The Gauss-Bonnet theorem characterizes the topology of surfaces by means of their Gaussian curvature. 


Do there exist results characterizing the topology of surfaces embedded in $\mathbf{R}^3$ via their mean curvature? 


For example, I find it hard to imagine how a topological 2-sphere could be embedded in $\mathbf{R}^3$ with, for example, everywhere negative mean curvature. However, I am not familiar with any easy results which immediately rule this out. The Gauss-Bonnet theorem seems insufficient here, because the trace of the shape operator is a priori unrelated to its determinant.



Topology of surfaces and mean curvature
1. Mean curvature depends on the [choice of the "unit normal"](http://en.wikipedia.org/wiki/Mean_curvature#Surfaces_in_3D_space). If you change the orientation (choose the other unit normal), the computed mean curvature for the sphere is everywhere negative.
2. In any case. Fix $x\_0\in \mathbb{R}^3$. Since the embedding $S$ of the two sphere is compact, there is a point $y$ at which $|y - x\_0| = \sup\_{x\in S} |x - x\_0|$. Thus $S\subset \overline{B\_{x\_0}(|y - x\_0|)}$ and $S$ is tangent to $\partial B\_{x\_0}(|y-x\_0|)$ at $y$. So trivially you have a lower bound on the Hessian and that the mean curvature must be positive at $y$ (for the "correct" choice of orientation).
3. In fact, this has nothing to do with topology. Let $\Sigma$ be any compact surface with no boundary embedded in $\mathbb{R}^3$, the same argument as above shows that both its mean curvature (with appropriate choice of orientation) and scalar curvature must be positive at some point.


