# A step that is not obvious in Moser's trick for proving Darboux Theorem

This is a step on pg 80 of Jost Riemannian Geometry and Geometric Analysis on Darboux Theorem proof which I do not find obvious. I could go through some calculation to find the relation. However, it seems that everyone finds that equation obvious as none of notes says anything on this equation.

Let $$\omega$$ be a closed 2 form on $$2d$$ smooth manifold. The point is to show $$\omega=dp_i\wedge dq_i$$ holds locally where $$p_i,q_i$$ are coordinates.

Set $$\omega_0=dp_i\wedge dq_i$$ locally around $$0$$ and it is clear $$\omega_0$$ is closed.

Define $$\omega_t=\omega_0+t(\omega-\omega_0)$$. The point is to interpolate between $$\omega_0$$ and $$\omega$$ via a family of diffeomorphism. Since $$d(\omega-\omega_0)=0$$ by individually closed, Poincare gives 1-form $$d\eta=\omega-\omega_0$$. Set $$i_{X_t}(\omega_t)=-\eta$$ by solving $$X_t$$ vector field which can be done by matching coefficient of chosen basis for $$\eta$$. Set $$\phi_t$$ as the flow determined by $$X_t$$.

$$\textbf{Q1:}$$ Why do I want to set $$i_{X_t}(\omega_t)=-\eta$$ here?(It is obviously used to cancel the term like $$t(\omega-\omega_0)$$ part but I do not see why I want to use interior product.) Is this merely coincidence? What is the reason behind this? I could see computation checks out but I could not find this step truly natural or I am missing something in picture.

Then use cartan formula $$L_{X_t}=i_{X_t}d+di_{X_t}$$ to deduce $$L_{X_t}\omega_t=di_{X_t}(\omega_t)=-d\eta=-(\omega-\omega_0)=-\frac{d}{dt}\omega_t$$. Then apply diffeo $$\phi_t^\star$$ on both sides. This will transform the equation into $$\phi_t^\star(L_{X_t}\omega_t+\frac{d}{dt}\omega_t)=\frac{d}{dt}(\phi_t^\star\omega_t)=0$$.

$$\textbf{Q2:}$$ Why is $$\phi_t^\star(L_{X_t}\omega_t+\frac{d}{dt}\omega_t)=\frac{d}{dt}(\phi_t^\star\omega_t)$$ obvious? I could check computation as the following. $$\frac{d}{dt}(\phi_t^\star(\omega_0+t(\omega-\omega_0))=L_{X_t}\omega_0+\frac{d}{dt}(t\phi_t^\star(\omega-\omega_0))$$. Then use product rule of $$\frac{d}{dt}$$ to second term and recombine terms to see it matches.

The answer to your first question is the Cartan formula computation you did right afterwards. You want to end up with a "total derivative" $$\mathscr L_{X_t}\omega_t + \frac d{dt}\omega_t$$. So it's typical math exposition, where you make the magic definition that makes things work, rather than discovering it naturally "along the way."
The usual method to prove "product rule" formulas like Q2 is to introduce an extra variable $$s$$ and then ultimately use the chain rule: Consider $$F(s,t) = \phi_s^*\omega_t,$$ compute $$\dfrac{\partial F}{\partial s}$$ and $$\dfrac{\partial F}{\partial t}$$, and then deduce the derivative of $$F(t,t)$$ by applying the chain rule. You don't need to break down $$\omega_t$$ into its definition.