On the Geroch's argument During the study of Geroch's argument to prove positive mass theorem, I faced a problem explained below:
Suppose $(M,g_{\mu \nu})$ is a four dimensional Lorentzian Manifold and $\Sigma$ is a submanifold of $M$ with induced metric $h_{ab}$ and extrinsic curvature $K_{ab}$. 
We introduce a function $\tau$ on $\Sigma$ such that the two dimensional surfaces  $\tau= \text{constant}$ in $\Sigma$ are
nested topological 2-spheres with the innermost surface reducing to a point. For each value
of $\tau$ let us assume that $S\subset \Sigma$   is one such surface and $\eta_a = \nabla_a\tau$ defines the normal to $S$ . The
unit normal is then given by $n_a = (\eta.\eta)^{-1/2}\eta_a$. Let $\xi^a:=un^a$ has the property that $\xi^a\nabla_a\tau = 1$.
Let $v^2:= \eta^a\eta_a$. So we have $\xi^a\nabla_a\tau = uv = 1$. So $u =1/v = (\eta^a\eta_a)^{-1/2}$. Next we consider the function $C(\tau)$ which for each value of $\tau$ is defined as
$$C(\tau):=\int_{S\subset \Sigma}(2\mathcal{R}-k^2)dA$$
where the integration extends over the surface $S$ and $\mathcal{R}$ and $k$ denote the scalar curvature and
the trace of the extrinsic curvature of the surface $S$ as a submanifold of $\Sigma$, respectively.
We note that the Gauss-Bonnet theorem implies that
$$\int_{S\subset \Sigma}\mathcal{R}dA=8\pi.$$
The trace of the extrinsic curvature $k$ of $S\subset \Sigma$   is defined as
$$k=\nabla_an^a.$$
The rate of change of any quantity with respect to $\tau$ is its Lie derivative by $\xi^a$. The rate of change of $k$ with respect to $\tau$ is then given by
$$\frac{\partial}{\partial \tau}‎‎k=\xi^b\nabla_b(\nabla_an^a)=\xi^b ‎\nabla_a\nabla_bn^a-\xi^bR^a_{\;mab}‎n^m=\xi^b ‎\nabla_a\nabla_bn^a-uR_{mb}n^bn^m.$$‎‎‎
Next after a calculation using the Gauss-Codazzi equation we arrive at the result that
$$‎‎‎\frac{\partial}{\partial \tau}‎C(\tau)=‎\int_{S‎\subset ‎\Sigma}(2kD^aD_au+ukk^{ab}k_{ab}-uk\mathcal{R}+ukR)dA.‎$$
where $D_a$ is the covariant derivative operator on the 2-surface $S$ with respect to the induced metric.
Question1: I don't understand why the bold statement is true. Can someone help me?
Question2: My efforts for deriving the last equation are failed. Can someone point me in the right direction? It is very important for me.
Thanks in advance for your time.
 A: I think that I misunderstood the notation in Question 1. The bold statement is incorrect. Consider the function $\tau(x) = 2x$ on $\mathbb{R}$ and the function $k(x) : = x$ on $\mathbb{R}$. Then $S_t = \{t/2\}$. So, at time $t$, the value of $k(x)$ on $S_t$ is $t/2$. So, the rate of change is $1/2$. On the other hand $\xi$ would be $\frac{\partial}{\partial x}$, so $\mathscr{L}_\xi k = 1$. 

As for question 2, I am not going to do the computation for you, but let me explain in modern notation how to understand. I will write $S_c:=\{\tau=t\}$ and $H$ for the mean curvature of $S_t$. Consider the quantity
$$
\int_{S_t} H^2 d\mu_t
$$
We let $u = |\nabla \tau|^{-1}$ as above. This is often called the lapse function. Now, how do we compute the rate of change of this integral?
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \frac{d}{dt} H^2 d\mu_t + \int_{S_t} H^2 \frac{d}{dt} d\mu_t
$$
(here, I am thinking of integrating over a fixed, abstract sphere, where the function $H$ and measure $\mu_t$ are time dependent).
What is the first term? To differentiate $H$, we must use the second variation formula,
giving
$$
\frac{d}{dt} H = -\Delta_{S_t} u -(Ric(\xi,\xi)+\Vert h \Vert^2)u.
$$
Here, $h$ is the second fundamental form. To differentiate the second term, one should use the first variation formula, giving
$$
\frac{d}{dt} d\mu_t = uH d\mu_t.
$$
Thus, putting these together yields
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \left(-2H \Delta u - 2H(Ric(\xi,\xi) + \Vert h\Vert^2)u + uH^3 \right) d\mu_t
$$
Now, using the Gauss equations, we have that 
$$
2(Ric(\xi,\xi) + \Vert h\Vert^2) = R-\mathcal{R}+\Vert h \Vert^2 + H^2 
$$
Inserting this into the above equation yields
$$
\frac{d}{dt}\int_{S_t} H^2 d\mu_t = \int_{S_t} \left(-2H \Delta u - uHR +uH\mathcal{R} -uH\Vert h\Vert^2  \right) d\mu_t
$$
This is exactly your equation.

I'll mention that an apt choice is $u=\frac{1}{H}$. This yields the so-called inverse mean curvature flow. It would be very instructive for you to plug this in and try to find a nice differential inequality for what you call $C(\tau)$ assuming that $R\geq0$. 
Of course, Geroch is unconcerned with the existence of such a function $\tau$. This turns out to be a very serious problem, and was only recently solved in the beautiful work of Huisken--Ilmanen in their proof of the Penrose inequality: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1090349447. In particular, the computation I have just done is contained in this article for the special case of inverse mean curvature flow (as this is the only case that is really important)
