ADM Formulation in General Relativity In the ADM(Arnowitt – Deser – Misner) formulation, we can foliate a globally hyperbolic spacetime by spacelike hyper-surface(Cauchy surface) $\Sigma_{t}$, which parametrised by global time function $t$. Therefore, in each point of spacelike hyper-surface, we can let $n^{a}$ to be a future -directed timelike unit vector field normal to the hyper surface $\Sigma_{t}$, which satisfies $n^{a}n_{a} = -1 $ and $n_{a} \propto \nabla_{a}t$ ($\nabla_{a}$ is associated with spacetime metric. $g_{ab}$). Therefore, the spacetime metric $g_{ab}$ induces a spatial metric $\gamma_{ab}$ which is defined as 
\begin{align}
\gamma_{ab} = g_{ab} + n_{a}n_{b}
\end{align}
My difficulties are that why the spacetime metric induce such spatial metric $\gamma_{ab}$ and how can I understand the above equation.  
 A: Consider a vector of the form, 
$$ v^a = \alpha \ \sigma^a + \beta \ n^a,$$
where $\sigma^a$ is a vector orthogonal to $n^a$, i.e, $\sigma_a n^a=0$. Here $\alpha$ and $\beta$ are just some constant scalars. 
Now lets contract our vector $v^a$ with the induced metric $\gamma_{ab}$. 
$$\gamma_{ab} v^a = \gamma_{ab} ( \alpha \sigma^a + \beta n^a) $$
$$ = (g_{ab} + n_a n_b) ( \alpha \sigma^a + \beta n^a) $$
$$ = \alpha\  g_{ab} \sigma^a + \beta \ g_{ab} n^a + \alpha \ n_b (n_a \sigma^a) + \beta \ n_b (n_a n^a)  $$
The ordinary metric just lowers the indices at this step. We also apply $\sigma^a n_a = 0$ and $n_a n^a = -1$. 
$$ = \alpha\ \sigma_b + \beta \ n_b + \alpha \ n_b (0) + \beta \ n_b (-1)  $$
$$ = \alpha\ \sigma_b + \beta \ n_b +  \beta \ n_b (-1)  $$
$$ = \alpha\ \sigma_b  $$
When $\gamma_{ab}$ is contracted against vectors parallel to the surface $\Sigma$ it lowers their indices as expected (by parallel I mean "orthogonal to $n^a$"). When $\gamma_{ab}$ is contracted against vectors normal to the surface the result is $0$. $\gamma_{ab}$ then acts as a projection onto the hypersurface and as a metric within that surface. 

Lets consider a concrete example in Euclidean space. We will be working in $\mathbb{R}^3$ using cartesian coordinates; which means our metric will be $g_{ij} = \delta_{ij}$. We will foliate the space with spheres centered on the origin. Each sphere will have a radius, $r=\sqrt{x^2+y^2+z^2}$. 
Our normalized unit vector will be $\hat{n} = \nabla r / \| \nabla r \|$. However this has a norm which is positive $\| \hat{n} \| = 1$, so to obtain $\gamma_{ij}$ we will need to subtract it from $g_{ij}$ rather than add it. 
$$\boxed{ \gamma_{ij} = g_{ij} - n_i n_j }$$
Now the following are the normalized basis vectors (vector fields really) for spherical coordinates. 
$$ \hat{e}_r = \frac{\nabla r }{\| \nabla r\|} = \begin{bmatrix} x/r \\ y/r \\ z/r \end{bmatrix}$$
$$ \hat{e}_\phi = \frac{\nabla \phi }{\| \nabla \phi\|} = \begin{bmatrix} -y/\sqrt{x^2+y^2} \\ x/\sqrt{x^2+y^2} \\ 0 \end{bmatrix}$$
$$ \hat{e}_\theta = \frac{\nabla \theta}{\| \nabla \theta \|} = \begin{bmatrix} \frac{zx}{r\sqrt{x^2+y^2}} \\ \frac{zy}{r\sqrt{x^2+y^2}} \\ -\frac{x^2+y^2}{r\sqrt{x^2+y^2}}\end{bmatrix}$$
Since these vectors form a basis any vector can be resoled in terms of them. 
$$ \vec{v} = \hat{e}_r (\hat{e}_r \cdot \vec{v}) + \hat{e}_\phi (\hat{e}_\phi \cdot \vec{v}) + \hat{e}_\theta (\hat{e}_\theta \cdot \vec{v})$$
$$ \vec{v} = \sum_{\alpha=r,\phi,\theta} \hat{e}_\alpha (\hat{e}_\alpha \cdot \vec{v})$$
$$ \vec{v} = \sum_{\alpha=r,\phi,\theta} \hat{e}_\alpha  (\hat{e}_{\alpha})_i v^i$$
$$ v_j = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i v^i$$
$$ g_{ij} v^i = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i v^i$$
Since the equality must hold for any vector $\vec{v}$ we can remove it from the equations and we obtain an identity.
$$g_{ij} v^i = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i v^i$$
$$\boxed{ g_{ij} = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i}$$
If you have experience in quantum mechanics you should compare this with $\sum |n \rangle \langle n| = 1$ for a complete set of states. Now lets return to our expression for $\gamma_{ij}$ in light of this identity. 
$$ \gamma_{ij} = g_{ij} - n_i n_j$$
$$ \gamma_{ij} = g_{ij} - (\hat{e}_r)_i (\hat{e}_r)_j$$
$$ \gamma_{ij} = \sum_{\alpha=r,\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i - (\hat{e}_r)_i (\hat{e}_r)_j$$
$$\boxed{ \gamma_{ij} = \sum_{\alpha=\phi,\theta} (\hat{e}_\alpha)_j  (\hat{e}_{\alpha})_i }$$
You can see we are literally subtracting out the part of the metric that has to do with $r$. 
You can use the explicit vectors that I wrote down for $\hat{e}_\phi$ and $\hat{e}_\theta$ to construct a matrix for $\gamma_{ij}$ in the Cartesian basis. If you diagonalize this matrix you will find that $\hat{e}_r$, $\hat{e}_\phi$, and $\hat{e}_\theta$ are eigenvectors where $\hat{e}_r$ will have an eigenvalue of $0$. The diagonal form will be the metric for a 2-sphere which you should recognize from your studies, along with a 0 in the diagonal entry corresponding to $r$.  
