Divergence on manifold

Divergence on manifold Suppose {(M,g)} is a Riemannian manifold. {X} is smooth vector field on it. {\{e_i\}} is a frame at {p}. Suppose {X=X^ie_i}, then define

\displaystyle \text{div} X=\sum_iX^i_{,i}

Under a local coordinates near {p}, choose {\{\frac{\partial}{\partial u^i}\}} as the basis of tangent space, then

\displaystyle \text{div} X=\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}

where {G=\det(g_{ij})}. One way to prove is straightforward verification. Let us see a more intuitive way, from the divergence theorem

\displaystyle \int_{M}\text{div } XdV=\int_{\partial M}X\cdot \vec{n}\,dS

Choose {h\in C_c^\infty(U(p))}, where {U(p)} is a neighborhood of {p}, apply the divergence theorem to {hX}

\displaystyle \int_{U}\text{div } (hX)dV=0

which means

\displaystyle \int_U h\text{div}\,XdV=-\int_U X\cdot \nabla h \,dV=-\int_U X^i\frac{\partial h}{\partial u^i}\sqrt{G}du^1\wedge\cdots\wedge du^n

Denote {\phi:U\rightarrow\mathbb{R}^n} is the diffeomorphism and {\tilde{X}=X\circ\phi^{-1}}, {\tilde{h}=h\circ \phi^{-1}}. From Green’s formula

\displaystyle \int_U X^i\frac{\partial h}{\partial u^i}\sqrt{G}du^1\wedge\cdots\wedge du^n=\int_{\phi(U)}\tilde{X}^i\frac{\partial \tilde{h}}{\partial u^i}\sqrt{G}du^1\cdots du^n

\displaystyle =-\int_{\phi(U)}\tilde{h}\frac{\partial (\tilde{X}^i\sqrt{G})}{\partial u^i}du^1\cdots du^n=-\int_{U}\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}hdV

Combining the two equalities, we get that

\displaystyle \text{div} X=\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}

When you are on a hypersurface, suppose {F:U\rightarrow \mathbb{R}^{n+1}} is the embedding. Metric is {g_{ij}=\partial_iF\cdot\partial_j F}

Denote {e_i=\partial_iF}, {X} is any smooth vector field, not necessaritly tangent to {M}, then

\displaystyle \langle\nabla^M_{e_i}X,e_l\rangle=\langle\bar{\nabla}_{e_i}X,e_l\rangle=\partial_{i}X\cdot e_l

So {\nabla^M_{e_i}X=g^{jl}(\partial_iX\cdot e_l)e_j}

\displaystyle \text{div}_M\,X=X^i_{,i}=g^{il}\partial_iX\cdot e_l

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: