the exact form in a manifold

Question

Let $M$ be a compact manifold, $X$ is a vector field on $M$, $\alpha$ is a closed 2-form on $M$, $\phi: M\to M$ is a diffeomorphism such that $\phi^*\alpha=\alpha$, then I want to konw whether $$ i_{(\phi^*X-X)}~~\alpha $$ is an exact form. Thanks in advance.

Answer 1

No. If $M=S^1\times S^1=(\mathbb R/\mathbb Z)^2$, $x,y$ are the coordinates on $\mathbb R^2$, $\alpha=dx\wedge dy$, $X=\partial_x$, $\phi(x,y)=(y,-x)$ ($\alpha$, $X$ and $\phi$ are well defined on $M$) then $\phi_*X=\partial_y$ and so $i_{(\phi_*X-X)}\,\alpha=dy-dx$ which is not exact on the torus $M$.

BTW in general you'll get (from $d\alpha=0$) $d i_{(\phi_*X-X)}\,\alpha=(\phi^*-id)L_X\alpha$. There is no reason why it should be $0$ (so not even closed); in the example above I have $L_X\alpha=0$ (which you perhaps wanted to suppose but forgot to add).