COURSE NOTES ON QUANTIZATION AND COHOMOLOGY, FALL 2006

JOHN BAEZ NOTES BY APOORVA KHARE FIGURES BY CHRISTINE DANTAS

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA RIVERSIDE, CA 92521, USA

Contents

1. Preface 3

2. Oct 3, 2006: Introduction 4 2.1. Perspective 4 2.2. (Higher) cohomology and physics 5 2.3. Classical dynamics vs. open string statics 6 2.4. The quantum case 7

3. Oct 10, 2006: Lagrangian Mechanics 9 3.1. Introduction to the Lagrangian approach 9 3.2. Deriving the Euler-Lagrange equations 10 3.3. Physics notation 11 3.4. Example: A particle in a potential 12 3.5. “Sneak preview” 12

4. Oct 17, 2006: From Lagrangian to Hamiltonian Dynamics 14 4.1. Recap 14 4.2. A matter of notation 14 4.3. Switching to the Hamiltonian approach 15 4.4. Energy 15 4.5. Hamilton’s Equations 16

5. Oct 24, 2006: Hamiltonian Mechanics and Symplectic Geometry 17 5.1. Recap 17 5.2. Some musical operators 17 5.3. The Hamiltonian vector field 18 5.4. Homework 19 5.5. Coordinate-free formulations 20

6. Oct 31, 2006: More on the canonical 1-form 22 6.1. Reconciling with the coordinate-based definition 22 6.2. Symplectic manifolds 23 6.3. Digression on five-body systems 24 6.4. The 1-form and action 25

7. Nov 07, 2006: The Extended Phase Space 26 7.1. Aside 26 7.2. Bringing in spacetime 26 7.3. Hamilton’s equations and the conservation of energy 27 7.4. Digression of the day: LIGO 29 7.5. A look back at the special case t(s) = s 29

8. Nov 14, 2006: From particles to strings and higher membranes 31 8.1. More derivations 32 8.2. Generalizing the Lagrangian 33

9. Nov 28, 2006: More on particles “vs.” membranes 34 9.1. (Functorial) construction of the multivelocity alternating tensor 34 9.2. Volume forms 35 9.3. The canonical p-form 37

10. Dec 05, 2006: Phases and connections on bundles 40 10.1. When connections come in 40 10.2. Phases and relative phases 41 10.3. Example: Rigid rotor 42 10.4. Integral cohomology and Max Planck 43

1. Preface

These are lecture notes taken at UC Riverside, in the Tuesday lectures of John Baez’s Quantum Gravity Seminar, Fall 2006. The notes were taken by Apoorva Khare. Figures were prepared by Christine Dantas based on handwritten notes by Derek Wise. You can find the most up-to-date version of all this material here:

          http://math.ucr.edu/home/baez/qg-fall2006/

Related notes on classical mechanics can be found here:

           http://math.ucr.edu/home/baez/classical/

For the continuation of this seminar in Winter 2007, see:

         http://math.ucr.edu/home/baez/qg-winter2007/

If you see typos or other problems with any of these notes, please let John

Baez know (baez@math.ucr.edu).

2. Oct 3, 2006: Introduction

2.1. Perspective. Our aim in this course is to try and categorify “everything in the universe”. More precisely, we want to replace sets by 1categories, categories by 2-categories, and so on.

Analogy from physics:

(0) Initially, people studied particles (or static ones).

(1) Then, they went on to study the “legal paths” that such particles could take in the ambient space (legal according to the laws of physics). This led to particle dynamics, which extended the previous study of particle statics.

Figure 1.

(1′) Relatively recently, people have reformulated particles in motion as strings, which leads to calling particle dynamics as string statics. (Here, a string is merely a map from an interval into the ambient space.)

(2) This makes us want to consider string dynamics now. (2′) But this should be the same as 2-brane statics.

And so on...

In general, we can reinterpret p-brane dynamics as p+1-brane statics, for any p ≥ 0.

Mathematics comes in: What happens in mathematical notation, is that we take a space X (a manifold, perhaps), and form P X = path space of X, defined as {γ : [t_{0}, t_{1}] → X : γ smooth},

Figure 2.

P P X = path space of P X, …

As a variation, we can consider paths (open strings) vs. loops (closed strings). Thus, P X is the configuration space of open strings,; the analogous space for closed springs is the loop space LX. Thus, LX = {γ : S^{1} → X}, where elements are free loops (i.e. not based at a point).

Remark 2.1. Note that LX ≠ {γ ∈ P X : γ(t_{0}) = γ(t_{1})} because these loops might have a “corner” at the basepoint, whereas LX was the space of smooth loops.

We can now similarly form LX, LLX, …, and it is these spaces, that are related to the cohomology of X. The first cohomology group of a topological space X can be defined as [X, U(1)] := {homotopy classes of maps : X → U(1)}. (Here, the unitary group U(1) = S^{1} ⊂ ℂ^{∗} is the set of unit modulus complex numbers.) So, we’re now asking how LX is related to [X, U(1)] = [X, S^{1}].

2.2. (Higher) cohomology and physics. Let’s see how maps S : L^{p}(X) → U(1) show up in the physics of closed p-branes, for various p. Note that by a particle, we’ll simply mean (below) a point in a (configuration) space. For example, the position of n objects in X denotes a particle in the configuration space X^{n} (for any n ∈ ℕ).

Thus, we have a “configuration space” X, whose points x ∈ X are possible positions for our particle (or the position of a general classical system).

We may decide that it is possible for two such particles to have the same position (or we may decide to ban such things) by considering X × X (or X × X \ Δ_{X} respectively).

Now, often X is a manifold, and we choose a 1-form F on X, called the force field. Thus, F : X → T^{∗}X, or F (x) ∈ T_{x}^{∗}X, or F is a section of T^{∗}X (each formulation containing more information than the previous one).

Definition 2.2.

(1) The work done on the particle as it moves along a path γ : [t_{0}, t_{1}] → X is defined to be W (γ) = ∫_{γ} F ∈ ℝ.

(2) A particle x ∈ X is in equilibrium if F (x) = 0; that is, for all infinitesimal displacements v ∈ T_{x}X, we have F (x)(v) = 0. (In D’Alembert’s terminology, the virtual work F (x)(v) vanishes.)

Often, the 1-form F comes from the differential of a 0-form V, i.e. F = −dV for some function V : X → ℝ called the potential energy. (Here, the negative sign is convention.)

Then at any critical point x of V, there is an equilibrium, meaning that dV (x) = 0. This is often (misleadingly?) called the principle of least energy, since often - but not always - x is a minimum.

The other advantage of having such an F, is that the Fundamental Theorem of Calculus (essentially) says that (recall the negative sign for F)

W (γ) = V (γ(t_{0})) − V (γ(t_{1}))

2.3. Classical dynamics vs. open string statics. Morally, these are the same concept, except that we use P X instead of X, or more precisely, we use P_{x_{0}→x_{1}}X := {γ ∈ P X : γ(t_{i}) = x_{i}, i = 0, 1} for some x_{0}, x_{1} ∈ X.

The idea now is that a particle now chooses an “optimal” path to “go” from x_{0} to x_{1}, i.e. dS(γ) = 0, where S : P_{x_{0}→x_{1}}X → ℝ is called the action. The equation dS(γ) = 0 is called the principle of least action.

The string picture: To think in terms of strings, we think of P_{x_{0}→x_{1}}X as a configuration space of an (open) string, and S as a potential.

