# How to see the Phase Space of a Physical System as the Cotangent Bundle

Two things today motivated this question.

First, the professor said that in a lecture Thurston mentioned

Any manifold can be seen as the configuration space of some physical system.

Clearly we need to be careful here, so the first question is

1) What is a precise formulation and an argument to see why the previous statement is true.

Second, the professor went on to say that because of the Poisson Bracket, we see the phase space of a physical system as the Cotangent Bundle of a manifold. I understand that we associate a symplectic form to the cotangent bundle, and that we want to think of Phase Space with a symplectic structure, but my second question is

2) Could you provide an example of a physical system, give the associated "configuration manifold" show the cotangent space, and explain why this is the phase space of the system.

I pushed the lecturer quite a bit to get this level of detail, so pushing much further would of probably been considered rude. I should also mention he is speaking about these things because we want to quantize the geometry associated to this manifold. So he looks at the cotangent space which apparently has a symplectic structure, and sends the symplectic form to the Lie Bracket. If any of the things I have said are incorrect, please comment with corrections. I am just learning this material and trying to understand how it fits with my current understanding.

One last bonus question(tee hee),

If your manifold is a Lie Group, we get a Lie algebra structure on the Tangent Space. Is there a relationship between this Lie Algebra structure and the one you would get by considering the cotangent space and then quantizing in the fashion of above?

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I would like this question better if "The Professor" weren't a main character (use their name instead). –  Theo Johnson-Freyd Feb 26 '10 at 5:59
V.I.Arnold(en.wikipedia.org/wiki/Vladimir_Arnold) says: "An "abstract" smooth manifold is a smooth submanifold of a Euclidean space considered up to a diffeomorphism. There are no "more abstract" finite-dimensional smooth manifolds in the world (Whitney's theorem). Why do we keep on tormenting students with the abstract definition? Would it not be better to prove them the theorem about the explicit classification of closed two-dimensional manifolds (surfaces)?" So it is nothing surprising that every smooth manifold is just R^N with bonds together. Abstracts not always work well... –  kakaz Feb 26 '10 at 9:55
@kakaz: What is the point of your comment? An - as you call it - "abstract" manifold definition clearly has its advantages. For example it allows one to distinguish between intrinsic and extrinsic (i.e. embedding) properties. Now the question is: Does the symplectic structure of the cotangent bundle of a manifold depend on this embedding? The answer is: no. But this is not clear from the definition in your comment! That is also the reason why it is unwise to use this definition. (Teaching students is another matter - that is also what Arnol'd adresses!) –  Orbicular Feb 26 '10 at 10:36
@Theo I didn't feel it was appropriate to post his name since I don't know if he would approve of associating his name to some of these comments. In the worst case scenario of me saying something very dumb, and another professor associating my stupidity with him, I would feel very guilty. Since no comments have proved this question to be entirely nonsensical, I can say the professor was Louis Crane, and the class was on Quantum Gravity. If you want even more details, we are going through a paper of Rovelli that Louis found useful: arxiv.org/abs/0708.1236 –  B. Bischof Feb 26 '10 at 14:51

Let's start by answering the first question.

Let $M$ be any manifold. Consider a physical system consisting of a point-particle moving on $M$. What are the configurations of this physical system? The points of $M$. Hence $M$ is the configuration space.

Typically one takes $M$ to be riemannian and we may add a potential function on $M$ in order to define the dynamics. (More complicated dynamics are certainly possible -- this is just the simplest example.)

As an example, let's consider a point particle of mass $m$ moving in $\mathbb{R}^3$ under the influence of a central potential $$V= k/r,$$ where $r$ is the distance from the origin. The configuration space is $M = \mathbb{R}^3\setminus\lbrace 0\rbrace$.

Classical trajectories are curves $x(t)$ in $M$ which satisfy Newton's equation $$m \frac{d^2 x}{dt^2} = \frac{k}{|x|^2}.$$ To write this equation as a first order equation we introduce the velocity $v(t) = \frac{dx}{dt}$. Geometrically $v$ is a vector field (a section of the tangent bundle $TM$) and hence the classical trajectory $(x(t),v(t))$ defines a curve in $TM$ satisfying a first order ODE: $$\frac{d}{dt}(x(t),v(t)) = (v(t), \frac{k}{m|x(t)|^2})$$ This equation can be derived from a variational problem associated to a lagrangian function $L: TM \to \mathbb{R}$ given by $$L(x,v) = \frac12 m v^2 - \frac{k}{|x|}.$$ The fibre derivative of the lagrangian function defines a bundle morphism $TM \to T^*M$: $$(x,v) \mapsto (x,p)$$ where $$p(x,v) = \frac{\partial L}{\partial v}.$$

