Timeline for pull-back connection
Current License: CC BY-SA 2.5
26 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2021 at 16:52 | comment | added | Deane Yang | @Sherose, it suffices to show that given two different frames of the pullback bundle, the transition matrix is a smooth function of the domain. This follows from the fact that the transition matrix is a smooth function of the range. | |
| Nov 11, 2021 at 12:57 | comment | added | Sherose | @DeaneYang I somehow agree with your argument. One thing that is not clear enough (at least to me) is the well-definedness of (2), since a section of the pull-back bundle may admit different linear combinations. I believe we should prove that if $F^*(s)(x_0) = 0$ then $F^*\nabla(F^*s)(x_0)$ should also be zero. However I cannot prove this... | |
| Dec 31, 2020 at 2:53 | comment | added | peter | but if i view X as a single vector, i will get out a single element, what does composition mean for that? | |
| Dec 30, 2020 at 15:12 | comment | added | Deane Yang | @peter, without the $F^*$, the right side is an element in the bundle $E$, but the left side is an element in the bundle $F^*E$. You could also write the right side as $(\nabla_{F_*X}s)\circ F$. | |
| Dec 30, 2020 at 13:14 | comment | added | peter | i like the approach of viewing X as a single vector. but the rhs has to lose the F*. | |
| Sep 7, 2019 at 22:31 | history | bounty ended | Ali Taghavi | ||
| Sep 4, 2019 at 14:26 | comment | added | Willie Wong | ... for example, a theorem of Yano (final theorem of the paper) states that on a compact orientable Riemannian manifold, any infinitesimal affine transformation is necessarily an infinitesimal isometry. (Or that restricted to the connected component of the identity, the affine group is the same as the isometry group; this restriction to the connected component is necessary, as you can take two spheres of unequal sizes as your manifold, and make your map swap their metrics.) | |
| Sep 4, 2019 at 14:23 | comment | added | Willie Wong | @AliTaghavi: not in general. On $\mathbb{R}^n$ the mapping $x \mapsto Ax$ where $A$ is any invertible $n\times n$ matrix is a diffeomorphism. Unless $A\in O(n)$ the mapping is in general not an isometry. But as an affine mapping the connection is preserved. This notion of affine mappings generalize to arbitrary Riemannian manifolds (in fact, to arbitrary manifolds with affine connections). In general, the affine group is bigger than the isometry group. But there are times when the two can be identified. ... | |
| Sep 3, 2019 at 23:24 | comment | added | Ali Taghavi | @WillieWong For parallelizable riemannian manifold $M$ there is no any ambiguity of speaking about $TM$ and $f^* TM$. In this condition is the space of all diffeomorphisms preserving LC connections the same as the isometric group? | |
| Sep 3, 2019 at 20:04 | comment | added | Ali Taghavi | @WillieWong yes I see thank you. | |
| Sep 3, 2019 at 19:14 | comment | added | Willie Wong | @AliTaghavi: I think that "generate" is read with coefficients in the ring of smooth functions over the domain, not just with coefficients being the reals. (May be simpler to reduce your question more and look first at the case where you have $f(x) = 0$. The pullback sections are only the constant sections.) | |
| Sep 3, 2019 at 16:55 | comment | added | Ali Taghavi | I do not think that the space of pull back sections generate even locally the space of sections in the initial space $f:\mathbb{R}\to \mathbb{R}\quad f(x)=x^2$ pull back sections areeven function(assuming we have trivial line bundle). Could you please ellaborate your answer in particular point 1)? Thank you. | |
| Sep 10, 2018 at 4:06 | comment | added | Deane Yang | @seub, The pullback of a bundle $E$ by a smooth map $F: P \rightarrow B$ is defined as follows: The fiber over each $x \in P$ is $E_{F(x)}$ and the set of all smooth sections consists of the all smooth maps $s: P \rightarrow B$ such that $s(x) \in E_{F(x)}$. It is straightforward to show that if $T: O\times \mathbb{R}^k \rightarrow E$ is a local trivialization of $E$, then $S: F^{-1}(O)\times \mathbb{R}^k \rightarrow E$, where $S(x,v) = T(F(x),v)$ is a local trivialization of $F^*E$. After that, it is straightforward to define the pullback connection using the local trivialization. | |
