# Is the optimal transport of radially symmetric measures also radially symmetric?

Let $$\mu$$ and $$\nu$$ be radially symmetric probability measures on $$\mathbb R^d$$. Consider the Kantorovich optimal transport problem between $$\mu$$ and $$\nu$$, with convex, nonnegative cost. Suppose there exists at least an optimal transport plan between $$\mu$$ and $$\nu$$ with finite cost.

Question. Does it follow that any optimal transport plan $$\gamma \in \mathcal P (\mathbb R^d \times \mathbb R^d)$$ is radially symmetric, in the sense that $$\gamma$$ is concentrated on the set $$\{(x, \, y) \mid \ x, \, y \in \mathbb R^d, \ x/|x| = y/|y| \}?$$

• A symmetric optimal plan always exists; it can be obtained by symmetrizing a given optimal plan. Now, I do not know the definition of radial symmetry --- if you assume that central symmetry is radial, then the answer is "no"; it is easy to construct an example with few atoms. Apr 29, 2021 at 3:41
• If radial symmetry is defined by all rotations, then answer is "yes". It follows since the quotient map $\mathbb{R}^d\to [0,\infty)$ is short. Apr 29, 2021 at 3:43
• Yes I do mean invariant under all rotations, sorry. What is a short map? And how does it follow from this? Apr 29, 2021 at 3:47
• Short means distance-nonexpanding. The optimal plan on $[0,\infty)$ lifts uniquely to an optimal plan on $\mathbb{R}^d$ Apr 29, 2021 at 4:18
• @FrancescoPolizzi "any=every" Apr 29, 2021 at 6:45

The answer is "yes". It follows since the quotient map $$\mathbb{R}^d\to[0,\infty)$$ is a submetry. The optimal plan between pushforward measures on $$[0,\infty)$$ lifts uniquely to an optimal plan on $$\mathbb{R}^d$$