Remark 2.3. Note that the ends of the string are fixed here; to avoid this, we may move to a bigger space, namely P X. Or for closed strings, use LX. Or for based strings or loops, use P_{∗}X, L_{∗}X etc.

For a fixed basepoint ∗ ∈ X, the space L_{∗}X is also called ΩX := {γ ∈ LX : γ(t_{0}) = ∗}.

We can now repeat this procedure by going to higher and higher (dimensional) branes. We’ll thus need actions on these spaces, for example for p = 2, we need some function of the type

S : P_{γ_{0}→γ_{1}}P_{x_{0}→x_{1}}X → ℝ

where all γ’s start at x_{0} and end at x_{1}, etc. Thus, we have

Figure 3. worldsheet Σ between two paths in P_{x_{0}→x_{1}}X

and we are trying to find the worldsheet(s) Σ ∈ P_{γ_{0}→γ_{1}}P_{x_{0}→x_{1}}X, so that dS(Σ) = 0. (Note that P_{γ_{0}→γ_{1}}P_{x_{0}→x_{1}}X are just maps : [0, 1] × [0, 1] → X.)

D-branes: (The D stands for Dirichlet.) This means that we study branes with added boundary conditions. For insatnce, the Σ above fixes all initial points to be x_{0}, and all final points to be x_{1}. Thus, the boundaries are coupled to 0-submanifolds x_{0}, x_{1}.

In general, we can use submanifolds of dimension d, and look at p-branes coupled wth d-branes for boundary conditions.

2.4. The quantum case. The question: ”How does the particle know in advance which path to take, before it has reached the end?”, led to the study of the quantum versions of all these ideas. A possible explanation was first given by Richard Feynman, who said that the quantum dynamics of particles is also governed by S : P_{x_{0}→x_{1}}X → ℝ, but in a new way:

Instead of choosing the path γ with dS(γ) = 0, it chooses all paths with certain “amplitudes”, given by

e^{iS/ℏ} ∈ U(1)

where ℏ is Planck’s constant, in units of action. (We’ll often choose units with ℏ = 1.) This is how U(1) gets into the picture in physics - through quantization.

In retrospect, we can do classical dynamics of particles using not S, but e^{iS} = A, since dA(γ) = 0 makes sense (given A : P_{x_{0}→x_{1}}X → U(1), we can define a complex-valued form dA on P_{x_{0}→x_{1}}X), and moreover, dA(γ) = 0 ⇔ dS(γ) = 0.

Similarly, we can do classical statics of a particle, using not V : X → ℝ, but A = e^{iV} : X → U(1).

So instead of using dA = 0 or integrating over a single path, we integrate over all paths, weighted by the U(1)-valued function In short, we have statics of particles ⟷ A = e^{iV} : X → U(1), particle dynamics = statics of strings ⟷ A = e^{iS} : P X → U(1), …, statics of p + 1-branes ⟷ A : P^{p}X → U(1).

If we restrict to Ω^{p}X ⊂ P^{p}X, we get A : Ω^{p}X → U(1), and in fact, [Ω^{p}X, U(1)] ≅ H^{p+1}(X, ℤ)! (Well, not exactly isomorphic, but close to it at any rate.)

This is how (higher) cohomology comes into the picture involving strings and higher branes!

3. Oct 10, 2006: Lagrangian Mechanics

Here’s some homework to do, first of all. Work out the “statics of a spring in imaginary time”! The problem (and some notes on it) can be found at

http://www.math.ucr.edu/home/baez/classical/.

Think of a rock thrown away from the earth. It chooses a parabolic path to come back to the ground, because this path minimizes the action.

A static (or hung) spring is traditionally of zero length because this minimizes energy.

Figure 4. parabolic shapes - of rock trajectory and hung spring

In both case, the energy or motion or static state is affected by (or counters) gravity.

Moreover, the paths in these two cases are upside-down relative to one another because there is a sign change, which ultimately comes from i^{2} = −1; hence, the notion of a spring in imaginary time!

To do this homework, one needs to know what we talk about today: Lagrangian mechanics.

3.1. Introduction to the Lagrangian approach. Suppose X is a (finitedimensional) manifold, called the configuration space. We want a law of physics (which we will call the Euler-Lagrange equation) satisfied by paths γ : [t_{0}, t_{1}] → X.

To get this, we define P_{x_{0}→x_{1}}X := {γ : [t_{0}, t_{1}] → X, γ(t_{i}) = x_{i}} as we did last time. This is a (smooth) infinite-dimensional manifold in its own right. (Do we also want it to be a Frechet manifold, i.e. locally homeomorphic to Frechet space?)

Moreover, we also choose a smooth function S : P_{x_{0}→x_{1}}X → ℝ, called the action.

The Euler-Lagrange equation then (abstractly) says that

dS(γ) = 0

where dS ∈ Ω^{1}(P_{x_{0}→x_{1}}X). So, a particle follows a path that is the critical point of the action. Slightly more concretely, dS(γ) ∈ T_{γ}^{∗}(P_{x_{0}→x_{1}}X), and so we’re saying that

dS(γ)(δγ) = 0 ∀δγ ∈ T_{γ}(P_{x_{0}→x_{1}}X)

(In physics notation, δγ is called the “(infinitesimal) variation” in γ.)

Figure 5. the plane as a t vs. X plot; how δγ is a path

Thus, if we think of the point γ ∈ P_{x_{0}→x_{1}}X as a path γ, then the tangent vector δγ can be thought of as a path in the tangent bundle:

δγ(t) ∈ T_{γ(t)}X ⊂ T X

3.2. Deriving the Euler-Lagrange equations. In physics, we often have actions of the form

S(γ) = ∫_{t_{0}}^{t_{1}} L(γ(t), γ̇(t)) dt

where γ(t) stands for the position, γ̇(t) denotes the velocity, and L is the Lagrangian. Thus, L : T X → ℝ is a smooth function, where T X is the space of position-velocity pairs.

Remark 3.1. The path γ in X might go through several different coordinate charts/patches. However, we then break it up into small paths, each of which is in only one such chart. Thus, locally we work over a chart in X, so the situation is homeomorphic (diffeomorphic?) to working in ℝ^{n}, and the tangent bundle T X is then locally homeomorphic to T ℝ^{n} ≅ ℝ^{n} ⊕ ℝ^{n}.

Thus, our Lagrangian is written as L({x_{i}}, {y_{i}}) ∈ ℝ, where {x_{i}} are local coordinates for position vectors, and {y_{i}} are local coordinates on the tangent space.

We now derive the Euler-Lagrange equations in the above situation. Given S as above, what does dS(γ) = 0 mean? It means that for all δγ ∈ T_{γ}P_{x_{0}→x_{1}}X, we have 0 = dS(γ)(δγ). We now expand this, and use Einstein summation notation henceforth.

0 = dS(γ)(δγ) = ∫_{t_{0}}^{t_{1}}(∇L · δγ)(t) dt = ∫_{t_{0}}^{t_{1}} (\frac{∂L}{∂x^{i}} δγ^{i}(t) + \frac{∂L}{∂y^{i}} δγ̇^{i}(t)) dt

But δγ̇^{i}(t) = \frac{d}{dt}δγ^{i}(t). We now use integration by parts on the second term, to get

0 = dS(γ)(δγ) = ∫_{t_{0}}^{t_{1}} (\frac{∂L}{∂x^{i}} δγ^{i}(t) + \frac{∂L}{∂y^{i}} \frac{d}{dt}δγ^{i}(t)) dt
= ∫_{t_{0}}^{t_{1}} (\frac{∂L}{∂x^{i}} δγ^{i}(t) − (\frac{d}{dt} \frac{∂L}{∂y^{i}} ) δγ^{i}(t)) dt

This works because the boundary terms in our integration vanish here, because the sum telescopes across coordinate charts - and at the “global” endpoints, (the picture shows that) δγ(t_{0}) = δγ(t_{1}) = 0.