| Sep 9, 2018 at 23:11 | comment | added | seub | @DeaneYang: by the way, in 1), not only existence is not always true, but neither uniqueness I think when f is not locally surjective, which is another problem for the well-definedness of the pullback connection. Also I'm not sure what is gained by writing "generated by". Maybe the whole idea is just no good! | |
| Sep 9, 2018 at 23:06 | comment | added | seub | @DeaneYang: I'm afraid I don't have a good suggestion because I don't understand very well myself what's the right approach to define this pullback connection, I wish I did. Your answer seems like the write approach superficially, but the whole answer relies on 1) being true, which is not the case. That's a problem! If you think the right way to justify it is to write everything in coordinates, that's fine, but it's just a different answer. That's just my humble opinion. | |
| Sep 9, 2018 at 21:50 | comment | added | Deane Yang | @seub, to be honest, I'm not sure what you'd like me to add. Could you write a comment with whatever you'd like to see and I'll insert it into the answer (and crediting you). | |
| Sep 9, 2018 at 19:41 | comment | added | seub | @DeaneYang ah, I thought so. Maybe it's worth editing the answer to reflect that. | |
| Sep 9, 2018 at 17:44 | comment | added | Deane Yang | @seub, indeed it looks funny. I suggest writing it all out in local coordinates assuming $F$ is a local immersion and observing that it all still works if $F$ is not | |
| Sep 9, 2018 at 17:14 | comment | added | seub | Wait, really? 1) just seems clearly false to me, especially if $F$ is not locally injective. Am I missing something? | |
| Dec 14, 2010 at 2:35 | comment | added | BCnrd | Dear Deane: You're right, it is the same content, just done in a different order. I tend to prefer postponing the "pointwise" aspect until the very end because in the algebraic (and especially scheme setting, with families over a base that may not be reduced) it is not something we can always use in the characterization step (it is just a "bonus" after the fact). | |
| Dec 14, 2010 at 0:01 | comment | added | Deane Yang | Dear BCnrd, thanks for the clarification. But I don't think there's any additional work required to verify smoothness. Smoothness of the pull-back connection follows directly from the smoothness of the original connection, smoothness of $F$, linear pointwise dependence on the tangent vector, and the Leibniz formula. | |
| Dec 13, 2010 at 23:50 | comment | added | BCnrd | Dear Deane: I just mean that if we make a pointwise construction then we may need to do more work (an explicit calculations over a suitable small open domain, or something else) to verify that its output is smooth in local coordinates, whereas if we make construction in terms of local sections of bundles (or in other words, from the viewpoint of sheaves) then the smoothness of the output of the connection operator drops out automatically from the framework. | |
| Dec 13, 2010 at 20:16 | comment | added | Deane Yang | Dear BCnrd, could you clarify what smoothness aspect you are referring to? | |
| Dec 13, 2010 at 19:41 | comment | added | BCnrd | Dear Deane: Aha, I didn't think of the interpretation with $X$ as a single vector (since the notation $\nabla_X$ is usually used with $X$ a vector field). OK, that then makes sense, but working that pointwise then requires a separate argument to verify the smoothness aspect (i.e., recognizing how to bypass the single-vector formulation after all). If we use the 1-form formulation then we work with local smooth sections throughout, so smoothness "comes along for the ride", and the single-vector interpretation can be inferred afterwards. Well, six of one, half dozen of the other. :) | |
| Dec 13, 2010 at 19:06 | comment | added | Deane Yang | As an aside, I stumbled onto this, because the use of a pullback connection is implicit when analyzing variations of geodesics and Jacobi fields on a Riemannian manifold. I've never seen this discussed explicitly in any textbook or paper, but it's actually quite necessary to make all the arguments rigorous. | |
| Dec 13, 2010 at 18:45 | history | answered | Deane Yang | CC BY-SA 2.5 |