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Let $M$ be a 2-dimensional Riemannian manifold of non-positive curvature everywhere, of genus > 1. Let $\textbf{D} \subset \textbf{C}$ be the open unit disc in the complex plane, the universal cover of $M$. Let $\gamma \subset \textbf{D}$ be a curve representing a geodesic in $M$ which is entirely in a region of zero curvature. It seems to me, because the definition of geodesic is local, that unless $\gamma$ is tangent to some region $A \subset \textbf{D}$ of negative curvature, $\gamma$ will be a Euclidean line through $\textbf{D}$. Is this correct?

Secondly, does anybody have any references that I could peruse to learn how a geodesic $\gamma$ which does in fact pass tangent to some $A$ of negative curvature reacts to this region (will it turn into $A$, away from it, etc. and maybe some way of calculating the actual effect)?

Thank you.

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Could you formulate it clearly? – Anton Petrunin Aug 2 '10 at 20:16
Your question needs to be better formulated: let $\phi$ be an arbitrary smooth map from $\Omega\in\mathbb{R}^2$ to itself. Take the Euclidean metric $e$ on $\Omega$, and consider the pull-back metric $\phi^*e$. The curvature will still be zero, but geodesics no longer look like straight lines for $(\Omega,\phi^*e)$. The concept of Euclidean line really is chart dependent. Secondly, the condition of zero curvature is a closed condition. So $\gamma$ that is only tangent to $A$ in fact never passes through $A$. So geodesics should not be affected. – Willie Wong Aug 2 '10 at 20:20
I edited my question -- I had forgotten to mention that D was supposed to be the universal cover of M. Willie: I am pretty sure I understand what your'e saying; does this change anything? Also, let me just check: even though the geodesics on regions of zero curvature may not look like straight lines, they will still not be affected in the scenario I describe, right? So if they happen to look like straight lines, they will remain so? – Nicolas Fernandez-Arias Aug 2 '10 at 20:46
Maybe you mean to specify that the covering projection from the unit complex disk to the manifold is chosen to be conformal (w.r.t. the usual conformal structure of the complex numbers and the conformal structure determined by the given metric on the manifold). So that part of your question is, "If a region of the plane is given a metric in which right angles are what they appear to be and the curvature is zero, must geodesics be straight lines?" ? – Tom Goodwillie Aug 2 '10 at 21:48
Rephrasing Willie Wong's comment: The issue is that the universal cover of $M$ is not $D$. You can identify it with $D$ in some way, but there is no "one true way" to do so. And what is a straight line after one identification is not straight after another - they differ by a map $\phi$ that "changes coordinates". Even if you restrict yourself to conformal coverings (as Tom Goodwillie suggested) $\phi$ can be any conformal bijection from $D$ to itself, and can transform any straight line to a circle arc. – Sergei Ivanov Aug 2 '10 at 23:05

I am answering the question as clarified in your Aug 3 comment. No it is not always possible to identify the universal cover with $D$ so that all zero-curvature geodesics become straight lines.

Begin with a metric in a small region where some 9 zero-curvature geodesics intersect each other but violate the Desargues theorem. For example, begin with a standard Desargues configuration and introduce a bit of negative curvature so that one of the lines misses one of the intersection points. For each of these 9 geodesics, attach a long narrow planar strip to the boundary of the region: each end of the strip is attached to the place where the geodesics meets the boundary, so that the geodesic extended to the strip closes up (and has a neighborhood isometric to a straight cylinder which lies partly in the strip and partly in the original region).

Now we have a surface with boundary, and on this surface some 9 zero-curvature closed geodesics violate the Desargues theorem within a simply connected region. Fill the boundary components by surfaces of sufficiently large genus - so that they can be equipped with a nonpositively curved metric. Now we have a closed surface. In its universal cover, the lifts of the 9 geodesics still violate the Desargues theorem. Hence they cannot be represented by straight lines no matter how you identify the universal cover with $D$.

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Suppose that $S$ is your Riemannian surface and $X \subset S$ is a flat subsurface (that is, locally isometric to $\mathbb{R}$ with the usual metric). Let's suppose that $X$ has some nontrivial topology. For example, $X$ is a unit disk minus one-half of a unit disk, and the core curve of $X$ is essential in $S$.

You are correct in thinking that the universal cover of $S$ is homeomorphic to $\mathbb{D}$ the unit disk. However, it is impossible to choose this homeomorphism so that the universal cover of $X$, call it $\bar{X}$, embeds isometrically in $\mathbb{D}$. You cannot even arrange this up to homothety. To see this, choose isometric charts for $X$ that lift to give isometric charts for $\bar{X}$. After choosing where any one chart of $\bar{X}$ goes in $\mathbb{R}^2$ (isometrically!) the positions of all others will be determined. (This is the so-called "developing map" of $\bar{X}$ and you can think of it as being similar to the process of analytic continuation of a analytic function.) The point here is that the developing map will not be injective - in fact it will have image isometric to $X$ itself.

I can't think off hand of a reasonable reference (Thurston's book is perhaps an unreasonable reference). Think about it and ask any local geometers or post more questions on MO.

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