Timeline for Comparison of two monodromies
Current License: CC BY-SA 4.0
13 events
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Apr 19, 2021 at 8:05 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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Apr 18, 2021 at 3:54 | history | became hot network question | |||
Apr 18, 2021 at 0:08 | comment | added | Moishe Kohan | Oh, sorry, somehow I missed the surjectivity assumption. | |
Apr 17, 2021 at 22:45 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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Apr 17, 2021 at 22:06 | answer | added | Will Sawin | timeline score: 7 | |
Apr 17, 2021 at 21:51 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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Apr 17, 2021 at 21:41 | comment | added | Will Sawin | The inclusion in one direction is easy. The $G$-invariant classes in $H_1$ are pullbacks from $\Sigma_b$, and that pullback map is monodromy-invariant. The equality seems very hard to me. | |
Apr 17, 2021 at 21:24 | comment | added | Francesco Polizzi | Sorry, probably I do not understand. The fibre of $f \colon X \to \Sigma_b$ over a point $p \in \Sigma_b$ is the preimage in $X$, via $X \to \Sigma_b \times \Sigma_b$, of the corresponding fibre of $\Sigma_b \times \Sigma_b \to \Sigma_b$. Now, this preimage is the curve corresponding to the group homomorphism $$\psi_1 \colon \pi_1(\Sigma_b-\{p \}) \to G,$$ and this is connected because I am assuming that $\psi_1$ is onto. Since one fibre is connected, all of them are so. Where am I wrong? | |
Apr 17, 2021 at 21:22 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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Apr 17, 2021 at 20:44 | comment | added | Moishe Kohan | Yes, since you require connected fibers over $ \Sigma_b$. | |
Apr 17, 2021 at 20:38 | comment | added | Francesco Polizzi | @MoisheKohan: why? A connected $G$-cover $X \to Y$ is equivalent to the datum of a group epimorphism $\pi_1(Y-B)\to G,$ where $B$ is the branch locus. Or am I missing something? | |
Apr 17, 2021 at 20:31 | comment | added | Moishe Kohan | I think you need more conditions to ensure connected fibers. | |
Apr 17, 2021 at 19:53 | history | asked | Francesco Polizzi | CC BY-SA 4.0 |