Timeline for What is a higher genus analogue of the Pontryagin product?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2012 at 9:07 | vote | accept | Daniel Moskovich | ||
| Aug 6, 2012 at 18:04 | comment | added | Daniel Moskovich | Agol: True- for the fundamental group itself, it's not interesting. The interest comes when one looks at homomorphs G of pi_1(M). These might have non-trivian H_2, but still the image of the Pontryagin product of the images of x and y under the homomorphism will vanish (I think this is noted in "Homomorphs of Knot Groups" by Johnson, where the significance of this condition is discussed- Johnson-Livingston says if you add a condition, this characterizes knot homomorphs). I'm asking about the analogous statement for a higher genus surface. But you're right- it isn't clearly stated. | |
| Aug 6, 2012 at 17:03 | comment | added | Ian Agol | I don't really understand your question: if $M$ is e.g. a knot complement, then $H_2(\pi_1(M))=0$, so the Pontryagin product vanishes for any immersed torus. So this doesn't characterize the peripheral torus. | |
| Aug 6, 2012 at 9:31 | answer | added | Mark Grant | timeline score: 3 | |
| Aug 5, 2012 at 19:55 | answer | added | Ryan Budney | timeline score: 1 | |
| Aug 5, 2012 at 9:46 | comment | added | Daniel Moskovich | Question edited to clarify this point. | |
| Aug 5, 2012 at 9:45 | history | edited | Daniel Moskovich | CC BY-SA 3.0 |
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| Aug 3, 2012 at 9:33 | comment | added | Mark Grant | Isn't $\pi_1(M)$ non-abelian in general? If so, how do you define the Pontryagin product on its homology? It seems you are just looking at the image of the fundamental class of $\partial M$ under inclusion into $M$, is this true? | |
| Aug 3, 2012 at 9:13 | history | asked | Daniel Moskovich | CC BY-SA 3.0 |