Earth movers distance (EMD) between two multivariate normals. Is it negative definite distance? I was looking at the closed form formula for 2-Wassersteins distance for multivariate normal distribution on Wikipedia. https://en.wikipedia.org/wiki/Wasserstein_metric#Normal_distributions
It has a closed form given as follows:
$W_2(\mathcal{N}(m_1,C_1);\mathcal{N}(m_2,C_2))=\Vert m_1-m_2\Vert_2^2 +\mathrm{Tr}(C_1+C_2-2(C_1^{1/2}C_2C_1^{1/2})^{1/2})$
I could not find any references regarding positive definiteness or negative definiteness of the Wasserstein's distance between two normals. Is it an open question ?  I know that in general p-Wassersteins distances may not be negative definite (https://stats.stackexchange.com/questions/64126/generalized-rbf-kernels). But here since we have a closed form it could be a special case. I will appreciate if someone can point out reference(s).
 A: The Wasserstein distance, defined and here for $p \geq 1$ for measures $\mu,\nu$ with finite $p^{th}$ moment as
$$
W_p(\mu,\nu) = \left( \inf_{\pi \in \Pi(\mu,\nu)} \iint ||x-y||^p d\pi(x,y) \right)^{1/p}
$$
for $\Pi(\mu,\nu)$ is the set of joint measures on $R^d \times R^d$ with marginals $\mu,\nu$,  is a positive definite distance; For any measures with finite $p^{th}$ moment (in particular true for Gaussians), one has 
$$
W_p(\mu,\nu) \geq 0, \quad W_p(\mu,\nu)= 0 \; iif\; \mu = \nu
$$
The proof can be found in Villani, topics in optimal transport, chapter 7. 
Concerning positive definiteness:
It is clear from the formula above and $\pi \geq 0$ (being a probability measure in the product space) that
$$
W_p(\mu,\nu) \geq 0
$$
It is also clear that if $\mu=\nu$, then $\pi(x,y)=\mu(x) \delta_{y=x}$ is optimal with transport cost $0$.
It remains to show that if $W_p(\mu,\nu)=0$ then $\mu=\nu$. This is because 
$$
W_p(\mu,\nu) = 0 \implies x=y \text{ on the support of } \pi \text{ (a.e.)}
$$
and hence one has necessarily $\pi(x,y)=\mu(x) \delta_{y=x}$ (one has always the disintegration $\pi(x,y) = \mu(x)\pi(y|x)$).
This implies that $\nu(y) = \int \mu(x)\delta_{y=x} = \mu(y)$.
The Gaussian case
This implies in particular that for $m_1,m_2 \in R^d$ and $C_1, C_2$ positive definite matrices of $R^d$:
$$
||m_1-m_2||^2 + 2 Tr \left( \frac{C_1+C_2}{2} -(C_1^{1/2}C_2 C_1^{1/2})^{1/2} \right) \geq 0
$$
and $=0$ iff and only if $m_1=m_2$ and $C_1=C_2$.
This is true, regardless of any optimal transport consideration; the difficult term involving the trace of the covariance matrices is nonnegative thanks to an arithmetic/geometric mean (AGM) type of inequality (see [1] for instance), and the equality case only works if we have equality of the matrices.
