Flat regions on surfaces of genus greater than 1

Question

Here is a polygonal disk + gluing scheme model of a surface we are attempting to construct. We want the regions of the surface bounded by the two vertical, dotted lines $\alpha,\beta$ to have zero curvature throughout, while the rest of the surface should be "smoothed out" with some kind of metric which gives it negative curvature and thus allows the entire shape to satisfy the requirement that a two-holed torus have negative total curvature.

Assuming that we can adjust the circled (and, separately, squared) angles so that they add up to $2\pi$ and the relevant identifications (side $a$ with side $a$, $b$ with $b$) can be made smoothly while maintaining the zero curvature of the inner region, the question is the following. Will the smoothing out of edges $1$ through $4$ necessarily need to be continued into the region inside of which we want to have zero curvature, thus giving it non-zero curvature instead? In other words, we have to smooth out the triangled vertices to some extent because their corresponding angles add up to more than $2\pi$. But, thinking geometrically (we have a limited knowledge of the underlying Riemannian geometry here), it seems that we would need to smooth out neighborhoods of these vertices and of the edges $1$ through $4$. Will we be able to stop before reaching the region inside of $\alpha,\beta$?

Thank you. Any references that might be able to help us are welcome. We're pretty new to this but if the answer to these questions is that we can maintain zero curvature, we've made progress on our problem!

Answer 1

Let $S$ be the underlying surface and call the image of the dotted lines $\gamma$. If we cut $S$ along $\gamma$ then $S$ falls apart into two components $X$ and $Y$ (both tori with a single boundary component). Let $X$ be the component that contains the edges $a$ and $b$. If I read your question correctly, you want the metric inside of $X$ to be flat and are willing to accept any smooth, negatively curved metric on $Y$.

I'll assume that you also want the surface $S$ to be a Riemannian manifold at the end of the day. I'll also assume that you are insisting that the two edges of $\gamma$ (on its $X$ side) be straight (otherwise you can proceed as Daniel Mehkeri suggests -ie in this situation you allow $\gamma$ to "bow-in" to the $X$ side. Eg, you could take any flat torus and cut out a round disk to get $X$.)

If my assumptions are correct then the answer to your question is negative. This is because in a Riemannian manifold the exponential map is well defined -- however in your surface, the dotted edges are limits of geodesic segments and hence geodesic. Thus at the square vertex the upward pointing tangent vector has two geodesic continuations, a contradiction.

Answer 2

This is not an answer, only a partly baked idea. There is a type of converse to the Gauss-Bonnet theorem, which says, roughly: if you specify a curvature function, then there is a Riemannian metric that realizes it. A bit more precisely, if $K$ satisfies a sign condition that I believe holds in your situation, then there is a Riemannian metric having $K$ as its Gaussian curvature. For details, see Hermann Gluck's survey, "Manifolds with preassigned curvature," Bull. Amer. Math. Soc. Volume 81, Number 2 (1975), 313-329, and the literature he cites.

Your diagram has only three vertices, two of which you will assign zero curvature, and the third (your triangles) must have curvature $-4\pi$ (if I understand the situation correctly). It may be that you could add more vertices of curvature zero to "protect" your $\alpha-\beta$ region, or otherwise design $K$ to be zero wherever you so desire, and then use the converse G-B theorem to get a metric matching your preassigned curvatures.

Answer 3

The obstacle, if I understand the question right, is that the angles around the squared and circled vertices can't be adjusted as you say. No matter how you do it, one triangled vertex is already inside the alpha-beta region, or, at best, two of them are right on the border. This is just because of the sum of the angles around the polygon. As a result any neighbourhood of the (fused) triangled vertex will be partly inside the alpha-beta region, so there is no way to smoothly do this.

If allowed, you can do it by subdividing the a-b edges. Move the border for the alpha-beta region inward. Temporarily elide edges 3-4, so that there are only two remaining triangled vertices. With the edges 1-2, and the divided edges a-a'-b-b', make a regular hexagon with each edge subdivided. This is a completely flat (genus 1) surface with the triangled vertices outside the new alpha-beta region. Then break apart one triangled vertex to re-introduce the edges 3-4, corresponding to adding a handle to the surface. If you are allowed to make the handle small enough, it will be entirely outside the alpha-beta region.