Why does phase space require a symplectic geometry rather than a metric? Is there some scenario where a metric is unable to describe the notion of length in phase space, specifically in relation to the uncertainty principle?

  • 4
    $\begingroup$ The symplectic structure is not there to provide a notion of length, but to provide the notion of Hamiltonian flow (as is mentioned in the answer to one of your other questions), and arises naturally on cotangent bundles. I am not sure what it is you are asking, and what the uncertainty principle (which is emphatically not a concept of Hamiltonian mechanics) has to do with it. $\endgroup$
    – ACuriousMind
    Sep 18, 2014 at 22:40
  • $\begingroup$ @ACuriousMind A manifold only has meaning when we impose upon it a geometrical structure. If we used a metric then what would be the interpretation of $(\Delta q)^2+(\Delta p)^2$? I believe the answer has to do with the symmetry or indeed asymmetry of the tensor we use in both space time and phase space? Namely the symplectic-2-form $\Omega$ and the metric tensor $g$. Forgive me if my question was misleading :) $\endgroup$ Sep 18, 2014 at 22:48
  • $\begingroup$ I can see how the uncertainty principle is rather misleading but I'm just imagining it as a small shift in coordinates. That isn't the essence of my question. $\endgroup$ Sep 18, 2014 at 22:50
  • 2
    $\begingroup$ A manifold has perfect meaning without a metric. What would your reason (physical or mathematical) be to pick your metric over all the others you could endow the manifold with? Yet, the symplectic structure on the cotangent bundle is natural/canonical and does not require the underlying "physical world"-manifold to possess a particular metric (nor the bundle itself, for that matter). The reasons why phase space is symplectic are twofold, but simple: The structure is natural, and it gives the correct results. $\endgroup$
    – ACuriousMind
    Sep 18, 2014 at 22:57

2 Answers 2

  1. A metric structure $g$ and
  2. a symplectic structure $\omega$

are two very different structures, although sometimes they can co-exist in a compatible way.

Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure.

There exists a unique torsionfree metric connection $\nabla$ on a pseudo-Riemannian manifold $(M,g)$, known as the Levi-Civita connection, while there exist infinitely many torsionfree symplectic connections $\nabla$ on a symplectic manifold $(M,\omega)$.

Unlike a pseudo-Riemannian manifold $(M,g)$, there are no scalar curvature, or local invariants on a symplectic manifold $(M,\omega)$.

The group of local symplectomorphisms on a symplectic manifold $(M,\omega)$ is infinite dimensional, while the group of local Killing symmetries on a pseudo-Riemannian manifold $(M,g)$ is typically finite dimensional.

A metric structure on spacetime is important for general relativity (GR), but irrelevant for the Hamiltonian formalism.$^1$ If we artificially assign a metric on phase space, it would typically not be preserved under symplectomorphisms.

On the other hand, note that in the Hamiltonian formalism, the cotangent bundle of the configuration space is born with a canonical symplectic structure, cf. e.g. this Phys.SE post.


$^1$ Note in particular that in the Hamiltonian formulation of GR, known as the ADM formulation, the pertinent phase space is different from spacetime itself, and also different from the configuration space of all possible spacetime metrics. The configuration space of all possible spacetime metrics is endowed with the DeWitt metric.

  • $\begingroup$ Minor stuff: (a) isn't it spelt "torsion-free"? Now for the major question: (b) I have forgotten, so this may be stupid, but isn't the lack of local invariants on a Symplectic manifold responsible for Darboux's theorem (more or less)? $\endgroup$ Sep 19, 2014 at 15:11
  • $\begingroup$ @Alex Nelson: (b) At least morally Yes, but I prefer to phrase it the other way around: Due to Darboux's theorem, there are no local invariants. $\endgroup$
    – Qmechanic
    Sep 19, 2014 at 15:16
  • $\begingroup$ Could you please explain why the ADM phase space is not the same thing as the space of all possible spacetime metrics? $\endgroup$ Jan 25, 2018 at 17:17
  • $\begingroup$ For starters, the latter doesn't appear to have a symplectic structure. $\endgroup$
    – Qmechanic
    Jan 25, 2018 at 19:37

So, in classical mechanics, we know nothing of this strange "spacetime" and its metric. We know only that systems are described by $n$ continuous generalized coordinates $q^i$ with a certain range, and we take the manifold $\mathcal{M}$ consisting of all possible different $\vec q$ as our starting point. Note that there is no metric, no form, nothing on this.

We now include the generalized momenta $p_i$ at any point as elements of the cotangent space at that point, and henceforth call the cotangent bundle our phase space $\mathcal{P} := T^*\mathcal{M}$.

Now, on the cotangent bundle, there is the tautological one-form $\theta : \mathcal{P} \to T^*\mathcal{P}$. It is uniquely characterized to be the form that undoes the pullback $\beta^*$ by $\beta : \mathcal{M} \to T^*\mathcal{M}$ on the coordinate manifold by $ \beta^*\theta = \beta$. It is natural in the sense that it is characterized by an universal property and always exists uniquely. In coordinates, it is $\theta = p_i \mathrm{d}q^i$.

From this, the symplectic form simply arises as $\omega := \mathrm{d}\theta$. As $\theta$ was natural, so is $\omega$, and the proof that $\omega$ is symplectic is simply that it is locally given by $\omega = \mathrm{d}p_i \wedge \mathrm{d}q^i$, which is obviously a symplectic form on $\mathbb{R}^n \times \mathbb{R}^n$

So, we do not choose a symplectic form, there is one, and only one, natural candidate. Note that there is no such candidate for a metric.

If we are now given an "energy function" - the Hamiltonian $H$ - on $\mathcal{P}$, it turns out that demanding that $\theta(X_H)$ be the action reproduces Hamilton's equations, which, by their equivalence to Lagrangian/Newtonian mechanics, are the correct equations of motions. Furthermore, the symplectic form gives the correct way to implement the Poisson bracket (which is antisymmetric, and which a metric could never do) by

$$ \{f,g\} := \omega(X_f,X_g)$$

where $X_f$ is the vector field defined by a differentiable function $f : \mathcal{P} \to \mathbb{R}$ through $\mathrm{d}f \overset{!}{=} \omega(X_f,\dot{})$.

One can also show now that this is really a Poisson bracket and derive the time evolution equation on the phase space for observables, but I believe this is sufficient to be convincing that the symplectic structure is indeed the natural setting for Hamiltonian mechanics.

Lastly, and somewhat unrelated, we note that, on Riemannian manifolds, it is quite possible to consider the geodesics as a Hamiltonian flow, where the Hamiltonian is defined by the metric, but we still use the symplectic cotangent bundle to define the flow on.

  • $\begingroup$ Awesome summary of the symplectic potential and form and also of the their uniqueness in reproducing Hamilton's equations. I have only one slight criticism: you might say that $\beta$ is any one form on $\mathcal{M}$. If I wouldn't have seen these ideas before, I think I'd be left looking around for $\beta$'s definition. Since this is such a good summary of all these ideas, you might also include the property that $X_{\{f,\,g\}} = [X_f,\,X_g]$ (well, at least I for one find that this one helps intuition for the Poisson bracket). $\endgroup$ Jun 2, 2017 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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