Rovelli Quantum Gravity: Clarification on Symplectic Forms & Hamiltonian

[albedo]

Please refer to p. 99 and 100 of Rovelli’s Quantum Gravity book (here).

I wonder what is the signification of the “naturalness” of the definition of $\theta_0=p_idq^i$? If I take $\theta_0'=q^idp_i$ inverting the roles of the canonical variables and have the symplectic 2-forms of the Darboux’s theorem changing sign, $\omega_0=d\theta_0=-d\theta_0'=-\omega_0'$, I would get $(d\theta'_0)(X_0)=dH_0$ the opposite of eq. 3.7. Or maybe I am not allowed to invert the roles of the variables because I need to work in the cotangent space and the only "natural" 1-form $\theta_0$ must be $p_idq^i$ where first you have the covector $p_i$? What is the signification of this inversion of sign of the $dH_0$? Could this be related to the assumption that the energy of a system is in reality the difference with an energy taken as zero? I would like to understand better the symplectic formulation of mechanics as it looks crucial when you go from the classical to the general-relativistic Hamiltonian formulation.

Thank you in advance!

 

[fisicist]

Sorry, but somehow the page that you've linked can't be viewed to me.
But perhaps this is what you're looking for: There is a natural pairing between tangent and cotangent vectors. But in Hamiltonian mechanics, you're working on the cotangent bundle. Every tangent vector to the configuration space ($(q,v) = (q, v^i \partial_{i}) \in T_q (M)$) can be naturally treated as a tangent vector to the phase space $(q, p; v^i \partial_{q^i}) \in T(T^* (M))$. Conversely, every tangent vector to the phase space $z = (q, p; v^{q^i} \partial_{q^i} + v^{p_i} \partial_{p_i}$ has a component that can be treated as a tangent vector to the configuration space. If $\Pi: T^*(M) \to M$ be the canonical projection, then $v = T(\Pi )\cdot z$ (T(Pi) is the derivative, other common notations are Pi' or dPi). So you can assign to every point of the phase space a cotangent vector that, paired with a tangent vector to the phase space, gives the pairing of the momentum component of that point with the tangent vector, treated as a tangent vector to the configuration space. This is exactly the canonical 1-form. $\Theta_{(q,p)} (z) := (p \vert T(\Pi)\cdot z)$.
Of course, $q^i\,dp_i = d(q^i\,p_i) - p_i\,dq^i$.
Hope this helps.

 

[albedo]

If the google books links does not work, you can also see at p. 70 and the paragraph just before the eq. 3.7 of the DRAFT version of the book:

http://www.google.ch/url?sa=t&rct=j...=IAzDVQxjqUFVfADJnO6Sag&bvm=bv.92885102,d.bGQ

Thank you for your reply!