Lie derivative of a one form in $S^1$ I'm reading section 4 in the article Cohomologie équivariante et Théorème de Stokes, it says the following:

The circle $S^1$ is parametrized by the angle $\theta$. The action of $S^1$ on $S^1$ is induced by the vector field $\frac{\partial}{\partial \theta}$. Then $J= -\frac{\partial}{\partial \theta}$. A function $f$ such that $\mathcal{L}(J)f=0$ is constant. A 1-form $gd\phi$ is annihilated by $\mathcal{L}(J)$ if $g$ is constant. So we have the set $\lbrace \alpha , \mathcal{L}(J)\alpha=0\rbrace$  is equal to $\mathbb{C}{1}_{S^1} \oplus \mathbb{C} d \phi$.

I have  two questions, the first one is why one the author want to choose a one form on $M$, he choose it to be of the form $gd\phi$, is every one form is of this form?
My second question is why does the equation $\mathcal{L}(J)(gd\phi)=0$ implies that $g$ is constant. My attempt in this question was first to use the fact that the Lie derivative is a derivation, so $\mathcal{L}(J)(gd\phi)=0$ implies $(\mathcal{L}(J)g)d\phi + g(\mathcal{L}(J)d\phi)=0$, and then to use the Cartan formula, I get that the last equation implies $(\mathcal{L}(J)g)d\phi + g(d \iota(J)d\phi)=0$. From this equation I don't see how to deduce that $g$ is constant?
 A: First of all, it is important to notice that $\theta$ and $\varphi$ stand for the same thing here. In her paper, the authors refers to $\theta$ when $\Bbb S^1$ is thought of as the coordinate on the group $\Bbb S^1$, acting on a manifold $M$. She refers to $\varphi$ when it is thought of as the manifold $\Bbb S^1$. For the action by translation of the circle on itself, they coincide and hence, $J = -\frac{\partial}{\partial \theta} = - \frac{\partial}{\partial \varphi}$.

is every one form is of this form?

$\Omega^1(\Bbb S^1)$ is a $\mathcal{C}^{\infty}(\Bbb S^1)$-module of rank $1$. Since $d\varphi$ is nonzero everywhere, $\{d\varphi\}$ is a basis of $\Omega^1(\Bbb S^1)$, that is, any differential form of degree $1$ on the circle is of the form $gd\varphi$.

why does the equation $\mathcal{L}(J)[gd\varphi ]=0$ implies that $g$ is constant

Since $\mathcal{L}(J)$ is a derivation, $\mathcal{L}(J)[g d\varphi] = (\mathcal{L}(J)g)d\varphi + g \mathcal{L}(J) d\varphi$. You can check that $\mathcal{L}(J) d\varphi = 0$ since the action of $\mathbb{S}^1$ onto itself is by translation and hence the coordinate $\varphi$ is invariant under this action. Therefore, $\mathcal{L}(J)[gd\varphi] = 0 \iff \mathcal{L}(J)g = 0$. The problem reduces to solving the equation $\mathcal{L}(J)g = 0$, which is in fact just the equation $dg = 0$, and since $\Bbb S^1$ is connected, its solutions are constant maps.
Comment The author isn't a 'he', but a 'she', since Michèle Vergne is a woman although she's a mathematician ;)
