On page 15 of his introduction to the WZW model Gawedzki states the following:
… a special case of the general, very useful, geometric identity: $$ \delta \int f^* \alpha = \int \mathcal{L}_{\delta f} \alpha$$
where $\mathcal{L}_X$ is the Lie derivative.
I don't understand what is meant by $\mathcal{L}_{\delta f}$ and what the domain of integration is supposed to be here.
Here is what I've tried so far:
Let $M$ and $N$ be smooth manifolds, $m := \operatorname{dim}(M) \leq n := \operatorname{dim}(N)$, $f:M \longrightarrow N$ smooth and $\alpha \in \Omega^m(N)$ a smooth $m$-form on $N$. Assume further that $M$ is compact.
If we consider the 1-parameter family of funtions $f_t:M \longrightarrow N$ given by $f_t= \Phi_t \circ f$, where $\Phi$ is a flow on $N$ whose generator is $Y \in \mathfrak{X}(N)$, then
$$\left.\frac{\partial}{\partial t} \right|_{t=0} f_t^*\alpha=
\left.\frac{\partial}{\partial t} \right|_{t=0}f^*\Phi^*_t\alpha =
f^*\left.\frac{\partial}{\partial t} \right|_{t=0}\Phi^*_t\alpha
= f^*\mathcal{L}_Y\alpha$$
implies
$$\left.\frac{\partial}{\partial t} \right|_{t=0} \int_M f^*_t\alpha =
\int_M \left.\frac{\partial}{\partial t} \right|_{t=0} f^*_t\alpha =
\int_M f^*\mathcal{L}_Y\alpha.$$
So if I interpret $\delta f$ as a vector field on $N$ it seems to me that this identity should read something like:
$$ \delta \int f^* \alpha = \int f^* \mathcal{L}_{\delta f} \alpha$$
What am I missing? Any help would be greatly appreciated.
Best Answer
I'd say your formula is the more precise one, and that the author was just trying to be brief. There are two special cases where the given formula can be interpreted as literally true: