Timeline for Automatic transfer of pointwise metric computations to bundle computations
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2017 at 6:58 | vote | accept | Asaf Shachar | ||
| S Apr 27, 2017 at 10:29 | history | bounty ended | Asaf Shachar | ||
| S Apr 27, 2017 at 10:29 | history | notice removed | Asaf Shachar | ||
| Apr 25, 2017 at 4:23 | comment | added | David Roberts♦ | I'm not saying it's stacks...but it's stacks. | |
| Apr 25, 2017 at 3:35 | answer | added | Deane Yang | timeline score: 2 | |
| Apr 24, 2017 at 15:11 | answer | added | Asaf Shachar | timeline score: 0 | |
| Apr 24, 2017 at 15:10 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
edited body; edited tags
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| Apr 24, 2017 at 12:44 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
focused the question more narrowly around bundles.
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| Apr 24, 2017 at 1:21 | answer | added | Deane Yang | timeline score: 1 | |
| Apr 23, 2017 at 23:41 | comment | added | Deane Yang | I'm pretty sure there is a meta-theorem, even in this case, assuming you're writing everything invariantly and using the naturally induced connections. You should be able to reduce it to a statement about a bundle map between vector bundles $E$ and $F$ with inner products and compatible connections. Here, $E = T_*\mathcal{M}$, $F=f^*T_*\mathcal{N}$. | |
| Apr 23, 2017 at 16:54 | comment | added | Deane Yang | Yes, you do have two different vector spaces. So you have to use an invariant definition of the determinant of a linear map between two different vector spaces and differentiate that. Overall, it's a little more complicated to work with a map between two manifolds instead of bundles over a single manifold. I think that's the real difficulty. | |
| Apr 23, 2017 at 16:29 | comment | added | Asaf Shachar | Your question is essentially what's bothering me: I "feel" the two results are really close to each other, so there is supposed to be a way to go from one to the other, immediately, by a clever application of the chain rule or some other change of perspective. The analogy $A \to df,B \to \nabla_Vdf$ is very appealing. | |
| Apr 23, 2017 at 16:29 | comment | added | Asaf Shachar | @DeaneYang Well, a naive application of the chain rule won't do it, right? Fix $p \in M$, and take a path $\alpha(t)$,$\alpha(0)=p,\dot \alpha(0)=V(p)$. Then $\big( V\det(df) \big)(p)=\frac{d}{dt}_{t=0}det(df_{\alpha(t)})$. But the problem is that this is not the derivative of a the determinant of a linear map between to fixed vector spaces. (The tangent spaces in the domain and codomain change with $t$ of course). As I said, one could take orthonormal frames and use representing matrices, but then we need to get back to the invariant interpretation somehow. | |
| Apr 23, 2017 at 16:08 | comment | added | Deane Yang | Doesn't the manifold identity follow directly from the pointwise identity combined with the chain rule? | |
| Apr 23, 2017 at 9:17 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
Elaborated on one approach for "solving" the problem.
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| S Apr 23, 2017 at 9:09 | history | bounty started | Asaf Shachar | ||
| S Apr 23, 2017 at 9:09 | history | notice added | Asaf Shachar | Draw attention | |
| Apr 19, 2017 at 7:28 | history | asked | Asaf Shachar | CC BY-SA 3.0 |