The inverse of the induced metric on a hypersurface which foliates spacetime can be expressed as


To my knowledge in terms of the lapse and shift

$$g^{00}=-\frac{1}{N^{2}}\quad\text{and}\quad g^{i0}=\frac{N^{i}}{N^{2}}.$$

Also $n^{0}$ equals $\frac{1}{N}$ and $n^{a}n_{a}=-1$. Now I believe that $h^{00}=0$ and $h^{i0}=0$. Thus its follows that $h^{i0}=g^{i0}+n^{i}n^{0}$ and $n^{i}n^{0}\neq0$. However if $h^{ii}=g^{ii}$ then $n^{i}n^{i}=0$ which makes no sense. I should mention i varies from 1 to 3 here.

I'm sure I have misunderstood something about the first fundamental form, the induced metric and the unit normal vector $n^{a}$ to the hypersurface. Can someone please explain what have I misunderstood which leaves me getting this nonsensical result that $n^{i}n^{0}\neq0$ while also $n^{i}n^{i}=0$.



Let $\Sigma$ be a spacelike hypersurface in the Lorentzian manifold $M$.

Define $h_{ab}=g_{ab}+g_{ac}g_{bd}n^cn^d$. It can be seen case-by-case that

  • if $X$ and $Y$ are any vectors tangent to $\Sigma$, then $h(X,Y)=g(X,Y)$
  • if $X$ is tangent to $\Sigma$, then $h(X,n)=0$
  • $h(n,n)=0$

However, given any local coordinate system for $M$, the matrix $[h_{ab}]$ is not invertible. For instance, in a local orthonormal frame $e_0,e_1,e_2,e_3$ where $e_1,e_2,e_3$ are tangential to $\Sigma$, the matrix $[h_{ab}]$ is diagonal with components $0,1,1,1$, and hence not invertible.

However $h$ is a 2-tensor on $M$, and so its indices can be raised by using $g$, i.e. one may define $h^{ab}$ to be given by $g^{ap}g^{bq}h_{pq}$. Following this definition, one has $h^{ab}=g^{ab}+n^an^b$. But it is not the case that the matrix $[h^{ab}]$ is the inverse of the matrix $[h_{ab}]$, as neither is even invertible.

Edit: on the ADM formalism, let $x_0,x_1,x_2,x_3$ be some local coordinates, and consider a $\{x_0=\text{constant}\}$ hypersurface. A normal vector is given by $g^{0a}$, since $g_{ab}g^{0a}X^b = X^0$, which vanishes if $X$ is tangent to the hypersurface. The length of this normal vector is $g_{ab}g^{0a}g^{0b} = g^{00}$. So one has that $\frac{g^{0a}}{\sqrt{-g^{00}}}$ is a unit normal vector to the hypersurface. And so $$h^{ab}=g^{ab}-\frac{g^{0a}g^{0b}}{g^{00}}.$$ So $h^{a0}=g^{a0}-\frac{g^{0a}g^{00}}{g^{00}}=0$ for any $a=0,1,2,3.$

Second edit, answering a question in the comments: in this context one has, following the definition at the beginning of my answer, $$h_{ab}=g_{ab}+g_{ac}g_{bd}\frac{g^{0c}}{\sqrt{-g^{00}}}\frac{g^{0d}}{\sqrt{-g^{00}}}=g_{ab}-\frac{\delta_a^0\delta_b^0}{g^{00}}$$ so the matrix $[h_{ab}]$ is identical to the matrix $[g_{ab}]$ except in its first diagonal entry. This is consistent with the matrix $[h^{ab}]$ not being identical to the matrix $[g^{ab}]$ in any entry, since schematically one gets $[h^\bullet]$ as $[g^{-1}][h_\bullet][g^{-1}]$, a matrix multiplication which scrambles all of the components of $h_\bullet$ together unless $[g^{-1}]$ happens to be of a special form.

Put a bit more directly, $n_a$ (in this specific context) is of the form $(\ast,0,0,0)$ but $n^a$ is of the form $(\ast,\ast,\ast,\ast).$

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    $\begingroup$ Thanks a lot for the input Quarto I appreciate it. Ah yes I was aware that in coordinate form $h_{a b}$ has no inverse because its determinant vanishes due to zeros in its time components, but when you only consider the subspace of M, where $h_{i j}$ acts on only spatial 3 vectors tangent to $\Sigma$ then you can define an inverse. I'm still unclear though on why it appears the unit normal $n^{i}$ has this weird behavior that I detailed in my question. $\endgroup$ – Topology21 Mar 23 at 2:29
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    $\begingroup$ I've added a part to my answer. I was a bit confused by the context of your question, so I've taken the specific context that the hypersurface is locally defined as a $\{x^0=\text{const}\}$ via some local coordinates. $\endgroup$ – Quarto Bendir Mar 23 at 3:17
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    $\begingroup$ I am certainly understanding it better now. I was mistaken in thinking that $g^{a 0} = N^{a}$ the shift which is a tangent vector to $\Sigma$. However I'm still confused, if $n^{a}=\frac{g^{0 a}}{\sqrt{g^{00}}}$ and if $g^{a 0} \neq 0$ then it follows that $n^{a}$ and $n^{0}$ does not equal zero as well. However if we let i and j range from 1 to 3 and if $h^{i j} = g^{i j}$, then $h^{i j}= g^{i j} +\frac{ (g^{0 i}g^{0 j})}{g^{0 0}}$, which results in $\frac{ (g^{0 i}g^{0 j})}{g^{0 0}}=0$ which doesn't makes sense if $\frac{ (g^{0 i}g^{0 0})}{g^{0 0}} \neq 0$. Thanks a lot for the help so far. $\endgroup$ – Topology21 Mar 23 at 22:16
  • $\begingroup$ added another edit $\endgroup$ – Quarto Bendir Mar 23 at 23:24

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