I have some trouble with refreshing ADM split and Hamilton formalism of GR in context of introducing Wheeler-de-Witt equation.

Having Lagrangian in form: $$\mathcal{L}_{ADM}=\sqrt{h}N(G^{abcd}K_{ab}K_{cd}+ {^3R})$$

where $$G^{abcd}=\frac{1}{2}(h^{ac}h^{bd}+h^{ad}h^{bc}-2h^{ab}h^{cd})$$ is supermetric, $$K_{ab}$$ is extrinsic curvature in form $$K_{ab}=\frac{1}{2N}(\dot{h}_{ab}-D_a\mathcal{N}_b-D_b\mathcal{N}_a)$$ ($$D$$'s are 3d covariant derivatives).

In order to obtain Hamiltonian constraint variation w.r.t lapse function $$N$$ has to be performed: $$\delta\mathcal{L}_{ADM}=\sqrt{h}\Big[(\delta N)(G^{abcd}K_{ab}K_{cd}+ {^3R})+N\delta(G^{abcd}K_{ab}K_{cd}+ {^3R})\Big].$$

First term will be trivial $$(=1)$$, while varying second term only extrinsic curvatures depend on $$N$$ and has to be considered, thus:

$$\delta(G^{abcd}K_{ab}K_{cd}+ {^3R})=G^{abcd}(\frac{\delta K_{ab}K_{cd}}{\delta N})=-\frac{2}{N} G^{abcd}K_{ab}K_{cd}$$

which will give after summing with first term and from requirement $$\delta (...)=0$$: $$G^{abcd}K_{ab}K_{cd}-{^3R}=0$$

Now, i have problem with obtaining variation w.r.t shift $$\mathcal{N}_a$$ to obtain momentum constraint which should be:

$$\mathcal{H}^b=-2\sqrt{h}D_a(K^{ab}-h^{ab}K)=-2\sqrt{h}D_A(G^{abcd}K_{cd})=0.$$

I dont know how to do it, $${K_{ab}}$$ part depends on shift only from covariant derivatives and i dont know how to get rid of them (integration by parts? but without any integral in expression?), 3d ricci scalar $${^3R}$$ and 3d metric $$h$$ are independent of shift.

section 20.6

Im also using Bojowalds Canonical GR and Quantum Gravitation The Feynman Path Integral Approach by H. Hamber but they are less useful (many things are ad hoc posted)

In your notation $$\mathcal{L}_{ADM}$$ is Lagrange density, i.e., what you have to play with is $$L_{ADM}=\int_\Sigma \mathcal{L}_{ADM} \, d\mu_h \,,$$ where $$d\mu_h$$ is the volume form on the hypersurface $$\Sigma$$ induced by the metric $$h$$. Keeping this in mind, and from the fact that $$\frac{\delta L_{ADM}}{\delta N} = 0\,, \quad \frac{\delta L_{ADM}}{\delta \mathcal{N}^a} = 0\,,$$ (since lapse function and shift vector are Lagrange multipliers) one gets the Hamiltonian constraint and momentum constraint equations, respectively. Now, let us consider the second equation and denote by $$\delta_{\mathcal{N}^c}$$ the variation w.r.t. the shift vector. Then, after some straightforward calculation you will get \begin{align} \delta_{\mathcal{N}^c} L_{ADM} &= 2 \int_\Sigma (D_a \delta \mathcal{N}_b K^{ab} - K h^{ab} D_a \delta \mathcal{N}_b ) \, d\mu_h \\ &= - 2 \int_\Sigma \delta \mathcal{N}_b D_a(K^{ab} - h^{ab} K) \, d\mu_h \,, \end{align} where I used integration by part and the fact that the shift vector has compact support in the second line. Using the fact that $$\delta_{\mathcal{N}^c} L_{ADM} = 0$$ will give the result.