Let $\pi: E \rightarrow B$ be a fibration over a Riemannian manifold $B$, with $\pi^{-1}[b]$ homeomorphic to $\mathbb{R}$.


More precisely:


* I want each fiber $\pi^{-1}[b]=Im(f\_b)$ for some $C^{\infty}$-curve $f\_b:\mathbb{R}\rightarrow \mathbb{R}$.
* If $(f\_b(x),b),(f\_b(y),b)$ lie on the same fiber (above the point $b\in B$) then: I would like the distance between them to be defined as:
\begin{equation}
d\_b((f\_b(x),b),(f\_b(y),b)):=\frac{\int\_x^y f\_b(u) du}{f\_b(1)=f\_b(0)}. 
\end{equation}
*(that is I want geodesics connecting two points on the curve $f\_b$ above $b$ to be the corresponding segment of that curve. )*
* I would like the distance between two points on the same "level" to but corresponding to different parameters $b,b'$'s distance to be measured purely in terms of B's distance, that is:
\begin{equation}
d((f\_{b'}(x),b'),(f\_b(x),b))=d\_B(\pi(f\_{b'}(x),b'),\pi(f\_b(x),b))=d\_B(b',b). 
\end{equation}
*(that is I want geodesics connecting two points on different curves $f\_b(x)$ and $f\_{b'}(x)$ on the same level (x) to correspond to the distance between the parameters $b$ and $b'$ in $B$. )*


**Notes on Notation:**
Where $d\_b$ is the Riemmanian metric existing on a specific fiber, $d\_B$ is the Riemmanian Metric on $B$ and $d$ is the metric I am looking for on $E$. 




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**Question:**


Then can I make $E$ into a manifold and it with a Riemannian metric satisfying the above such that moreover:


each fiber $\pi^{-1}[b]$ is a geodesic in $E$?




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**Nutshell Resume of Goal:**


(essentially I want to be able to move from $f\_b(x)$ to $f\_{b'}(y)$ in the shortest way given the above constraints, and explicitly calculate that distance) 



Construction of fibration over Riemannian Manifold
Let's consider the case $B = \mathbb R^n$ and $E = B \times \mathbb R$, with bundle map $\pi(x,t) = x \in B$. 


This isn't much of a simplification as your fibre bundle is locally diffeomorphic to such models.


In this model you could define the Riemann metric on $E$ to be the product metric of $B$ and $\mathbb R$. Your integral condition is basically saying what the metric on the $\mathbb R$ factor should be. 


But the problem is the ambiguity in the choice of these local models. 


Such product decompositions (fixing the base) correspond to the space of maps $B \to Diff(\mathbb R)$. Up to deformation there are only two such choices,but this is assuming the simplified model. The actual space of choices is infinite-dimensional.


If we use your condition involving $f$, the above becomes the space of maps $B \to O\_1 \ltimes \mathbb R$. While much simpler,it is still infinite dimensional. And in general it sounds like you do *not* have a model with a canonical element in it, like this (the zero section).


The thing you have to decide, which would fix the metric and allow you to use a formula would be an orthogonal complement to your fibres $\pi^{-1}(b)$. In the situations that you care about,is it clear what those should be? 