In this example, $p = mv$. The Legendre transform of the lagrangian function $L$ gives a hamiltonian function $H$ on $T^*M$, which in this example is the total energy of the system: $$H(x,p) = \frac{1}{2m}p^2 + \frac{k}{|x|}.$$

The equations of motion can be recovered as the flow along the hamiltonian vector field associated to $H$ via the standard Poisson brackets in $T^*M$:

$$\frac{dx}{dt} = \lbrace x,H \rbrace \qquad\mathrm{and}\qquad \frac{dp}{dt} = \lbrace p,H \rbrace.$$

Being integral curves of a vector field, there is a unique classical trajectory through any given point in $T^*M$, hence $T^*M$ is a phase space for the system; that is, a space of states of the physical system. Of course $TM$ is also a space of states, but historically one calls $T^*M$ the phase space of the system with configuration space $M$. (I don't know the history well enough to know why. There are brackets in $TM$ as well and one could equally well work there.)

Not every space of states is a cotangent bundle, of course. One can obtain examples by hamiltonian reduction from cotangent bundles by symmetries which are induced from diffeomorphisms of the configuration space, for instance. Or you could consider systems whose physical trajectories satisfy an ODE of order higher than 2, in which case the cotangent bundle is not the space of states, since you need to know more than just the position and the velocity at a point in order to determine the physical trajectory.

It's late here, so I'll forego answering the bonus question for now.

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This was really great, thanks a lot. I had considered your example myself for the statement of Thurston, but I dismissed it as non-physical due to dimension. The example you provided is pretty easy to follow, but I am confused on a point, could you say a little more about the fibre derivative of the Lagrangian function? I don't think I see why passing from the Lagrangian to the Hamiltonian is a mapping of the tangent space to the Cotangent Space. Also, thank you again for your last comment about examples when it is not a cotangent bundle, this was a question I decided not to post. :) Thanks! –  B. Bischof Feb 26 '10 at 1:27
@José Figueroa-O'Farrill: I know that you wanted to keep it simple, but choosing a Riemannian metric somehow kills the point. The point is that the (total space of the) cotangent bundle of any manifold is a symplectic manifold. Any function on it induces a dynamical system. Choosing a Riemannian metric boils down to choosing an identification of the tangent and cotangent bundle, inducing a dynamical system on the tangent bundle. That's also the reason why "Hamiltonian dynamics" is quite useless unless you're concerned with quantization! –  Orbicular Feb 26 '10 at 10:20
@ B. Bischof: Why does the Legendre transform map to the cotangent bundle? Well, this comes from the transformation behaviour under coordinate changes. The easiest was to see this is to look at the formula p=dL/dv, which says that p is a "differential", i.e. a one form. –  Orbicular Feb 26 '10 at 10:23
@Orbicular: I realise this, but I was trying to motivate Hamiltonian dynamics from a Newtonian point of view. Once you see this for a simple lagrangian of the form "kinetic - potential", then the generalisation is clear, but to write down the kinetic term one needs a metric (or at least a quadratic form) on the configuration space. –  José Figueroa-O'Farrill Feb 26 '10 at 12:09
QB. Bischof: I used somewhat compact notation, but if you write the "kinetic" term in the lagrangian as $\frac{1}{2} m \langle v,v\rangle$, then the derivative with respect to $v$ is something of the form $m \langle v,-\rangle$, which is a 1-form, as in Orbicular's comment. –  José Figueroa-O'Farrill Feb 26 '10 at 12:11

This is just to add to what the previous posters have said so much better. It is a wonderful "fact" that whatever manifold you choose, there will always be some mechanical system that can be encoded as geodesic flow on this space (relative to some metric). Here are some examples:

1. $SO(3)$ : rigid body dynamics. The elements of $SO(3)$ are rotation matrices, determining the orientation of the rigid body.

2. $\mathrm{Diff}_{vol}(M)$, the group of volume-preserving diffeomorphisms of a manifold $M$: incompressible fluid dynamics. A diffeo $\varphi: M \to M$ tells you the following: a fluid particle initially at $X \in M$ ends up at $x = \varphi(X)$.

