0
$\begingroup$

Picture below is from 1706th page of Zhao Liang's The first eigenvalue of Laplace operator under powers of mean curvature flow.

$g_{ij}$ is Riemannian metric, $h_{ij}$ is second fundamental form, and $H=g^{ij}h_{ij}$. $\nabla$ is Riemannian connect, and $\nabla^pH=g^{pl}\nabla_lH$. $h$ is a constant independ to $x$.

For getting the below equation, I get stuck at $$ g^{ij}g^{pl}H^k\nabla_lh_{ij}-2g^{ij}g^{pl}H^k\nabla_ih_{jl}=-H^k\nabla^pH $$ How should I to do it ?

enter image description here

$\endgroup$
2
$\begingroup$

Note

$$g^{ij} \nabla_l h_{ij} = \nabla_l (g^{ij} h_{ij}) = \nabla _l H$$

as $\nabla_l g^{ij} = 0$. Thus

$$g^{ij}g^{pl}H^k\nabla_lh_{ij}=g^{pl} H^k\nabla_l H = H^k\nabla^pH$$

by definition of $\nabla ^p = g^{pl}\nabla_l$.

For the next term, it is important to remark that one is working on hypersurfaces in $\mathbb R^{n+1}$. In particular the ambient curvature is zero and the Codazzi equation gives

$$\nabla_i h_{lj} - \nabla_l h_{ij} =0.$$

Thus we have $$-2g^{ij}g^{pl}H^k\nabla_ih_{jl}=-2g^{ij}g^{pl}H^k\nabla_lh_{ij}=-2g^{pl}H^k\nabla_lH = -2H^k\nabla^pH.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.