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.



On the Geroch's argument
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)