3. $\mathrm{Diff}(S^1)$, ordinary diffeos on the circle: geodesic flow is nothing but the Burgers' equation.

4. $\mathrm{Diff}(S^1) \times \mathbb{R}$, the Bott-Virasoro group (where the multiplication is defined using the Bott-Virasoro cocycle). Geodesic flow here gives you the Korteweg-de Vries equation, which is Burgers' equation with a nonlinearity. The nonlinearity in the equation stems directly from the BV cocycle.

Anyway, the point is that you can have a good time just writing down a manifold with a Riemannian metric and writing down the geodesic equations. More often than not, the resulting system will be out there somewhere in the guise of some famous differential equation, and in this way, you can often relate the geometry of the configuration space to the properties of the equation.

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Do you have a reference for this? –  Dinakar Muthiah Feb 26 '10 at 22:44
The observations on $SO(3)$ and $\mathrm{Diff}_{vol}(M)$ are due to Arnold in his paper "Sur la géométrie différentielle des groupes de Lie de dimension infinie et ses applications a l’hydrodynamique des fluides parfaits", Ann. Inst. Fourier (Grenoble) (1966) 16, 319–361, but in fact this goes back to Euler. The link between the BV group and KdV is due to Ovsienko and Khesin (1987), "The super Korteweg-de Vries equation as an Euler equation", Funktsional. Anal. i Prilozhen. 21, 81–82. A good book that collects all this is Marsden et al. "Hamiltonian Reduction by stages", LNM 1913. –  jvkersch Feb 27 '10 at 8:48

Well, I'll not say anything deeply new here - just a (hopefully correct) summary.
Suppose you are given your favorite manifold, say $Q.$ Then its cotangent bundle $M=T^* Q$ comes equipped with some canonical structure (Keep in mind that the cotangent bundle $T^* Q$ and the tangent bundle $TQ$ are isomorphic as vector bundles over $Q$ (in particular their total spaces are diffeomorphic, but not canonically so), but for some miraculous reason the cotangent bundle has "more" structure). Denote by $\pi:T^* Q\rightarrow Q$ the projection, associating to a covector its basepoint. Differentiating this yields the tangent map of $\pi$, $T\pi: T(T^* Q)\rightarrow TQ$. With its help one can define a one-form $\theta$ on $T^* Q$ (usually called "canonical" or Liouville one-form). It is defined via
$\theta_\alpha:T_\alpha (T^* Q)\rightarrow \mathbb{R},$ $v\mapsto \alpha(d\pi(\alpha).v)$
for any point $\alpha\in T^* Q$. To explain why the definition makes sense: $v$ is an arbitrary element of $T_\alpha (T^* Q)$, the differential of $\pi$ at $\alpha,$ $d\pi(\alpha)$ is a linear map from $T_\alpha (T^* Q)$ to $T_{\pi(\alpha)} Q$. Furthermore $\alpha$ can be interpreted as a linear form on the tangent space of its base point; consequently it makes sense to evaluate it on $d\pi(\alpha).v$.
Now the symplectic form $\omega\in \Omega^2(T^*Q)$ is defined as the exterior differential of $\theta$ (some authors prefer to smuggle a minus sign in). Notice $\omega$ is defined purely intrinsically (no choice of coordinates for instance, even though one often sees expressions like $\theta=p_idq^i$).
That's the reason why you can associate a symplectic manifold (aka phase space) to any manifold (aka configuration space). But to elaborate on what José Figueroa-O'Farrill already said: there are symplectic manifolds which are not of the form $T^* Q$ for some $Q$. Probably the easiest example are closed symplectic manifolds, i.e. compact ones without boundary (they occur for instance as surfaces with a fixed volume form or as nonsingular complex projective varieties together with the Fubini-Study form). You can easily show that their symplectic form can not be exact (that is, of the form $d\theta$ for some one-form $\theta$) unless they are zero-dimensional. Because if it were exact, by Stokes' theorem the integral $\int \omega^{\wedge (dim M/2)}$ would have to be zero. And it is difficult to find manifolds with zero total volume!

In order to answer the second question: the Lie algebra structure on $\mathfrak{g}=Lie(G)$ induces a Poisson structure on its dual $\mathfrak{g}^*,$ as for instance explained in http://en.wikipedia.org/wiki/Poisson_manifold#Example.

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I was just recently trying to understand a similar lecture by a colleague at my university. He's using Folland's "Quantum field theory" as source; you may find it helpful.

