Timeline for Proof of Ehresmann's theorem
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 11, 2022 at 12:31 | comment | added | Tom | @AndreasBlass, I think the condition $a_1<a_2<\dots$ approach a limit $b$ is not easy to achieve, we still need to cover the compact interval $[a,b]$ by finitely many open intervals and apply the local triviality. | |
| May 10, 2022 at 14:34 | comment | added | Andreas Blass | Maybe I'm being silly, but this situation looks easier to me. Consider the bad case, where $X_0$ is diffeomorphic to $X_{a_1},X_{a_2},\dots$ and the points $a_1<a_2<\dots$ approach a limit $b$, with $X_b$ not diffeomorphic to $X_0$. Then the argument used for $X_0$ could be applied to $X_b$ in the reverse direction to get $X_b$ diffeomorphic to $X_c$ for any $c<b$ that is sufficiently close to $b$. But that would include some $a_n$ as a possible $c$. | |
| May 10, 2022 at 10:55 | comment | added | Tom | @BenMcKay, thanks, I know where my problem lies in now: local triviality is transitive from 0 to 1 since any compact set can be covered by finitely many open sets, while local stability is not necessarily transitive between arbitrarily two points in a line. | |
| May 10, 2022 at 8:21 | history | edited | Tom | CC BY-SA 4.0 |
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| May 9, 2022 at 15:29 | comment | added | Ben McKay | To prove the result stated, that all fibers are $C^{\infty}$ diffeomorphic, his proof is fine. He splits up his interval into finitely many subintervals, and proves diffeomorphisms one step at a time through each of these subintervals. What I should have written is that the stronger result, which is what is usually called Ehressmann's theorem, states that the map is a $C^{\infty}$ fiber bundle, for which we need to ensure the local triviality. | |
| May 9, 2022 at 14:37 | comment | added | Tom | @BenMcKay, so you mean Huybrechts' proof contains a gap and it makes his proof invalid? | |
| May 9, 2022 at 14:28 | comment | added | Ben McKay | it is not enough to prove that the fibers are diffeomorphic, as far as I can see, because we need to prove that there is a trivialization simultaneously of all fibers to prove triviality. Think of a nontrivial bundle, like the Moebius strip. But also, you are right that a sequence of fibers of a map can be all diffeomorphic without the limit fiber being diffeomorphic to any of them. | |
| May 9, 2022 at 12:33 | comment | added | Tom | @BenMcKay, but I don't think it suffices to prove $X_0$ and $X_{\tau}$ are diffeomorphic for any $\tau>0$, for example, if we have proved $X_0\cong X_{\frac{1}{2}}\cong X_{\frac{3}{4}}\cong\cdots\cong X_{1-\frac{1}{2^n}}$, then we can't get the conclusion $X_0\cong X_1$, right? | |
| May 9, 2022 at 11:53 | comment | added | Ben McKay | Yes, that is right. But $F_t$ need not be holomorphic, since the vector field is produced using a partition of unity, so only $C^{\infty}$. | |
| May 9, 2022 at 11:04 | comment | added | Tom | @BenMcKay, for the diffeomorphism $F:X_0\times I\to \mathcal X;(x,t)\mapsto F(x,t)$, let $F_t(x):=F(x,t)$, then $F_t:X_0\to X_t$ provides a diffeomorphism between $X_0$ and $X_t$, which implies all the fibers over $B$ are diffeomorphic, isn't it? | |
| May 9, 2022 at 9:56 | comment | added | Ben McKay | Since all fibers are diffeomorphic, all fibers are diffeomorphic to a Kaehler manifold just when one fiber is a Kaehler manifold. But they are only diffeomorphic as real manifolds, not as complex manifolds. The complex structure can vary, and does in many elementary examples, already on complex surfaces. | |
| May 9, 2022 at 9:54 | comment | added | Ben McKay | It is not enough to show that they are diffeomorphic. He is showing that we can provide a fiber preserving diffeomorphism between $X_0 \times I$ and an open set of $X$ containing $X_0$, $I$ an interval. So $X$ is locally trivial, a fiber bundle over the interval. | |
| May 9, 2022 at 8:01 | history | asked | Tom | CC BY-SA 4.0 |