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 
 A: 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 $(\ast,\ast,\ast,\ast).$
