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:
$$\lambda(p)=0 \\ \frac{dx^{\mu}}{d\lambda}(\lambda=0)=k^{\mu}$$
(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 $s<<S_M$. 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 $p' = exp_p(v)$, the coordinates for $p'$ will be just $(x+v_x, y+v_y, z+v_z)$. Then $v$ can be interpreted as the "displacement vector" connecting $p$ and $p'$.

 

[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).