Continuing with the calculations,

∫_{t_{0}}^{t_{1}} (\frac{∂L}{∂x^{i}} − \frac{d}{dt} \frac{∂L}{∂y^{i}} ) δγ^{i}(t) = 0 ∀δγ^{i}

But if we have smooth functions φ_{i}, so that ∫_{t_{0}}^{t_{1}} φ_{i}f^{i} = 0 for all smooth functions f^{i}, then φ_{i} ≡ 0 for all i. Thus, the previous equation gives us the Euler-Lagrange equations:

\frac{∂L}{∂x^{i}} = \frac{d}{dt} \frac{∂L}{∂y^{i}} ∀i  (EL)

or, more pedantically,

\frac{∂}{∂x^{i}} L(γ(t), γ̇(t)) = \frac{d}{dt} \frac{∂}{∂y^{i}} L(γ(t), γ̇(t)) ∀i

3.3. Physics notation. Physicists don’t write {x^{i}, y^{i}} as coordinates on T X; they use {q^{i}, q̇^{i}}, even though q̇^{i} here is not the time derivative of anything. They also write q : [t_{0}, t_{1}] → X instead of γ : [t_{0}, t_{1}] → X, which makes the notation q̇^{i} ambiguous. (And no one cares!)

Therefore the physicists’ version of the Euler-Lagrange equations looks like

\frac{∂L}{∂q^{i}} = \frac{d}{dt} \frac{∂L}{∂q̇^{i}}

3.4. Example: A particle in a potential. Let X = ℝ^{n} and consider L : T X → ℝ given by

L(q, q̇) = \frac{1}{2}m||q̇||^{2} − V (q)

where q ∈ X, q̇ ∈ T_{q}X, and the terms on the right-hand side are the kinetic and potential energies. Thus, m > 0 is the mass, and V : X → ℝ is the potential.

This Lagrantian is “weird”; it’s not the kinetic plus potential energies (which is the total energy), but rather, the kinetic energy (i.e. how much is happening) minus potential energy (how much could - potentially! - happen, but is not happening). This we call the total “happening-ness” ¨⌣. (The reason, of course, is that the Lagrangian is not the energy; that’s what comes from the Hamiltonian approach!)

Nature usually tries to minimize the integral of this over time.

Let us now carry out the computations. The Euler-Lagrange equations now say:

−\frac{∂V}{∂q^{i}} = \frac{∂L}{∂q^{i}} =^{EL} \frac{d}{dt} \frac{∂L}{∂q̇^{i}} = \frac{d}{dt}mq̇_{i} = mq̈_{i}

or, in Newton’s words, F = ma, where F_{i} = −\frac{∂V}{∂q_{i}} is the force, and a_{i} = q̈_{i} is the acceleration.

Definition 3.2. For any (smooth) Lagrangian function L : T X → ℝ, we define the momentum to be p_{i} := \frac{∂L}{∂q̇^{i}}, and the force to be F_{i} := \frac{∂L}{∂q^{i}}.

The Euler-Lagrange equations now say that \frac{d}{dt}p_{i} = F_{i}.

3.5. “Sneak preview”. To relate this to cohomology, let’s step back for a moment: we have derived classical mechanics from the principle of least action, based on S : P_{x_{0}→x_{1}}X → ℝ, or more generally, S : P X → ℝ.

It would be nice (at least, from the point of view of de Rham cohomology), if there were a 1-form α ∈ Ω^{1}(X) so that S(γ) = ∫_{γ} α.

Also note that the action S in our example is not of this form (because paths can be reparameterized - whereby the integral above remains the same - but this action is not independent of the parametrization). For instance, a ball rolling along at constant speed, does not possess the same action function as a ball going one way, then reversing, and then going back again the “correct way” to the end.

In other words, S(γ) depends here on the parametrization of γ. So, we need to write S(γ) as the integral of some 1-form over a path (that we cook up from γ) in some other space.

Question. What is this other space? Hint. For any manifold M, the cotangent bundle T^{∗}M has a God-given 1-form on it, called the canonical 1-form.

Answer. We’ll use this to get the job done, but with M = X × ℝ. In other words, M = X ×ℝ stands for space-time, so to speak, and we need the extra dimension to get the reparametrization-invariance.

This new manifold M is also known as the extended configuration space. Thus, we now reparametrize both space and time in terms of some other, arbitrary parametrization, and this helps us achieve S = ∫_{γ} α.

4. Oct 17, 2006: From Lagrangian to Hamiltonian Dynamics

4.1. Recap. We want a description of classical mechanics where the action is the integral of some 1-form along a path. This path will lie not in X = configuration space ∋ position, nor in T X ∋ (position, velocity), nor in T^{∗}X = phase space ∋ (position, momentum), but in T^{∗}(X × ℝ) = extended phase space ∋ (position, momentum, time, energy).

(Thus, energy : time :: momentum : position.)

To get there, we first study the phase space T^{∗}X and energy (also called the Hamiltonian). We start with a Lagrangian L : T X → ℝ, and get the Euler-Lagrange equations

\frac{d}{dt} \frac{∂L}{∂q̇^{i}} = \frac{∂L}{∂q^{i}}

where \frac{∂L}{∂q̇^{i}} = p_{i} is the momentum, and \frac{∂L}{∂q^{i}} = F_{i} is the force.

Notation: We use subscripts for cotangent vectors, and superscripts for tangent vectors.

4.2. A matter of notation. The first question we ask, is: Why is momentum a cotangent vector?

Figure 6. X vs. T X plane (e.g. X = ℝ)

Here, \frac{∂L}{∂q̇^{i}} describes the derivative of L in the vertical direction (i.e. along the fiber, or tangent space). In other words, {vertical vectors} = T_{(q,q̇)}T X.

Moreover, we have that the set of vertical vectors is precisely the kernel of the map dπ : T_{(q,q̇)}T_{q}X → T_{q}X is the differential of the projection π : T X → X ((q, q̇) ↦ q).

But for any (real finite-dimensional) vector space V, we have T_{v}V ≅ V for all v ∈ V. So T_{(q,q̇)}T_{q}X ≅ T_{q}X, whence momentum is really the derivative of L : T X → ℝ, but only in the “vertical” directions.

Now, the derivative of L is the 1-form dL(q, q̇) : T_{(q,q̇)}T X → ℝ, and the vertical vectors in the domain are just T_{q}X. So the momentum is a linear map p : T_{q}X → ℝ, i.e. p ∈ T_{q}^{∗}X is a cotangent vector.

4.3. Switching to the Hamiltonian approach. We will now switch from the “Lagrangian approach”, based on (q, q̇) : T X → ℝ, to the “Hamiltonian approach”, based on (q, p) ∈ T^{∗}X = phase space. We’ll do this using the Legendre transform λ : T M → T^{∗}M, that takes (q, q̇) ↦ (q, p), with p_{i} = \frac{∂L}{∂q̇^{i}}. (Here, λ is defined using L.)

