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 Sci Advisor P: 1,593 I can't figure out what George is getting at, sorry. The induced metric on should only be 3x3. Generically what you're doing is you have some global time function $t$ with gradient $dt$. You want to find the induced metric on the level sets of $dt$. The level sets of $dt$ are generated by a triplet of linearly-independent vector fields X, Y, Z such that $$dt(X) = dt(Y) = dt(Z) = 0.$$ In order that each level set be a surface, this set of vector fields needs to be integrable; that is, the set should be closed under the Lie bracket. This should hold automatically, given that $t$ is a global time function, and X, Y, Z are everywhere perpendicular to $dt$. Then the induced metric $h$ can be given by (where X and Y are some vectors within the level surface) \begin{align}h(X,Y) &= g(X,Y) = g_{tt} dt(X) dt(Y) + g_{ti} \Big( dt(X) dx^i(Y) + dt(Y) dx^i(X) \Big) + g_{ij} dx^i(X) dx^j(Y) \\ &= 0 + 0 + g_{ij} X^i Y^j. \end{align} So perhaps this is what George means by "think of $h$ as 4x4". Note that I'm assuming the vectors X and Y are already tangent to the level surfaces of $dt$. One can imagine instead a 4x4 metric on general vectors that includes some extra terms to project those vectors onto the level surfaces of $dt$. I think that is what George wrote down. But I wouldn't call that the "induced metric", since it acts on a vector space of the wrong dimension. I also see that I haven't used the covariant derivative $\nabla_\mu t$ anywhere; I'm not sure exactly why this is needed. Probably the method I am outlining above is a different route to the same result, rather than the method Wald (?) is using.