TITLE: Total mean curvature of an immersed torus.
QUESTION [9 upvotes]: How to prove that the total mean curvature of an immersed torus of $R^3$ such that has nontrivial self-intersection must $> 8 \pi$? The definition of total mean curvature is the integral of $H^2$ over the immersed manifold, where H is the mean curvature.
I don't have any idea about it.

REPLY [2 votes]: Li and Yau (1982): If  is not embedded, then $\mathcal{W(\Sigma)}\geq{8\pi}$.
Fix $v\in\mathbb{R}^4$ and consider the vector field $x=v^\perp(x)$, given by tangential
projection on $T_x\mathbb{S}^3$. 
It is a conformal killing vector field in $\mathbb{S}^3$ and generates
a flow $\phi_t$ of centered dilations. By conformal invariance of Willmore energy,
$$\mathcal{W(\Sigma)}=\mathcal{W(\phi_t(\Sigma))}=\int_{\phi_t(\Sigma)}(1+H^2_{\phi_t(\Sigma)})d\phi_t(\Sigma)\geq\hbox{area}(\phi_t(\Sigma))$$
If we choose $v\in\Sigma$ and let $t\longrightarrow\infty$, we have
$\mathcal{W(\Sigma)}\geq{4}\pi$. In case $F:\Sigma\longrightarrow{\mathbb{S}^3}$ is immersion and $p\in\Sigma$ satises that #$F^{-1}(p)=k$, then $\mathcal{W(\Sigma)}\geq{4\pi}$k.