Timeline for Is canonical class a topological invariant?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 14, 2012 at 5:45 | vote | accept | Yuchen Liu | ||
| Jul 15, 2011 at 15:44 | answer | added | Tim Perutz | timeline score: 13 | |
| Jul 15, 2011 at 15:18 | comment | added | diverietti | You are so kind! <3 hehe | |
| Jul 15, 2011 at 15:02 | comment | added | Francesco Polizzi | Diverietti, ok! I erased mine, too :-) | |
| Jul 15, 2011 at 14:52 | comment | added | diverietti | Francesco, you are absolutely right! I was confused since I have in mind the case of deformations. Thank you, I think it's better if I erase my comment, then! | |
| Jul 15, 2011 at 14:28 | comment | added | Mike Usher | For the "diffeomorphism" version of the question, Dmitri seems to have settled the matter in complex dimension larger than 2. I'm not sure of the answer in dimension 2, but if one asks the more general question where the manifold is just symplectic (and so still has a canonical class) there was a flurry of examples around the turn of the century showing that the answer is no; see MR1739225 (McMullen-Taubes), arXiv:math/0005195 (LeBrun), math.SG/0012096 (Smith). | |
| Jul 15, 2011 at 14:10 | comment | added | Mike Usher | For a simple counterexample to the "homeomorphism" version of the question, take, say, a K3 surface (which of course has canonical class zero) which elliptically fibers over $\mathbb{P}^1$, and then do a logarithmic transformation of multiplicity $p>1$ to one of the fibers. You won't have changed the homeomorphism type and the resulting manifold will still admit Kahler forms, but the result will now have nontrivial canonical class. (with divisibility $p-1$ in $H^2(M;\mathbb{Z})$). Of course these examples are all mutually nondiffeomorphic. | |
| Jul 15, 2011 at 14:07 | answer | added | Dmitri Panov | timeline score: 16 | |
| Jul 15, 2011 at 13:53 | answer | added | Francesco Polizzi | timeline score: 13 | |
| Jul 15, 2011 at 13:26 | comment | added | Francesco Polizzi | Of course in the last line I meant $X'$ and not $Y'$. I hope that some of the prominent topologists/Seiberg Witten experts who are on MO can give an answer more exhaustive than my comment... | |
| Jul 15, 2011 at 13:23 | comment | added | Francesco Polizzi | The answer is yes if the diffeomorphism is induced by a deformation equivalence; this is an easy consequence of Ehresmann theorem. Otherwise, I think that the answer is not at all obvious. For instance, using Seiberg Witten theory, one proves that any diffeomorphism $\phi \colon X \to X'$ between smooth $4$-manifolds (for instance, algebraic surfaces) maps $K_X$ either into $K_{X'}$ or into $-K_{X'}$, and the second case may occur. I do not know, however, if there are examples where the second case occurs and $X$, $Y'$ are both smooth and projective. | |
| Jul 15, 2011 at 13:17 | comment | added | Yuchen Liu | @diverietti's comment: Yes, your statement is Poincare dual to my question. | |
| Jul 15, 2011 at 12:57 | comment | added | diverietti | Can I ask you if I understand correctly your question? I see it as follows: Let $X$ be a compact complex manifold and $K_X$ its canonical bundle. Let $c_1(K_X)=-c_1(X)$ its first Chern class in $H^2(X,\mathbb Z)$. Suppose we are given another complex structure let's say $X'$ on $X$ and consider the corresponding canonical bundle $K_{X'}$. Then, is it true that $c_1(K_X)=c_1(K_{X'})$ in $H^2(X,\mathbb Z)=H^2(X',\mathbb Z)$? | |
| Jul 15, 2011 at 12:33 | comment | added | Yuchen Liu | It may seems that the case "homeomorphism" and "diffeomorphism" have many differences, but I don't know whether the differences will make the answer different. However, I want to know the answers both in "homemorphism" and "diffeomorphism" cases. | |
| Jul 15, 2011 at 12:25 | comment | added | user5117 | Just to be sure, does "topological invariant" here mean "diffeomorphism invariant"? | |
| Jul 15, 2011 at 12:11 | history | asked | Yuchen Liu | CC BY-SA 3.0 |