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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 $E$.


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 $E$?


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)

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    $\begingroup$ Yes, you can do it in a variety of ways. Are there any more constraints you would like on the connection? The ones you give are pretty mild. If your fibration is a vector bundle, and you reduce to structure group to $\mathbb Z_2$ then there is a fairly canonical choice. $\endgroup$ – Ryan Budney Feb 16 '16 at 4:14
  • $\begingroup$ Basically I want each fiber $\pi^{-1}[b]$ to be a curve smoothly parameterized $b\in B$ and the minimal path connecting two points on the fiber $\pi^{-1}[b]$ curve to be part of the fiber itself. $\endgroup$ – AIM_BLB Feb 16 '16 at 4:22
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    $\begingroup$ Ultimately, choosing a connection is going to amount to choosing a sense of "uniform speed parametrization" for the fibers. So you either have to make such a choice beforehand, or you have to make the choice when defining the connection. Either way, there is a choice to be made. Are you hoping for a formula? Or just a "yes you can do it" statement? $\endgroup$ – Ryan Budney Feb 16 '16 at 5:19
  • $\begingroup$ I was thinking of defining length on points on each of the curves $f_a$ (above $a\in A$) by using the formula: $d_a(x,y):=\frac{\int_x^y f_a(u)du}{f_a(1)-f_a(0)}$ and then choosing the Levi-Civita connection associated to this metric. All in all, I was seeking an explicit formula since I will need to make calculations of certain geodesics on the resulting manifold. $\endgroup$ – AIM_BLB Feb 16 '16 at 6:05
  • $\begingroup$ What's $f$? It does not appear in your original question. If you have some extra assumptions it might help. With the way your question is written right now, I do not believe there is any canonical choice. So one will have to make choices.. If you are in a particular formalism or a more special-case situation it might make the choice easier. $\endgroup$ – Ryan Budney Feb 16 '16 at 6:09
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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?

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  • $\begingroup$ I believe I can build an orthogonal complement, because I'm taking my curves to live in a certain Hilbert space (the space of all real-analytic functions with inner product $<f,g>:=\int f'g'$, so for each $\pi^{-1}[b]$, I guess I can build the bundle of all (norm 1) functions $g_a$ such that $<f_a,g_a>=0$; would this bundle help? (So i would look both bundles as part of the Whitney sum bundle they form). How can I use this to get a metric then? Ps.: thank you so much for all your help so far :) $\endgroup$ – AIM_BLB Feb 16 '16 at 14:27
  • $\begingroup$ @CSA: could you provide a more concrete example. Is $B$ the curve? $\endgroup$ – Ryan Budney Feb 16 '16 at 22:14
  • $\begingroup$ for example $B=\mathbb{R}^2$ and $f_b(x)=e^{b_1x}+b_2x$ ($x\in [0,1]$). $\endgroup$ – AIM_BLB Feb 17 '16 at 0:08
  • $\begingroup$ What is the Hilbert space? I'm not seeing one. You are integrating functions but where are these functions defined, and are they related to your $f_b$ functions? $\endgroup$ – Ryan Budney Feb 17 '16 at 18:25

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