Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The formula seems slightly confusing. $\phi^*(\omega)$ is a form on $S^1\times X$, but $\phi_t^*(\omega)$ is a form on $X$ (depending on your choice of $t$). There must be an abuse of notation somewhere, but I'm not quite sure where it is. –  Aaron Jan 27 '12 at 19:58
    
Let $x_1,...,x_n$ be local coordinates for $X$. Any $k$-form $\alpha$ on $S^1 \times X$ can be written as $\alpha= \alpha' + dt \wedge \alpha''$ where $\alpha'$ is a $k$ form not involving $dt$, i.e. it can be written as a sum $\alpha '^I dx_i_1 \wedge ... \wedge dx_i_k$ and $\alpha''$ is a $k-1$ form not involving $dt$, i.e. it can be written as a sum $\alpha''^J dx_j_1 \wedge ... \wedge dx_j_{k-1}$. The formula says that $\alpha'=\phi_t^*(\omega)$ and $\alpha''=\iota_T(\phi_t^*(\omega)$. –  Manuel Jan 27 '12 at 20:54
    
So given a form $\omega$ on $X$, you are identifying it with $\pi_X^*(\omega)$ where $\pi_X:S^1\times X \to X$ is the natural projection map. However, there is still the notational problem that $\phi_t$ depends on a fixed value of $t$, and $dt$ is something else entirely (unrelated to the other $t$'s in the equation). With imprecise notation, you might be able to "prove" the identity in coordinates, but you won't understand it in any real sense unless you straighten out exactly what is going on. The left hand side has no dependence on $t$, while the right side seems to. Why? –  Aaron Jan 27 '12 at 21:14
    
Well, following my notation above, $\alpha'$ and $\alpha''$ do depend on $t$ in the sense that they can be seen as functions of $t$ with values on $\Omega^*(X)$ since we can write $\alpha^'= \sum \alpha'^{i_1,...,i_k}(t,x_1,...,x_n)dx_{i_1} \wedge... \wedge dx_{i_k}$ where $\alpha'^{i_1,...,i_k}$ are real valued functions on $S^1 \times X$, so fixing the $t$ in $\alpha'^{i_1,...,i_k}(t,x_1,...,x_n)$ gives a function on $X$. Similarly, $\phi_t^*(\omega)$ is a function of $t$ with values on $\Omega^*(X)$. Does this make sense? –  Manuel Jan 28 '12 at 2:10
    
In other words, for each $t\in S^1$, by restricting each $\alpha'$ and $\alpha''$ to $t \times X$ we get forms in $X$. –  Manuel Jan 28 '12 at 13:51

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.