Notation : $\#$ is the pushforward measure and $X$ is the whole space on which $\mu,nu$ are deffined.
In optimal transport, a transport plan between two probability measures $\mu$ and $\nu$ always exists, namely the plan given by the product measure $\mu\otimes\nu$. But what does this plan actually mean. Is it something like redistributing the mass $\mu$ uniformly or something like that?
In particular, in the following quote
Let $\pi$ be the transference plan obtained by keeping fixed all the mass shared by $\mu$ and $\nu$, and distributing the rest uniformly: this is $\pi = (Id , Id )_\# (\mu \wedge \nu) +\frac{1}{a}(\mu −\nu)_+ \otimes ( \mu − \nu)_−$ where $\mu \wedge \nu = μ − (μ − ν)_+$ and $a = (μ − ν)_− [X] = (μ − ν)_+ [X]$.
I don't understand why $\frac{1}{a}(\mu −\nu)_+ \otimes ( \mu − \nu)_−$ distributes the rest uniformly.
I use the following intuition for the independent transport plan $\pi:=\mu\otimes\nu$: For each location $x\in \mathrm{supp} \mu$ you take the mass $d\mu({x})$ located on $x$ and then transport $d\pi(x, y) = d\mu({x})d\nu({y})$ of it to $y$ for every location $y \in \mathrm{supp} \nu$.
So yes, for each $x\in \mathrm{supp}(\mu)$ the mass located on $x$ is always distributed to the points of $\mathrm{supp} \nu$ with the same proportions given by $\nu$.