From now on, assume that L is strongly regular, i.e. λ : T M → T^{∗}M is a diffeomorphism.

Example: We once again look at our familiar example of a particle moving on a Riemannian manifold (X, g), in a potential V. The Lagrangian is

L(q, q̇) = \frac{m}{2}g(q̇, q̇) − V (q)

where m > 0 and V : X → ℝ. (Thus, the components on the righthand side are the kinetic and potential energies.) In what follows, not that g(q̇, q̇) = g_{ij}q̇^{i}q̇^{j}, whence p_{i} = \frac{∂L}{∂q̇^{i}} = mg_{ij}q̇^{j}. For this, we have used the coordinate-dependent form of the metric:

g(q̇, −) = g_{ij}q̇^{j} ∈ T_{q}^{∗}X

So λ is a diffeomorphism, since g is nondegenerate (i.e. q̇ ↦ g(q̇, −) is a bijection). In other words, L is strongly regular, as claimed. □

4.4. Energy. To translate the Euler-Lagrange equations into equations satisfied by q, p, we need the concept of energy.

Theorem 4.1 (Conservation of Energy). Given any Lagrangian L : T X → ℝ, and q : [t_{0}, t_{1}] → X satisfying the Euler-Lagrange equations, the function E(q(t), q̇(t)) is independent of t, where q̇(t) = \frac{d}{dt}q(t), and E : T X → ℝ is given by E(q, q̇) = p_{i}q̇^{i} − L(q, q̇) = \frac{∂L}{∂q̇^{i}} q̇^{i} − L(q, q̇).

Proof. We compute:

\frac{d}{dt}E(q, q̇) = \frac{d}{dt} (\frac{∂L}{∂q̇^{i}} ) q̇^{i} + \frac{∂L}{∂q̇^{i}} \frac{d}{dt}q̇^{i} − \frac{∂L}{∂q^{i}} \frac{d}{dt}q^{i} − \frac{∂L}{∂q̇^{i}} \frac{d}{dt}q̇^{i}
= \frac{d}{dt} (\frac{∂L}{∂q̇^{i}} ) q̇^{i} − \frac{∂L}{∂q^{i}} \frac{d}{dt}q^{i}

since two terms cancel. But now, \frac{d}{dt}q^{i} = q̇^{i}, and we now apply the Euler-Lagrange equations to make the entire expression vanish, as claimed. □

Back to our example: For a particle on (X, g) as above, with L(q, q̇) = \frac{m}{2}g_{ij}q̇^{i}q̇^{j} − V (q), we have

E(q, q̇) = p_{i}q̇^{i} − L(q, q̇) = mg_{ij}q̇^{j}q̇^{i} − \frac{m}{2}g_{ij}q̇^{i}q̇^{j} + V (q) = \frac{m}{2}g_{ij}q̇^{i}q̇^{j} + V (q)

So the energy is indeed the sum of the kinetic and potential energies, just as the Lagrangian was their difference.

4.5. Hamilton’s Equations. Using our diffeomorphism λ : T X → T^{∗}X, we can define the Hamiltonian to be

H = E ∘ λ^{−1} : T^{∗}X → ℝ

In other words, H(q, p) = E(q, q̇), since λ(q, q̇) = (q, p). Let us now figure out Hamilton’s equations, that describe the time evolution of (q, p) = λ(q, q̇), given that (q, q̇) satisfy the Euler-Lagrange equations. Here’s how: compute dH in two different ways.

Method 1. H : T^{∗}X → ℝ, and T^{∗}X has local coordinates (q^{i}, p_{i}) coming from local coordinates q^{i} on X. So we get

dH = \frac{∂H}{∂q^{i}} dq^{i} + \frac{∂H}{∂p_{i}} dp_{i}

Method 2. But we can also compute dH using the coordinates (q^{i}, q̇^{i}). These are really local coordinates on T X coming from local coordinates q^{i} on X, but they become coordinates on the cotangent bundle, using λ : T X →^{∼} T^{∗}X. Thus, we get

dH = d(p_{i}q̇^{i} − L(q, q̇)) = (dp_{i})q̇^{i} + p_{i}dq̇^{i} − \frac{∂L}{∂q^{i}} dq^{i} − \frac{∂L}{∂q̇^{i}} dq̇^{i}

The second and fourth terms cancel by definition of momentum, and we now compare with the expression from the previous method, equating the coefficients for dp_{i}, dq̇^{i}. This gives us

q̇^{i} = \frac{∂H}{∂p_{i}} , −\frac{∂H}{∂q^{i}} = \frac{∂L}{∂q^{i}}

But the last term above equals \frac{d}{dt}p_{i}, by the Euler-Lagrange equations. Therefore, given a path q satisfying the Euler-Lagrange equations, we get Hamilton’s equations:

q̇^{i} = \frac{∂H}{∂p_{i}} ,  ṗ_{i} = −\frac{∂H}{∂q^{i}}

5. Oct 24, 2006: Hamiltonian Mechanics and Symplectic Geometry

5.1. Recap. We have seen that any Lagrangian L : T X → ℝ gives Euler-Lagrange equations

\frac{d}{dt} \frac{∂L}{∂q̇^{i}} = \frac{∂L}{∂q^{i}}

describing a flow on T X, i.e. describing evolution (at least, such a global solution exists if L is well-behaved). We also get a Legendre transform λ : T M → T^{∗}M, and if L is strongly regular (i.e. λ is a diffeomorphism), then we get a flow on T^{∗}M, describing the time evolution of position q and momentum p, which satisfies Hamilton’s equations

\frac{d}{dt}q^{i} = \frac{∂H}{∂p_{i}} , \frac{d}{dt}p_{i} = −\frac{∂H}{∂q^{i}}

where H : T^{∗}M → ℝ is the Hamiltonian, given by

H(q, p) = p_{i}q̇^{i} − L(q, q̇) = p_{i}q̇^{i} − L(λ^{−1}(q, p))

Remark 5.1. Note that the Euler-Lagrange equations were “one” secondorder equation, whereas the Hamiltonian equations are “two” first-order equations. This is the same, because there is a standard way to make an nth order equation into n first-order equations, by introducing auxiliary (“intermediate”) variables.

Also recall our indexing notation: subscripts are used for cotangent vectors, and superscripts for tangent vectors.

5.2. Some musical operators. Suppose (X, g) is a Riemannian manifold, and V : X → ℝ is the “potential energy”. Then the (nonrelativistic) Hamiltonian for a particle of mass m > 0 is the sum of the kinetic and potential energies, namely,

H(q, p) = \frac{|p|^{2}}{2m} + V (q)

Here, Hamilton’s equations say

\frac{d}{dt}q^{i} = \frac{∂H}{∂p_{i}} , q̇^{i} = \frac{p^{i}}{m}

Question. We only saw p_{i}’s above. What is p^{i}? Answer. p^{i} = g_{ij}p_{j}, where g is the Riemannian metric. Reason. Force must be a cotangent vector, because the integral of force is work, which indicates that we are essentially integrating a 1-form on a path to get a number. But 1-forms come from the cotangent bundle.

Now, the time derivative of the momentum is force, so momentum should also be a cotangent vector! But momentum is related to velocity (i.e. d(position)), so this appears to be a tangent vector! What’s wrong?

