Dimensionality and determinant of the first fundamental form

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The first fundamental form is related to the metric tensor of the manifold as follows:

$$h_{ab} = g_{ab} - \sigma \text{ } n_a n_b$$

Where $\sigma$ is +1 or -1 depending on normalization of the normal vector, and $n_a$ is the component of the normal vector. As I understand it, the first fundamental form is of the same dimensionality as the metric tensor with each of the indices running from 1 to n. However, some people refer to the first fundemental form as the pullback of the metric on to the surface. This implies that the form is of dimensionality n-1. What is the relation between these two definitions?

The second question is maybe more of a general relativity question than mathematics but if we define the lapse function and the shift vector as the following respectively:

$$N^a = h^a_bt^b$$ and $$N = -g_{ab}t^an^b$$ Where $n^b$ is the normal to the hypersurface, and $t^a$ is some vector field on the manifold. How do we arrive at the following relation between the determinants of the metrics?

$$\sqrt{-g} = N\sqrt{h} $$ Where N is the lapse function and $g$ and $h$ are the determinants of the metric and the first fundemental form respectively.

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The first fundamental form may be thought of as either an $(n - 1)$ dimensional object (the pullback of the ambient metric) or as an $n$ dimensional object, by acting as zero on any vector which is not tangent to the hypersurface. By the way, it looks like you're considering a spacelike hypersurface; the following is not true for a null hypersurface, as in this case you cannot split the ambient tangent space into tangent space of the hypersurface direct sum its orthogonal complement.

Your formula formally gives $h$ as an $n$ dimensional object, but note that if you input to $h$ a vector which lies only along $n$, you get zero. The form $h$ does not "see" orthogonal components, and so may be thought of as intrinsic to the hypersurface, and $(n - 1)$-dimensional.

The second question: if it is coordinate-invariant, it is easiest to pick the shift to be zero and write down the matrix for $g$ in coordinates at a point, in which case it is a block-diagonal matrix with $-N$ in the top-left entry and $h$ in the bottom $3\times 3$ entry.