I was working on some mathematics of Wasserstein GAN and found out a seemingly interesting research problem but I am not quite sure whether it has already been studied in some recent literature of Optimal Transport Theory (As far as I know, it hardly is)

The observation is as follows (which has been validated with some toy simulations).


Problem (not rigorously stated)

Let $\mu, \nu$ probabilistic measures on regular manifolds $\mathcal{M}, \mathcal{N}$, $C^{\infty}(\mathcal{M}, \mathcal{N})$ the set of continuous mapping from $\mathcal{M}$ to $\mathcal{N}$, and $\Pi(\mu,\nu)$ the set of measures on $\mathcal{M}\times\mathcal{N}$ s.t. its marginal distributions are respectively $\mu, \nu$.

Consider the following optimization problem

$$ (*) = \min_{T\in{C^{\infty}(\mathcal{M}, \mathcal{N})}} \inf_{\gamma\in\Pi(\mu,\nu)} \int d^{2}(T(p), q) d\gamma(p,q) $$ where the cost function can be considered as the $L_2$ distance on $\mathcal{N}$'s total space as a real vector space.

My question is that whether $(*) \propto h(\chi(\mathcal{M}), \chi(\mathcal{N}))$ where $h$ is certain metric function and $\chi(\cdot)$ denotes Euler characteristic. Generally, would it be possible that the minimum cost of the Wasserstein game is deeply related with the difference between some topological invariants of underlying manifolds'?


Look forward to any feedbacks and welcome discussions and potential references :D. I am willing to provide details of my toy experiments if one is interested in this problem.

  • $\begingroup$ (1) Which cost function are you thinking off? (2) As there is lots of choice for $T$ the answer seems to be $(*) = \inf_{S:\mathcal{N}\to\mathcal{N} \int c(S(y),y)d\nu(y)$ if the dimensions of the manifolds agree. $\endgroup$ – Martin Kell Nov 29 '18 at 13:24
  • $\begingroup$ For (1), em.. I am thinking about a general form of $h$ since I am not sure if anyone has seen a similar formulation before. Concretely, we can view $c$ as the geodesic distance on $\mathcal{N}$ if it is Riemannian. $\endgroup$ – Morino_Hikari Nov 29 '18 at 13:29
  • $\begingroup$ Your problem does not demand continuous transport of $\mu$ to $\nu$. Any coupling and any continuous/smooth $T$ would do. Hence let $T(r,s)=s$ where $D^1$ is the unit disk and $S^1$ is parametrized by $[-1,1)$. The push-forward is a absolutely continuous measure on the circle with density having a bump at $0$ and being zero at $\pm 1$. It's even possible to find a continuous map such that the push forward is exactly the uniform measure on $S^1$. Choose a coupling concentrated on $\{ (r,s,s) | r,s \in [-1,1]\}$. Problems arise if $\dim \mathcal{M} < \dim \mathcal{N}$. $\endgroup$ – Martin Kell Nov 29 '18 at 13:52
  • $\begingroup$ Oh, I see. Thanks for your swift feedbacks!And I have also considered a non-uniform case on two-dimension where $\mu = p\delta_{0} + (1-p)\text{Unif}(D^{1})$ and $\nu = \text{Unif}(S^{1})$ as I thought the negligible measure at point $(0,0)$ would make the conjectured 'topological obstruction' covered up by integral. And with experiments, I found out that even a small $p$ would make the transportation cost un-vanishing. By the way, have you ever read about some similar issues in recent literature? Since as far as I known, mathematician would not include the $T$ term in their setting. $\endgroup$ – Morino_Hikari Nov 29 '18 at 14:02
  • $\begingroup$ And I thought the simplification you made in the first reply is based on the existence of T that push $\mu$ forward to $\nu$, which is however not obvious to me... $\endgroup$ – Morino_Hikari Nov 29 '18 at 14:20

As for me, two trivial cases currently can be safely stated.

1. If $\mathcal{M}$ is $C^{\infty}$-diffeomorphic to $\mathcal{N}$, then $(*) = 0$.

2. In the toy case between $D^{1}$ and $S^{1}$, where $d\mu = p\delta_{0} + (1-p)d\text{Unif}_{D^{1}}$ and $d\nu = d\text{Unif}_{S^{1}}$, $ (*) = p\inf\int{d^{2}(y_{0}, y)}d\nu(y)$ exactly, since one can always construct a $C^{\infty}$-diffeomorphism from $D^{1}\backslash{\{(0,0)\}}$ to $S^{1}$.

Since $\beta_{1}(D^{1}) = 0$, $\beta_{1}(S^{1}) = 1$, $\beta_{1}(D^{1}\backslash{\{(0,0)\}}) = 0$, is there some possible relation between the different betti numbers and the non-vanishing $(*)$ in the toy case?


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