Solution. Absolutely nothing is wrong! However, this phenomenon (and the p_{i} ↔ p^{i} issue) achieves consistency only when we introduce an isomorphism that we need, between T_{q}X and T_{q}^{∗}X. This is the lowering index operator (that takes p^{i} to p_{i}), or the flat operator

♭ : T_{q}X →^{∼} T_{q}^{∗}X ∀q ∈ X

And a Riemannian metric gives it to us! Since the metric is nondegenerate, hence we get an isomorphism ♭, given by

♭(v) = g_{q}(v, −), v ∈ T_{q}X

We use this isomorphism - or more precisely, its inverse map, called the sharp operator ♯ (“natural”ly (pun intended) ¨⌣!). This turns the cotangent vector p into a tangent vector ♭^{−1}(p) = ♯(p) = p^{♯}.

Thus, if g_{q}^{ij} is the inverse of (g_{ij})_{q} = g_{q}, then

p^{♯} = (p^{i})_{i} = (g^{ij}p_{j})_{i}

This equation, which had only to do with the kinetic energy, is “boring”; it is the other one (among Hamilton’s equations), which deals only with the potential energy, that is interesting:

\frac{d}{dt}p_{i} = −\frac{∂H}{∂q^{i}} ⇒ ṗ_{i} = −\frac{∂V}{∂q^{i}} ⇒ ma = F

5.3. The Hamiltonian vector field. Let us now seek a coordinate-free formulation of Hamilton’s equations. These give a vector field v_{H} on T^{∗}X, describing how (q(t), p(t)) moves around. v_{H} is called the Hamiltonian vector field:

\frac{d}{dt}(q(t), p(t)) = v_{H}(q(t), p(t)) ∈ T (T^{∗}X)

so M = T^{∗}X always. Moreover, v_{H} = d(q, p), so

v_{H} = \frac{dq^{i}}{dt} \frac{∂}{∂q^{i}} + \frac{dp_{i}}{dt} \frac{∂}{∂p_{i}}

Here, \frac{∂}{∂q^{i}} and \frac{∂}{∂p_{i}} are a basis of vector fields on some open set in T^{∗}X, coming from local coordinates (q^{i}, p_{i}) on T^{∗}X.

But now, Hamilton’s equations allow us to rewrite v_{H} as

v_{H} = \frac{∂H}{∂p_{i}} \frac{∂}{∂q^{i}} − \frac{∂H}{∂q^{i}} \frac{∂}{∂p_{i}}

Example: Say X = ℝ^{n}, so T^{∗}X ≅ ℝ^{2n}. Then, as the picture below shows, if we identify these cotangent vector fields dH, v_{H} with tangent vectors using the metric, then they are orthogonal at each point. This is because

dH = (\frac{∂H}{∂q^{i}} , \frac{∂H}{∂p_{i}} ) , v_{H} = (\frac{∂H}{∂p_{i}} , −\frac{∂H}{∂q^{i}} )

(so one can easily verify that their dot product vanishes). Moreover, the gradient dH is always normal to the level curves, so (at least if n = 1) we get that v_{H} must lie along the vector field!

Figure 7. level curves for H, and how dH ⊥ v_{H} on a p-q plot

Furthermore, we can recover v_{H} from the flow, and H from v_{H} (upto a scalar), on any connected component of the manifold.

Back to our example, in the graph above, dH is really a 1-form, but in ℝ^{2}, we can use the metric to turn it into a vector field ∇H, and rotate this 90^{∘} clockwise to get v_{H}. So, any solution of Hamilton’s equations moves along the level curves of H, and conservation of energy follows automatically!

But back to our original motivation: to seek a coordinate-free description of Hamilton’s equations. Recall from above, that one uses Hamilton’s equations to rewrite v_{H}. Thus, we seek a coordinate-free description of how to turn the 1-form dH into the vector field v_{H}.

One way to turn 1-forms into vector fields is to use the Riemannian metric: use ♭, ♯. But, ♯(dH) = ∇H is not parallel (proportional), but perpendicular (normal) to the level curves of H!

So, instead of a metric, we would need to use an anti-symmetric nondegenerate bilinear form ω on M = T^{∗}X.

The good news is that every cotangent bundle M = T^{∗}X is automatically equipped with such an ω : T M ×T M → ℝ. This is called a symplectic form.

5.4. Homework. In our example, we have ω = dq^{i}∧dp_{i} (summing over i, of course). Show that with this 2-form on M = T^{∗}X, we have ω(v_{H}, −) = dH.

Proof. This is an exercise in wedge calculations. As a warmup, we note some random facts about wedges. For instance, f ∧ dx = f dx, and

dx ∧ dy ∧ … (f \frac{∂}{∂x}, −) = f dy ∧ …
dx ∧ dy ∧ … (f \frac{∂}{∂y} , −) = −f dx ∧ …
dp_{i}(\frac{∂}{∂q^{j}}) = dq^{j}(\frac{∂}{∂p_{i}}) = 0 ∀i, j
dp_{i}(\frac{∂}{∂p_{j}}) = dq^{j}(\frac{∂}{∂q^{i}}) = δ_{i}^{j} ∀i, j

where δ_{i}^{j} is the Kronecker delta. We now show the result. Given H : T^{∗}X = M → ℝ, we have

v_{H} = \frac{∂H}{∂p_{i}} \frac{∂}{∂q^{i}} − \frac{∂H}{∂q^{i}} \frac{∂}{∂p_{i}} ∈ T M

We also have dH ∈ T^{∗}M, and ω = dq^{k} ∧ dp_{k}. We now compute, using the “warmup facts” above:

ω(v_{H}, −) = \frac{∂H}{∂p_{i}} δ_{i}^{k}dp_{k} − dq^{k} ∧ \frac{∂H}{∂q^{i}} dp_{k}(\frac{∂}{∂p_{i}}) = \frac{∂H}{∂p_{i}} dp_{i} + δ_{k}^{i} \frac{∂H}{∂q^{i}} dq^{k}
= \frac{∂H}{∂p_{i}} dp_{i} + \frac{∂H}{∂q^{i}} dq^{i} = dH
□

Another way of stating this result, is to define ♭ : T M → T^{∗}M, given by v ↦ ω(v, −). We once again say that ω is nondegenerate if ♭ is an isomorphism. If this happens, then the inverse ♯ of ♭ takes T^{∗}M to T M, and we can set v_{H} = ♯(dH) now.

5.5. Coordinate-free formulations. Getting back to our original problem: we wanted to write Hamilton’s equations in a coordinate-free way. To do this, we needed to write v_{H} in a coordinate-free way. By the above homework problem, we only need to write ω in a coordinate-free way! And we do this now.

This is, moreover, good from another point of view: the definition of ω then stays the same under more symmetries - not just only those that fix a particular choice of coordinates.

So, how do we write this? In fact, ω = −dα, where α = p_{i} dq^{i} is a 1-form on T^{∗}X, called the canonical 1-form, which can be defined without using coordinates! (We show this next time.) First check this:

dα = d(p_{i} dq^{i}) = dp_{i} ∧ dq^{i} + p_{i}(−1)^{0}d(dq^{i}) = −ω

since p_{i} is a 0-form for all i.

Now, how do we define α without using coordinates? Keep in mind that α likes to eat tangent vectors to M = T^{∗}X, e.g. v ∈ T_{(q,p)}(T_{q}^{∗}X), and spit out numbers.

