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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 $$\{(x, \, y) \mid \ x, \, y \in \mathbb R^d, \ x/|x| = y/|y| \}?$$

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  • $\begingroup$ 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. $\endgroup$
    – Anton Petrunin
    Apr 29, 2021 at 3:41
  • $\begingroup$ 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. $\endgroup$
    – Anton Petrunin
    Apr 29, 2021 at 3:43
  • $\begingroup$ Yes I do mean invariant under all rotations, sorry. What is a short map? And how does it follow from this? $\endgroup$
    – Nate River
    Apr 29, 2021 at 3:47
  • $\begingroup$ Short means distance-nonexpanding. The optimal plan on $[0,\infty)$ lifts uniquely to an optimal plan on $\mathbb{R}^d$ $\endgroup$
    – Anton Petrunin
    Apr 29, 2021 at 4:18
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    $\begingroup$ @FrancescoPolizzi "any=every" $\endgroup$
    – Anton Petrunin
    Apr 29, 2021 at 6:45

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This is my commnent (which nobody sees).

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 $\mathbb{R}^d$

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  • $\begingroup$ (I use that the cost function is a strictly increasing function of distance.) $\endgroup$
    – Anton Petrunin
    Apr 29, 2021 at 19:36
  • $\begingroup$ I see your comment sir. $\endgroup$
    – Nate River
    Apr 30, 2021 at 1:39

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