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Apr
8
accepted Fundamental group of a manifold with an $S^1$-action
Apr
8
comment Fundamental group of a manifold with an $S^1$-action
Igor, I will accept your answer. I think indeed, that explicit counterexamples can be constructed for example as follows. Take the linear S^1 action on S^3 that fixes $S^1$. Then the quotient is $D^2$. Now take as many $S^1$ fibers in $S^3$ as you want and make a Dehn surgery at these fibres. Then the quotient will become a disk with orbi-points, and I guess the obrifold fundamental group of the disk will be related to the fundamental group of the surgered $3$-manifold. Would you agree with this?
Apr
8
comment Fundamental group of a manifold with an $S^1$-action
Thank you Igor! I will check the paper once I wake up :)
Apr
8
comment Fundamental group of a manifold with an $S^1$-action
Francesco, thank you for this example. I amended the question to ask if at least the kernel of the map is finite. This is what I really want...
Apr
8
revised Fundamental group of a manifold with an $S^1$-action
added 173 characters in body
Apr
8
comment Fundamental group of a manifold with an $S^1$-action
Thank you, indeed, this is a counterexample. I amended the question, because I was asking for more than I really need.
Apr
7
asked Fundamental group of a manifold with an $S^1$-action
Mar
27
awarded  Necromancer
Mar
5
awarded  Yearling
Feb
26
comment Image of a hypersurface under a map $\mathbb CP^n\to \mathbb CP^n$
Jason, thank you for this comment, you are right of course.
Feb
26
revised Image of a hypersurface under a map $\mathbb CP^n\to \mathbb CP^n$
added 12 characters in body
Feb
25
asked Image of a hypersurface under a map $\mathbb CP^n\to \mathbb CP^n$
Feb
21
comment Easy to state applications of dimension theory in algebraic geometry
ACL, sure I had in mind the non-algebraic action. I can not yet grasp why dimension theory is doing the job in the algebraic case (while it fails in the analytic case)...
Feb
21
comment Easy to state applications of dimension theory in algebraic geometry
Daniel, thank you for this answer. I am not quite sure though that I understand the phrase in grey. For example $\mathbb C^*$ is acting on an any elliptic curve over $\mathbb C$, and the action is without fixed points.
Feb
21
asked Easy to state applications of dimension theory in algebraic geometry
Feb
21
accepted Extending the tautological bundle of $G(1,3)$?
Feb
21
comment Extending the tautological bundle of $G(1,3)$?
Alex, I think your reasoning is correct. Could you give me a reference to some nice book that explains that tautological Chern classes generate $H^*$ of the Grassmanian?
Feb
21
revised Extending the tautological bundle of $G(1,3)$?
edited body
Feb
21
asked Extending the tautological bundle of $G(1,3)$?
Feb
14
comment Sufficient conditions for a divisor to be connected on a K3 surface
In fact after I wrote this answer I realised that the first comment of Artie solves the problem.