Let $(X,d)$ be a (separable, complete) metric space with uniformly strictly non-positive curvature in the sense of Alexandrov, i.e. $(X,d)$ satisfies a $CAT(K)$ inequality for some $K<0$.


Does it hold that the 2-Wasserstein space on $(X,d)$ has strictly non-positive curvature?


I suspect this is false, but haven't found this result anywhere. Note that the analogous statement for $CAT(0)$ is definitely false, because the 2-Wasserstein space on $\mathbb{R}^d$ is positively curved (see for example Remark 2.10 in this paper by Bertrand and Kloeckner: <https://arxiv.org/pdf/1010.0590.pdf>).



Wasserstein space with strictly non-positive sectional curvature
Yes, it is false. Note that the space $W\_2(X)$ contains the space $\tfrac1{\sqrt{n}}\cdot X^{\times n}/S\_n$ where the group $S\_n$ permutes the $X$-factors. While space $X^{\times n}$ is $\mathrm{CAT}(0)$, the quotient $X^{\times n}/S\_n$ is not --- for example, take $X=$ Lobachevsky plane.


