What does $g_x^{-1}$ mean where $g$ is a Riemannian metric?

Question

I am reading this paper about the Wasserstein-Fisher-Rao distance and they define the Wasserstein-Fisher-Rao distance on a manifold as follows (page 9):

Here, $M$ is a compact Riemannian manifold, $\mathcal{M}_+(M)$ is the set of positive finite Radon measures on $M$, and $\mathcal{M}(M)$ is the set of signed Radon measures on $M$.

Definition 7. (Dynamical formulation of WFR metric). Let $\rho_0,\rho_1\in \mathcal{M}_+(M)$, the WFR metric is defined by $\begin{equation}\textrm{WFR}^2(\rho_0,\rho_1) = \inf_{\rho,m,\mu}\mathcal{J}(\rho,m,\mu) \tag*{} \end{equation} $ where $\begin{equation} \mathcal{J}(\rho,m,\mu) \\= \displaystyle a^2 \int_{0}^{1}\int_{M} \frac{g_x^{-1}(\tilde{m}(t,x), \tilde{m}(t,x))}{\tilde{\rho}(t,x)}d\nu(t,x) + b^2 \int_{0}^{1}\int_{M} \frac{\tilde{\mu}(t,x)^2}{\tilde{\rho}(t,x)}d\nu(t,x) \tag*{} \end{equation}$ over the set $(\rho, m, \mu)$ satisfying $\rho\in \mathcal{M}_+([0,1]\times M), m\in (\Gamma_M^{0}([0,1]\times M, TM))^*$ which denotes the dual of time dependent continuous vector fields on $M$ (time dependent sections of the tangent bundle, $\mu\in \mathcal{M}([0,1]\times M)$... Moreover, $\nu$ is chosen such that $\rho,m,\mu$ are absolutely continuous with respect to $\nu$ and $\tilde{\rho},\tilde{m},\tilde{\mu}$ denote their Radon-Nikodym derivative with respect to $\nu$.

My question is: what is $g_x^{-1}$ here? It should not be a reciprocal of the Riemannian metric because that goes against the definition of WFR in the Euclidean setting.

I am also confused about what "Radon-Nikodym derivative" means for $m$, since it is not really a measure. My guess is as follows:

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In geometric measure theory, the duals of $k$-forms are called $k$-currents, and since $1$-forms can be identified as vector fields through Riemannian metric, the dual space of smooth vector fields can be seen as the space of 1-currents. A 1-current (with good enough condition) can be represented by an integral (copied from Geometric Measure Theory, A Beginner's Guide by Frank Morgan):

$\displaystyle T(\phi) = \int \langle \vec{T}(x),\phi(x)\rangle d\|T\| \tag*{}$

where $\langle,\rangle$ is an inner product (the book only talks about the Euclidean case, but the above has obvious generalization in the case of Riemannian manifold), $\vec{T}$ is a vector field such that $(\vec{T},\vec{T})=1$, and $\| T\| $ is a Radon measure. This theorem can be seen as a decomposition theorem of a current into a "vector part" $\vec{T}$ and a "measure" part $\|T\|$, and using this decomposition, we can define the Radon Nikodym derivative of $T$, a 1-current on $M$, by $\nu\in \mathcal{M}_+(M)$, as

$\displaystyle \frac{dT}{d\nu} = \vec{T} \frac{d\|T\|}{d\nu}\tag*{}$

which is well-defined since $\|T\|$ is a genuine measure.

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The above is a guess, so I don't know if it is a good definition. I thought maybe it is a dual metric defined here, but it does not make sense if I took my definition of Radon-Nikodym, since $\frac{dT}{d\nu}$ is an element of $TM$ and not $T^*M$.

