13
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I don't know if this will help (I found it incomprehensible so, if it makes sense to you, please share...): math.stackexchange.com/q/252180 $\endgroup$ – Alfred Centauri Dec 4 '13 at 2:30
  • $\begingroup$ I'm thinking this should be posted on Math.SE. $\endgroup$ – joshphysics Dec 4 '13 at 4:15
  • $\begingroup$ supposedly, it has to do with solder forms and the tautological 1-form of the frame bundle (that one is tautological because it's a $R^n$-valued tensorial 1-form that corresponds to the identity of the tangent bundle understood as an associated vector bundle); however, I was never able to make the definitions match $\endgroup$ – Christoph Dec 4 '13 at 12:13
  • $\begingroup$ For logistics, mathworld.wolfram.com/Tautology.html tautological meant the same an equivalent relation. So it sounds like people were trying to say that this was a definition, but it's equivalent to some other generic definitions. "Liouville" and "Poincaré" meant either those people had something to do with it, or their work was vastly connected with it. "canonical" and "symplectic" meant "physics" and "math". It sounded like a naming strategy by mathematician: "tautological."...."oh!" $\endgroup$ – ShoutOutAndCalculate Apr 21 at 0:47
10
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Damn. Re-reading the question, I see that you already understand why it's "canonical." It might still be worthwhile to leave this answer up for others who (like me) didn't understand even this much. At worst, it's no less informative than MBN's answer. $\endgroup$ – A_P Apr 14 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.