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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!

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Just a comment, not an answer, because I don't know if I've understood your question, but you might want to look up "translation surfaces." –  JSE Jul 20 '10 at 3:06
I've embedded the image and copyedited accordingly. –  Victor Protsak Jul 20 '10 at 21:34
I cannot really comprehend the question, but it appears to me that you want to put non-positively curved metric structure on the cell complex in the picture so that (a) the middle hexagonal cell is flat (Euclidean) (b) the glued surface is smooth at the vertices (i.e. the angles at cone points add up to $2\pi$). Is this correct? Do you insist on Riemannian structure, or would metric structure be enough? Can you relax one of the requirements (a), (b)? This doesn't appear to be a translation surface because e.g. pairs of edges labeled 3 and 4 aren't parallel. –  Victor Protsak Jul 20 '10 at 21:53
It is supposed to be a compact, closed manifold. I don't know if that relaxes any requirements...I've never taken a course in Riemannian geometry, just learned a little here and there for this project! I definitely can't relax (a). Thanks! –  Nicolas Fernandez-Arias Jul 21 '10 at 21:59
What I was trying to say is that perhaps for your problem Riemannian structure is not crucial. In the past 20 years or so there's been a resurgence of interest to metric geometry point of view (see e.g. Bridson and Haefliger's book). Specifically, you can try to construct a topological surface by gluing polygons with metrics of constant curvature (e.g. a flat hexagon in the middle and two hyperbolic triangles in the wings, making sure that the lengths of the corresponding edges match). It's a bit more general than the usual setup, where the pieces are of the same type (PE, PH or PS complexes) –  Victor Protsak Jul 23 '10 at 4:17
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3 Answers

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.

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Sam, So you are saying that one does not necessarily obtain a Riemannian 2-manifold from a polygonal disk with a gluing scheme, because of the geodesic "splitting" at the square vertex? Wouldn't this maybe just reflect the fact that, in order to identify edges in the diagram, we also have to identify some subsets of the regions they enclose? Specifically, at least the paths of the "splitting" geodesic (which soon afterwards reunite!). If I am wrong, are there any other conditions I should know about? Any references? Thank you so much! –  Nicolas Fernandez-Arias Jul 20 '10 at 18:09
Before we discuss that, could I ask you to clarify your question by telling me if my two assumptions are correct? 1) Do you want $S$ to be a Riemannian manifold? 2) Do you want the dotted edges to be straight? (Well, at least on the "flat side", where straight makes sense.) –  Sam Nead Jul 20 '10 at 19:53
Oh, sorry. 1) Yes, it has to be a Riemannian manifold. 2) Yes - we are trying to find a non-periodic geodesic on regions of curvature zero, and the ones we were considering in the above diagram were in X and parallel to the dotted lines as they are drawn in the picture, i.e. vertical. As I commented below, we have realized that the above doesn't work because we can't adjust the angles appropriately. But yeah, they should be vertical. –  Nicolas Fernandez-Arias Jul 20 '10 at 20:46
Ok, the issue is not really one of angle (see my comment on Daniel's answer). Instead there is an issue of smoothness. Imagine cutting open the $z$ plane along the real axis and gluing in a "cusp": in the $w = u + iv$ plane this is the region with $u$ positive and $|v| \leq u^2$. At the origin of this new "glued" space there is angle $2\pi$. However, this can't be a Riemannian manifold as there is a tangent vector at the origin that doesn't have a unique geodesic running through it. –  Sam Nead Jul 21 '10 at 11:09
Reference for the claims about geodesics - See Chapter VII, Remark 5.7 of Boothby's "An introduction to differentiable manifolds and Riemannian geometry". --- "2) Yes - we are trying to find a non-periodic geodesic on regions of curvature zero" If those regions are supposed to be open subsets of some larger Riemannian surface then I will make the guess that this cannot happen. –  Sam Nead Jul 21 '10 at 11:24
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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.

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Super helpful, thanks a bunch for the paper. –  Nicolas Fernandez-Arias Jul 20 '10 at 18:10
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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.

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This isn't quite right - you need to rule out the possibility that edge 1 curves in to become tangent to the dotted line at the circled vertex (and similarly for edge 2). –  Sam Nead Jul 20 '10 at 13:23
Dan, Soon after I posted this I realized that it is indeed impossible to adjust the square and circle angles to add up to 360 without making the alpha and beta lines the actual boundaries of the polygon, because the sum of the interior angles of a hexagon is 720 (and because a-a are parallel, b-b are parallel). So we are definitely going to have to come up with a different figure to solve our actual problem; but these responses have been quite helpful as we try to better understand how this works. –  Nicolas Fernandez-Arias Jul 20 '10 at 18:13
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