first variation of area

Question

Does anyone know any good references or notes for proving the first variation of area under inverse mean curvature flow? I know what the result should be, but I dont understand the proofs really, and I havent learned the Lie derivative. There seems a nice proof on the bottom of page 10 here http://www.math.zju.edu.cn/swm/RG_Section_4.pdf but I dont know how they get the evolution equation for the metric. Any help would be appreciated.

Answer 1

The link you've given looks like a fine reference. Let us derive the evolution equation for the metric. To make the answer self-contained, lets recall the setup in your linked document. The symbol $M^n$ denotes an $n$-dimensional Riemannian manifold, and we consider a smooth, one-parameter family of maps $x_t : S \to M,$ so that $x_t(S) \subset M$ is a hypersurface in $M$ for each $t$. This family of maps satisfies the evolution equation

$$\frac{dx}{dt} = \beta\nu,$$ where $\beta$ is a function on $x_t(S)$. Thus, for inverse mean curvature flow, we have $\beta = 1/H$, where $H$ is the mean curvature of $x_t(S)$ in $M$. The evolution equation for the metric is $$\frac{\partial}{\partial_t}g_{ij} = 2\beta h_{ij},$$ where $h_{ij}$ is the second fundamental form of $x_t(S)$, i.e. $h(X,Y) = g(\nabla_X \nu, Y),$ and we write $h_{ij} = h(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j})$ for the coefficients of $h$ in a coordinate basis.

From the definition of Lie derivative, we have that $$L_{\nu}g(X,Y) = 2h(X,Y)$$ and moreover, in a coordinate basis $(x^1, \ldots, x^{n-1}, \nu)$, we get $$\nu g(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}) = L_{\nu}g(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}) = 2h(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j}).$$ on the other hand, $\nu = 1/\beta \frac{\partial}{\partial t},$ which when combined with the previous equation gives the result that $$ \frac{\partial}{\partial t}g_{ij} =2\beta h_{ij}$$ as desired.