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Let $X$ be a Riemann surface of genus $g\ge 2$ and $q$ a non-vanishing holomorphic quadratic differential on $X$. Denote by $\mathbb{H}$ the universal cover of $X$ and by $\widetilde q$ the pullback of $q$ to $\mathbb{H}$. The space $\mathbb{H}$ endowed by the flat singular metric $|\widetilde q|$ is a Cat(0) metric space.

For every $x,y,z\in (\mathbb{H},|\widetilde q|)$ denote by $\Delta$ the corresponding geodesic triangle (i.e. the subset of $(\mathbb{H},|\widetilde q|)$ composed by $x,y,z$ and the three geodesics connecting them). It seems to me that it makes sense to talk about the interior of $\Delta$: it is the bounded region of $(\mathbb{H},|\widetilde q|)$ delimited by $\Delta$ (and of course if can by empty). Denote by $\Delta'$ the union of $\Delta$ with its interior.

Is there a nice characterization of such $\Delta'$? In particular, is it true that they are always isometric to polygons with possibly one dimensional components (it seems to me that there can not be zeroes of $\widetilde q$ in the interior of $\Delta$)?

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Triangles do not have singular points in their interior ⟺ curvature of any singular point is at least π.

In this case each triangle admits a length-preserving immersion into the plane. (The interior admits an locally isometric immersion since it is locally flat, then you can extend it to the closure of the interior in the solid triangle and then add line segments for each vertex with overlapping adjacent sides.)

The total negative turn of sides can not exceed π. It is sufficient to conclude that the length-preserving immersion is actually an embedding.

Does it answer you question?

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  • $\begingroup$ I do not completely follow your reasoning, are you saying that geodesic triangles in $(\mathbb{H},\widetilde q)$ do not have singular points in their interior and consequently can be embedded in the plane? From this it should follow that they are isometric to polygons with possibly one dimensional components, right? $\endgroup$
    – user101163
    Nov 7, 2017 at 18:51
  • $\begingroup$ @user389604 yes, but isometric here means with respect to the intrinsic metric of polygon. (I have added bit more.) $\endgroup$ Nov 7, 2017 at 19:09
  • $\begingroup$ I am sorry to bother you again, but lately I have thought again about this question and there is something I do not understand. You said that the interior of the geodesic triangle admits a locally isometric immersion: how do you know that this is in fact an immersion? Are you sure that the fact that $\Delta'$ is locally flat is enough? When developing $\Delta'$ into the real plane couldn't one have overlaps? $\endgroup$
    – user101163
    Feb 18, 2018 at 19:00
  • $\begingroup$ @user389604 the sum of angles of triangle + the total curvature of sides $=\pi$. But to have a self intersection the total curvature must be strictly larger than $\pi$. $\endgroup$ Feb 19, 2018 at 6:18

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