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As @peek-a-boo wrote in one of his answer, "the word "momentum" gets thrown around more often than candy during Halloween".

I found two definitions of momentum generalized coordinates I want to reconcile one way or another. We go with the usual adapted coordinate charts on a manifold $M$: the ones on $TM$ are noted $(q, \dot{q})$ and the ones on $T^*M$ are noted $(q, p)$.

The first definition is given in the Wikipedia article about Canonical coordinates as a function on the set of vector fields on $M$ to the set of functions on $T^*M$ by $$X\to \mu_X(p)=p(X(\pi(p))$$ where $p\in T^*M$ and $\pi: T^*M\to M$ the projection.

The second definition is a function on the set of functions on $TM$ to the set of one-forms on $TM$ given by

$$f\to\Theta_f=(\text{F}f)^*\alpha$$ where $\alpha$ is the tautological one form defined on $T^*M$, and $\text{F}f$ the Legendre transform of $f$.

We get $\mu_{\frac{\partial}{\partial q}}=p$ and I want to find a way to get $p$ using the second moment map $\Theta_{\dot{q}}$ if possible (here I have edited my question following the correction of @peek-a-boo in his comment). How can I do it?

brunoh
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  • @peek-a-boo I hope you do not mind being quoted … – brunoh Jan 15 '23 at 09:44
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    Calling both $\mu$ is obviously a recipe for confusion. Anyway, the first is just a dressed up version of the simple double-duality from linear algebra, and as a consequence it says that there is a (bijective) correspondence between (local) vector fields on the base and (local) fiberwise linear maps on the cotangent bundle. If in particular you choose the the coordinate-vector fields on the base, then you get the adapted coordinates for the cotangent bundle. Nothing more, nothing less. – peek-a-boo Jan 15 '23 at 13:12
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    The second (let us call it $\Theta_f=(Ff)^*\theta$ instead to avoid confusion) tells you how to go from functions on $TM$ to 1-forms on $TM$ (which are not just the exterior-derivative of that function). Also, $\Theta_{\dot{q}^i}=dq^i$, so if you evaluate on the vector field $\frac{\partial}{\partial q^j}$, then you get $\delta^j_i$. I don’t see where you’re getting $p$ from. – peek-a-boo Jan 15 '23 at 13:16
  • @peek-a-boo your first comment is clear and illuminating. – brunoh Jan 15 '23 at 17:02
  • @peek-a-boo your second comment is absolutely right. Actually what I wanted to do was to recover $p$ from the function $\dot{q}$ on the tangent bundle using the second definition of the moment. What I wrote had no meaning, and I will edit my question accordingly. – brunoh Jan 15 '23 at 17:13
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    Anyway, we have the formula $p_i\circ Ff=\frac{\partial f}{\partial \dot{q}^i}$. But unless your Legendre transform is a diffeomorphism, you can't invert this to explicitly get $p_i$. So, as you can see, a lot of things naturally want to live on the cotangent bundle. – peek-a-boo Jan 15 '23 at 17:22
  • @peek-a-boo if you want to put the spirit of your comments in an answer I will accept it. I think my own question was not very clear in my mind. I was trying for the different definitions of the moment maps to go back and forth between canonical coordinates of the tangent and cotangent bundles using them. For instance, is it right to say $p_i=\theta(\frac{\partial}{\partial q_i})$, where $\frac{\partial}{\partial q_i}$ is actually considered as an element of tangent bundle to the cotangent bundle using the zero section? – brunoh Jan 15 '23 at 18:33
  • @peek-a-boo is there a more elegant way of writing the formula in my last comment to see precisely how the tautological one form solder the $\dot{q}$ to the corresponding $p$? – brunoh Jan 15 '23 at 18:38
  • Let us continue this discussion in chat. – peek-a-boo Jan 15 '23 at 19:31

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Actually, after chatting with @peek-a-boo, I discovered that what was bothering me was the Wikipedia glose in the article about the Tautological one-form where it is said that

The tautological one-form assigns a numerical value to the momentum $p$ for each velocity $\dot {q}$ and more: it does so such that they point "in the same direction", and linearly, such that the magnitudes grow in proportion. It is called "tautological" precisely because, "of course", velocity and momenta are necessarily proportional to one-another. It is a kind of solder form, because it "glues" or "solders" each velocity to a corresponding momentum. The choice of gluing is unique; each momentum vector corresponds to only one velocity vector, by definition. The tautological one-form can be thought of as a device to convert from Lagrangian mechanics to Hamiltonian mechanics.

I was trying to make sense of this paragraph, and finding a formula linking the two. The only formula I found was $$p=\alpha(\frac{\partial}{\partial q})$$ where where $\alpha$ is the tautological one form, and $\frac{\partial}{\partial q}$ is considered as an element of $TT^*M$ using the push forward by the zero section $s$ of the projection $\pi: T^*M\to M$, that is to say we abusively write $\frac{\partial}{\partial q}$ for $s_*(\frac{\partial}{\partial q})$, but it does not link $\dot{q}$ to $p$.

@peek-a-boo convinced me that using only the tautological one form to connect the two should not be possible without using an extra structure like a Legendre transform or a riemannian metric directly connecting the two tangent / cotangent bundles because it would amount to find a natural way to connect the two bundles in general.

Therefore since the evaluation map is the only natural way to connect a vector space and its dual, what Wikipedia meant was the fact that the Tautological one-form was precisely doing the same at the level of the bundle, not less, no more.

brunoh
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