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Tangent vectors

  1. Oct 15, 2005 #1
    Here's the question. Say you've got some manifold M and at a point p in M you want the tangent space. I don't quite understand why equivalence classes of paths through p need to be considered. That is, what goes wrong if you don't?

    Kevin
     
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  3. Oct 15, 2005 #2

    HallsofIvy

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    I'm not sure I understand. How do you define tangent vectors if not as equivalence groups of paths? Two such paths are "equivalent" of course, if their derivatives are the same at the given point.
     
  4. Oct 15, 2005 #3

    Hurkyl

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    Well, you need a definition for the tangent vectors, and an equivalence of paths is certainly a definition.

    This particular definition has the advantage of being constructive: it's built out of concepts you already understand.

    It also has the advantage of being directly related to the concept it's trying to capture: tangent vector, by this definition, is precisely the class of things that are going in the "same direction" at that point.


    Of course, in the end, it doesn't really matter if you choose to use this definition, or some other equivalent definition. It's the properties that tangent vectors have that matters.
     
  5. Oct 15, 2005 #4
    At the risk of sounding increasingly stupid I'll carry on...

    Let's see if I can word my question better. So at some point p in your manifold, why not let the tangent space be the collection of all linear combinations of derivatives of curves passing through p (and evaluated at p)? I realize that this is just one definition (I like the derivation definition better), but I'm just not sure why we bother to form equivalence classes. However, I'm sure that there's a good reason, something fairly obvious most likely.

    Thanks for your patience,

    Kevin
     
  6. Oct 15, 2005 #5

    Hurkyl

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    So you're fixing a coordinate chart, and then defining the tangent space at a point P to be the tangent space at the corresponding point in R^n, right?

    (Tangent space has a more obvious definition when we're dealing with a vector space, so I'll assume we don't need to define that)
     
  7. Oct 16, 2005 #6
    Well I don't see any reason to bring R^n into the discussion yet.
     
  8. Oct 16, 2005 #7

    Hurkyl

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    Then how do you give meaning to the phrase "derivative of a curve"?
     
  9. Oct 16, 2005 #8
    Right sorry. So for our point p in our manifold M, we have a path g:[a,b]->M in M that goes through p. But, you're right, in order to differentiate g, we'll need to get its coordinate rep, i.e. x(g(t)) where x is a chart that has p in its domain. So yes we're now looking at the image of g under x in an open subset of R^n and we're computing the derivative of x(g(t)) at x(p).
     
  10. Oct 16, 2005 #9

    HallsofIvy

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    Hurkyl's point is that your definition of "tangent vector" depends on a specific choice of coordinate system. If I were to choose a different coordinate system for the same manifold, I would get different vectors. Yes, it would be easy to show they were isomorphic but the point is that by defining the vectors themselves to be "equivalence classes of paths through p that have the same derivative is some coordinate system", the vectors themselves do not depend on the coordinate system.
     
  11. Oct 16, 2005 #10

    Hurkyl

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    Now homology, consider this:

    You've defined "the tangent space at the point P in the coordinate chart A".

    But that's not good enough, because now you'll have to carry the baggage of coordinates around with you wherever you go, which is a bad thing.

    So now, how do you plan to proceed to get a definition of "the tangent space at the point P"?
     
  12. Oct 16, 2005 #11
    Hmmm, the light is slowly creeping through the dense jungle that protects my brain from learning too quickly...

    So, let's see what I've learned. a point p in a manifold sits in all kinds of coordinate chart domains. And the images of those domains don't necessarily intersect in R^n. So when we define the tangent vector of a path g, at p, then we'll get different vectors (in R^n) depending on which chart we use. This of course sucks, because the tangent of a curve at a point ought to be independent of the coordinates. Okay, so two curves (at p) are equivalent (make the same tangent vector in T_pM) if their images have the same tangent vector in R^n under some chart.

    So two curves, h and g at p are equivalent if there is some chart x, such that the derivative of x(g) = x(h) at some p.

    Is that okay?

    Again, thanks,

    Kevin
     
  13. Oct 16, 2005 #12

    Hurkyl

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    I remember when I first saw that definition, I thought to myself "I don't like that definition. I'll work out my own"... then once I started getting happy with my definition, I realized that I had essentially just replicated that definition. :smile:


    Incidentally, you could carry your approach further:

    You could look at all of the (x, v) pairs where x is a coordinate chart and v is a tangent vector at p in x. You could then say (x, v) ~ (y, w) iff the change of coordinate map from x to y would map v to w. Then, you could take these equivalence classes, and define that to be the tangent space at P.

    You'd still have to prove this is an equivalence relation, but the work is roughly equivalent to proving that the usual definition is well-defined.

    In summary, you could let the tangent vector v be the set of pairs of coordinate charts with v's coordinate representation, instead of letting v be the set of curves to which its tangent.


    I just cracked open Spivak's Differential Geometry and skimmed the section on the tangent bundle.

