# Does the Hamiltonian formulation of classical mechanics require an inner product on physical space?

The Hamiltonian formulation of classical mechanics is quite broad and flexible; one of the only nontrivial physical assumptions that need to be made is that the degrees of freedom are continuous rather than discrete.

In the Hamiltonian formalism (working on flat space in Cartesian coordinates), the positions $$r^i$$ transform as vectors, while the canonical momenta $$q_j := \frac{\partial \mathcal{L}}{\partial \dot{r}^j}$$ transform as one-forms.

The symplectic manifold of phase space is equipped with a symplectic form, i.e. a closed nondegenerate 2-form over the manifold. But since it's a two-form, it can only input pairs of vectors, not one-forms. Therefore, it isn't obvious that the symplectic form can directly input canonical momenta, which are one-forms. Moreover, I don't think you can directly take the wedge product of a position and a momentum, since one is a vector and one is a one-form, so their tensor product is a type-(1,1) tensor, and it doesn't make any sense to talk about pairs of indices of opposite variance being symmetric or antisymmetric.

If you have a (symmetric) inner product on physical space (not phase space), then this isn't an issue, because you can just use the inner product to convert the position vector into a one-form before taking the wedge product.

But does it make sense to talk about wedge products, Poisson brackets, symplectic forms, etc. for a physical space that's a smooth manifold without a metric tensor? (Of course, in practice the Lagrangian will typically contain such a metric tensor anyway, but I'm just talking about the general canonical formalism, independent of any particular choice of Lagrangian.) If so, then what does it even mean for a symplectic form or Poisson bracket to be "antisymmetric" under the interchange of a tensor input and a one-form input? They seem "apples to oranges", so it isn't obvious to me what it even means to interchange them.

This Wikipedia page seems to imply that there is a way to define symplectic forms on smooth manifolds without a metric structure, but it's way over my head and I don't understand it (I have no idea what a "solder form" is). It would be great if someone could explain this using more elementary language.

• I think you may be confusing objects in physical space vs on the phase space manifold. For example, indeed in flat space with Cartesian coordinates position transform as vectors and momentum as a one-form. But in phase space $q^i$ and $p_i$ become rather coordinates in a Darboux chart. You never need the sympletic form to act on one-forms, it acts on the associated coordinate basis vector fields.
– Gold
Commented Feb 11, 2023 at 18:44
• The $q_i$ and $p_i$ are generalized coordinates and momenta (coordinates in a chart). They are neither vectors nor forms. Instead, $dq_i$ and $dp_i$ are 1-forms. Commented Feb 11, 2023 at 18:57
• @GiorgioP So $q^i$ is a vector but $dq_i$ is a one-form? Commented Feb 11, 2023 at 19:23
• In the phase space manifold $q^i$ is not a vector, nor $p_i$ is a one-form. They are coordinate functions in a local chart. Then $dq^i$ and $dp_i$ are one-forms as usual and $\frac{\partial}{\partial q^i}$ and $\frac{\partial}{\partial p_i}$ are vector fields.
– Gold
Commented Feb 11, 2023 at 19:33
• Also $p_i dq^i$ is not a scalar, it is a one-form in the phase space manifold often called the pre-sympletic potential.
– Gold
Commented Feb 11, 2023 at 20:00

In many of our classical textbook toy problems (mass moving in a certain potential), yes there is a ‘hidden’ Riemannian metric (because we’re talking about massive particles) which is how we formulate the kinetic term. But, abstractly, no it is not necessary to have a Riemannian metric in order to define a symplectic form or to formulate Hamiltonian mechanics.

You write

In the Hamiltonian formalism (working on flat space in Cartesian coordinates), the positions $$r^i$$ transform as vectors, while the canonical momenta $$q_j := \frac{\partial \mathcal{L}}{\partial \dot{r}^j}$$ transform as one-forms.

Well, that’s just confusing. The $$r^i$$ are not vectors or covectors or anything. They are coordinate functions. I have written about this in several answers, so you may want to read some of them and go down the linked rabbit hole starting from Help with geometric view of conjugate momenta and Legendre transformation (which has all the relevant links inside).

Next, regarding your middle paragraphs, you’re right, you can’t talk about symmetry or skew-symmetry without first fixing an isomorphism between the vector space and its dual. But this is not relevant in this discussion, and I think you’re only bringing it up out of confusion.

