What happens to time component of metric in Wheeler-de-Witt equation? In the Wheeler-DeWitt equation, space-time is "foliated" and the metric $g_{\mu\nu}$ is decomposed into a metric on the surface of a 3D slice of space-time. The Wheeler-DeWitt equation is then written in terms of this intrinsic metric.
What happens to the $g_{00}$ component of the metric? Why does it get "thrown away"? i.e. the metric perpendicular to the slice of space-time which gives the time scale. Surely the time scale is very important!
 A: No components are thrown away.  When you go to a 3+1 slicing, the $00$ and the $0i$ components go into the lapse function and the shift vector.  A lot of times lapse is denoted $N$ and the shift vector $N^i$, although that's not universal.  On the page that you linked, as it currently shows, the lapse is $N$ and the shift is $\beta^i$.
This is a generic feature of 3+1 splitting, not specific to Wheeler-DeWitt.  It falls into the general category of ADM formalism. (https://en.wikipedia.org/wiki/ADM_formalism)
EDIT
In response to the comments, I'll continue the explanation a bit further, with the caveat that I'm coming at it from the classical GR side, not the quantum side.
The lapse and the shift represent your freedom to choose any coordinate system that you like.  Roughly, the lapse is the amount of coordinate time passing per unit of proper time along a geodesic from slice to slice.  The shift measures how much your coordinate position changes from slice to slice as you move along a geodesic.  On any single slice, they don't have much meaning.
In purely classical GR, there are four constrains that are related to the coordinate freedom (known usually as gauge freedom).  One Hamiltonian constraint that comes from variation with respect to lapse and a 3-vector momentum constraint that comes from variation with respect to shift.  These are, specifically, $H=0$ and $P^i=0$.  Those are satisfied on each time slice. They do not themselves contain lapse and shift if you look up their formulae, but the lapse and shift still matter because, if you take any physical slice you like, the constraints will be satisfied and you'll have $H_{,t}=0$ and $P^i_{,t}=0$ since those are the necessary conditions that the constraints are also satisfied on the "next" slice.  When you compute those time derivatives, the lapse and shift do enter.
I didn't work out the terms so somebody tell me if I'm wrong, but I believe the same must be true for Wheeler-DeWitt equation.  The equation $\hat{H} \left| \psi \right> = 0$ holds on any slice you like, where you specify the 3-metric and slice as an "initial value" to the variational calculus problem.  Then you should be able to show that it continues to hold on future (or past) slices by computing time derivatives, which will include lapse and shift terms.
