Timeline for Monodromy action on the local system $R^2\phi_*\mathbb{Z}$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2013 at 16:55 | answer | added | Donu Arapura | timeline score: 1 | |
| Jun 26, 2013 at 16:21 | history | edited | michael waltz | CC BY-SA 3.0 |
deleted 245 characters in body
|
| Jun 26, 2013 at 16:13 | comment | added | michael waltz | yes, both of you are right, i'm sorry, i will edit the question | |
| Jun 26, 2013 at 15:06 | comment | added | Serge Lvovski | To what Jason said, I would add that monodromy does not act on the cohomology of $X_0$ (that is, fiber over zero, which contains critical points of $f$). It acts on the cohomology of $X_c=f^{-1}(c)$, where $c\ne 0$. | |
| Jun 26, 2013 at 14:47 | comment | added | Jason Starr | I am not sure I understand the question, and I wonder if there might be a mistake in the formulation. What do you mean by "family of complex manifolds". Are you assuming that the morphism $\phi$ is proper and submersive (i.e., "smooth" in the language of algebraic geometry)? If so, and if $B$ is contractible, then the monodromy action is trivial, just as you say. However, if the morphism is only smooth over $B\setminus \{0\}$, then there can be a nontrivial action of $\pi_1(B\setminus\{0\},b)$. | |
| Jun 26, 2013 at 11:38 | history | edited | Qfwfq | CC BY-SA 3.0 |
(typo)
|
| Jun 26, 2013 at 11:33 | history | asked | michael waltz | CC BY-SA 3.0 |