Well, we have the projection π_{1} from T^{∗}X, taking (q, p) to q. Differentiate π_{1} to get dπ_{1} : T (T^{∗}X) → T X. Thus, we can form dπ_{1}(v) ∈ T_{q}X (which is the analogue of dq^{i}). But we also have p ∈ T_{q}^{∗}X, which is the second (component of the first) projection π′_{2} : T (T^{∗}X) → T^{∗}X. So we get

α(v) = p_{i} dq^{i}(v) = p(dπ_{1}(v)) ∈ ℝ

Conclusion: Therefore, in coordinate-free terms,

α(v) = π′_{2}(v)(dπ_{1}(v)), ω = −dα

Next time we will see how this agrees, given a choice of coordinates p_{i}, q^{i}, with the usual formula α = p_{i} dq^{i} (note that some people write this as p_{i} ∧ dq^{i} too).

6. Oct 31, 2006: More on the canonical 1-form

Let us draw a picture to illustrate the coordinate-free definition of α from last class. Note that in it, q ∈ X, p ∈ T_{q}^{∗}X, v ∈ T_{(q,p)}T_{q}^{∗}X, and dπ_{1}(v) is the “shadow” or the projection, of v.

Figure 8. X vs. T_{q}^{∗}X plot, to show how dπ_{1}(v) is the “shadow”

This picture doesn’t make it clear, though, that the vertical fibers are dual to tangent vectors.

6.1. Reconciling with the coordinate-based definition.

Theorem 6.1. If q^{i} are any local coordinates on X, and (q^{i}, p_{i}) the corresponding local coordinates on T^{∗}X, then α = p_{i} dq^{i}.

Proof. Choose any v ∈ T_{(q,p)}T_{q}^{∗}X. Then

v = ∑_{i} a^{i} \frac{∂}{∂q^{i}} + b_{i} \frac{∂}{∂p_{i}}

for some choice of scalars a^{i}, b_{i}, say. Evaluating at p_{i} dq^{i}, we have

p_{i} dq^{i}(v) = p_{i} dq^{i}(a^{k} \frac{∂}{∂q^{k}} + b_{k} \frac{∂}{∂p_{k}})

Since dq^{i} kills all \frac{∂}{∂p_{k}} and almost all \frac{∂}{∂q_{k}} ’s too, we just end up with a^{i}p_{i}.

On the other hand, α(v) = π′_{2}(v)(dπ_{1}(v)) = p(a^{j} \frac{∂}{∂q^{j}}) because dπ_{1} removes the vertical components \frac{∂}{∂p_{j}}. Now, recall that in α, the q^{i} is already a coordinate on M = T^{∗}X. But the dq^{i}’s form coordinates on T^{∗}X because q^{i}’s are coordinates on X itself! So there’s some sort of ambiguity here, but this is resolved if we remember that dπ_{1} takes one of these q’s to the other.

In other words, in coordinates, it just throws away the vertical part, so we really are fine, in using q^{i}’s in both instances.

This means that p = p_{i} dq^{i}, whence α(v) = p_{i} dq^{i}(a^{j} \frac{∂}{∂q^{j}}) = a^{i}p_{i} once again.

Thus, both definitions agree on every v, whence they are equal. □

6.2. Symplectic manifolds. There are various related ways in which α shows up in classical mechanics. We’ve now seen that dα = dp_{i} ∧ dq^{i} = −ω, where ω is a symplectic structure on M = T^{∗}X.

Note that in general, more than just cotangent spaces have symplectic structures; these are precisely what allow us to treat M as a “phase space” (space of states of our system). That is, we have (position, momentum). Such M’s are called symplectic manifolds, and it essentially means that we can do classical mechanics on them!

Definition 6.2. A symplectic structure on a (finite-dimensional) manifold M (which is then called a symplectic manifold) is a 2-form ω on M, such that

(1) ω is closed (i.e. dω = 0).

(2) ω is nondegenerate (i.e. the map ♭ : T_{m}M → T_{m}^{∗}M, given by v ↦ ω(v, −), is one-to-one - and since M is finite-dimensional, onto as well).

Note that this ω is skew-symmetric.

Remark 6.3. Every M = T^{∗}X has such a form on it. In fact, if ω = −dα as earlier on M = T^{∗}X, then we claim that ω is one such. For

dω = d(dq^{i} ∧ dp_{i}) = 0

by the Leibnitz rule, so that ω is closed. Moreover, to see that ω is nondegenerate, we need to show that ♭ is one-one. So suppose that ♭ kills a^{i} \frac{∂}{∂q^{i}} + b_{i} \frac{∂}{∂p_{i}}. Using our homework problem and the warmup random facts

mentioned therein, we evaluate both sides at \overline{a^{i}} \frac{∂}{∂p_{i}} − \overline{b_{i}} \frac{∂}{∂q^{i}} (write a^{i}, b_{i} ∈ ℂ). This gives us that

∑_{i} |a^{i}|^{2} + ∑_{i} |b_{i}|^{2} = 0

so that a^{i} = b_{i} = 0 ∀i, as required. □

How does ω on M allow us to do classical (Hamiltonian) mechanics?

As we said above, roughly speaking, the nondegeneracy of ω lets us define a Hamiltonian vector field v_{H} (given H : M → ℝ) on M, by: v_{H} = ♯(dH) = ♭^{−1}(dH).

This now lets us write Hamilton’s equations, describing the time evolution of states.

Definition 6.4. Given x(t) ∈ M for t ∈ [t_{0}, t_{1}], we say that it satisfies Hamilton’s equations, if \frac{dx(t)}{dt} = v_{H}(x(t)).

What does the closedness of ω do for us? Well, under “mild assumptions”, we get a flow on M.

Definition 6.5. A flow on M is a smooth(?) map F : ℝ×M → M, denoted by (t, x) ↦ F_{t}(x) ∈ M, so that

(1) F_{t}F_{s} = F_{t+s} for all t, s ∈ ℝ.

(2) F_{0} is the identity map.

In particular, F_{−t} = F_{t}^{−1}, whence F is a group homomorphism from ℝ to the diffeomorphism group of M.

How do we get this flow? It describes time evolution, and satisfies Hamilton’s equations:

\frac{d}{dt}F_{t}(x) = v_{H}(F_{t}(x))

So: use this equation to define the flow as the integral curve (solution) to a differential equation. Then the closedness of ω can be used to show that: Given any H, the corresponding flow F_{t} preserves ω.

Formalism: How does F_{t} act on ω? In general, given a “good” (or smooth) φ : M → N, we get dφ : T M → T N, and one can now define the pullback map φ^{∗} : Ω^{p}(N) → Ω^{p}(M) for any p ≥ 0, by

(φ^{∗}ω)_{m}(v_{1}, …, v_{p}) := ω_{φ(m)}(dφ(v_{1}), …, dφ(v_{n}))

where v_{i} ∈ T_{m}M and dφ(v_{i}) ∈ T_{φ(m)}N for all i.

6.3. Digression on five-body systems. The “mild assumptions” that allow us to derive a flow on M, rule out some physics “real-life” situations. For instance, the n-body problem: describe the time evolution of a system of n particles that interact gravitationally. For n = 5, it has been shown that they split off into two pairs and a fifth

[figure: two pairs orbiting each other, fifth particle orbits around]

where the paired objects “revolve” around each other, and the fifth particle orbits around both systems. As time passes, the distance d(t) between the two paired systems increases, and the energy (Hamiltonian) involves 1/d(t)^{2}, so the mild assumptions don’t work here! Moreover, d(t) → ∞ as t goes to some finite time point, and as d(t) increases, the potential energy decreases, so the kinetic energy increases. That is, the fifth particle travels faster on its (larger and larger) orbit. Eventually, in finite time, its speed becomes infinite!