Answer 1

$\newcommand{\dive}{\operatorname{div}}$$\newcommand{\R}{\mathbb R}$There is a slight (but common) confusion here: as a rule of thumb, momenta are always cotangent vectors, not tangent vectors. For example in classical mechanics the very notation for (generalized) momentum $p=\frac{\partial L}{\partial\dot q}$ should make it clear: the derivative (or rather, the Fréchet differential) of the Lagrangian with respect to the variable $\dot q\in T_q M$ is of course acting on tangent vectors, i-e a cotangent vector. In your specific optimal transport framework, you probably realized that the momentum $m$ plays a role in the continuity equation $\partial_t\rho +\dive_M m=\mu$. But, in manifolds just as in $\R^d$, the divergence operator $\dive_M$ is the dual of the gradient (a.k.a. nabla) operator $\nabla_M$. So, since $\nabla_M$ maps $C^1(M)$ to the tangent bundle $TM$ (if $\phi(x)$ is a function then $\nabla \phi(x)$ is a tangent vector field), the dual $\dive=\nabla^*$ maps $TM^*$ to $(C^1)^*$. Hence $\dive$ indeed acts on cotangent fields, not on tangent fields. And since you are feeding $m$ to the divergence in the continuity equation, this momentum should indeed be a cotangent field (all regularity issues left aside). Let me illustrate this even further: in $\R^d$, in the weak (distributional) formulation of the continuity equation, one usually writes down $$ \int_{\R^d} (\dive m)\phi=-\int_{\R^d} m\cdot \nabla\phi. $$ Here you see that indeed the dual $\nabla$ appears. However, there is a slight abuse in what the meaning of the dot means in $m\cdot \nabla\phi$ in the context of manigolds. Indeed, the right interpretation would be that it is NOT the scalar product between $m$ and $\nabla\phi$, but rather the dual $T^*T$pairing, so in manifolds one should rather write down $$ \int_M(\dive m)\phi=-\int_M\langle m,\nabla\phi\rangle_{T^*_x,T_x}. $$ So, if you accept the idea that $m\in TM^*$ is indeed a cotangent field, indeed it only makes sense to use the inverse metric tensor $g_x^{-1}$.

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Let me be more precise: you copy-pasted from [Frank Morgan] the definition of vector-measure decomposition of 1-currents in $\R^d$, and correctly pointed out that it admits an immediate generalization to manifolds. However, in $\R^d$ there is this usual ambiguity between tangent and cotangent vectors. The right decomposition of a 1-current $T$ over a manifold should read, modulo reasonable smoothness, $$ T(\xi) = \int_M \langle \bar{T}(x),\xi(x)\rangle_{T_x^*M,T_xM} |T|(d x) $$ for any tangent field $\xi\in TM$. The point is that in defining this decomposition one should primarily use the canonical $T^*T$ pairing, not directly the scalar product (both are of course equivalent modulo standard musical isomorphisms, but the primordial definition is really that). Here $|T|\in\mathcal M_+(M)$ is a nonnegative measure, and $\bar T(x)\in T^*_xM$ is a cotangent field. Then, if $\nu\in\mathcal M_+(M)$ we say that the current $T$ is absolutely continuous with respect to $\nu$ if $|T|\ll\nu$ (in the usual sense of measures), and we simply write $$ \frac{dT}{d\nu}(x)=\bar T(x)\frac{d|T|}{d\nu}(x)\in T_x^*M. $$ This is again a cotangent vector field, as the pointwise multliplication of the cotangent vector $\bar T(x)$ by the scalar $\frac{d|T|}{d\nu}(x)$ above every $x\in M$.

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Just to wrap it up: In your particular Wasserstein-Fisher-Rao case the Radon-Nikodym derivative $\tilde m(x)=\frac{dm}{d\nu}(x)$ is indeed a cotangent vector, and therefore it only makes sense to evaluate $g_x^{-1}(\tilde m(x),\tilde m(x))$ using the inverse metric, not the metric itself. Formally, one expects as in classical optimal transport that $m(x)=\rho(v)v(x)$, where $v(x)$ is now a tangent vector (if you know a little bit of optimal transport, $v(X_t)=\frac{d}{dt}X_t$ is indeed the time-derivative of Lagrangian particles, so it really is a tantent vector $v(x)\in T_xM$). To be more rigorous, but discarding regularity issues, one should use the musical isomorphisms to write $m=(\rho v)^{\flat}$. The kinetic energy is then indeed, by definition of the $\flat$-isomorphism, \begin{multline*} \rho(x) g_x(v(x),v(x)) \\ =\frac{1}{\rho(x)}g_x(\rho v(x),\rho v(x)) =\frac{1}{\rho(x)}g_x^{-1}((\rho v)^\flat(x),(\rho v)^\flat(x)) \\ =\frac{1}{\rho(x)}g_x^{-1}(m(x),m(x)). \end{multline*} In $\R^d$ one usually dispenses from writing $\flat$ and simply abuses notations to write $m=\rho v$, and one moreover uses the same notation $|\cdot|^2$ undistinctly for the tangent metric and cotangent inverse metric. Hence the confusion.