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I am trying to prove the invariance of the ADM formalism under (infinitesimal) diffeomorphisms. I have checked Wald and other textbooks on the subject but have been unable to find expressions for how the lapse, shift and extrinsic curvature transform under diffeomorphisms. If anyone knows the relations, or could point me to a reference where they are presented I would be extremely grateful.

What I am looking for is of the form

\begin{align} N\longrightarrow N'=F_{1}(N,\xi)\\ N_{i}\longrightarrow N_{i}'=F_{2}(N_{i},\xi)\\ K_{ij}\longrightarrow K_{ij}'=F_{3}(K_{ij},\xi) \end{align}

Where $N$ is the lapse, $N_{i}$ the shift, $K_{ij}$ the extrinsic curvature and $\xi$ an arbitrary infinitesimal diffeomorphism. This would effectively be similar to the $A_{\mu}(x)\longrightarrow A_{\mu}(x)+\partial_{\mu}\phi(x)$ from EM (where $\phi(x)$ is an arbitrary scalar field).

I know that the ADM 3-metric $\gamma_{ij}$ varies as $\gamma_{ij}\longrightarrow \gamma_{ij}+\xi_{i|j}+\xi_{j|i}$, where $|$ indicates covariant derivation using the 3-metric. It should be possible to construct the required transformations above from this, but I am still unsure about the manipulations, for example I have not been able to figure out how $\det\gamma$ varies, given the variation of $\gamma_{ij}$.

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  • $\begingroup$ Any given metric, in any given coordinate system, can be organized into a lapse, shift, and spatial part. Since you know how a generic metric transforms (just like any other rank-2 tensor), you can just transform the metric and then re-write the new one in terms of a lapse, shift, and spatial part. By comparing these to the ones in the original metric, you can read of how they transformed. (They don't transform like tensors, because their definitions are not coordinate-independent.) Does this address your question? $\endgroup$ – Chiral Anomaly Feb 15 at 21:43
  • $\begingroup$ Unless your differomorphism respects the slicing, I doubt that you are going to get some very insightfull results. $\endgroup$ – magma Feb 16 at 5:17
  • $\begingroup$ @DanYand I'm not sure whether the following is correct but I thought I'd give it a go based on your comment. Given the 4D metric $g_{\mu\nu}$, which transforms to $g^{\prime}_{\mu\nu}=g_{\mu\nu}+\xi_{\mu;\nu}+\xi_{\nu;\mu}$, the lapse is defined as $N:=(-g^{00})^{-1/2}$ so it should transform to $N'=(-g^{00}-2\xi^{0;0})$? The shift is defined as $N_{i}=g_{0i}$, so it should transform to $N^{\prime}_{i}=g_{0i}+\xi_{0;i}+\xi_{i;0}$. The problem for me is the extrinsic curvature, which has a more complicated formula, as, for example, I don't know how $\det\gamma$ transforms. $\endgroup$ – Carlo Palazzi Feb 17 at 8:47
  • $\begingroup$ Yes, that's exactly what I had in mind. I haven't worked it out myself, but what you wrote looks correct, with the understanding that you're considering infinitesimal transformations. To handle $\det\gamma$, you can use this identity: $\delta \det\gamma=(\det\gamma)\gamma^{jk}\delta \gamma_{jk}$. (Hopefully I got the details right; it's something like that.) I can post an answer with the derivation of that identity, if that would be helpful. $\endgroup$ – Chiral Anomaly Feb 17 at 16:27
  • $\begingroup$ @DanYand, your first comment makes it sound as though all rank-2 tensors should transform in the same way under infinitesimal diffeomorphisms. This question and Weinberg reference seem to say the same thing physics.stackexchange.com/questions/174651/…. So should $K_{ij}$ just transform to $K^{\prime}_{ij}=K_{ij}+\xi_{i|j}+\xi_{j|i}$? $\endgroup$ – Carlo Palazzi Feb 18 at 12:00

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