Identity Involving Lie Derivative and Local Flows

Question

I'm trying to show,

$$ \frac{d}{dt} \varphi_t^* \omega = \varphi_t^* \left( \mathcal{L}_{X_t} \omega \right)$$

but I have another question as well. Every case in which the lie derivative is mentioned, that notation $\mathcal{L}_{X_t}$ has never been given interpretation and so I would like to understand this notation first.

Given $\frac{d}{dt}\bigr|_{t=0} \varphi_t = X_p = X_{\varphi_0(p)}$ and so $\frac{d}{dt} \varphi_t = X_{\varphi_t(p)}$ which we can call $X_t$. And so,

$$\mathcal{L}_{X_t} = \mathcal{L}_{\frac{d}{dt} \varphi_t}$$

Is this correct?

Update: attempt to prove identity

Answer 1

Edit: The following answer assumes $\varphi_t$ is the associated flow of $X,$ but for a general flow it is incorrect.

Edit 2: It seems like OP was asking about this case, in which case my answer would apply.

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If $\omega$ is some tensor field on $M$ and $X$ is a vector field, the Lie derivative is usually defined as, $$ \mathcal{L}_X\omega|_p = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_t^* \omega|_p = \lim_{t \rightarrow 0} \frac1t\left(\varphi_t^*(\omega_{\varphi(p)}) - \omega_p \right), $$ where $\varphi_t$ is the associated flow of $X.$

From this, the identity follows essentially by definition, noting you can commute pullbacks with limits. Indeed assuming the flow exists up to some time $t_0,$ we have,

$$\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=t_0} \varphi_t^*\omega = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_{t_0}^* \varphi_t^*\omega = \left.\varphi_{t_0}^* \frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \varphi_t^*\omega = \varphi_{t_0}^* \mathcal{L}_{X}\omega. $$ I'll leave you to check the details on a pointwise level.

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Now what I wrote isn't the same as what you asked, since I don't have $X_t$ appearing in my expression. I think this is because of how I interpret the pullback, namely I define, $$ \varphi^*\omega|_p = \varphi^*|_{\varphi(p)} \omega_{\varphi(p)} \in E_p,$$ where $E$ is whatever bundle $\omega$ is a section of. If you define the pullback differently however, then you may find you'll need to differentiate with respect to $X_t$ when checking the above pointwise.

In general there's nothing special about $X_t = \varphi_{-t}^*X$ however, all you're really doing is shifting the points over so everything lands where they should.

Answer 2

In order to prove the identity: $$\frac{\mathrm{d}}{\mathrm{d}t}{\varphi_t}^*\omega={\varphi_t}^*(\mathcal{L}_{X_t}\omega),$$

1. Prove that it holds for functions ($0$-forms), it is just the chain rule,

2. Prove that if it holds for $\omega$, then it holds for $\mathrm{d}\omega$, use that the exterior derivative commutes with the pullback and the Lie derivative,

3. Prove that if it holds for $\omega$ and $\omega'$, then it holds for $\omega\wedge\omega'$, use that the pullback of a wedge product is the wedge product of their pullbacks,

4. Conclude using that the algebra of differential forms is locally generated as an algebra by functions and differentials of functions.

Full computations can be found in Appendix B of An introduction to Contact Topology by H. Geiges.