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Let $pxy$ be a triangle in a CAT(0) space $X$, and $p' x' y'$ be a triangle in $\mathbf{R}^2$ such that the lengths $|px|=|p'x'|$, $|py|=|p'y'|$ and the angle $\angle(xpy)=\angle(x'p'y')$. Let $z\in xy$ and $z'\in x'y'$ be points with $\angle(xpz)=\angle (x'p'z')$. Does it follow that $|pz|\leq |p'z'|$?

Using basic trigonometry, one can check that this holds when $X$ is the hyperbolic space $\mathbf{H}^2$. But I think there should be a better way to show this using the comparison results for triangles, hinges or quadrilaterals in CAT(0) spaces. I am mainly interested in the case where $X$ is a Cartan-Hadamard manifold.

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2 Answers 2

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No, it is not true.

Suppose $X$ glued from two solid plane triangles $\blacktriangle p x \bar y$ and $\blacktriangle y x\bar y$, the angle at $y$ is obtuce and the total angle at $\bar y$ is reflex. Then the inequality has opposite sign.

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  • $\begingroup$ Thanks Anton, are the triangles glued only along the side $x\overline{y}$? Are there such examples which are Cartan-Hadamard manifolds, or at least geodetically complete? $\endgroup$
    – Mohammad Ghomi
    Commented Jan 30 at 23:32
  • $\begingroup$ @MohammadGhomi yes and yes; this example is isometric to a solid quadrangle in a 2-dimensional cone with nonpositive curvature. The latter can approximated by a smooth metric on the plane with curvature supported in a small neighborhood of one point. $\endgroup$
    – Anton Petrunin
    Commented Jan 31 at 3:29
  • $\begingroup$ Thanks again, I got it now. $\endgroup$
    – Mohammad Ghomi
    Commented Jan 31 at 3:39
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This is just a bit more elaboration on Anton's nice example, and subsequent comment. Here is the picture of the two triangles:

enter image description here

Note that $p'x'y'$ is obtained from $pxy$ by rotating the side $\overline{y}y$ about $\overline{y}$ until the interior angle at $\overline{y}$ reaches $\pi$. Since the angle at $y$ is obtuse, the points on the side $xy$ move closer to $p$. Thus we obtain $|p'z'|<|pz|$.

The quadrilateral $p\overline{y} y x$, which forms the space $X$, is CAT(0) because each of the subtriangles $p\overline{y}x$ and $\overline{y}yx$ is a convex subset of $\textbf{R}^2$ and a convex subset of a CAT(0) space is CAT(0). Furthermore, gluing two CAT(0) spaces along isometric convex subsets (in this case the common edge $x\overline{y}$) yields a CAT(0) space. The quadrilateral $p\overline{y} y x$ may be regarded as a geodesic triangle $pxy$ in $X$.

Note that $X$ may be extended to a geodesically complete CAT(0) space, since we may regard $X$ as a geodesic triangle in a cone with vertex at $\overline{y}$. Setting the total angle of the cone at $\overline{y}$ to twice the interior angle of the quadrilateral at $\overline{y}$ makes sure that $py$ is indeed a geodesic. Smoothing the cone near $\overline{y}$ yields a Cartan-Hadamard manifold, i.e., a complete simply connected manifold of nonpositive curvature, where the comparison property in the question I had asked (which holds in $\mathbf{H}^2$) does not hold.

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