Wasserstein distance to the set of Gaussians ; Boltzman dissipation rate I am interested in the $2$-Wasserstein distance for probabilities over ${\mathbb R}^n$,
$$W_2(\mu,\nu)=\left(\inf\int_{{\mathbb R}^n\times{\mathbb R}^n}|w-v|^2\pi(v,w)\right)^{1/2}$$
where the infimum is taken over the probabilities $\pi$ having marginals $\mu$ and $\nu$.
Let $GP$ denote the subset of probabilities that have a Gaussian density. Is there any close formula, or approximate formula, for the distance (in terms of $W_2$) from a given $\mu$ to $GP$ ? How can it be related to the entropy dissipation rate encountered in the Boltzman equation, where one integrate the non-negative quantity
$$\left(f(v^*)f(v'^*)-f(v)f(v')\right)\left(\log\big(f(v^*)f(v'^*)\big)-\log\big(f(v)f(v')\big)\right)$$
where $f$ is the density of the velocity distribution and the arguments are constrained by (conservation of momentum and energy)
$$v+v'=v*+v'^*,\qquad\|v\|^2+\|v'\|^2=\|v^*\|^2+\|v'^*\|^2.$$
Notice that, for consistency, the Wasserstein distance should be applied to measures of the form
$$\frac1\rho\,f(v)\,dv,\qquad\rho:=\int f(v)\,dv.$$
 A: I'm not sure about the dissipation rate part, but what I can tell you is that the projection of any probability density $\mu$ (with second moments) onto $GP$ should be the Gaussian density with the same mean and (I'm almost sure, see below for details) variance.
Here's an idea of the proof. From Brenier's theorem, the optimal transport problem from any absolutely continuous density has a Monge solution that is the gradient of a convex function, so there exist functions $\phi_{m,\Sigma}$ whose gradient map any Gaussian density $N(m,\Sigma)$ to $\mu$.
$$\nabla \phi_{m,\Sigma} \# N(m,\Sigma) = \mu .$$
In Wasserstein-2 space, the optimal map between two Gaussians is the affine map; therefore we can derive a formula for $\nabla \phi_{m,\Sigma}$ in terms of, say, $\nabla \phi_{0,I}$ (the map corresponding to the central normalized Gaussian).
Using it gives a formula for the Wasserstein distance, that can be expressed as an integral against $N(0,I)$.
$$W_2^2(N(m,\Sigma),\mu) = \int |\sqrt{\Sigma}x + m - \nabla\phi_{0,I}(x)|^2 dN(0,I)(x).$$
Then we can compute the first-order variation of the distance in terms of $m,\Sigma$. Checking when it is zero yields $m = \mathbb{E}(\mu)$, and another condition.
I think this other condition should lead to $\Sigma = \mathbb{V}(\mu)$, although I was unable to finish the computation.

Update: My conjecture was wrong: the variance of the closest Gaussian may be different from the variance of $\mu$.
Here is an example where they differ.
Let $\phi(x)=|x|$ be the "absolute value" function on $\mathbb{R}$. Consider the distribution defined by $\mu = \nabla\phi_{\#} N(0,1)$. It is actually composed of two Diracs: $\mu = \frac{1}{2}(\delta_{-1} + \delta_{1})$.
After the first trick, the Wasserstein distance from $\mu$ to $N(m,\sigma^2)$ is
$$W_2^2(N(m,\sigma^2),\mu) = \int (\sigma x + m - \nabla\phi(\sigma))^2 dN(0,1)(x)$$
Minimizing in terms of $\sigma$ yields
$$
0=\sigma-\int |x| dN(0,1)(x).
$$
And so $\sigma = \sqrt{\frac{2}{\pi}}$, and not 1 as I thought.
