I'm reading the first lines of page 5 in the article which called "Cohomologie équivariante et théorème de Stokes", which says

Let $M$ be a smooth manifold on which $S^1$ acts. Let $J$ be the vector field generated by the rotations of a circle.

We denote by $\mathcal{L}(J)$ the lie derivative in the direction of $J$.

If $\xi$ is a vector field which vanishes at a point $p$, then $\mathcal{L}(J)$ induces an invertible transformation on $T_pM$.

What does it mean that $J$ is the vector field generated by rotations of a circle?


1 Answer 1


Given a point $p \in M$, the action of $\mathbb{S}^1$ yields a path $\gamma_p(t) = e^{it}\cdot p $ which is a loop in $M$ with basepoint $p$. It induces a vector field $J(p) = \frac{d}{dt}|_{t=0} \gamma_p(t) = \gamma_p'(0)$.

Edit after reading the article online, it seems you are actually talking about page 6 (not 5) and just a few words after the first sentence "Soit $M$ une variété$\ldots$", the precise definition of $J$ is stated. It is opposite to my answer: the author's choice is $J_x = -\gamma'_x(0)$, which is detailed by its action on smooth functions.


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