Let $\phi$ be a homogeneous symplectomorphism of tangent bundle $\dot{T}^*M=T^*M-0_M$ and let $\alpha_M$ be the canonical Liouville 1-form of $\dot{T}^*M$. Then is it true that $\phi^*\alpha_M=\alpha_M$?
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3$\begingroup$ If you mean that $\alpha_M$ is equivariant with respect to the $\mathbb R^\times$-action of rescaling the fibers, this follows from $\alpha_M = \iota_{E}\omega$, where $E$ is the generator of this action and therefore preserved by any equivariant map. $\endgroup$– Bertram ArnoldCommented Sep 30, 2019 at 9:00
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$\begingroup$ @Bertram Arnold: You mean that $\phi$ is equivariant with..., not $\alpha_M$ (otherwise you are right of course). $\endgroup$– abxCommented Sep 30, 2019 at 19:00
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$\begingroup$ @Bertram Arnold: I mean that $\phi$ is $\mathbb{R}^{\times}$-equivariant with respect to the action and I don't know whether $\alpha_M$ is $\mathbb{R}^{\times}$-equivariant. $\endgroup$– SoYuCommented Oct 1, 2019 at 1:46
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$\begingroup$ you mean 'cotangent' instead of 'tangent' $\endgroup$– sanetteCommented Oct 1, 2019 at 20:15
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$\begingroup$ @sanette: Oh, thanks. I corrected. $\endgroup$– SoYuCommented Oct 4, 2019 at 1:06
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1 Answer
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Yes this is true. You can prove it in Darboux coordinates $(x,\xi)$. Let $\phi=(\phi_1,\phi_2)$ be the symplectomorphism. Since it is homogeneous, $\phi(x,t\xi) = (\phi_1(x,\xi), t\phi_2(x,\xi))$ and differentiating wrt $t$ you get
$$ \partial_\xi\phi_1 \cdot \xi = 0$$ and $$ \partial_\xi\phi_2 \cdot \xi = \phi_2$$ (these are Euler identities).
On the other hand, $\phi$ is a symplectomorphism, so $d\phi_2 \wedge d\phi_1 = d\xi \wedge d x$. Applying these 2-forms to a pair of vectors $(U,V)$ of the form $$U=(0,\xi) \text{ and } V=(v,0)$$ you get $$d\xi \wedge d x (U,V) = \xi\cdot v $$ while $$d\phi_2 \wedge d\phi_1 (U,V) = (\partial_\xi \phi_2 \cdot \xi)(\partial_x \phi_1 \cdot v) - (\partial_\xi \phi_1 \cdot \xi)(\partial_x\phi_2\cdot v).$$ The equality between these 2-forms, combined with the Euler identities, gives $$\phi_2 \partial_x\phi_1\cdot v = \xi\cdot v,\quad \forall v,$$ which is the same as $$\phi^* (\xi d x) = \xi d x.$$ So the Liouville 1-form $\xi d x$ is preserved.
Alternative, coordinate free proof: you want to use Bertram Arnold's claim, which can be proved as follows: (since you were asking whether $\alpha$ was homogeneous).
We denote the dilation action on the cotangent fibers by $h_t(\beta) =e^t \beta$, where $\beta$ is a 1-form on $M$. Differentiating with respect to $t$, you see that $h_t$ is the flow of the Euler vector field, which is tangent to the vertical fibers: $E(\beta) = (0,\beta)$. Hence $\iota_E \alpha = 0$.
Now there is a nice characterization of the Liouville 1-form by the tautological formula $\beta^* \alpha = \beta$, for any 1-form $\beta$ on $M$ viewed as a map $M\to T^*M$. Hence $\beta^*(h_t^*\alpha) = (h_t\circ\beta)^*\alpha = h_t\circ\beta = e^t\beta$, which implies that $h_t^*\alpha = e^t \alpha$.
You differentiate this last identity at $t=0$ to obtain $$\mathcal{L}_E\alpha = \alpha.$$ The Cartan formula together with (1.) yields $\iota_E d\alpha = \alpha$.
Let $\phi$ be a homogeneous symplectomorphism: $\phi\circ h_t = h_t\circ \phi$, and $\phi^*(d\alpha) = d\alpha$. From the first equality you obtain that the operators $\iota_E$ and $\phi^*$ commute. Hence
$$ \phi^*\alpha = \phi^*(\iota_E d\alpha) = \iota_E \phi^* (d\alpha) = \iota_E d\alpha = \alpha. $$
