I am trying to show that any geodesic quadrangle $Q$ in any CAT($0$) space $X$ has a comparison quadrangle in $\mathbb{R}^2$ (same definition as for triangles). One can split $Q$ in two triangles $T_1$ and $T_2$ and consider the quadrangle $\overline{Q} \subseteq \mathbb{R}^2$ obtained gluing the comparison triangles $\overline{T_1}$ and $\overline{T_2}$ along the "diagonal", which we call $\overline{d}$ (note that, in general, we would not obtain the same quadrangle $\overline{Q}$ if we had chosen the other "diagonal" in $Q$). Pick $p,q \in Q$, wlog one in $T_1$ and one in $T_2$. If the straight line segment between $\overline{p}$ and $\overline{q}$ crosses $\overline{d}$, then we call the intersection point $\overline{g}$, its "preimage" $g$, and we obtain the inequality $$d_X(p,q) \le d_X(p,g) + d_X(g,q) \le d_{\mathbb{R}^2}(\overline{p},\overline{g}) + d_{\mathbb{R}^2}(\overline{g},\overline{q}) = d_{\mathbb{R}^2}(\overline{p},\overline{q}),$$ and everything is fine. If, however, $\overline{Q}$ is not so nice and we do not get the intersection point $\overline{g}$, then the above reasoning does not work. I would appreciate any hint on how to adapt the proof for the general case.

  • $\begingroup$ In fact, more is true: A polygon in $R^2$ with side-lengths $r_1,...,r_n$ exists if and only if each $r_i$ is $\le$ the sum of the rest of the side-lengths. Thus, CAT(0) spaces are irrelevant. $\endgroup$ – Moishe Kohan Jul 27 '16 at 13:49
  • $\begingroup$ By "comparison quadrangle" I mean one such that $d_X(p,q) \le d_{\mathbb{R}^2}(\overline{p},\overline{q})$ holds for all $p,q \in Q$. $\endgroup$ – 57Jimmy Jul 27 '16 at 14:11

You can do this using the CAT(0) 4-point condition, which all CAT(0) spaces satisfy, namely for any four points $x_1,y_1,x_2,y_2$ in your CAT(0) space $X$ there are points $\bar{x}_1,\bar{y}_1,\bar{x}_2,\bar{y}_2$ in $\mathbb{R}^2$ with $d(x_i,y_j)=d(\bar{x}_i,\bar{y}_j)$ and $d(x_1,x_2)\leq d(\bar{x}_1,\bar{x}_2)$ and $d(y_1,y_2)\leq d(\bar{y}_1,\bar{y}_2)$; see https://www.math.bgu.ac.il/~barakw/rigidity/bh.pdf p164 for the proof (1.11 is stated for complete metric spaces but the proof of the CAT(0) 4-point condition doesn't use this).

We can assume that $\bar{x}_1,\bar{y}_1,\bar{x}_2,\bar{y}_2$ (in that order) form a convex quadrilateral $Q$ (if WLOG $\bar{x}_1$ is in the interior of the triangle of the other three points then reflect it in the line through $\bar{y}_1,\bar{y}_2$). I claim that $Q$ is the desired comparison quadrangle.
First consider comparison points on adjacent sides. If $z_1\in [x_1,y_1]$ and $z_2\in[x_2,y_1]$ then their comparison points $\bar{z}_1,\bar{z}_2$ will satisfy $d(z_1,z_2)\leq d(\bar{z}_1,\bar{z}_2)$; indeed if $\hat{x}_1,\hat{y}_1,\hat{x}_2$ is a comparison triangle for $x_1,y_1,x_2$, and $\hat{}$ denotes comparison points in this triangle, then $d(z_1,z_2)\leq d(\hat{z}_1,\hat{z}_2)\leq d(\bar{z}_1,\bar{z}_2)$ - the latter inequality due to $\angle_{\hat{y}_1}(\hat{x}_1,\hat{x}_2)\leq\angle_{\bar{y}_1}(\bar{x}_1,\bar{x}_2)$.
Now consider comparison points on opposite sides. If $z_i\in[x_i,y_i]$ for $i=1,2$, then by the above we have $d(x_2,z_1)\leq d(\bar{x}_2,\bar{z}_1)$ and $d(y_2,z_1)\leq d(\bar{y}_2,\bar{z}_1)$. Now let $x'_2,y'_2,z'_1$ be a comparison triangle for $x_2,y_2,z_1$ and let $'$ denote comparison points in this triangle. Then $d(z_1,z_2)\leq d(z'_1,z'_2)\leq d(\bar{z}_1,\bar{z}_2)$ - the latter inequality because $d(x'_2,y'_2)=d(\bar{x}_2,\bar{y}_2)$, $d(x'_2,z'_1)\leq d(\bar{x}_2,\bar{z}_1)$ and $d(y'_2,z'_1)\leq d(\bar{y}_2,\bar{z}_1)$ (I won't write out a proof for this, but it is elementary euclidean geometry using the cosine rule).


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