How does a spatial covariant derivative act on tensors that are not purely spatial? I have a possibly dumb question on ADM formalism. Starting with a metric in ADM form
\begin{equation}
 ds^2 = -N^2dt^2 + q_{ij}(dx^i + N^idt)(dx^j + N^jdt)
\end{equation}
where $i,j$ only run over the spatial components. I'm worried that I've misunderstood how to use the spatial covariant derivative $D_i$ compatible with the spatial metric $q_{ij}$. The sources that I read seemed to imply it could be formed entirely from the spatial metric $q_{ij}$ and its inverse $q^{ij}$ using the usual symmetric connection:
\begin{equation}
 ^{(n - 1)}\Gamma^i_{jk} = \frac{1}{2}q^{im}(\partial_jq_{mk} + \partial_kq_{jm} - \partial_mq_{jk})
\end{equation}
Then for a spatial covariant derivative acting on a purely spatial vector we would have
\begin{equation}
 D_kV^m = \partial_kV^m + ^{(n-1)}\Gamma^m_{kn}V^n
\end{equation}
So what happens when a spatial covariant derivative acts on a vector that is not purely spatial? How would it act on the time component? Would it reduce to a partial derivative, such as 
\begin{equation}
D_r\xi^t \rightarrow \partial_r\xi^t
\end{equation}
Sorry for the probably dumb question but I really want to make sure I understand the formalism.
 A: An equivalent way of defining the 3-covariant derivative $D$ is to say it projects the 4-covariant derivative $\nabla$ onto the hypersurface. This is worked out explicitly in an appendix to Carroll's Spacetime and Geometry, where he writes
$$ D_\mu V^\nu = P^\alpha{}_\mu P^\nu{}_\beta \nabla_\alpha V^\beta $$
(D.48), given the projection tensor
$$ P_{\mu\nu} = g_{\mu\nu} + n_\mu n_\nu $$
(D.37). Here $\vec{n}$ is the unit timelike normal to the surface. (If it were spacelike, the "$+$" would become a "$-$" in the projection operator.)
Right away you can see we can actually take any 4-vector $\vec{V}$, apply the 3-derivative, and end up with a rank $(1,1)$ tensor that in principle lives in the full spacetime (or rather its tangent+cotangent bundle).
But given the use of projection operators (or, using the other definition, that fact that no covariant derivative should produce objects living outside the submanifold in question, given that they start there), one might expect there to be some zeros. Indeed, using the relations
\begin{align}
n_0 & = -N & n^0 & = 1/N \\
n_i & = 0 & n^i & = -N^i/N
\end{align}
we find
\begin{align}
P^0{}_\mu & = 0 \\
P^i{}_0 & = N^i \\
P^i{}_j & = \delta^i_j.
\end{align}
Thus
$$ D_i V^0 = P^\alpha{}_i P^0{}_\beta \nabla_\alpha V^\beta = 0. $$
