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).

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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'?

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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.

  • (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. – Martin Kell Nov 29 at 13:24
  • 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. – Morino_Hikari Nov 29 at 13:29
  • 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 $\dim \mathcal{M} < \dim \mathcal{N}$. – Martin Kell Nov 29 at 13:52
  • 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. – Morino_Hikari Nov 29 at 14:02
  • 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... – Morino_Hikari Nov 29 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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