Mapping Tangent Space to Manifold - Questions & Answers

[guitarphysics]

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

From this, we can then construct a map, call it \exp_p: T_p\to M such that \exp_p(k)=x(\lambda=1)
Where x(\lambda=1) is the point in M belonging to the parametrized path introduced earlier (the geodesic for k) evaluated at \lambda=1. Now, my question is: why are we evaluating at \lambda=1? 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 \exp_p maps to the neighborhood of p; 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 \lambda=1, 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" S_M, and then redefine the map so that \exp_p(k)=x(\lambda=s), where s&lt;&lt;S_M. For an example of what I mean by this "size", maybe we could say S_M=2\pi R for S^2? (The problem is I doubt this could be done in general :c ).

Many thanks in advance!

 

[Orodruin]

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.

 

[stevendaryl]

From this, we can then construct a map, call it \exp_p: T_p\to M such that \exp_p(k)=x(\lambda=1)
Where x(\lambda=1) is the point in M belonging to the parametrized path introduced earlier (the geodesic for k) evaluated at \lambda=1. Now, my question is: why are we evaluating at \lambda=1?

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

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

F(p, v, t) = F(p, tv, 1)

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

The choice of \lambda = 1 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 p has coordinates (x,y,z) and v has components (v_x, v_y, v_z). Then if p&#039; = exp_p(v), the coordinates for p&#039; will be just (x+v_x, y+v_y, z+v_z). Then v can be interpreted as the "displacement vector" connecting p and p&#039;.

 

[guitarphysics]

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).