Agol | CC BY-SA 3.0 |