# 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.

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)

• Thanks for the answer. Can you explain me why it isn't true that: the rate of change of any quantity with respect to $τ$ is its Lie derivative by $n^a$? – Sepideh Bakhoda Dec 7 '13 at 7:00
• sorry, I got confused with the notation in the answer for the question 1. what you have bolded is false. what you write in the comment is true. I updated my answer. – Otis Chodosh Dec 7 '13 at 7:20
• Please see Geroch's paper here. I thought what I write in the comment is true, but during the study of this paper I surprised and got confused. Geroch has written what I have bolded. – Sepideh Bakhoda Dec 7 '13 at 7:35
• Please read Geroch's paper more carefully, you have made a mistake in transcribing what he writes. – Otis Chodosh Dec 7 '13 at 16:29
• I don't copy and paste Geroch's notation in the post. Geroch has written A dot, affixed to a quantity, will denote its rate of change with respect to t (i.e., its Lie derivative by $\phi \xi^a$), while in his work $\xi^a$ is unit normal, not $\phi \xi^a$! – Sepideh Bakhoda Dec 7 '13 at 16:46