# Formula for variation of pullback

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.

• if $$M=N$$ and $$f_0$$ is the identity; or
• if $$f_0$$ is an embedding, we can interpret it as $$\delta \int_M f^\star \alpha = \int_{f_0(M)} \mathcal{L}_{\delta f} \alpha.$$