    He starts by defining, for R^n (and its subspaces), the tangent space at a point p.

    He then defines the notion of a vector bundle.

    He then states a theorem asserting the existance of a vector bundle with special properties (which we call the tangent bundle). The proof he uses is basically:

    (1) Use the above idea to construct the tangent spaces at points p. (and not using an equivalence class of curves)

    (2) Assemble all of the tangent spaces into the tangent bundle.

    (3) Prove that it really is a vector bundle satisfying the special properties.


    However, once he's done this, he then goes on to mention the equivalence classes of curves definition, asserting that it is "really the same" to the tangent bundle. He gives another construction as well.


    P.S. it does sound to me like you've got the equivalence classes of curves definition. Just don't forget someone needs to prove that definition is independent of the choice of charts!
     
    Last edited: Oct 16, 2005
  14. Oct 17, 2005 #13

    George Jones

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    Here's what Chris Isham has to say in the preface of his very nice book Modern Differential Geometry for Physicists:

    "In particular, I have laid considerable stress on the basic ideas of 'tangent space structure', which I develop from several different points of view: some geometric, some more algebraic. My experience in teaching this subject for a number of years is that a firm understanding of the various way of describing tangent spaces is the key to attaining a grasp of differential geometry that goes beyond just a superficial aquiescence in the jargon of the subject."

    Isham decribes tangent vectors as both as equivalence classes of curves, and as derivations. He shows that the equivalence class definition gives a nice conceptual picture of a push-forward ("linearization") of a map between manifolds.

    Regards,
    George
     
  15. Oct 17, 2005 #14
    that's a very nice quote. I would daresay that it applies to virtually every mathematical idea. A key to understanding any mathematical discipline is to be able to non-trivially describe any of its components in more than one way.
     
  16. Oct 19, 2005 #15
    okay so

    So I'm stuck on transitivity. Clearly our relation is reflexive and symmetric. But let's say we have three paths, a, b, c and in our manifold M. and that

    a ~b and b~c

    So there exists a chart x such that the tangent of a and the tangent of b are equal under x and there exists a chart y such that the tangents of b and c are equal under y.

    Now I suspect I may be making this too hard, but I can't see how to get a chart for a to be equivalent to c. I've stuck in x^-1 composite x and played with the change of coordinates diffeo and play with chain rule, but I end up some derivative stuff I can't get rid of. Any ideas?

    Thanks,

    Kevin
     
  17. Oct 19, 2005 #16

    George Jones

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    Suppose you want to find the tangent space at p.

    The definition of tangency at p of 2 curves is chart-independent. Pick a chart that has p in its domain and use this for all 3 curves.

    Regards,
    George
     
  18. Oct 19, 2005 #17
    Thanks George, but that's what I'm trying to prove. That the relation amoung curves is an equivalence relation.
     
  19. Oct 19, 2005 #18

    George Jones

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    Everything is at p.

    If a is tangent to b in one chart, a is tangent to b in all charts.

    To show ((a tangent to b) and (b tangent to c)) implies (a tangent to c),
    it suffices to show that this statement is true in one chart.

    Maybe the definitiion of tangency is the thing that's giving you problems.

    Regards,
    George
     
  20. Oct 19, 2005 #19

    Hurkyl

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    homology is essentially trying to prove that tangency has the property that you state, so it's not fair game to use that property. :smile:


    I assume you meant to compose y with x^-1! I'd like to suggest simply calling the function f, and forget that it's actually the composition of two functions. Anyways, it would be easier to help if we saw your work. :smile:
     
  21. Oct 20, 2005 #20
    Okay so let p be a point in a manifold M and let a, b, and c be paths in M. Now suppose that a~b and that b~c.

    Then we know that there exists charts x and y such that:
    (I'm going to use @ as composition)

    (1) d(x@a)/dt = d(x@b)/dt
    (2) d(y@b)/dt = d(y@c)/dt

    I would like to find a chart z, such that

    (3) d(z@a)dt = d(z@c)/dt

    Well one thing I thought of doing is to put x^-1@x into the RHS of (2) to get

    (4) d{(y@x^-1)@(x@b)/dt

    You're right Hurkyl we could just call y@x-1 f, its nice and differentiable. So we'd have

    (5) d{f@(x@b)}/dt

    So carrying out this differentiation, f maps R^n to R^n, and x@b maps R to R^n so we if we let Jf be the jacobian of f we have, upon rewriting (2)

    (6) Jf(d(x@b)/dt = d(y@c)/dt

    Now using (1), we have

    (7) Jf(d(x@a)/dt = d(y@c)/dt

    Now at this point, its probably my weak calc.3. This jacobian is an isomorphism (because f is diffeo) from the tangent space at x(p) to the tangent space at y(p), and...I want to say that the Jacobian takes tangents of curves to tangents of image curves, but that fact may rely on what I'm trying to prove. I'm going to work on that aspect for a bit. This is what I have so far. Thanks for your help.
    Kevin
     
    Last edited: Oct 20, 2005
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