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Jan 1, 2018 at 19:00 comment added Deane Yang Asaf, I consider anything I write on MathOverflow to be informal communications which can be used by anyone without acknowledgement. If you want, I am happy to have you thank me in the acknowledgments, but it's not necessary.
Jan 1, 2018 at 17:41 comment added Asaf Shachar I wanted to thank you again for your proposals in this question. You have suggested a bundle version statement regarding the gradient of the determinant of a bundle map. I am going to use it in a paper, as a minor lemma in deriving an Euler-Lagrange equation. Of course, I am going to give you credit in the paper. If you have any objections, or would like explicit citations, please tell me. BTW, I have now learned enough about bundles, so I fully understand your meta-theorem here. It's really nice.
Apr 25, 2017 at 3:29 comment added Deane Yang Yes, what you stated in your answer is pretty close to what I have in mind.
Apr 24, 2017 at 13:59 comment added Deane Yang I'll try to state a meta-theorem, when I get a chance. It's essentially the chain rule. And it's useful.
Apr 24, 2017 at 6:43 comment added Asaf Shachar Perhaps one can try to state a meta-theorem that says "every finite dim result carries to the manifold/bundle context", but the phrasing would probably be very abstract, and not very helpful(?) (This was the essence of the question actually, I was hoping someone might come with such an approach).
Apr 24, 2017 at 6:37 comment added Asaf Shachar Indeed, the "same" proof should work (I demonstrated one possible approach) but some (admittedly not very hard) justifications need to be addressed (where the metricity come in etc). An ideal situation, from my perspective would be to find a way to view the bundle calculation as a derivative of a the determinant between fixed vector spaces (then the result would really "transfer automatically"). However, I am becoming more and more convinced that this is not possible in any "natural way", so in some sense, there are no "further shortcuts", and my/our approach is "the best" that can be done.
Apr 24, 2017 at 6:37 comment added Asaf Shachar Yes, I agree with everything you said. Working with a map between different manifolds is not a problem, because we can indeed view $df$ as a map between bundles over $M$, and use the induced connection. (Of course, this is what I did in my proof, it was just the special case where $A=df$ ($A$ is the bundle map in your notation). My point is that even this reduction to maps between bundles (with metric and connection structures) is not a trivial consequence from the finite dimensional statement, where the two vector spaces, while different, are fixed.
Apr 24, 2017 at 1:21 history answered Deane Yang CC BY-SA 3.0