$\newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\R}{\mathbb R} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\C}{\mathbb C} \newcommand{\Pb}{\mathbb P} \newcommand{\D}{\mathcal{ D}} \renewcommand{\P}{\mathcal{P}} \newcommand{\E}{\mathbb E} \newcommand{\dx}{\mathrm{d}} $I am currently reading about Optimal Transport. I am interested in the Otto Calculus and the corresponding Benamou Brenier Formula. I encountered a problem in my understanding. The Otto Calculus induces a formal (infinite dimensional) riemannian structure $g$ on a smooth manifold $M$. On the one hand the corresponding riemannian/geodesic distance is given by $$ d(\mu_0,\mu_1) = \inf_{\mathcal{CE}(\mu_0,\mu_1)} \mathcal{L}(\gamma) $$ where $$\mathcal{L}(\gamma) = \int_0^1 \sqrt{g_{\gamma(t)}( \gamma'(t), \gamma'(t)) }\mathrm{d} t$$ denotes the length of a smooth enough curve and $\mathcal{CE}$ a set of smooth curves connecting $ \mu_0,\mu_1$. On the other hand we know the benamou brenier formula which says $$W_2(\mu_0,\mu_1) = \sqrt{ \inf_{\mathcal{CE}(\mu_0,\mu_1)} \left\{ \int_{0}^{1} g_{\gamma(t)}( \gamma'(t), \gamma'(t)) \dx t \right\} }.$$ It is stated that the riemannian/geodesic distance and the Wasserstein distance $W_2$ coincide. However the position of the square root in each expression is different. This leads to my following question.
How can one resolve the square root problem? I have seen that one can define the riemannian/geodesic distance different. See for example here where $$ d(\mu_0,\mu_1) = \inf_{\mathcal{CE}(\mu_0,\mu_1)} \sqrt{ \int_0^1 g_{\gamma(t)}( \gamma'(t), \gamma'(t)) \mathrm{d} t}.$$ Why is this possible? Is there a relation to geodesics?
Here the named equality is stated in defintion 1.5.
PS. I think this questions is more about riemannian geometry and the associated distance than about optimal transport. Therefore i missed out on many details to narrow down my problem. However, if i am wrong please tell me and i will edit for more details.
The functional $$E[\gamma] \doteq \int_0^1 g_{\gamma(t)}(\gamma'(t),\gamma'(t))\,{\rm d}t,$$perhaps with a $1/2$ multiplying factor, is called the energy functional associated to the metric $g$. Geodesics are precisely the critical points of $E$. Also, it follows from Cauchy-Schwarz that ${\cal L}[\gamma]^2 \leq E[\gamma]$.
If I understand it correctly, there is no problem with the square root whatsoever. The relevant result is Proposition 1.3 in your last link, which establishes the desired equality.