# Derivation of equation 3.10 from ADM paper

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I am currently reading the  ADM paper  https://arxiv.org/abs/gr-qc/0405109 on General Relativity and I am having trouble deriving equations 3.10-3.11 from 3.9 a). I did get to the equation $$^4g^{0i}(N^2({^4g^{0k}g_{ik}})-N_i)=0$$ which holds if 3.11 a) is true. However I don't know how to derive it directly since the implicit sum over $i$ does not allow one to write $N^2{(^4g^{0k}g_{ik})}-N_i=0$. Any help or hint would be much appreciated!

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I eventually found a derivation. The issue is finding the right relations to use for the metric tensor since they are so many. Instead of the notation used in the paper I used $q_{ij}$ to denote the 3-metric and $g_{\mu\nu}$ for the 4-dimensional one. Here it is in case someone else needs it:

From 3.9 a) we immediately obtain $g^{00}=-\frac{1}{N^2}$.

From the orthogonality relation $g^{\mu0}g_{\mu0}=1$, we obtain

g^{00}g_{00}+g^{i0}g_{i0}=1\implies -\frac{1}{N^2}g_{00} + g^{i0}N_i=1

\implies g_{00}=-N^2(1-g^{i0}N_i).

We now use $g_{\mu i}g^{\mu 0}=0$ to get

g_{0i}g^{00}+g_{ji}g^{j0}=0\implies N_i(-\frac{1}{N^2})+q_{ij}g^{j0}=0.

Multiplying by the inverse 3-metric $q^{ik}$ and summing over $i$ gives

N^k(-\frac{1}{N^2})+\delta^k_jg^{j0}=0

where $N^k=q^{ik}N_i$. Then

N^k(-\frac{1}{N^2})+g^{k0}=0

\implies g^{k0}=\frac{N^k}{N^2}.

g_{00}=-(N^2-N^iN_i)

Further, we have

g^{\mu j}g_{\mu i}=\delta^j_i \implies g^{0j}g_{0i}+g^{kj}g_{ki}=\delta^j_i \implies g^{0j}N_i + g^{kj}q_{ki}=\delta^j_i

\implies g^{kj}q_{ki}=\delta^j_i - g^{0j}N_i

Multiplying by $q^{il}$ and summing over $i$ gives

g^{kj}\delta^l_k=q^{jl}-g^{0j}N^l

and we finally get, relabelling the indices $l \to i$ and using the symmetry property of the metric tensor,

g^{ij}=q^{ij}-\frac{N^iN^j}{N^2}.

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