Question on the first fundamental form 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?
In addition, 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 fundamental form respectively.
 A: As a matrix, $h_{\mu\nu}$ has the same dimensionality as $g_{\mu\nu}$, but it's rank is $n-1$. To see this, we note
$$
h_{\mu\nu} n^\nu = n_\mu - \sigma n^2 n_\mu = 0
$$
since $n^2 = \sigma$. The first fundamental form is strictly defined as
$$
h_{ab} = e_a^\mu e_b^\nu h_{\mu\nu}  , \qquad e_a^\mu =\frac{\partial x^\mu}{\partial y^a} . 
$$
where $y^a$ is any intrinsic choice of coordinates on the hypersurface $\Sigma$. Notice here that $\mu,\nu=0,\cdots,d-1$ whereas $a,b=1\cdots,d-1$. $h_{\mu\nu}$ is not an invertible matrix, but $h_{ab}$ is!
The lapse and shift is defined as
$$
N^a = h^{ab} e^\mu_b t_\mu , \qquad N = - n^\mu t_\mu.
$$
In terms of the lapse and shift, the metric takes the form
$$
ds^2 = - N^2 dt^2 + h_{ab} ( dy^a + N^a dt ) ( dy^b + N^b dt ).
$$
So in matrix form, we have
$$
g_{\mu\nu} = \begin{pmatrix}
- N^2 + h_{ab} N^a N^b  & h_{bc} N^c \\ h_{ac} N^c & h_{ab}
\end{pmatrix}
$$
To find the determinant, we can use row operations on this matrix. We replace the first row with $R_0 \to R_0 - N^1 R_1 - N^2 R_2 - \cdots - N^{d-1} R_{d-1}$. The determinant is invariant under row operations so we have
\begin{align}
\det ( g_{\mu\nu} ) &= \det \begin{pmatrix}
- N^2 + h_{ab} N^a N^b - N^a ( h_{ac} N^c )  & h_{bc} N^c - N^a ( h_{ab} ) \\ h_{ac} N^c & h_{ab}
\end{pmatrix} \\
&= \det \begin{pmatrix}
- N^2  & 0 \\ h_{ac} N^c & h_{ab}
\end{pmatrix}
\end{align}
We can further simplify (though its not really necessary) by doing column operations $C_0 \to C_0 - N^1 C_1 - N^2 C_2 - \cdots - N^{d-1} C_{d-1}$. Then,
\begin{align}
\det ( g_{\mu\nu} ) &= \det \begin{pmatrix}
- N^2  & 0 \\ 0 & h_{ab}
\end{pmatrix} = - N^2 \det (h_{ab} ) .
\end{align}
We therefore have
$$
g = - N^2 h \quad \implies \quad \sqrt{-g} = N \sqrt{h}.
$$
A: Suppose that $X$ is an $n$ dimensional spacetime manifold with metric tensor $g$, and $\Sigma$ is a hypersurface ($n-1$ dimensional embedded submanifold) in $X$ with $\phi:\Sigma\rightarrow X$ being the embedding. We further suppose that $\Sigma$ is pseudo-Riemannian in the sense that the pullback metric $h=\phi^\ast g$ is nondegenerate.
The usual argument shows that there exists (we assume both $X$ and $\Sigma$ are orientable) a normal 1-form $n$ along $\Sigma$ which satisfies $g(n,n)=\sigma$ and is unique up to sign. If $n^\sharp$ is the (contravariant) vector field along $\Sigma$ that corresponds to $n$ (in coordinates $n^\sharp=n^\mu\partial_\mu$ while $n=n_\mu dx^\mu$) by metric duality, then $n^\sharp$ is nowhere tangential.
We can make the following definition. Let $T$ be a tensor field of type (r,s) along $\Sigma$ (that is a smooth map $T:\Sigma\rightarrow T^{r,s}X$ which satisfies $\pi\circ T=\phi$, where $\pi:T^{r,s}X\rightarrow X$ is the bundle projection and $T^{r,s}X$ is the type (r,s) tensor bundle of $X$). We say that $T$ is tangential if it is true that any contraction of $T$ with $n$ vanishes. Let $$ T^{r,s}_\mathrm{tan}(\Sigma,X) $$ denote the subbundle of the pullback bundle $\phi^\ast T^{r,s}X$ that consists of tangential type (r,s) tensors. That is $T^{r,s}_\mathrm{tan}(\Sigma,X)$ is a vector bundle over $\Sigma$ whose elements are tangential type (r,s) tensors of $X$ whose basepoint is on $\Sigma$.
Let $T^{r,s}\Sigma$ denote the vector bundle of type (r,s) tensors of $\Sigma$.
Theorem: The vector bundles $T^{r,s}_\mathrm{tan}(\Sigma,X)$ and $T^{r,s}\Sigma$ are isomorphic.
This isomorphism is described in coordinates as follows. Let $x^\mu$ be a coordinate system in $X$ and $y^a$ a coordinate system in $\Sigma$. With respect to these charts the embedding functions take the form $$ x^\mu=\phi^\mu(y^1,...,y^{n-1}). $$ Let us use the notation $$ \phi^\mu_a:=\frac{\partial\phi^\mu}{\partial y^a}. $$ Using the normal we can also define a "pseudo-inverse" of this matrix as $$ \phi^a_\mu:=h^{ab}g_{\mu\nu}\phi^\nu_b. $$
if $T^{a_1...a_r}_{b_1...b_s}$ are the components of a type (r,s) tensor field defined in $\Sigma$, we define the corresponding tangential tensor field (of $X$) along $\Sigma$ by $$ T^{\mu_1...\mu_r}_{\nu_1...\nu_s}=\phi^{\mu_1}_{a_1}...\phi^{\mu_r}_{a_r}\phi^{b_1}_{\nu_1}...\phi^{b_s}_{\nu_s}T^{a_1...a_r}_{b_1...b_s}. $$ Conversely, if $T^{\mu_1...\mu_r}_{\nu_1...\nu_s}$ is a tangential tensor field of $X$ along $\Sigma$, we can define the corresponding intrinsic $\Sigma$-tensor field by $$ T^{a_1...a_r}_{b_1...b_s}= $\phi^{a_1}_{\mu_1}...\phi^{a_r}_{\mu_r}\phi^{\nu_1}_{b_1}...\phi^{\nu_s}_{b_s}T^{\mu_1...\mu_r}_{\nu_1...\nu_s}.$$
In this scheme, the projection tensor $$ h_{\mu\nu}=g_{\mu\nu}-\sigma n_\mu n_\nu $$ is the tangential tensor field along $\Sigma$ corresponding to the intrinsic $\Sigma$-tensor field $$ h_{ab}=\phi^\mu_a\phi^\nu_b g_{\mu\nu}. $$
