Action of $S^1$ on itself

Question

I'm reading some lecture notes about equivariant cohomology. It says that:

Consider the action of the circle on itself by rotations. The circle $S^1$ is parametrized by the angle $\phi$. The action of $S^1$ on $S^1$ is generated by the vector fields $\frac{\partial}{\partial \phi}$.

Please someone explain to me why is this action generated by that vector field.

Ps: the lecture notes I am reading are written in french, so please excuse me if my translation to the quote is bad !

Answer 1

You can see this action as follows: $G=SO(2)=\Big\{R(\phi)=\left(\begin{smallmatrix}\phantom{-}\cos\phi&\sin\phi\\-\sin\phi&\cos\phi\end{smallmatrix}\right),\ \phi\in[0,2\pi]\Big\}\cong S^1$, $M=S^1=\{(x,y)\in\mathbb{R}^2,\ x^2+y^2=1\}$ and the action is by rotation: $$\phi\cdot\left(\begin{smallmatrix}x\\ y\end{smallmatrix}\right):=R(\phi)\left(\begin{smallmatrix}x\\ y\end{smallmatrix}\right).$$ Equivalently, the vector field $\partial_{\phi}$ is $X=-y\partial_x+x\partial_y$ (in cartesian coordinates). The flow of $X$ acts on $S^1$ and it's of the form $$\exp tX=R(t)\ \text{rotation}.$$

Answer 2

The vector field $\frac\partial{\partial\phi}$ is rotation invariant, hence the action generated by it must be rotation invariant. The only ortation invariant actions on $S^1$ are - rotations.