Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Map from tangent space to manifold

  1. May 17, 2016 #1
    Hi all, this might be a silly question, but I was curious. In Carroll's book, the author says that, in a manifold [itex] M [/itex], for any vector [itex] k [/itex] in the tangent space [itex] T_p [/itex] at a point [itex] p\in M [/itex], we can find a path [itex] x^{\mu}(\lambda) [/itex] that passes through [itex] p [/itex] which corresponds to the geodesic for that vector ([itex] k [/itex] being the tangent vector to the path). Two conditions for this path are:
    [tex] \lambda(p)=0 \\ \frac{dx^{\mu}}{d\lambda}(\lambda=0)=k^{\mu}
    (And, of course, it must satisfy the geodesic equation.)

    From this, we can then construct a map, call it [itex] \exp_p: T_p\to M [/itex] such that [tex] \exp_p(k)=x(\lambda=1) [/tex]
    Where [itex] x(\lambda=1) [/itex] is the point in [itex] M [/itex] belonging to the parametrized path introduced earlier (the geodesic for [itex] k [/itex]) evaluated at [itex] \lambda=1 [/itex]. Now, my question is: why are we evaluating at [itex] \lambda=1 [/itex]? Not only does this seem arbitrary- it also seems completely independent of all aspects of the manifold. What I mean by this is that we could pick any parameter, big or small, for our geodesic (since we're working with an affine parameter, I think). Given this, it seems like we lose our ability to say that [itex] \exp_p [/itex] maps to the neighborhood of [itex] p [/itex]; it could map to faraway places in the manifold, given the right parameter. So given this,

    1) Why was this chosen? Arbitrary convention?
    2) Is what I pointed out above problematic, or is it a non-issue? (Regarding the fact that we can map to faraway places.)
    3) Could something be chosen instead of [itex] \lambda=1 [/itex], that sort of characterizes a "small scale" in the manifold? This is related to something I'm not very sure about- is there a way to establish the "physical size" of a manifold? Throwing rigor out the window, what I mean is this: maybe we could say that a manifold has "size" [itex] S_M [/itex], and then redefine the map so that [itex] \exp_p(k)=x(\lambda=s) [/itex], where [itex] s<<S_M [/itex]. For an example of what I mean by this "size", maybe we could say [itex] S_M=2\pi R [/itex] for [itex] S^2 [/itex]? (The problem is I doubt this could be done in general :c ).

    Many thanks in advance!
  2. jcsd
  3. May 17, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    The value of ##\lambda## is one because you can always select the tangent vector in such a way that you get a different parametrisation of the same curve.
  4. May 17, 2016 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    I would say that it's just a convention, but it's convenient. Here is a way to think about it:

    The most general way to specify how to travel from a point [itex]p[/itex] along a geodesic would be given by a three-parameter function [itex]F(p,v,t)[/itex], the meaning of which is: Find a geodesic [itex]\mathcal{P}(\lambda)[/itex] such that [itex]\mathcal{P}(0) = p[/itex] and [itex]\frac{d\mathcal{P}}{d\lambda}|_{\lambda=0} = v[/itex]. Then let [itex]F(p,v,t)[/itex] be the point [itex]\mathcal{P}(t)[/itex]. So it makes sense that the way you specify a destination is to give a starting point, [itex]p[/itex], a direction to travel, [itex]v[/itex], and a specification of how far to travel along the geodesic, [itex]t[/itex].

    However, since a vector has both a direction and a magnitude, we can absorb the parameter [itex]t[/itex] into [itex]v[/itex] as follows:

    [itex]F(p, v, t) = F(p, tv, 1)[/itex]

    So in [itex]F(p,v,t)[/itex] you can just fix the third parameter to be 1, and allow the [itex]v[/itex] parameter specify both the direction and how far to go. Carrol's [itex]exp_p(v)[/itex] is just my [itex]F(p,v,1)[/itex].

    The choice of [itex]\lambda = 1[/itex] is pretty much arbitrary. However, it is convenient, because in a small region of space, you can approximate space by flat 3D space (or 4D, if you're talking about spacetime). Then you can use local Cartesian coordinates, so that [itex]p[/itex] has coordinates [itex](x,y,z)[/itex] and [itex]v[/itex] has components [itex](v_x, v_y, v_z)[/itex]. Then if [itex]p' = exp_p(v)[/itex], the coordinates for [itex]p'[/itex] will be just [itex](x+v_x, y+v_y, z+v_z)[/itex]. Then [itex]v[/itex] can be interpreted as the "displacement vector" connecting [itex]p[/itex] and [itex]p'[/itex].
  5. May 17, 2016 #4
    Awesome, thank you both! (Funnily enough, it looks like Orodruin's answer is an abstract for stevendaryl's :D; same concepts, essentially, just different amount of detail hehe).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted