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Let $Q$ be a smooth manifold, $\alpha\in\Omega^1(T^*Q)$ the tautological $1$-form and $F:T^*Q\to T^*Q$ a symplectomorphism such that $F^*\alpha=\alpha$ ($F$ could be the cotangent lift of some diffeomorphism $f:Q\to Q$, for example). Let $X\in\mathfrak{X}(T^*Q)$ be the Euler vector field, which is defined as $X:=\sum_{i=1}^n\xi_i\frac{\partial}{\partial \xi_i}$ in local coordinates $(x_1,...,x_n,\xi_1,...,\xi_n)$. Prove that $dF_m(X_m)=X_{F(m)}$.

I have an initial idea: $\alpha$ is given locally by $\alpha=\sum_{i=1}^n\xi_i dx_i$. If $F$ takes coordinates $(x_1,...,x_n,\xi_1,...,\xi_n)$ into coordinates $(x'_1,...,x'_n,\xi'_1,...,\xi'_n)$, the condition $F^*\alpha=\alpha$ basically means $\sum_{i=1}^n\xi_i dx_i=\sum_{i=1}^n\xi'_i dx'_i$. Writing $dx'_i$ in terms of $dx_i$ and $d\xi_i$, I've concluded that: $$\sum_{i=1}^n\xi'_i\frac{\partial x'_i}{\partial x_j}=\xi_j$$ $$\sum_{i=1}^n\xi'_i\frac{\partial x'_i}{\partial \xi_j}=0$$

for every $j=1,...,n$. This seemed like a good start, but I'm not being able to conclude $dF\left(\sum_{i=1}^n\xi_i\frac{\partial}{\partial \xi_i}\right)=\sum_{i=1}^n\xi'_i\frac{\partial}{\partial \xi'_i}$. To arrive at that, I would need: $$\sum_{i=1}^n\xi_i\frac{\partial x'_i}{\partial \xi_j}=0$$ $$\sum_{i=1}^n\xi_i\frac{\partial \xi'_i}{\partial \xi_j}=\xi_j$$ but I don't know how to get there. Any hints?

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The canonical symplectic form $\omega$ identifies $\mathfrak{X}(T^*Q)$ and $\Omega^1 (T^*Q) $ by taking a vector field $Z$ to the $1$-form $Z_\flat =\omega(Z,\cdot)$. The idea is that the Euler field corresponds to the tautological form $\alpha$, perhaps modulo sign.

By the expression $\omega=\sum_{i=1}^n {\rm d}x^i\wedge {\rm d}\xi_i$, we get that $$\omega(X,\cdot) =\sum_{i=1}^n \begin{vmatrix} {\rm d}x^i(X) & {\rm d}x^i \\ {\rm d}\xi_i(X) & {\rm d}\xi_i\end{vmatrix}=\sum_{i=1}^n \begin{vmatrix} 0 & {\rm d}x^i \\ \xi_i& {\rm d}\xi_i\end{vmatrix} = -\sum_{i=1}^n \xi_i\,{\rm d}x^i =-\alpha.$$So $F^*\alpha = \alpha$ reads ${\rm d}F(X) = X\circ F$.


Further clarification: let $m\in T^*Q$ and $Y\in \mathfrak{X}(T^*Q)$. Then $F^*(-\alpha)_m (Y_m)=-\alpha_m (Y_m) $ becomes $$\begin{align} \omega_{F(m)}(X_{F(m)},{\rm d}F_m(Y_m)) &= -\alpha_{F(m)}({\rm d}F_m (Y_m)) \\ &= -\alpha_{m}(Y_m)\\ &=\omega_m (X_m,Y_m) \\ &= \omega_{F(m)}({\rm d}F_m (X_m), {\rm d}F_m(Y_m)).\end{align}$$Non-degeneracy of $\omega$ gives $X_{F(m)} = {\rm d}F_m (X_m)$.

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  • $\begingroup$ I was aware that $i_X\omega=-\alpha$, but I still don't see how $F^*\alpha=\alpha$ "reads" $dF (X)=X\circ F$ $\endgroup$
    – rmdmc89
    Jul 11 '18 at 12:30
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    $\begingroup$ I added the rest of the computation. See if it helps $\endgroup$
    – Ivo Terek
    Jul 11 '18 at 14:15
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    $\begingroup$ That's excellent, thank you $\endgroup$
    – rmdmc89
    Jul 11 '18 at 16:07
  • $\begingroup$ just another question. This exercise was written in the following way: "defining $X$ as the only field such that $i_X\omega=-\alpha$, check that $X=\sum_i\xi_i\frac{\partial}{\partial \xi_i}$ and prove $dF(X)=X_F$". So I was a little embarrassed that my attemp with coordinates didn't take me anywhere, while your coordinate-free approach worked out so elegantly. Either the hint in the exercise was misleading, or I missed something very simple. What do you say? $\endgroup$
    – rmdmc89
    Jul 11 '18 at 20:38
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    $\begingroup$ It is important to know how to describe things locally, so I'd interprete it as having two parts: (a) prove that $dF(X) = X\circ F$; (b) check that $X = \xi_i \partial/\partial \xi_i$. One might be tempted to think that (b) always follows from (a), but that's not always the case. That said, I agree that the exercise could have been stated in a better way... $\endgroup$
    – Ivo Terek
    Jul 11 '18 at 20:55

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