$p\ \text dq$ is the "tautological" one-form?

Question

The one-form $$\theta=\sum_i p_i\, \text dq^i$$ is a central object in hamiltonian mechanics. It has a bunch of applications: $\omega=\text d\theta$ is the symplectic structure on phase space, $S=\int\theta$ is the classical action, and so on and so forth. It is associated with the names Liouville one-form, Poincaré one-form, canonical one-form, and symplectic potential, none of which surprises me, but its Wikipedia entry informs me that the preferred^{[by whom?]} name for it is actually "tautological" one-form, on the grounds that 'canonical' (which would be my natural choice) is 'already heavily loaded', and because of the risk of confusion with some algebraic thingammy.

This name completely mystifies me. Why was the name "tautological" chosen for this object? When, where, and by whom? Or was this name chosen because that's its name?

Answer 1

The name seems appropriate if consider that it probably comes from the case when the manifold is the cotangent bundle of a manifold. Then a point on $T^*M$ is a pair $(x,\alpha)$, where $x$ is a point on $M$ and $\alpha$ a one form. The definition of the tautological one form is: the value of the form at a point $(x,\alpha)$ on a tangent vector is obtained by projecting the vector to a tangent vector on $M$ and evaluating (at $x$) the form $\alpha$. In other words for $v\in T(T^*M)$ the value is $\theta_{(x,\alpha)}(v):=\alpha_x(d\pi(v))$, where $\pi : T^*M \rightarrow M$ is the projection.

In a way it is pretty tautological.

Answer 2

I) On a general symplectic manifold $({\cal M},\omega)$ (typically called phase space by physicists), one can locally choose a symplectic potential $\theta\in \Gamma(T^{*}{\cal M}|_{\cal U})$, which is a one-form such that

$$\tag{1} \mathrm{d}\theta~=~\omega,$$

cf. Poincare Lemma. Here ${\cal U}\subseteq {\cal M}$ denotes a local neighborhood.

Note that the symplectic potential $\theta$ is never unique (or 'canonical') in the sense that

$$\tag{2} \theta^{\prime}~=~\theta+\mathrm{d}F$$

would also be a symplectic potential, if $F$ is a zero-form (aka. a function).

For a general symplectic manifold $({\cal M},\omega)$ there may not exist a globally defined symplectic potential $\theta$.

Darboux' theorem states that any $2n$-dimensional symplectic manifold $({\cal M},\omega)$ is locally isomorphic to the cotangent bundle $T^*(\mathbb{R}^n)$ equipped with the canonical symplectic two-form.

II) Consider next the special case where the symplectic manifold ${\cal M}=T^{*}M$ happens to be a cotangent bundle

$$\tag{3} {\cal M}~=~T^{*}M~\stackrel{\pi}{\longrightarrow}~ M $$

equipped with the canonical symplectic two-form $\omega$, which in local coordinates reads

$$\tag{4} \omega|_{\pi^{-1}(U)} ~=~\sum_{i=1}^n\mathrm{d}p_i\wedge \mathrm{d}q^i.$$

Here $U\subseteq M$ denotes a local neighborhood of the base manifold $M$. (The base manifold $M$ is typically called the configuration space by physicists.) Moreover, $q^i$ are local coordinates on the base manifold $M$, and $p_i$ are local coordinates of the cotangent fibers.

Then there always exists a globally defined symplectic potential $\theta\in \Gamma(T^{*}{\cal M})$ that in local coordinates reads

$$\tag{5} \theta|_{\pi^{-1}(U)}~=~\sum_{i=1}^n p_i ~\mathrm{d}q^i.$$

Since the globally defined one-form (5) comes for free on a cotangent bundle ${\cal M}=T^{*}M$ for any manifold $M$, it is 'tautological' in that sense. But wait, there there is more: It can be defined in a manifestly coordinate-independent way, cf. Wikipedia & MBN's answer.

Answer 3

Here's how it finally clicked for me. Let $\phi: M \rightarrow N$ be a smooth map, and $\alpha$ be a 1-form on $N$. Recall that the pullback of $\alpha$ by $\phi$ is a 1-form $\phi^*\alpha$ on $M$ defined by

$$(\phi^*\alpha)_x(v) = \alpha_{\phi(x)}(d\phi_x(v))$$

where $d\phi$ is the pushforward of $\phi$.

When $M$ is $T^*N$, something nice happens. We find that a point $(x, \alpha$) on $T^*N$ itself gives us a 1-form ($\alpha$) to pull back, and the canonical projection $\pi$ gives us something obvious to pull back by. ($\alpha$ isn't actually a 1-form, as another answer says, but a single covector. Nonetheless, it can be pulled back in the same way.) So we get a 1-form

$$\theta_{(x, \alpha)}(v) = \alpha(d\pi_{(x,a)}(v))$$

or in other words

$$\theta_{(x, \alpha)} = \alpha \circ d\pi_{(x,a)}$$

without having to make any additional choices. It is in this sense canonical (though why they call it "tautological" I don't know).