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let $U\subset \mathbb R^n$ open and $\omega_t\in \Omega^2(U)$ a differential 2-form with a parameter $t\in \mathbb R$.

Let $v_t\in \mathcal X(U)$ a vector field and $\phi_t$ its flow.

I don't understand why we have $$\frac{d}{dt}\phi_t^*\omega_t=\phi_t^*(\mathcal L_{v_t}\omega_t+\frac{d}{dt}\omega_t)$$

I would like to use my definition which is that $\mathcal L_v\omega=\lim_{t\to 0}\frac{\phi_t^*\omega-\omega}{t}$ But I can't make the link unfortunately...

Thank you very much in advance.

Stylel
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  • Mimic the product rule and pass the pullback through the limit. – Matematleta Nov 24 '20 at 19:46
  • @Matematleta Thank you for your answer. However in the link you've provided I do not understand the last equality in Jordan Payette's answer. It seems to me that $\psi_{t_0}^{\ast} \left\lbrack \lim_{t \to t_0} \frac{(\psi_{t_0}^{\ast})^{-1}\psi^{\ast}{t} - Id}{t-t_0} \right\rbrack \cdot \lim{t \to t_0} \omega_t=\psi_{t_0}^{\ast} \left\lbrack \lim_{t \to t_0} \frac{(\psi_{t_0}^{\ast})^{-1}\psi^{\ast}{t} - Id}{t-t_0} \right\rbrack \cdot \omega{t_0}=\mathcal L_{X_{t_0}}\omega_{t_0}$ instead of having $=\psi^{\ast}{t_0} \mathcal{L}{X_{t_0}}\omega_{t_0}$ because of my definition which is – Stylel Nov 25 '20 at 17:10
  • @Matematleta which is $\mathcal L_{X_{t_0}}\omega_{t_0}=\lim_{t\to t_0}\frac{\psi_t^\omega_{t_0}-\psi_ {t_0}^\omega_{t_0}}{t-t_0}$ – Stylel Nov 25 '20 at 17:12

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