= Planar rotation as a one-parameter subgroup
{title2=$R_\theta=\exp(\theta J)$}
Differentiate the planar <rotation matrix> at angle zero:
$$
J=\left.\frac{dR_\theta}{d\theta}\right|_{\theta=0}=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad J^2=-I.
$$
The even and odd powers in the <matrix exponential> therefore give
$$
\exp(\theta J)=\sum_{n=0}^\infty\frac{\theta^nJ^n}{n!}=\cos\theta\,I+\sin\theta\,J=R_\theta.
$$
This exact formula gives the whole <one-parameter subgroup>. Taking $X=J$ means unit <angular velocity>: at $t=\pi/2$, $\exp(tJ)=J$ sends $(1,0)^T$ to $(0,1)^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi/4$. Changing the generator changes the speed, while preserving the circular path.
\Image[/lie-rotation-exponential.png]
{title=The generator J gives a tangent velocity, while its exponential gives finite rotations}
For a moving <vector> $v(t)=R_t v(0)$, the <ordinary differential equation> $v'(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant <matrix> specifies every angle of this endless family of <rotations>.
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