You write

Therefore, it isn't obvious that the symplectic form can directly input canonical momenta, which are one-forms

and

This Wikipedia page seems to imply that there is a way to define symplectic forms on smooth manifolds without a metric structure, but it's way over my head and I don't understand it…

So, it seems to me you’re misunderstanding the spaces things are defined on. What we’re saying is you start with a smooth manifold $$Q$$. Then, on $$T^*Q$$, there is a natural 1-form $$\theta$$ (note this is a much more complicated beast than a 1-form on $$Q$$. We have $$\theta\in \Gamma(T^*(T^*Q))$$, i.e it is a mapping $$\theta:T^*Q\to T^*(T^*Q)$$ such that composing with the projection $$\pi_{T^*(T^*Q)\to T^*Q}$$ gives you the identity mapping on $$T^*Q$$). I emphasize again, $$\theta$$ is NOT a 1-form on $$Q$$, which is what your wording seems to suggest your interpretation is, instead it is a 1-form on the larger space $$M:=T^*Q$$. The canonical symplectic form is then given by the exterior derivative $$\omega:=d\theta$$. This is a 2-form on $$M=T^*Q$$, giving us the symplectic manifold $$(M=T^*Q,\omega)$$.

Now, I should mention that not every symplectic manifold arises as the contangent bundle of some manifold. For example, the complex projective spaces $$\Bbb{C}P^n$$ have standard symplectic structures (by using the standard one, $$\sum_{j=1}^{n+1}dx^j\wedge dy^j$$, on $$\Bbb{C}^{n+1}=\Bbb{R}^{2n+2}$$, pulling it back to the sphere $$S^{2n+1}$$, and then noting that it is invariant under the $$U(1)$$-action, and hence descends to a closed 2-form, i.e a symplectic form, on the quotient $$S^{2n+1}/U(1)=\Bbb{C}P^n$$). Complex projective spaces are compact, while cotangent bundles are non-compact, hence they provide examples of symplectic manifolds which are not cotangent bundles. By the way, a concrete special case is when $$n=1$$, in which case one has the well-known diffeomorphism $$\Bbb{C}P^1\cong S^2$$, meaning that the sphere $$S^2$$ has a symplectic structure. This can even be described explicitly as follows: \begin{align} \omega_{S^2}=\iota^*(x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy), \end{align} where $$(x,y,z)$$ are the usual Cartesian coordinate functions on $$\Bbb{R}^3$$, and $$\iota:S^2\hookrightarrow \Bbb{R}^3$$ is the canonical injection. In other words, if you let $$(\xi,\eta,\zeta)$$ be the functions obtained by restricting the functions $$(x,y,z)$$ from $$\Bbb{R}^3$$ to $$S^2$$, then $$\omega_{S^2}=\xi\,d\eta\wedge d\zeta+\eta\,d\zeta\wedge d\xi+\zeta\,d\xi\wedge d\eta$$. This is none other than the area 2-form of the sphere (so it is also not exact, i.e this $$\omega$$ doesn’t arise from a $$\theta$$ like in the cotangent bundle case).

As you can see from these concrete examples, I never used any Riemannian metric to define the $$\omega$$ (I didn’t assume any Riemannian metric on $$Q$$ or $$T^*Q$$, yet I defined $$\omega$$ on $$T^*Q$$. Likewise, I defined $$\omega$$ on $$\Bbb{C}P^n$$, and especially on $$S^2$$, all without referencing a Riemannian metric).

Next, if you have a Poisson manifold $$(P,\{\cdot,\cdot\})$$, for which symplectic manifolds $$(M,\omega)$$ serve as basic examples, this again has nothing to do with a Riemannian metric anywhere. The bracket is a skew-symmetric $$C^{\infty}(P)$$-bilinear mapping $$C^{\infty}(P)\times C^{\infty}(P)\to C^{\infty}(P)$$, and hence it gives rise to a bivector field $$\pi$$ on $$P$$, i.e $$\pi$$ is a section of the bundle $$\bigwedge^2(TP)$$, i.e $$\pi:P\to\bigwedge^2(TP)$$ such that it projects back to the identity on $$P$$ (you can think of it as a $$(2,0)$$ tensor field on $$P$$ satisfying a bunch of properties). But again, this section has been defined without referencing any Riemannian metric.

