Skip to main content

Timeline for Comparison of two monodromies

Current License: CC BY-SA 4.0

13 events
when toggle format what by license comment
Apr 19, 2021 at 8:05 history edited Francesco Polizzi CC BY-SA 4.0
added 553 characters in body
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
edited body
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
edited body
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
added 4 characters in body
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