Mean Curvature flow: Evolution equation of any invariant symmetric homogeneous polynomial with input the Weingarten map.

Question

I have the following evolution equations realted to mean curavture flow, with the induced metric $g=\{g_{ij}\}$, measure $d\mu$ and second fundamental form $A=\{h_{ij}\}$:

1)$\frac{\partial}{\partial t}g_{ij}=-2Hh_{ij}$

2)$\frac{\partial}{\partial t}d\mu=-H^2d\mu$

3)$\frac{\partial}{\partial t}h_{ij}=\Delta h_{ij}-2Hh_{il}h^l_j+|A|^2h_{ij}$

Now we consider the Weingarten map $W:T_pM\to T_pM$ associated with $A$ and $g$, given by the matrix $\{h^i_j\}=\{g^{il}h_{lj}\}$, and let $P$ be any invariant symmetric homogeneous polynomial.

Then I wish to prove the result.

If $W=\{h^i_j\}$ is the Weingarten map and $P(W)$ is an invariant polynomial of degree $\alpha$, i.e. $P(\rho W)=\rho^{\alpha}P(W)$, then

1) $\frac{\partial}{\partial t}h^i_j=\Delta h_j^i+|A|^2h_j^i$

2)$\frac{\partial}{\partial t}P=\Delta P- \frac{\partial^2 P}{\partial h_{ij}\partial h_{pq}}\nabla_lh_{ij}\nabla_lh_{pq}+\alpha|A|^2P$

I'm able to prove the first part as follows,

$\frac{\partial}{\partial t}h^i_j=\frac{\partial}{\partial t}(g^{ik}h_{jk}) =\frac{\partial}{\partial t}(g^{ik})h_{jk}+\frac{\partial}{\partial t}(h_{jk})g^{ik}$

$=2Hg^{is}h_{sl}g^{lk}h_{jk}+(\nabla h_{jk}-2Hh_{jl}h^l_k+|A|^2h_{jk})g^{ik}$

$=\Delta h_j^i+|A|^2H_j^i$.

To work out $\frac{\partial}{\partial t}g^{ij}$ you just need to use the fact that $g^{ij}g_{ij}=1$

I think I'm missing a simple useful result to get part 2) out?

Answer 1

I'm guessing the equality you need is

$$h^i_j \frac{\partial P}{\partial h^i_j} = \alpha P.$$

To see that this is true, note that the LHS is the derivative of $P$ in the outwards radial direction; i.e.

$$ h^i_j \frac{\partial P}{\partial h^i_j} = \frac{d}{dt}\Big|_{t=0} P((1+t)W).$$

If we apply the $\alpha$-homogeneity of $P$ to the RHS we get $$\frac{d(1+t)^\alpha}{dt}\Big|_{t=0} P(W)=\alpha P(W).$$

Answer 2

I've found this to be the best way to do it. I won't post a full solution but a rough sketch,

$\partial_tP=\sum(h\cdots h)_t$

$=\sum\sum(h_t h\cdots h)$

$=\sum\sum((\Delta h) h\cdots h)+\sum\sum|A|^2h\cdots h)$

$=\sum\sum((\Delta h)h\cdots h)+\alpha|A|^2P$

Now $\Delta P=g^{ij}(\sum h\cdots h)_{ij}$

$=g^{ij}(\sum\sum h_{ij} h\cdots h +\sum\sum h_ih_jh\cdots h)$

$=\sum\sum(\Delta h)h\cdots h+\frac{\partial^2 P}{\partial h \partial h}\nabla h\nabla h$.

Putting this all together gives the required result.