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How to compute the exponential map

  1. Apr 22, 2012 #1
    I need help calculating the exponential map of a general vector.

    Definition of the exponential map
    For a Lie group [itex]G[/itex] with Lie algebra [itex]\mathfrak{g}[/itex], and a vector [itex]X \in \mathfrak{g} \equiv T_eG[/itex], let [itex]\hat{X}[/itex] be the corresponding left-invariant vector field. Then let [itex]\gamma_X(t)[/itex] be the maximal integral curve of [itex]\hat{X}[/itex] such that [itex]\gamma_X(0)=e[/itex]. Then the exponential map [itex]\mbox{exp}:\mathfrak{g} \to G[/itex] is [itex]\mbox{exp}(A) = \gamma_A(1)[/itex].
    _______________________________________________________________________

    It can be shown that the exponential map when [itex]A[/itex] is a matrix is just the 'exponential taylor series' in matrix form.

    However, how do you actually compute the exponential map for a general vector that isn't a matrix?

    Say, for example, we have the Lie group [itex]\mathbb{R}^2 - \lbrace (0,0)\rbrace[/itex] with binary operation [itex](a,b) *(c,d) = (ac-bd,ad+bc)[/itex], with identity [itex](1,0)[/itex] and basis [itex]\left(\frac{\partial}{\partial x}\bigg|_{(1,0)},\frac{\partial}{\partial y}\bigg|_{(1,0)}\right)[/itex]. What steps are required to compute [itex]\mbox{exp}\left(\frac{\partial}{\partial x}\bigg|_{(1,0)}\right)[/itex] here?
     
  2. jcsd
  3. Apr 22, 2012 #2

    quasar987

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    Well, since exp(X) is the time 1 map of integral curve of the left-invariant vector field on G induce by X, then it might be a good idea to figure out first what is this induced vector field on G. If you know the group action explicitely, then this is easy: [itex]\hat{X}_g=(\theta_g)_*X[/itex], where [itex]\theta_g:M\rightarrow M [/itex] is the map "act by g": [itex]\theta_g(p)=g\cdot p [/itex]. So once you have [itex]\hat{X}[/itex], it remains to solve the first order ODE which defines its integral curve.
     
  4. Apr 22, 2012 #3
    Thanks... so for the example I gave, if [itex]\theta_{(x,y)}(a,b) = (xa-yb,xb+ya)[/itex], then
    [tex](\theta_{(x,y)})_* = \left[\begin{matrix} x & -y \\ y & x \end{matrix}\right][/tex]
    so
    [tex]X_{(x,y)} = \left[\begin{matrix} x & -y \\ y & x \end{matrix}\right]\left[\begin{matrix} 1 \\ 0\end{matrix}\right] = x\frac{\partial}{\partial x}\bigg|_{(1,0)}+y\frac{\partial}{\partial y}\bigg|_{(1,0)}[/tex]

    Then if we let [itex]\gamma = (\gamma_1,\gamma_2)[/itex] be the integral curve we should have:

    [tex]\gamma' (t) = \gamma_1'(t)\frac{\partial}{\partial x}+\gamma_2'(t)\frac{\partial}{\partial y}\ \ \ =\ \ \ \gamma_1(t)\frac{\partial}{\partial x}+\gamma_2(t)\frac{\partial}{\partial y} = X_{\gamma(t)} [/tex]

    Then equate and solve... is that right?
     
  5. Apr 22, 2012 #4

    quasar987

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    Yep! (except in the second displayed equation, (1,0) should be (x,y))
     
  6. Apr 23, 2012 #5
    Cheers quasar :)
     
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