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I would like to see a "metric proof" that if two metric spaces $X$ and $Y$ are CAT($\kappa$) for some $\kappa \ge 0$, then so is their product. I would be satisfied to see a proof for $X=Y=S^2$.

By "metric proof" I mean one which does not rely on Riemannian geometry, but rather only uses Alexandrov (metric) geometry. I already understand the case where $\kappa \le 0$ (where in fact the product will only be CAT($\text{max}(0,\kappa)$).

Delfador Logalmier
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  • What is ${\rm CAT}[k]$ and their product ? – HK Lee Sep 06 '18 at 11:36
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    A CAT($\kappa$) space is a geodesic space all of whose geodesic triangles of perimeter less than $2D_\kappa$ satisfy the CAT$(\kappa)$ inequality. Here $D_\kappa= +\infty$ if $\kappa \leq 0$ or $\frac{\pi}{\kappa}$ if $\kappa > 0$. One version of the CAT$(\kappa$) inequality says that if one takes such a triangle in $X$ and a comparison triangle in the model plane of curvature $\kappa$, then the Alexandrov angles in the triangle in $X$ are no greater than the corresponding angles of the comparison triangle. – Delfador Logalmier Sep 06 '18 at 12:17
  • The product of two metric spaces $X\times Y$ has the usual cartesian product $X \times Y$ as its underlying set, and the metric is defined in the expected way: $$ d_{X \times Y}((x_1,y_1),(x_2,y_2))^2 = d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2 $$ – Delfador Logalmier Sep 06 '18 at 12:18
  • Sorry, in the first comment, $ \frac{\pi}{\kappa}$ should have been $\frac{\pi}{\sqrt{\kappa}}$. $X$ is the CAT$(\kappa)$ space under discussion. – Delfador Logalmier Sep 06 '18 at 12:24

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In ${\rm CAT}\ [k]$-space, two points of distance smaller than $\frac{\pi}{\sqrt{k}}$ have unique geodesic.

So if $\mathbb{S}^2\times \mathbb{S}^2$ is ${\rm CAT}\ [k]$-space s.t. $0<k<1$, then assume that $(p,q),\ (-p,q_\epsilon)$ s.t. $|q-q_\epsilon|=\epsilon$ and $\sqrt{ \pi^2+\epsilon^2 }<\frac{\pi}{\sqrt{k}}$. Note that there are infinitely many shortest paths between them.

HK Lee
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