Graduate Mapping Tangent Space to Manifold - Questions & Answers

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SUMMARY

The discussion centers on the mapping of tangent space to a manifold using the exponential map, specifically the function exp_p: T_p → M. Participants clarify that evaluating at λ=1 is a convention that simplifies the representation of geodesics, allowing for the tangent vector to encapsulate both direction and distance. The choice of λ=1 is deemed convenient for approximating local geometry as flat space, facilitating the use of Cartesian coordinates. The conversation also addresses concerns about the potential for mapping to distant points in the manifold, ultimately concluding that this is a non-issue due to the nature of affine parameters.

PREREQUISITES
  • Understanding of differential geometry concepts, particularly manifolds and tangent spaces.
  • Familiarity with geodesics and their properties in curved spaces.
  • Knowledge of affine parameters and their role in parametrizing curves.
  • Basic grasp of local Cartesian coordinates and their application in manifold theory.
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  • Explore the properties of geodesics in Riemannian geometry.
  • Study the implications of the exponential map in differential geometry.
  • Investigate the concept of affine connections and their role in defining geodesics.
  • Learn about local coordinate systems and their applications in manifold analysis.
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Mathematicians, physicists, and students of differential geometry seeking to deepen their understanding of manifold theory and the geometric interpretation of tangent spaces.

guitarphysics
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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!
 
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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.
 
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guitarphysics said:
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;.
 
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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).
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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