I am reading about differential forms on manifolds with group actions and there is an 'obvious' formula which I don't quite understand.
Let $X$ be a manifold with a smooth circle action, that is a smooth one parameter group of diffeomorphisms $\phi_t: X \to X$ with period 1. Let $T$ be the infinitesimal generator of this circle action, i.e. the vector field tangent to the $S^1$ orbits. Let $\iota_T: \Omega^*(X)\to \Omega^{*-1}(X)$ denote interior multiplication with the vector field $T$.
Let $\phi: S^1 \times X \to X$ be the map $\phi(t,x)=\phi_t(x)$.
How can I prove the following formula for any differential form $\omega \in \Omega^*(X)$:
$\phi^*(\omega)=\phi_t^*(\omega) +dt \wedge \iota_T(\phi_t^*(\omega))$ ?
Thanks.
One way of showing this is to see what $\phi^* \omega$ is when applied to vectors in $T(S^1 \times X)$. Note that $T(S^1 \times X) = TS^1 \oplus TX$. A first step then is to understand what $\phi_* \partial_t$ and $\phi_* v$ are where $v \in T_x X$. We have $\phi_* \partial_t = T$ and $\phi_* v = {\phi_t}_* v$.
The only case we have to look at is to see what happens when we apply $\phi^* \omega$ to a collection of vectors that starts with $\partial_t$, say $\partial_t, v_1, v_2, \dots$. Then, \[ \phi^* \omega( \partial_t, v_1, \dots) = \omega( T, (\phi_t)_* v_1, \dots) \] Now, using the final observation that $T = (\phi_t)_* T$, we get the identity you wanted.
An exercise for you, so you make sure you understand what's going on, is to chase through at which point each of these vector fields is evaluated.