# Flat embedded surfaces in arbitrary 3-manifolds.

It is generically known that any surface (at least embedded in $\mathbb{R}^3$ - and see edit below) can be deformed so that it is flat - i.e. has a metric of zero curvature - as long as one adds some conical singularities (for reference, here's an overview by Zorich (link) with the main results I am referring to being Troyanov (link)).

My question is would this still be true for a surface embedded in an arbitrary 3-manifold? The Zorich paper is unclear, and the Troyanov results seem to be for a Riemann surface without explicit embedding. It seems a bit too much to hope for, but the basic technique is by solving some differential equations, which are local. Anyone know anything about this?

EDIT: It's not true that "any surface" can be flattened in this matter - there are some conditions, which vary depending on which result we are looking at. For instance, genus $g>1$ is a classic result, although in that Troyanov paper he proves the following must be true: $$2\pi\chi(S)+\sum(\theta_i-2\pi)<0$$ for a set of singular points $p_i$ with angles $\theta_i$.

• Dear levitopher, I don't have an answer to your question, but here is an example you might be able to visualize, and see what conclusions you come to: consider $S^2 \times S^1$, which you can visualize as a spherical shell $S^2 \times [0,1]$ with the outer and inner boundary identified, and take the surface to the be any of the spheres $S^2 \times \{t\}$ for some $t \in [0,1]$. It's not immediately obvious to me that there is room to flatten this sphere in $S^2 \times S^1$, since it is "wrapped around" in a non-trivial way. Regards, – Matt E Sep 1 '13 at 2:06
• What's a zero curvature embedding of $S^2$? Isn't the Troyanov theorem only for surfaces of genus $2$ or larger? – Ryan Budney Sep 25 '13 at 19:52
• There are a variety of results (including one for $g>1$) but the specific condition I am referring to is $2\pi\chi (S)+\sum (\theta_i-2\pi)<0$ for a set of singularities with angles $\theta_i$. In any case, your statement about the spheres is certainly helpful. – levitopher Sep 26 '13 at 4:01
• But even with those conditions satisfied, there isn't an embedding of the sphere in $\mathbb R^3$, is there? I'm not seeing one. – Ryan Budney Sep 26 '13 at 8:47
• I see - I've added an edit for some clarification. – levitopher Sep 26 '13 at 13:53