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

Tangent vectors as equivalence classes of curves

  1. Aug 4, 2011 #1
    defining a tangent vector v as the equivalence class of of curves: [tex]v = [\sigma] = \left. \frac{df(\sigma)}{dt} \right|_{t=0}[/tex], i want to show that this definition is independent of the member of the equivalence class that i choose.

    where [tex]\sigma[/tex] represents a function from the reals to the manifold and f is a coordinate function from the manifold to the reals.

    so i am starting with [tex]\sigma_{1} (0) = \sigma_{2}(0)[/tex]since i am picking two curves that go through the same point. i then define the functions [tex]F = \phi o \sigma_{1}[/tex] and [tex] G = \phi o \sigma_{2}[/tex] and since these are both functions from R to R my goal is to now show that F'(0) = G'(0). i also know that f(0) = g(0) since [tex]\sigma_1(0) = \sigma_2(0)[/tex] but now i am stuck and can't figure out the next step. can someone give me a few hints in the right direction? thanks.
     
  2. jcsd
  3. Aug 4, 2011 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hey. So you've defined a relation on the set of smooth curves passing through a given point p on the manifold by declaring equivalent two curves iff they tangent vector at p with respect to some chart about p are the same. Presumably, you've shown that this is an equivalence relation, and you then call such an equivalence class a "tangent vector at p".

    What is there to show? What is ambiguous with this definition?

    The next step however would be to show that the set of all tangent vectors is a vector space. And then, there will be ambiguities and it will be necessary to check that the vector space operations are well defined (independant of the class representatives).
     
  4. Aug 17, 2011 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    given this definition,check that two curves are equivalent iff they have the same tangent vector in one and hence all coordinate charts., therefore it is a vector space, oops once you check the isomorphism for different coordinate charts is linear (chain rule). this stuff seems trivial, i.e. a good exercise if you don't agree.
     
  5. Aug 20, 2011 #4

    Landau

    User Avatar
    Science Advisor

    I think he wants to show that this equivalence relation (or rather, 'being tangent') is chart-independent, i.e. that if two curves are tangent in some chart, then also in any other chart. This follows directly from the chain rule.
     
  6. Aug 21, 2011 #5
    the book i am using says to show that this equivalence class is independent of the representative i choose but i guess that follows from the set of curves being an equivalence class to begin with. as for showing that the equivalence classes are coordinate chart independent i know how to do that as well. i think my question was a bit confusing, but thanks for everyone's input.
     
  7. Aug 21, 2011 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not a fan of this notation, or the precise statement you made. If there's an f on the right, there obviously needs to be one on the left as well. Why "d/dt" when there's no "t" in the expression that's supposed to tell us what function you're taking the derivative of? (The latter is an abuse of notation that seems to be somewhat standard). Also, f isn't a coordinate function. I think it should be any smooth function defined on an open neighborhood of the relevant point in the manifold.

    I'd write [itex][\sigma]f=(f\circ\sigma)'(0)[/itex] or [itex]\phi([\sigma])f=(f\circ\sigma)'(0)[/itex]. The point of this definition is to define a function from the set of equivalence classes of curves through p to the set of derivations at p, so it might be useful to choose a symbol for it, in anticipation of the next step, which is to prove that this function is a vector space isomorphism. The equality you posted doesn't define the tangent vectors or the equivalence relation. It just defines the isomorphism.

    Isham?

    If follows from the definition of this particular equivalence relation.

    A couple of latex tips: If you use tex tags and you want a comma or a period at the end, you need to put it before the closing tex tag, not after it. I also recommend that you use itex tags instead of tex tags when it makes more sense for the math expression to appear in the middle of a line of text rather than indented on a line of its own.
     
  8. Aug 22, 2011 #7
    yes i am using Isham's book which may not be the best book for a first exposure to differential geometry. i often get confused about what the notations are trying to say in the definitions. as i understand to show that the function taking a set of equivalence classes to a set of derivations is a vector space isomorphism, i need to show that it is a linear transformation and that it is both one to one and onto. i seem to be having trouble showing first that it is a linear transformation.
     
  9. Aug 22, 2011 #8

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I like the book, especially for the stuff about tangent spaces, Lie groups, Lie group actions and the introduction to fiber bundles. But it wasn't the first book I read on differential geometry. I don't really like his notation, but I don't really like anyone's notation but my own. I also felt that he was rushing things towards the end. I had to work much harder to understand each page once he started talking about connections on principal bundles. It was also weird that the book ended where it did. He could have included another chapter or two about integration on manifolds, the Yang-Mills lagrangian, etc.

    "Introduction to smooth manifolds" by John M. Lee, is a better book for most purposes. Its biggest flaw is that it doesn't cover connections and curvature. The best place to read about those things is an earlier book by the same author, "Riemannian manifolds: an introduction to curvature".

    The definition of linear combinations of equivalence classes of curves makes it complicated. Let's see if I can remember how he did that. Hm, I'm thinking that he probably just defined [itex]a[\sigma]+b[\tau]=[a\sigma+b\tau][/itex], where the linear combination of the curves on the right is defined in terms of a coordinate system x, by [tex]a\sigma+b\tau=x^{-1}\circ(a(x\circ\sigma)+b(x\circ\tau)).[/tex] If this is right, then it would take some work just to show that this makes sense and that the result doesn't depend on the coordinate system x. I'm not going to do that now. Assuming that things are defined as above, [tex]
    \begin{align} &\phi(a[\sigma]+b[\tau])f=\phi([a\sigma+b\tau])f =(f\circ(a\sigma+b\tau))'(0) =(f\circ x^{-1}\circ(a(x\circ\sigma)+b(x\circ\sigma)))'(0)\\
    &=(f\circ x^{-1})_{,i}(ax(\sigma(0)))+bx(\tau(0))) \ (a(x\circ\sigma)+b(x\circ\tau))^i'(0)
    \end{align}
    [/tex] The second factor is [tex]=(a(x^i\circ\sigma)+b(x^i\circ\tau))'(0) =a(x^i\circ\sigma)'(0)+b(x^i\circ\tau)'(0) =a(x\circ\sigma)^i'(0)+b(x\circ\tau)^i'(0).[/tex] Should be easy enough from here. Insert this result into the main calculation and use the chain rule again (this time in the other direction).

    If the way I'm writing the chain rule looks weird to you, see this post for an explanation.
     
    Last edited: Aug 22, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tangent vectors as equivalence classes of curves
  1. Tangent vectors (Replies: 21)

Loading...