The problem is that using point particles allows us to get infinity as a limit in finite time. For instance, if both particles were points, they could “superimpose”, resulting in the realisation of an infinite amount of potential energy.

In reality, such a scenario only occurs for black holes. Thus, there are problems in such situations, which is why we need the theory of relativity etc.

6.4. The 1-form and action. There is another (related) way in which α ∈ Ω^{1}(T^{∗}X) shows up in physics. We can use it to describe the action of a path q : [t_{0}, t_{1}] → X, our configuration space, as follows: We get (q, q̇) : [t_{0}, t_{1}] → T X, whence γ : [t_{0}, t_{1}] → T^{∗}X, our phase space. This is given by t ↦ λ(q(t), q̇(t)) = (q(t), p(t)), if we choose a Lagrangian L : T X → ℝ, and use it to define the Legendre transform λ : T X → T^{∗}X.

In this case, set H = p_{i}q̇^{i} − L, α = p_{i} dq^{i}. Then the action of our path q is

S(q) = ∫_{t_{0}}^{t_{1}} L(q, q̇) dt = ∫_{t_{0}}^{t_{1}} p_{i}(t)q̇^{i}(t) dt − ∫_{t_{0}}^{t_{1}} H(q(t), p(t)) dt

But the first term is the integral of p_{i}(t)\frac{dq^{i}(t)}{dt}dt = α, so we conclude that

S(q) = ∫_{γ} α − ∫_{t_{0}}^{t_{1}} H(q(t), p(t)) dt

(Thus, the integral along a path is “almost” the action of that path; so, we’re amost doing Lagrangian mechanics.) In particular, we see that the Principle of Least Action

δ(S(γ)) = 0

is equivalent to the principle δ(∫_{γ} α) = 0, as long as we let γ vary only over the paths which conserve energy! (Because we would need ∫_{t_{0}}^{t_{1}} H(q, p) to be just a number.) Therefore we only consider the paths in the set

M_{E}^{1} := {γ : [t_{0}, t_{1}] → T^{∗}X : q(t_{i}) = q_{i}, H(q(t), p(t)) = E ∀t}

and then we would get the action to be

S(γ) = −(t_{1} − t_{0})E + ∫_{γ} α ∀γ ∈ M_{E}^{1}

7. Nov 07, 2006: The Extended Phase Space

7.1. Aside. As we saw last time, the canonical 1-form α on a symplectic manifold M = T^{∗}X occurs in classical mechanics in two ways:

(1) We can write down Hamilton’s equations.

(2) We can look at the action along a path, so long as we only consider paths γ so that H ≡ E ∈ ℝ on γ.

(In this second case, S(γ) = ∫_{γ} α − E(t_{1} − t_{0}), as we saw last time.)

Putting these together, we can give a physical interpretation of − ∫_{Σ} ω, where Σ is a surface with boundary γ for some loop γ. Stokes’ Theorem says that

− ∫_{Σ} ω = ∫_{Σ} dα = ∫_{∂Σ} α = ∫_{γ} α = S(γ) + ∫_{t_{0}}^{t_{1}} H dt

which is just S(γ) in the limit t_{1} → t_{0}, where this amounts to reparametrizing γ suitably (i.e. covering it in smaller and smaller time).

Figure 9. γ bounds Σ

Thus, the job of ω is to tell us how much action it costs, to run around the surface Σ, by:

ω ↦ − lim_{t_{1}→t_{0}} ∫_{Σ} ω = S(∂Σ)

7.2. Bringing in spacetime. Let us now consider the extended phase space T^{∗}(X ×ℝ). Thus, our configuration space X gets replaced by space-time, or the extended configuration space X ×ℝ, and a point (q, t) in it says precisely where and when the system is. In the extended phase space, we thus add two coordinates from the phase space: time and energy.

The notation used is (q, t, p, p_{0}) ∈ T^{∗}(X × ℝ), with the terms denoting (position, time, momentum, negative energy(!)) respectively. Thus, the energy is given by E = −p_{0}.

In these terms, any path γ : [t_{0}, t_{1}] → T^{∗}X gives rise to γ̃ : [t_{0}, t_{1}] → T^{∗}(X × ℝ), given by

γ̃ : t ↦ (q(t), t, p(t), −H(q(t), p(t)))

so note that we still do need a Hamiltonian H : T^{∗}X → ℝ.

The other thing to note, is that later on, we will replace time by any old parameter s, so that time would be a function t(s). This would make the entire procedure independent of parametrization!

Now, the action is

S(γ) = ∫_{t_{0}}^{t_{1}} p_{i}q̇^{i} − H(q, p) dt = ∫_{t_{0}}^{t_{1}} p_{i} dq^{i} − ∫_{t_{0}}^{t_{1}} H(q, p) dt = ∫γ̃ α̃

where (comparing with the original 1-form α) α̃ is the canonical 1-form on T^{∗}(X × ℝ):

α̃ = α + p_{0} dt = p_{i} dq^{i} + p_{0} dt

(so we will use the notation t = q^{0}!) because we had cleverly chosen the p_{0}-coordinate of the path γ̃ to be −H(q(t), p(t)).

(In other words, α̃ is canonical, but H is not. However, the “noncanonical-ness” is built into the choice of γ̃, and then α̃ just gets γ̃ into S(γ)!)

7.3. Hamilton’s equations and the conservation of energy. Carlo Rovelli (see link on webpage!) has reformulated this, to apply not just to particles, but strings (and higher-dimensional “branes”). This requires a canonical 2-form (or higher forms).

To do this, let us allow γ̃ to be more general: γ̃ : [s_{0}, s_{1}] → T^{∗}(X × ℝ) is given by

s ↦ (q(s), t(s), p(s), −H(q(s), p(s)))

where H : T^{∗}X → ℝ is as before. In other words, γ̃ : [s_{0}, s_{1}] → Y ⊂ T^{∗}(X × ℝ), where Y “contains all information about H”:

Y = {(q, t, p, p_{0}) : p_{0} = −H(q, p)}

This is a codimension-1-submanifold, hence is odd-dimensional, hence not a symplectic manifold. But it has a 1-form on it, namely, α̃|_{Y} = i^{∗}α̃, the pullback of the inclusion i : Y ↪ T^{∗}(X × ℝ). Similarly, we also have a 2-form on Y, namely, ω̃|_{Y} = i^{∗}ω̃.

Aside: This kind of situation is called a contact manifold, where this is an odd-dimensional manifold, with a foliation whose leaves are symplectic (codimension 1) submanifolds. Here, we foliate along the time coordinate...

Why is this approach nice? For one thing, we have

Proposition 7.1. Assume \frac{dt}{ds} ≠ 0. Then a curve γ̃ : [s_{0}, s_{1}] → Y gives a solution to Hamilton’s equations, if and only if its tangent vector γ′(s) satisfies ω̃|_{Y} (γ′(s), −) ≡ 0. Moreover, if this holds, then conservation of energy automatically holds.

Remark 7.2.

(1) This is analogous to ω(v_{H}, −) = dH in the Hamiltonian approach above. But ω̃ is nondegenerate, so should it imply that γ′(s) = 0? No, because ω̃|_{Y} may well be degenerate!

