The simplest objects in differential geometry attain some geometric intuition for definition of the Lie derivative:

  1. Let $X$ be the vector field on the manifold $\mathcal{M}$ and $f:\mathcal{M} \rightarrow \mathbb{R}$ is some smooth function. Then Lie derivative can be defined as follows: \begin{equation} \mathcal{L}_Xf|_p := \lim\limits_{t\rightarrow 0} \dfrac{\phi_t^*f(p) - f(p)}{t}. \end{equation} This somehow reflects the definition that scalar function $f$ is pullbacked by group of diffeomorphisms $\phi_t$ generated by X.
  2. The vector field $Y$ along $X$ attains also limit-style definition: \begin{equation} \mathcal{L}_X Y|_p := \lim\limits_{t\rightarrow 0} \dfrac{Y|_p-\phi_t^*Y|_p}{t}, \end{equation} which relfects the fact that vector field on manifold is pushed forward under action of group $\{\phi_t\}$.

For differential forms the definition is analogous to the case (1), for tensors of type $(r,0)$ it is similiar to case (2), for tensors of type $(0,s)$ the definitions are of case (1). This reflects the ideology that push-pulls of tensors can be defined via push-pulls of their arguments.

However, for the tensor of general type $(p,q)$ it is less clear for me how to get this limit-style definition since $p$ - arguments of tensor are pushed forward by $\{\phi_t\}$ and $q$ arguments are pushed backwards by the same group.

Also I want to note that it is easy to write the Lie derivative of mixed-tensor $T$ (of type (1,1) for example) via algebraic axioms: \begin{equation} \mathcal{L}_XT(Y,\omega) = X (T(Y,\omega)) - T(\mathcal{L}_XY, \omega)- T(Y, \mathcal{L}_X\omega), \end{equation} where $X$ is a vector field and $\omega$ is a differential form.

So the question is: Do you know the limit-style definition for mixed type tensor via diffeomorphisms group?

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    $\begingroup$ It would be helpful to notice that for diffeomorphisms, the pullback is just the pushforward of the inverse. So essentially there's no need to distinguish pullbacks and pushforwards; you just need to care where your tensors lie over, and then there's only one way to identify two different fibers by a diffeomorphism. $\endgroup$ – cjackal Feb 9 '17 at 0:51
  • $\begingroup$ I think I understand what you are saying. That push-pulls are not so different (as you said). It is obvious on the example of the group of diffeomorphisms and the only question is then at which point I want to do the comparison -- this makes the fix of diffeomorphisms and fixes whether it will be a push or a pull. Thanks for the comment, the last sentence you said is very good. $\endgroup$ – Fedor Goncharov Feb 9 '17 at 9:03

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