The two terms you wrote down are roughly the horizontal and vertical parts of the metric. Roughly speaking, the first part gives the metric for the first factor $T_pM$ of $T_{p,v}TM$. The second part gives the metric for the second factor $T_vT_pM$.
By the definition of the projection operator $\pi$ and the definition of the covariant derivative on $(M,\langle\cdot\rangle)$, it is clear that the expression you wrote down is coordinate independent. There are several things to check to make sure that it is a metric
- It is positive definite
- It is bilinear
- It is tensorial
Since we already have coordinate independence, it is most convenient to work over a fixed coordinate system.
Let $\{x^1,\ldots,x^n\}$ be a system of coordinates for $M$; this can be extended to a system of local coordinates $\{x^1,\ldots,x^n;y^1,\ldots,y^n\}$ for $TM$ where $(x,y)$ corresponds to the point $(p,v)\in TM$ with $p$ the point in $M$ specified by $x$, and $v\in T_pM$ given by $\sum y^i\partial/\partial x^i$.
At a fixed point $(p,v)$, an element of $T_{p,v}TM$ can be then described by
$$ V = \sum \xi^i \frac{\partial}{\partial x^i} + \sum \zeta^i \frac{\partial}{\partial y^i}$$
with the projection
$$ d\pi(V) = \sum \xi^i \frac{\partial}{\partial x^i} $$
Express the curve $\alpha(t) = (p,v)(t)$ in this coordinates we have that the condition $\alpha'(0) = V$ is simply the statement that $\frac{d}{dt}p^i(0) = \xi^i$ and $\frac{d}{dt}v^i(0) = \zeta^i$.
With this we can compute
$$ \frac{D}{dt}v^i(0) = \frac{d}{dt} v^i(0) + \Gamma^i_{jk}\left(\frac{d}{dt}p^j(0)\right)\left(v^k(0)\right) $$
using the definition, and where $\Gamma$ is the Christoffel symbol of the Riemannian metric on $M$. This we immediately see to be
$$ \frac{D}{dt}v^i(0) = \zeta^i + \Gamma^i_{jk}v^k(0)\xi^j $$
which in fact is a linear map from $T_{p,v}TM$ to $T_pM$.
Now the three properties are easily checked:
- It is tensorial because the expressions are completely independent of which curve $\alpha$ is chosen, as long as $\alpha'(0)= V$.
- It is bilinear because $T_{p,v}TM\ni V\mapsto (d\pi(V),\frac{D}{dt}v(0))\in T_pM \oplus T_pM$ is linear, and the Riemannian metric on $M$ is bilinear
- Since the Riemannian metric induced on $T_pM\oplus T_pM$ is positive definite, to prove that the new object is also positive definite, it suffices to check that the map $V\mapsto (d\pi(V),\frac{D}{dt}v(0))$ is injective. But this is true by direct inspection.
Yes, this is basically fine. One small correction: You wrote
Because of this $|X_1| = \sqrt{|g(X_1,X_1)|}$ is a non-vanishing
smooth vector field on $U_1$, ...
But $|X_1|$ is not a vector field. It should say
Because of this $|X_1| = \sqrt{|g(X_1,X_1)|}$ is a non-vanishing
smooth function on $U_1$, ...
Best Answer
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.
(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.