# First fundamental form, induced metric and unit normal vector in general relativity

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

$$h^{ab}=g^{ab}+n^{a}n^{b}.$$

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 $$n^{i}n^{i}=0$$.

Thanks

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 $$(\ast,\ast,\ast,\ast).$$

• 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. – Topology21 Mar 23 at 2:29
• 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. – Quarto Bendir Mar 23 at 3:17
• 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. – Topology21 Mar 23 at 22:16
• added another edit – Quarto Bendir Mar 23 at 23:24