How to compute the exponential map

[Identity]

I need help calculating the exponential map of a general vector.

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

It can be shown that the exponential map when A 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 \mathbb{R}^2 - \lbrace (0,0)\rbrace with binary operation (a,b) *(c,d) = (ac-bd,ad+bc), with identity (1,0) and basis \left(\frac{\partial}{\partial x}\bigg|_{(1,0)},\frac{\partial}{\partial y}\bigg|_{(1,0)}\right). What steps are required to compute \mbox{exp}\left(\frac{\partial}{\partial x}\bigg|_{(1,0)}\right) here?

 

[quasar987]

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: \hat{X}_g=(\theta_g)_*X, where \theta_g:M\rightarrow M is the map "act by g": \theta_g(p)=g\cdot p. So once you have \hat{X}, it remains to solve the first order ODE which defines its integral curve.

 

[Identity]

Thanks... so for the example I gave, if \theta_{(x,y)}(a,b) = (xa-yb,xb+ya), then
(\theta_{(x,y)})_* = \left[\begin{matrix} x & -y \\ y & x \end{matrix}\right]
so
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)}

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

\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)}

Then equate and solve... is that right?

 

[quasar987]

Yep! (except in the second displayed equation, (1,0) should be (x,y))

 

[Identity]

Cheers quasar :)