The canonical formula for the tautological 1-form

Question

In classical momentum-phase space the tautological 1-form $ \theta \in \Omega^1 (T^*M) $ is (by definition)

$ \theta_p = p^{-1} \circ \pi_{*p} $

With $ T^*M \ni p : T_pM \rightarrow \mathbb{R} $ and $ \pi_p : T^*M \rightarrow M $

In clasical mechanics for $ n $-dimensional space with canonical Darboux coordinates $ (q^i, p_i) $we have the formula

$ \theta = \sum_{i=1}^n p_idq^i $

How is this formula achieved via the definition of the tautological 1-form?

Answer 1

First, recall the following basic fact about expanding relative to the dual basis of a given basis.

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Preliminaries.

Your definition of the tautological 1-form shouldn’t have the inverse (you can’t invert the linear map $p$ in all but the most trivial cases), and it’s also confusing because you’re overusing the letter $p$ (an element of $T^*M$ and also for the induced coordinates… also, $\pi_p:T^*M\to M$ does not make sense).

The tautological 1-form is a $1$-form on $T^*M$, meaning a mapping $\theta:T^*M\to T^*(T^*M)$. To define this, fix a covector $\lambda\in T^*M$, and let $a=\pi(\lambda)$ be its base point (here $\pi:T^*M\to M$ is the canonical projection of the cotangent bundle). Then,

• $\pi_{*,\lambda}\equiv (T\pi)_{\lambda}: T_{\lambda}(T^*M)\to T_{\pi(\lambda)}M=T_aM$ is a linear map.
• $\lambda\in T_a^*M$, meaning $\lambda:T_aM\to\Bbb{R}$ is a linear map
• hence the two can be composed to give \begin{align} \theta_{\lambda}:=\lambda\circ (T\pi)_{\lambda},\tag{!} \end{align} which is a linear map $T_{\lambda}(T^*M)\to\Bbb{R}$.

Now, let $(U, x=(x^1,\dots, x^n))$ be a coordinate chart on $M$, and consider the induced coordinate chart on $T^*U$ given by $(T^*U, (q,p)= (q^1,\dots, q^n, p_1,\dots, p_n))$. I explained this in detail in this MSE answer but let me conform to the present notation and repeat myself:

• $q^i: T^*U\to\Bbb{R}$ is a coordinate function defined explicitly as $\pi^*(x^i):= x^i\circ\pi$. It is by an abuse of notation that people call both of these the same symbol $q^i$ (or $x^i$ depending on context).
• $p_i:T^*U\to\Bbb{R}$ is defined by setting for each $\lambda\in T^*U$, \begin{align} p_i(\lambda):= \lambda\left(\frac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)}\right)\tag{!!} \end{align} In other words, $p_i(\lambda)\in\Bbb{R}$ is the unique number in the expansion of $\lambda$ relative to the basis $\{dx^1|_{\pi(\lambda)},\dots, dx^n|_{\pi(\lambda)}\}$ of $T_{\pi(\lambda)}^*M$, i.e making the following equation true: \begin{align} \lambda&= p_{i}(\lambda)\,(dx^i)|_{\pi(\lambda)}. \end{align} Here, both sides are elements of $T_{\pi(\lambda)}^*M$ (review the link at the top).

Now, we’re almost ready to prove the desired claim. The final ingredient is the following calculation which I leave to you to verify: \begin{align} (T\pi)_{\lambda}\left(\frac{\partial}{\partial q^i}\bigg|_{\lambda}\right)&= \frac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)}, \quad \text{and}\quad (T\pi)_{\lambda}\left(\frac{\partial}{\partial p_i}\bigg|_{\lambda}\right)=0.\tag{$*$} \end{align}

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The proof.

Now, recalling the very first link about expanding relative to the dual basis, and keeping in mind the definitions (!), (!!), and the easy calculation (*), we have \begin{align} \theta_{\lambda}&=\theta_{\lambda}\left(\frac{\partial}{\partial q^i}\bigg|_{\lambda}\right)\cdot (dq^i)|_{\lambda}+ \theta_{\lambda}\left(\frac{\partial}{\partial p_i}\bigg|_{\lambda}\right)\cdot(dp_i)|_{\lambda}\\ &=\lambda\left(\frac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)}\right)\cdot (dq^i)|_{\lambda}+ 0\\ &= p_i(\lambda)\cdot (dq^i)|_{\lambda}. \end{align} Since this is true for all $\lambda\in T^*U$, it follows that $\theta|_{T^*U}= p_i\,dq^i$.

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Extra reading.

You may also want to read the following to understand how this picture in the cotangent bundle is ‘translated’ on the tangent bundle side (i.e the relationship between the Lagrangian and Hamiltonian formulations of things)

• Help with geometric view of conjugate momenta and Legendre transformation
• Help pulling back the tautological one-form on the cotangent bundle $T^*M$ to the tangent bundle $TM$ using a Riemannian metric on $M$,

and all of the tons of sublinks.