2
$\begingroup$

If $(M,g)$ is a 2-dimensional surface with boundary, now we have a deformation of metric $(g_t)$, suppose $\frac{dg_t}{dt}\big|_{t=0}=h$ which is a symmetric $(0,2)$ tensor, then the derivative of volume is $$\frac{dV(g_t)}{dt}\big|_{t=0}=\frac{1}{2}\int_M tr_g(h) dv_{g}$$ The volume of boundary is $$A(g_t)=\int_{\partial M}dv_{\tilde{g}_t}$$ But I don't know how to calculate the derivative of this, is it same as above?

Now if $M$ is n-dimensional Riemannian manifold with boundary, I think the derivative of volume is still same as above, but what happens to boundary now?

$\endgroup$
1
$\begingroup$

It is just $$\frac{dA(g_t)}{dt}\big|_{t=0}=\frac{1}{2}\int_{\partial M}h(T,T)dV_{\bar{g}}$$ where $T$ is the tangent vector field on the boundary. Since for orthonormal basis $\{e_i\}$ we have $$tr_gh=g^{ij}h_{ij}=h(e_i,e_i)$$ but on the boundary, the basis is just the tangent vector $T$.

$\endgroup$

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.