First I think we need some definitions. A phase space is a symplectic manifold, which in turn is a pair $$(\Gamma,\Omega)$$ where $$\Gamma$$ is an even dimensional smooth manifold and $$\Omega$$ is a closed and non-degenerate two-form in $$\Gamma$$ called symplectic form. A Darboux coordinate system in $$(\Gamma,\Omega)$$ is a local chart $$(q^i,p_i)$$ on some open subset $${\cal U}\subset \Gamma$$ such that the two-form $$\Omega$$ written in this chart has the form $$\Omega=dp_i \wedge dq^i\tag{1}.$$

In that case we call $$(q^i,p_i)$$ canonical coordinates on $$(\Gamma,\Omega)$$ and say that $$p_i$$ and $$q^i$$ are canonically conjugate variables. The one-form defined $$\theta = p_i dq^i$$ is called sympletic potential because $$\Omega = d\theta$$ by definition of exterior differentiation. This is valid locally on $${\cal U}$$, but such sympletic potential may not be globally-defined if $$H^2(\Gamma)\neq 0$$.

Now, in all this discussion you should observe the following: $$q^i$$ here are not components of a vector, they are coordinates in a smooth manifold $$\Gamma$$. Likewise, $$p_i$$ are not components of a one-form, they are coordinates in a smooth manifold as well. The basis differentials $$dq^i$$ and $$dp_i$$ are then one-forms in $$\Gamma$$ while the coordinate basis $$\frac{\partial}{\partial q^i}$$ and $$\frac{\partial}{\partial p_i}$$ are vector fields. The sympletic potential $$\theta=p_i dq^i$$ is not a scalar, but rather a one-form in $$\cal U$$. Very important to the question, all of this is occurring inside the manifold $$\Gamma$$.

There are many cases in which one constructs $$(\Gamma,\Omega)$$ directly without talking about any other space before. And many cases in which $$(q^i,p_i)$$ are not actual position and momenta, but some other variables obeying canonical commutation relations. Such an example is readily provided in field theory, in the context of the covariant phase space. We start from a Lagrangian describing the dynamics of some fields and we give a prescription to directly build $$(\Gamma,\Omega)$$.

Nevertheless, in classical mechanics, it often happens to be the case that $$(\Gamma,\Omega)$$ is constructed out of some other space. In particular, we often have a configuration space $$Q$$ and then Lagrangian mechanics takes place in $$TQ$$ while Hamiltonian mechanics takes place in $$T^\ast Q$$.

So let $$\pi:TQ\to Q$$ be the canonical projection and suppose that $$q^i$$ are coordinates on the base configuration manifold over some open subset $$U\subset Q$$. These will naturally lift to coordinates $$(q^i,\dot{q}^i)$$ on $$\pi^{-1}(U)\subset TQ$$ which are defined so that if $$v\in \pi^{-1}(U)$$ lives over the point $$a\in U$$ then $$q^i(v)$$ are the coordinates $$q^i(a)$$ of the base point, plus $$\dot{q}^i(v)$$ are the components of $$v$$ in the coordinate basis: $$v = \dot{q}^i(v)\frac{\partial}{\partial {q}^i}\bigg|_{a}.\tag{2}$$

Again observe that the $$q^i$$ are not vector components, but they are just coordinate functions, be in $$Q$$ or $$TQ$$. The same goes for $$\dot{q}^i$$. Nevertheless, given the Lagrangian $$L:TQ\to \mathbb{R}$$ you may construct the one-form

$$\dfrac{\partial L}{\partial \dot{q}^i}dq^i\tag{3},$$

and you might define those coefficients as $$p_i$$. In that case the $$p_i$$ so-defined are the components of a one-form over $$TQ$$. But when we go to $$T^\ast Q$$ the $$p_i$$ become just coordinate functions as in the general symplectic manifold analysis. Using the same notation is in fact an abuse of notation since $$p_i: TQ\to \mathbb{R}$$ and $$p_i:T^\ast Q\to \mathbb{R}$$ cannot possibly be the same function since they even have different domains.

The bottom line is that you must be aware on which space you are working on:

1. The $$q^i$$ are never vector components really, be in $$Q$$, $$TQ$$, $$T^\ast Q$$ or in a generic Darboux chart in a symplectic manifold $$(\Gamma,\Omega)$$. It just so happens that for the manifold $$\mathbb{R}^D$$, if $$q^i$$ are Cartesian coordinates, you can identify the point with a vector at the origin through the identification between $$\mathbb{R}^D$$ and any of its tangent spaces. But that is a very special case that doesn't generalize - curved spaces have no position vectors really.

