Timeline for Isomorphism of cotangent bundles..
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2010 at 22:26 | answer | added | Misha Verbitsky | timeline score: 2 | |
| Nov 15, 2010 at 12:21 | comment | added | Spiro Karigiannis | Yes, Michael is correct. I wasn't thinking it through completely. I noticed that Anton Deitmar posted the solution as an answer. | |
| Nov 13, 2010 at 14:47 | answer | added | user1688 | timeline score: 3 | |
| Nov 11, 2010 at 15:09 | comment | added | Michael Bächtold | @Spiro: but any 1-form is a linear combination of exact ones. | |
| Nov 11, 2010 at 14:08 | comment | added | Spiro Karigiannis | Johannes: $phi$ is a vector bundle isomorphism, so on each fibre of $T^* M$ it is a linear map, but the induced map on $\Lambda^0 T^* M = M \times \mathbb R$ is necessarily the identity map. What you have shown is that for $\phi$ to commute with $d$, it must be the identity on exact $1$-forms. But that does not imply that it is the identity on all $1$-forms. | |
| Nov 11, 2010 at 13:21 | comment | added | Johannes Ebert | $\phi$ commutes with $d$ if and only if $\phi$ is the identity. Proof: If $\phi$ commutes with $d$, then $\phi(df)=d (\phi f)=d (\phi (f\cdot 1))$. Since $\phi$ is a bundle homomorphism, $d (\phi (f\cdot 1)) = d (f \phi (1))$. Since $\phi(1)=1$, this is $df$. Or did I miss something in your question? | |
| Nov 11, 2010 at 13:17 | comment | added | Yunhyung Cho | I edited my question to clarify what I mean. For example, If $g_t$ is a 1-parametrized family of Riemannian metric ($t \in [0,1]$) on $M$, then we can construct a bundle automorphism $\phi$ of $TM$ such that $\phi^* g_1 = g_0$. Of course there are many such kind of automorphisms. I want to know when the induced cotangent bundle automorphism preserves the differential $d$. | |
| Nov 11, 2010 at 13:10 | history | edited | Yunhyung Cho | CC BY-SA 2.5 |
added 63 characters in body
|
| Nov 11, 2010 at 12:34 | comment | added | Spiro Karigiannis | Above, $\Lambda T*M$ should of course be $\Lambda T^*M$. I still don't know how to edit my own comments. Is that possible? Also, I just realized the actual question was not about whether $\phi$ would always commute with $d$, so I shouldn't have said the answer to the question is no. (It's 7am for me, and my brain doesn't work at these hours...) | |
| Nov 11, 2010 at 12:32 | comment | added | Spiro Karigiannis | You're right, I used the wrong word. It is indeed part of the data of a bundle map. I just understood the question, actually. In general, of course, the answer is no. Consider for example, the bundle map which is multiplication by -1 on each fibre of $T^* M$. Then the induced map on $\Lambda T*M$ will be +1 on even forms and -1 on odd forms. Since $d$ changes degree, we have $d \phi \alpha \neq \phi d \alpha$ for any $k$-form $\alpha$. I would have to think about it for a general criterion. It's not immediately obvious. | |
| Nov 11, 2010 at 11:55 | comment | added | Andrew Stacey | Spiro: your choice of wording bothers me ("induces"). I would say that part of the data of a bundle isomorphism is a diffeomorphism of the base and that since the exterior derivative always commutes with a pullback by a diffeomorphism, you can use the base map to reduce the problem to that of an isomorphism of $T^* M$ covering the identity. But there's still a question to answer there since there can be several of those. | |
| Nov 11, 2010 at 11:46 | comment | added | Spiro Karigiannis | Maybe I am confused by your question, but a bundle isomorphism induces a diffeomorphism $f: M \to M$ over the base, and the exterior derivative $d$ always commutes with pullback by a diffeomorphism. Is there something else that you mean here? | |
| Nov 11, 2010 at 11:36 | history | asked | Yunhyung Cho | CC BY-SA 2.5 |