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When I try to compute $g^{ij}g^{pq}\nabla_{i,j}^2h_{pq}=\Delta tr_gh$, I need $g^{ij}=\delta^{ij},\nabla_i g_{kl}=0$,i.e, it's normal coordinate. But $g(t)$ will change with $t$, only for a $t_0$, I can use normal coordinate. So, whether $t=0$ is missed in the proof in picture below ?

The picture is from 59th page of this paper.

enter image description here

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    $\begingroup$ Not completely sure I understand your question, but if I'm not mistaken, you should be able to fix a point $p$ arbitrarily, then use normal coordinates at $p$ to show the indicated PDE holds at $p$ and at $t = 0$, and finally conclude the PDE holds in the large for arbitrary time because the terms have coordinate-independent meaning. $\endgroup$ Commented Nov 20, 2015 at 11:29
  • $\begingroup$ @AndrewD.Hwang Yes, your understand is right. I am fuzzy with coordinate-independent.I mean that I can't understand the terms have coordinate-independent.Is it because they are tensor ? But how to deal with time ? Could you give a detail answer ? Thanks. $\endgroup$
    – Farmer
    Commented Nov 20, 2015 at 11:36
  • $\begingroup$ No time at the moment, but if I can put together a convincing argument in the next day or two I'll write it up. On the other hand, I'll be Perfectly Happy if someone posts a clear, concise explanation in the meantime. :) $\endgroup$ Commented Nov 20, 2015 at 11:47
  • $\begingroup$ @AndrewD.Hwang Thanks :) $\endgroup$
    – Farmer
    Commented Nov 20, 2015 at 12:04

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I think you are missing the essence of using normal coordinate, if you are able to derive $$g^{ij} g^{pq} \nabla^2_{ij} h_{pq} = \Delta \text{tr}_g h$$ using normal coordinate, then the above equation holds under ANY coordinate, as both sides of the equations are tensor fields. So to derive that equality, fix a $p$ and a $t$, then assume $g_t$ is the normal coordinate at $p$ and compute.

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  • $\begingroup$ In fact ,I don't know why tensor is independent to coordinate. But $g^{ij} g^{pq} \nabla^2_{ij} h_{pq} = \Delta \text{tr}_g h$ still be a question for me .Only under normal coordinate I can get it ... $\endgroup$
    – Farmer
    Commented Nov 20, 2015 at 13:22
  • $\begingroup$ I might just give an example: If $A_{ij} = B_{ij}$ under some coordinates, then under any other coordinate, $$A_{\alpha\beta} = F^i_\alpha F^j_\beta A_{ij} = F^i_\alpha F^j_\beta B_{ij} = B_{\alpha\beta}$$ @lanse2pty $\endgroup$
    – user99914
    Commented Nov 20, 2015 at 13:33
  • $\begingroup$ I get it ,because tensor is multilinear ,so ,as your example, it it independent to coordinate. $\endgroup$
    – Farmer
    Commented Nov 21, 2015 at 2:15

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