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This question asks about the time-dependent case: Lie derivative along time-dependent vector fields

I am totally confused with even the time-independent case:

$\mathcal L_v \omega = \frac{d}{dt} (\textrm{exp } t v)^* \omega|_{t=0}$.

(Here $\omega $ is a $k$-form on the manifold $M$ and $v$ is a vectorfield on $M$.)

For each $t\in \mathbb R$, $\textrm{exp } t v$ is a function from $M$ to $M$ which is the identity at $0$, so we can pullback $\omega$ at each point (to the same point when $t=0$) to get a new $k$-form on $M$. But what does it mean to take the derivative with respect to $t$ here? I mean I could think of $(\textrm{exp } t v)^* \omega$ as a function on $(TM)^k$, but that seems wrong...very confused. Maybe I am not understanding notation. Help!

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Ashley
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  • You can subtract, scale and take limits of differential forms, so the usual limit definition of the derivative makes sense. – Anthony Carapetis Feb 23 '17 at 02:05
  • But usually you don't get the same type of object w a derivative, so is it somehow clear here that you get a k form? – Ashley Feb 23 '17 at 04:16
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    For a function of one variable taking values in a linear space you get the same type of object. Think of the $k$-form at each point as an element of the vector space $T^* M^{\otimes k}$, so that you're just differentiating a curve in a vector space to get its velocity vector. – Anthony Carapetis Feb 23 '17 at 04:19
  • Ohhh I see. I'm fixing a point in M and pulling back this form in a time varying way so only the functions in front of my basis for the k forms at that point are changing. – Ashley Feb 23 '17 at 06:17

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