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Jul 11, 2018 at 22:00 comment added Deane Yang Note that, if all you want is the top order term, then you can ignore all terms involving the derivatives of the metric tensor (equivalently, the Christoffel symbols and their derivatives). You can also do the calculation at a single point and assume that $g_{ij} = \delta_{ij}$ at that point. This simplifies the calculations.
Jul 11, 2018 at 19:50 answer added Piotr Hajlasz timeline score: 5
May 23, 2018 at 20:37 comment added Felix Schlag If there is interest, I will post the proof, but these days I am too busy.
May 22, 2018 at 5:54 comment added Piotr Hajlasz Very good question. Actually I have been thinking about it recently since I need it in my research.
May 19, 2018 at 14:17 comment added Felix Schlag I think I proofed everything now, but my proof involves very shady business with the metric tensor.
May 19, 2018 at 13:20 history edited Felix Schlag CC BY-SA 4.0
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May 19, 2018 at 13:19 comment added Felix Schlag By hand, I could also veryfy all assertions for k=1
May 19, 2018 at 12:54 review First posts
May 19, 2018 at 13:13
May 19, 2018 at 12:53 history asked Felix Schlag CC BY-SA 4.0