## Tag Archives: self-shrinker

### Self-shrinker and polynomial volume growth

Proposition: If $M$ is an entire graph of at most polynomial volume growth and $H=\langle X,\nu\rangle$, namely $M$ is a self-shrinker. Then $M$ is a plane.

Proof: Suppose

$\displaystyle v=\frac{1}{\langle \nu, w\rangle}$

Then one can derive the following equation

$\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2$

Multiplying both sides by $e^{-|X|^2/2}$ and integration on $M$, which makes sense because of the polynomial volume growth, we get

$\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu$

However, integration by parts shows the LHS is zero. Thus $A\equiv v\equiv 0$, $M$ must be a plane.