ADM decomposition of the general scalar tensor theory Lagrangian

Question

I have question about ADM decomposition of some general scalar-tensor theory of gravity.
Starting with ADM form of the metric:
$$ds^2=-N^2dt^2+h_{ij}(dx^i+N^idt)(dx^j+N^jdt)$$
provided with extrinsic curvature:
$$K_{ij}=frac{1}{2N}(dot{h_{ij}}-D_iN_j-D_jN_i)$$,
where $N$ is lapse function and $N^i$'s are shift vector components. Timelike normal vector to the hypersurface is denoted as $n_a$.
I'm considering given action:
$$int d^4xsqrt{-g}[f(phi) R-nabla_muphinabla^muphi + U(phi)]$$
and I want to recast this Lagrangian in the 3+1 form which is suitable for discussing Hamiltonian formulation of this theory.
Kinetic term $(X=nabla_muphinabla^muphi)$ decomposes as:
$$
X=-A_*^2+D^iphi D_iphi
$$
where:
$A_*=n^munabla_mu=frac
{1}{N}(dot{phi}-N^iD_iphi)$ and $D^i$ is 3d covariant derivative associated with metric $h_{ab}$ ($h^a_bnabla_a=D_b$).
However, i have trouble with the first part of the action:
begin{align}
int d^4xsqrt{-g}f(phi) R=int dtint d^3x sqrt{-h}NBig[f(phi)&big(R^{(3)}+K_{ij}K^{ij}-K^2
&+2nabla_{mu}(n^munabla_nu n^nu-n^nunabla_nu n^mu)big)Big].
end{align}
First term is left as it is, while term involving derivatives of normal vector $n^mu$ needs to be integrated by parts (dropping boundary term) - normally in GR this is total divergence and is discarded (this is not the case here):
begin{align}
int dtd^3xsqrt{-h}N  f(phi) 2nabla_{mu}(n^munabla_nu n^nu-n^nunabla_nu n^mu) 
&=-2int dtd^3xsqrt{-h}N(n^munabla_nu n^nu-n^nunabla_nu n^mu)nabla_mu f(phi) 
&= -2int dtd^3xsqrt{-h}N(n^munabla_nu n^nu)f_phinabla_muphi
&quad+2int dt d^3xsqrt{-h}N(n^nunabla_nu n^mu)f_phi nabla_mu phi
&=-2int dt d^3xsqrt{-h}N(n^mu Kf_phinabla_muphi-n^nu(nabla_nu n^mu) f_phi nabla_mu phi)
&=-2int dt d^3xsqrt{-h}N (Kf_phi A_*-(nabla_nu n^mu )n^nu 
 f_phinabla_mu phi)
end{align}
where  I used identity $K=nabla_nu n^nu$ and $nabla_alpha f=f_phi nabla_alpha phi$ (chain rule).
I have trouble with the last part of the above equation - i have no idea how to simplify this expression and put it into 3+1 form.
Here are some references that I'm trying to follow:
https://arxiv.org/abs/1101.3403,
https://arxiv.org/abs/1708.02951,
https://arxiv.org/abs/1812.02667,
https://arxiv.org/abs/1512.06820.

haelewiin · Accepted Answer

Before we start
Let's take a few steps back. The ADM decomposition implies the existence of a globally defined scalar field $T$. The spatial hypersurfaces are then the level surfaces of $T$. The unit normal (co)vector $n_mu$  can be constructed as
$$
n_mu = -Nnabla_mu T.tag{1}label{normal}
$$
The lapse function $N$ shows up as a normalisation factor. The metric $g_{munu}$ then decomposes as
$$
g_{munu} = h_{munu} - n_mu n_nu.tag{2}label{metric}
$$
In this decomposition $h_{munu}$ is the embedding of your spatial 3-metric $h_{ab}$ in spacetime.
It is useful to define projectors parallel ($P_{parallel}{}^{alpha}_{mu}$) and orthogonal ($P_{perp}{}^{alpha}_{mu}$) to $n_mu$:
begin{align}
P_{parallel}{}^{alpha}_{mu} &equiv - n_mu n^alpha,
P_{perp}{}^{alpha}_{mu} &equiv delta_mu^alpha - P_{parallel}{}^{alpha}_{mu} = delta_mu^alpha + n_mu n^alpha.
end{align}
One can then construct spatial tensors by taking a tensor in your spacetime, and applying $P_{parallel}$ to each component. In particular, for the gradient of  a scalar field $f$ one has
$$
D_mu f = P_{perp}{}^alpha_mu nabla_alpha f.
$$
Now, $D_mu$ is the covariant derivative compatible with the spatial metric $h_{munu}$. You can check this by using the decomposition eqref{metric}.
The derivative $D_mu$ is equivalent to your $D_i$.
Regarding your question
The last term in your expression can be rewritten in terms of the quantities that I have defined above. With eqref{normal} one can show that
$$
n^nunabla_nu n_mu = P_{perp}{}^{nu}_{mu}nabla_nulog N = D_mulog{N}.
$$
Since this is a spatial vector, the only part of $f_phinabla_muphi$ that survives must also be spatial, to wit $D_mu f$. Your term, in full, is then
$$
2int {rm d}t{rm d}^3xsqrt{h}N D^mulog{N} D_mu f =
2int {rm d}t{rm d}^3xsqrt{h} left[D^mu(ND_mu f) - NDelta fright],
$$
where $Delta = D_mu D^mu$ is the Laplacian on the spatial hypersurface. The first term is a total spatial derivative and can be neglected, as you did before.
Note that $h_{munu}$ is positive definite, and so has a positive determinant.
References

Reference on the ADM decomposition: https://arxiv.org/abs/gr-qc/0703035
an example where this calculation is done: https://arxiv.org/abs/1712.08543

ADM decomposition of the general scalar tensor theory Lagrangian

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