2. If you have a configuration manifold $$Q$$ and you build $$TQ$$ and define $$p_i = \frac{\partial L}{\partial \dot{q}^i}$$ then the $$p_i$$ become components of a one-form in $$TQ$$ because of your definition. However, this should not be confused with the $$(q^i,p_i)$$ coordinates used in $$T^\ast Q$$, in which case both $$q^i$$ and $$p_i$$ are just coordinate functions. The key issue here is an abuse of notation of identifying $$p_i$$ in $$T^\ast Q$$ and $$p_i$$ in $$TQ$$.

3. In a general phase space with a Darboux chart $$(q^i,p_i)$$ are both just coordinate functions.

Apart from all that, a small comment is that there is no situation in which you need $$\Omega$$ to "act on coordinates and momenta". $$\Omega$$ acts on phase space vector fields and one of its main uses is to define a canonical transformation (symmetry) to be a transformation preserving $$\Omega$$ and also to define the Poisson brackets $$\{\}$$. It also establishes the connection between symmetries and conserved charges. If $$X$$ generates a canonical transformation, $$\Omega(X,\cdot)$$ is a closed form and the corresponding charge is, in principle, determined by the equation $$\Omega(X,\cdot)=-dH_X(\cdot)$$. There may be topological obstructions for the charge to be globally-defined though, since only when $$H^1(\Gamma)=0$$ we will have such $$H_X$$ defined on $$\Gamma$$ globally.

Finally, to answer your title question: no, no inner product is required in any of that, c.f. the definition of a symplectic manifold. I suggest you take a look at the book "Physics for Mathematicians" by Spivak, which has a nice account of Hamiltonian mechanics using differential geometry.

• A few comments regarding local vs global. First, “The pre-sympletic potential $\theta=p_i dq^i$ is not a scalar, but rather a one-form in $\Gamma$”, well, it is only a 1-form on an open subset of $\Gamma$, and in general cannot be defined globally (not all symplectic forms are exact, e.g the one on $S^2$). Next, same issue towards the end: ”act on a canonical symmetry generator $X$ and define the associated canonical charge by $\Omega(X,\cdot)=−dQ(\cdot)$”, here again all we know is the 1-form $\Omega(X,\cdot)$ is closed (also denoting $Q$ as the charge, if exact, is unfortunate notation). Commented Feb 12, 2023 at 3:25
• @peek-a-boo these are good points, thanks for bringing them up. I was indeed tacitly assuming $H^2(\Gamma)=0$ and $H^1(\Gamma)=0$ so that closed two-forms and closed one-forms are exact. I have edited the text with the more general situation.
– Gold
Commented Feb 12, 2023 at 12:27
• This is extremely helpful, thank you. The Wikipedia article on "tautological one-form" says that it "is used to create a correspondence between the velocity of a point in a mechanical system and its momentum, thus providing a bridge between Lagrangian mechanics with Hamiltonian mechanics" that "assigns a numerical value to the momentum $p$ for each velocity $\dot{q}$". ... Commented Feb 13, 2023 at 2:01
• ... Does this mean that to convert between the two senses of $p_i$ in your case #2, we first take the negative exterior derivative of the one-form $\theta$ to produce a two-form $\omega$, and then we feed into that two-form the velocity vector $\dot{q} \in TQ$ to get the corresponding momentum $p \in T^*Q$? Commented Feb 13, 2023 at 2:01
• So the tautological one-form essentially provides an alternate formulation of the physical mapping from velocities to canonical momenta that doesn't explicitly reference the Lagrangian? Commented Feb 13, 2023 at 2:11
1. Indeed a metric tensor is typically needed for a notion of kinetic energy (both in the Lagrangian and Hamiltonian formulation). See also e.g. this Phys.SE post.

2. However, there exist (Lagrangian and Hamiltonian formulations of) topological field theories that don't depend on a metric tensor or a volume form (although gauge-fixing terms therein typically need a metric tensor or a volume form).

3. As a simple example, think of a theory $$L=p_i\dot{q}^i$$ where the Hamiltonian $$H=0$$ vanishes identically.