The symplectic structure on the cotangent bundle to $M$ is given by the differential $\Omega$ of the 1-form $\omega$ defined by $\omega_{v^*}(v) = v^*(\pi_*(v))$ for $v^* \in T^*M$ and $v \in T_{v^*}(T^*M)$. (I couldn't tell from your question whether or not this was already clear. Thanks to Orbicular for pointing out that my original version depended on a choice of coördinates.)

The following is a direct quote from Folland:

… we start with a configuration space $N$, which is taken to be a manifold and is supposed to be a description of the possible “positions” of the system. There is quite a lot of flexibility here. For example, if the system consists of $k$ particles moving in $\mathbb R^3$ …, $N$ will be $\mathbb R^{3k}$, or perhaps $\{(\mathbf x_1, \ldots, \mathbf x_k) \in \mathbb R^{3k} : \mathbf x_i \ne \mathbf x_j\text{ for }i \ne j\}$. If the motion is subject to some constraints, $N$ could be a submanifold of $\mathbb R^{3k}$ instead. For an asymmetric rigid body, however, the appropriate configuration space is $\mathbb R^3 \times \operatorname{SO}(3)$: three linear coordinates to give the location of the center of mass (or some other convenient point in the body), and three angular coordinates to give the body's orientation in space.

Velocities are taken to be tangent vectors to the configuration space, so the position–velocity state space is $T N$. On the other hand, the appearance of Poisson brackets in the Hamiltonian formalism [earlier] leads us to take the position–momentum state space to be the symplectic manifold $T^*N$ …. [T]he mass matrix $m$ should be interpreted as a Riemannian metric on $N$ that mediates between vectors and covectors …. $T^*N$ is traditionally called the phase space of the system. (The origin of the name lies in statistical mechanics.)

Thus, it seems to me that it is the configuration, not phase, space that is meant to be physically obvious; and that the designation of the cotangent bundle as the phase space is mostly a matter of terminology.

Once we have the symplectic structure, we use it to equip, not $N$ or $T^*N$, but $C^\infty(T^*N)$, with a Poisson bracket, with respect to which it is a Lie algebra. However, this is a really immense (hugely infinite dimensional) Lie algebra, so it is hard to believe that it has anything to do with the much smaller Lie algebra $T_1G$ when $N = G$ is a Lie group.

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A remark on your last sentence. You can define a differential on $\bigwedge^* \mathfrak{g}^*$ making it a cochain complex. The de Rham cochains on $G$ are the hugely infinite dimensional space of all $k$-forms $\Omega^*(G)$, so you might think the two couldn't be related. But $\bigwedge^* \mathfrak{g}^*$ sits inside $\Omega^*(G)$ as the left-invariant forms, giving a map from $H^*(\mathfrak{g}):=H^*(\bigwedge^* \mathfrak{g}^*)$ to $H^*(G;\mathbb{R})$, and for $G$ compact this is even an isomorphism. I do agree that in this particular case there is probably no relation though. –  Tom Church Feb 26 '10 at 3:47
You cannot split the tangent space of a point in the total space of the cotangent bundle the way you indicated unless you choose a connection. But the symplectic form is independent of such choices, rendering your definition somewhat obscure (not to say wrong). –  Orbicular Feb 26 '10 at 14:43
Tom, you are quite right that “one is bigger” is far from “they are unrelated”—thanks! Orbicular, I'm afraid I have to reveal my near-ignorance of differential geometry: why not? I thought that, since the co-tangent bundle was locally free (or trivial, or whatever the term is), there was no harm in identifying, informally speaking, $T^*(M)$ with $M \cross T^*_p(M)$ “near $p$”. (Indeed, to me, that is the definition of local free-ness.) In that case, it is clear that $T_{(p, v^*)}(T^*(M)) \cong T_p(M) \cross T^*_p(M)$. What have I mis-understood? –  L Spice Feb 26 '10 at 17:56
Err, sorry, \cross is no good. I meant “… identifying, informally speaking, $T^*(M)$ with $M \times T^*_p(M)$ “near $p$” …. In that case, it is clear that $T_{(p, v^*)}(T^*(M)) \cong T_p(M) \times T^*_p(M)$.” –  L Spice Feb 26 '10 at 17:58
@L Spice: The point is that your splitting (or identification, as you call it) depends a priori on your choice of coordinates. So, why should that thing you wrote down make any sense on the manifold? This is unclear from your post (and - as you claim yourself - unclear to you). –  Orbicular Feb 26 '10 at 18:10