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Inverse in lie group, tangent space

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to prove, that when G is a Lie group, [itex]i:G\to G[/itex] is the inverse mapping [itex]i(g)=g^{-1}[/itex], then

    [tex]
    i_{*e} v = -v\quad\quad\forall \; v\in T_e G
    [/tex]

    where [itex]i_{*e}:T_e G \to T_e G[/itex] is the tangent mapping.

    2. Relevant equations

    I'm not sure how standard the tangent mapping notation or terminology is, so here's how it is defined on our course. If [itex]\phi:M\to N[/itex] is a smooth function between two manifolds, then for each [itex]p\in M[/itex]

    [tex]
    \phi_{*p}:T_p M\to T_{\phi(p)} N,\quad\quad (\phi_{*p} v) f = v(f\circ \phi)\quad
    \forall \; v\in T_p M \quad f:N\to\mathbb{R}
    [/tex]

    Quick googling revealed that this is often given other notations similar to derivative notations.


    3. The attempt at a solution

    We can choose some chart [itex]x:U\to\mathbb{R}^n[/itex], where [itex]U\subset G[/itex] is some environment of the [itex]e\in G[/itex], so that x(e)=0. After simply writing down some definitions, I was able to show that the result follows if we know

    [tex]
    \partial_j (x_k\circ i\circ x^{-1})(0) = -\delta_{jk}
    [/tex]

    Here [itex]x^{-1}:(x(U)\subset \mathbb{R}^n)\to U[/itex] is the inverse of the x, [itex]x_k:U\to\mathbb{R}[/itex] is a component mapping of the x, and [itex]\partial_j[/itex] is a normal partial derivative in an Euclidean space.

    This is where I ran out of ideas. I have no clue how the group structure gets related to what happens in the chart images. I know that [itex]g\mapsto g^{-1}[/itex] is supposed to be smooth, but this merely justifies the existence of the partial derivative in the above equation.

    (btw. I will not return any solution anywhere. I'm preparing to an exam by doing old exercises)
     
    Last edited: Dec 5, 2007
  2. jcsd
  3. Dec 5, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can think of tangent vectors as represented by differentiable curves through e, the group identity. So pick g(s) to be a representative of v, i.e. g(0)=e, g'(0)=v. Now g(s)*i(g(s))=e. Differentiate both sides, put s=0 and get t+i_*(t)=0.
     
    Last edited: Dec 5, 2007
  4. Dec 5, 2007 #3
    Your hint certainly pushed in the right direction, but I still have some problems with this. One way to prove

    [tex]
    i_{*e} v = -v
    [/tex]

    would be to show

    [tex]
    vf = -v(f\circ i)
    [/tex]

    for arbitrary smooth [itex]f:U\to\mathbb{R}[/itex] on some environment [itex]U\subset G[/itex] of [itex]e\in G[/itex].

    If [itex]g:I\to G[/itex] has been chosen so that [itex]0\in I[/itex], [itex]g(0)=e[/itex], and [itex]\dot{g}_0 = v[/itex], then the left side is

    [tex]
    \dot{g}_0 f = D_t (f\circ g)(t)\Big|_{t=0}
    [/tex]

    and the right side is

    [tex]
    -\dot{g}_0 (f\circ i) = -D_t(f\circ i\circ g)(t)\Big|_{t=0}.
    [/tex]

    According to you hint, it looks like I should start computing something like this

    [tex]
    0 = D_t f\big(g(t)\cdot (i\circ g)(t)\big)\Big|_{t=0} = \cdots
    [/tex]

    But I don't know how to do anything to that. How do you get the derivation inside the f?

    I would know how to compute

    [tex]
    D_t\Big((f\circ g)(t)\; (f\circ i\circ g)(t)\Big)\Big|_{t=0} =\cdots = f(e)\big( vf + (i_{*e} v) f\big)
    [/tex]

    for example, but for this there does not seem to be any obvious reason why this would be zero.
     
    Last edited: Dec 5, 2007
  5. Dec 5, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you think of the Lie group as a matrix group, then the argument I gave is not just a hint, it IS the proof. I'm having more than a little trouble wrapping my head around your more abstract formalism.
     
  6. Dec 6, 2007 #5
    I can see this.

    This is the rigor general formalism. In general Lie group is just a differentiable manifold with a group structure. Tangent vectors are linear mappings that map smooth functions into real numbers.

    In fact the problem comes down to this:

    If [itex]\gamma^1[/itex] and [itex]\gamma^2[/itex] are some smooth paths [itex]I\to G[/itex] on some Lie group, so that [itex]\gamma^1(0)=\gamma^2(0)=e[/itex], then we get a new smooth path

    [tex]
    t\mapsto \gamma^1(t)\cdot \gamma^2(t)
    [/tex]

    by multiplying the images.

    Is the equation

    [tex]
    \dot{(\gamma^1\cdot\gamma^2)}_0 = \dot{\gamma}^1_0 + \dot{\gamma}^2_0
    [/tex]

    true? (okey, that dot looks horrible, when its trying to be above both of the gammas...)

    It seems to be, but I don't know the proof. Once this is clear, the previous problem becomes solved as you showed.
     
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