I'm reading some lecture notes which discuss in a part of it some examples of the circle actions on manifolds. I hope that someone help me to understand the following:
Let $M$ be a a manifold provided with an action of the 1-sphere $S^1$ by a group of transformations with a parameter $g(\theta)$. We denote $J$ the vector field on $M$ defined by $$ (Jf)(x)= -\frac{d}{d\theta} {f(g(\theta)x)}_{\theta=0}, \qquad f \in C^\infty(M).$$
- Consider the action of $S^1$ on $\mathbb{R}^2$ by rotations. Why in this case the vector field $J$ is given by : $$J = x_2 \partial_1 - x_1 \partial_2 ? $$
- what is the importance of the vector field $J$, I mean what are the informations it gives us about the circle action ?
Any help would be greatly appreciated!
As Lee Mosher points out, your first bullet is solvable via direct calculation. You know what $g(\theta)$ is (it's the rotation matrix), so you know what the function $g(\theta)x$ is. Now write down the derivative of $f(g(\theta)x)$ using the chain rule, substitute $\theta=0$, and write the end result as $(\text{differential operator})f$.
The second bullet is a more worthy question.
Consider the circle acting on itself and the function $\exp{}\in S^1\to\mathbb{R}$. An action of $S^1$ shifts points in $S^1$…which multiplies the exponential of those points by a constant. So rather than talk about shifting points in $S^1$, we can talk about multiplying exponentials by a constant, and this is (in some sense) the same action. Of course, the same holds true for any function on $S^1$, not just exponentials; just the resulting action is much more complicated in that case.
More generally, your circle action on $M$ also gives a circle action on $M\to\mathbb{R}$ (as well as higher-regularity subspaces): rather than moving your points, you move the observables instead. $J$ is the infinitesimal generator of this dual action. That is, the effect of an angle $\theta$ on those functions is given by $\exp{(-\theta J)}$, where the exponential is in the sense of Lie theory. The intuition is that if $S^1$ were instead a finite cyclic group, generated by $g$, then each action is given by repeated actions of $g$; similarly, here the $S^1$ action is given by repeating a tiny neighborhood of $0\in S^1$.
To be precise: for all $\theta$ and $x$, $$\frac{d}{d\theta}f(g(\theta)x)=-(Jf)(x)$$ Thus for all $\theta$ and $\phi$, $$\frac{d}{d\theta}f(g(\theta+\phi)x)=\frac{d}{d\theta}f(g(\theta)g(\phi)x)=-(Jf)(g(\phi)x)$$ The only solution to this partial differential equation (with $x$ as the "spatial" variable and $\theta$ as "time") is $\exp{(-\theta J)}f$.