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 $\mbox{exp}(A) = \gamma_A(1)$.
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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 :)