Riemannian manifold of absolutely continuous functioins I´m struggling with a definition of a Riemannian metric proposed in the textbook "Functional and Shape Data Analysis" by Anuj Srivastava, Eric P. Klassen:
Let $\mathcal{F}$ be the set of absolutely continuous functions $f: [0,1] \rightarrow \mathbb{R}$ and let $\mathcal{F}_0 = \{ f \in \mathcal{F}|\, f'>0 \}$.
Definition: Let $f \in \mathcal{F}_0$ and $v_1, v_2 \in T_{f}(\mathcal{F})$, where $T_{f}(\mathcal{F})$ is the tangent space of $\mathcal{F}$ at $f$, the Fisher-Rao Riemannian metric is defined as the inner product
\begin{align} 
    \langle\langle v_1, v_2 \rangle \rangle_{f} = \frac{1}{4} \int_0^1 v_1'(t)  v_2'(t) \frac{1}{| f'(t)|}  dt. 
\end{align}
My questioin is: How do I check $\mathcal{F}_0$ is a Riemannian manifold in this context? The textbook does not give any comments on this and after several hours of thorough web searching I feel particularly lost.
 A: They are quite sloppy here. Caveat: I am a differential geometer, not an expert in data analysis, so take what will say here with a grain of salt.

*

*I think, they implicitly assume that the function $f$ not only has the property $f'>0$ (almost everywhere), but satisfies the condition
$$
\frac{1}{f'(t)}\in L^1([0,1]). 
$$
Without this assumption, the integral that they write may diverge.


*They completely neglect to indicate which topology they use on the space ${\mathcal F}$: Without such a specification, it is meaningless to talk about structure of an infinite-dimensional Riemannian manifold. There is a chance, they simply take the $L^1$ topology on the space of derivatives of functions $f\in {\mathcal F}$. (In fact, they'll need a bit more control to ensure openness of ${\mathcal F}_0$. I will refrain from speculating what conditions they would impose, maybe $f\in C^1$, or maybe they have in mind only constructing a metric space rather than an infinite-dimensional Riemannian manifold. In this case, all what they care about are lengths of paths. My guess is that it is the latter.) To make this work, they need to normalize functions $f\in {\mathcal F}$ so that one can uniquely recover $f$ from its derivative. For instance, one may require $f(0)=0$ or $\int_0^1 f(t)dt=0$.
I will be assuming some normalization like that in what follows (you will see why soon).
Then the space they are dealing with is a convex open subset in a Banach space, hence, a Banach manifold.


*Now, let's check at least some Riemannian properties of the F-R metric:

(a) Bilinearity and semipositivity of the form that they define is clear.
(b) Suppose that $\langle v, v \rangle_f=0$, i.e.
$$
\int_0^1 \frac{(v'(t))^2}{f'(t)} dt=0. 
$$
Since $f'>0$ a.e., the integrand has to vanish a.e., which implies that $v'(t)=0$ a.e., hence, $v$ is constant. This, of course, does not imply that $v\equiv 0$ as required by the definition of a Riemannian metric. Since I assumed (unlike the authors of the book) that elements of ${\mathcal F}$ are normalized, the tangent vectors $v\in T_f{\mathcal F}$ are also normalized, which implies that indeed $v\equiv 0$.
Lastly, as a Riemannian geometer, I would like to check that my Riemannian metric depends smoothly on $f\in {\mathcal F}$. However, since the authors do not compute connections or curvatures, they do not really need smoothness. All what they need is continuity, so that they can define lengths of paths in ${\mathcal F}$. I suspect they do this only for "nice" paths $\alpha: [0,1]\to {\mathcal F}_0$,
for which the integral
$$
\int_0^1 ||\alpha'(s)||ds
$$
converges. But this is something for you to check.
