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Tangent vector to a curve (Differential geometry/Lie theory).

  1. Nov 19, 2011 #1

    B L

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    1. The problem statement, all variables and given/known data

    Let [itex] c(s) = \left( \begin{array}{ccc}
    \cos(s) & -\sin(s) & 0 \\
    \sin(s) & \cos(s) & 0 \\
    0 & 0 & 1 \end{array} \right) [/itex] be a curve in SO(3). Find the tangent vector to this curve at [itex] I_3 [/itex].

    2. Relevant equations

    Presumably, the definition of a tangent vector as a differential operator would be useful here:
    If X is the tangent vector to [itex] c(s) [/itex] at [itex] I_3 [/itex] (i.e. [itex] s = 0 [/itex]), then for functions [itex] f: SO(3) \rightarrow \mathbb{R} [/itex],
    [itex] X\left[ f \right] = \frac{df\left(c\left(s\right)\right)}{ds}|_{s=0} [/itex]

    3. The attempt at a solution
    The fact that this problem is found in the section of the textbook (Nakahara, Geometry, Topology and Physics) dealing with Lie groups and Lie algebras, together with the fact that the problem asks for the tangent vector to c(s) at the identity leads me to think that we are supposed to use the Lie algebra in some way. However, I can't see how one would do that, so I've tried proceeding naively from the definition:
    [itex] X\left[ f \right] = \frac{df\left(c\left(s\right)\right)}{ds}|_{s=0} [/itex]
    [itex] = \frac{df}{dc} \frac{dc}{ds}|_{s=0} [/itex]
    [itex] = \frac{df}{dc} \left( \begin{array}{ccc}
    0 & -1 & 0 \\
    1 & 0 & 0 \\
    0 & 0 & 0 \end{array} \right) [/itex]

    I have no idea how to proceed from here.

    Thanks for any and all help!

    EDIT: I quickly realized how trivial this is.
     
    Last edited: Nov 19, 2011
  2. jcsd
  3. Nov 19, 2011 #2

    Dick

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    You are already done. Sort of. X[f] would indeed be df(c(s))/ds at s=0. But that would be a real number. Writing something like df/dc is a little misguided. That would be the derivative of a real function with respect to a matrix. It would be tough to clearly define that. I think all they really want is the matrix dc/ds at s=0. And that you already have.
     
  4. Nov 19, 2011 #3

    B L

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    Am I correct in thinking that the corresponding tangent vector (i.e. element of the Lie algebra [itex] so(3) \cong T_eSO(3) [/itex]) would be [itex] -\frac{\partial}{\partial x^{12}} +\frac{\partial}{\partial x^{21}} [/itex]?

    Thanks for the help.
     
  5. Nov 19, 2011 #4

    Dick

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    Yes, you could write it that way. That would tell you how X acts on a function f:SO(3)->R.
     
  6. Nov 19, 2011 #5

    B L

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    I know why I went wrong - what I didn't realize is that by differentiation wrt c, I really meant partial differentiation wrt c^ij (and differentiation of c^ij wrt s) As soon as you do that the problem is trivial.
     
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