(2) However, it does imply that some (most!) things about γ′(s) must vanish.

Proof. We compute:

γ′(s) = \frac{dq^{i}}{ds} \frac{∂}{∂q^{i}} + \frac{dp_{i}}{ds} \frac{∂}{∂p_{i}} + \frac{dt}{ds} \frac{∂}{∂t} + \frac{dp_{0}}{ds} \frac{∂}{∂p_{0}}

But as we are in Y, hence

\frac{dp_{0}}{ds} = −\frac{∂H}{∂q^{i}} \frac{dq^{i}}{ds} − \frac{∂H}{∂p_{i}} \frac{dp_{i}}{ds}

Moreover, ω̃ = dp_{i} ∧ dq^{i} + dp_{0} ∧ dt, so wedge computations give

ω̃|_{Y} (γ′(s), −) = −\frac{dq^{i}}{ds} dp_{i} + \frac{dp_{i}}{ds} dq^{i} − \frac{dt}{ds}dp_{0} − (\frac{∂H}{∂q^{i}} \frac{dq^{i}}{ds} + \frac{∂H}{∂p_{i}} \frac{dp_{i}}{ds} ) dt

Now, note that dp_{0} and dt are linearly independent on T^{∗}(X × ℝ), but they are dependent on Y! In fact, on Y, we have

\frac{dt}{ds}dp_{0} = −\frac{dt}{ds}dH = −\frac{dt}{ds} (\frac{∂H}{∂q^{i}} dq^{i} + \frac{∂H}{∂p_{i}} dp_{i})

We now put this back into the computation above, and collect coefficients. Thus, for ω̃(γ′(s), −) to vanish on Y, is equivalent to the dq^{i}, dp_{i}, dt-parts all vanishing. In other words, if and only if

\frac{dp_{i}}{ds} = \frac{dt}{ds} \frac{∂H}{∂q^{i}}
\frac{dq^{i}}{ds} = −\frac{dt}{ds} \frac{∂H}{∂p_{i}}
\frac{∂H}{∂q^{i}} \frac{dq^{i}}{ds} + \frac{∂H}{∂p_{i}} \frac{dp_{i}}{ds} = 0

Note that the left-hand side in the last equation is just \frac{dH}{ds}.

How do we get Hamilton’s equations (and conservation of energy) out of this? Well, if t is a locally invertible function of s, then the equations above can be converted, using the Chain Rule, to involve \frac{df}{dt} instead of \frac{df}{ds}. This gives us an equivalence with the new (modified) equations

\frac{dp_{i}}{dt} = \frac{∂H}{∂q^{i}}
\frac{dq^{i}}{dt} = −\frac{∂H}{∂p_{i}}
\frac{dH}{dt} = \frac{dH}{ds} /\frac{dt}{ds} = 0

In other words, ω̃|_{Y} (γ′(s), −) is identically zero if and only if Hamilton’s equations and conservation of energy hold.

The proof will be complete if we can show that the first two equations automatically imply the third. But this is obvious - we compute, using Hamilton’s equations (or the equations in terms of s instead of t, as above):

\frac{dH}{ds} = \frac{∂H}{∂q^{i}} \frac{dq^{i}}{ds} + \frac{∂H}{∂p_{i}} \frac{dp_{i}}{ds} = \frac{dt}{ds} · (\frac{∂H}{∂q^{i}} · −\frac{∂H}{∂p_{i}} + \frac{∂H}{∂p_{i}} · \frac{∂H}{∂q^{i}} ) = 0

□

Remark 7.3. As mentioned above, ω̃|_{Y} (γ′(s), −) ≡ 0 implies that “most components” of it vanish. In other words, there are a lot of constraints. More precisely, we can now say what constraints there are. They are exactly the Hamilton equations above.

7.4. Digression of the day: LIGO. Neutron stars and black holes colliding among themselves (or each other) are the only known cosmic events that radiate enough gravitational energy to be detectable on Earth. There is a project funded by NASA (with the largest funding, barring the expedition to send men to Mars!) which attempts to measure such effects. It is called LIGO.

However, the effects are so small compared to local gravitational effects (like people walking, even!) that the instruments (detectors) have to be extremely sensitive to measure this. Thus, there are L-shaped mirrors in rooms only four degrees above absolute zero; the mirrors themselves hang from sapphire wires, and so on. Attempts are being made to get everything as precise and sensitive as possible, but instead of the 10^{−18} order of magnitude, we are a few orders away from this.

7.5. A look back at the special case t(s) = s. Let us keep the notation and the setup as above - where we work with the extended phase space T^{∗}(X × ℝ). Earlier, we considered γ : [t_{0}, t_{1}] → X, giving rise to the phase space (points have coordinates of position and momentum here), together with the extra-special parametrization of the time function: t(s) = q^{0}(s) = s. We re-view the old theory now, from the point of view of the new one.

In general, the conservation of energy result in Lecture 3 holds in the extended phase space too, which means that the summation is starting from i = 0 (in q^{i}). Moreover, we keep our old Lagrangian L(q, q̇), which is independent of p_{0}, q^{0}.

Since q^{0} = t, hence q̇^{0} = 1, and the (extended) Hamiltonian is given by

H_{new}(q, p) = p_{0} · 1 + p_{i}q̇^{i} − L(q, q̇) = p_{0} + H_{old}(q, p)

Thus, classical mechanics (what we saw in earlier classes) comes under the case H_{new} = 0! Because once we have this, we get that p_{0} is independent of time, since conservation of energy tells us that H_{old}(q, p) was. Moreover, p_{0} = −E, where E = H_{old}(q, p).

Moreover, α̃ = α + p_{0}dq^{0} = α + p_{0} dt, and p_{0} is independent of time. Since t(s) = s, hence we are integrating on [t_{0}, t_{1}], and we get

S(γ̃) = ∫γ̃ α̃ = ∫_{γ} α+p_{0} ∫_{t_{0}}^{t_{1}} dt = ∫_{γ} α+p_{0}(t_{1}−t_{0}) = ∫_{γ} α−(t_{1}−t_{0})E = S(γ)

or, in other words, the old notion of action really is the integral of a 1-form on our path. We will see, in later classes, how this generalizes to higherdimensional membranes.

8. Nov 14, 2006: From particles to strings and higher membranes

Let us now generalize everything we did so far for point particles

γ : [s_{0}, s_{1}] → X × ℝ = M

(where M is “space-time”) to strings and higher membranes. Thus, we replace the one-dimensional [s_{0}, s_{1}] by Σ, where now Σ is a p-dimensional manifold with boundary (or with corners, like a p-(hyper)cube [0, 1]^{p}).

Moreover, we write γ : Σ → M (and also start to forget that we ever had the decomposition M = X × ℝ in an explicit way! This is what we mean when we talk about the “indistinguishability of the space and time directions”). The image of γ is now a p-dimensional membrane, which string theorists unfortunately call a ‘(p−1)-brane’, since they’re thinking about the ‘spatial’ dimensions of Σ instead. For example, in the case p = 2, we have a string, which they would call a 1-brane because it looks 1-dimensional at any moment of time, even though its ‘worldsheet’ (the image of γ) is 2-dimensional.

In the particle case, the canonical 1-form α̃ on T^{∗}M played a key role in defining the action, Hamiltonian etc. We had defined S(γ̃) = ∫γ̃ α̃, where γ̃ is a path in T^{∗}M. More precisely,

γ̃ : [s_{0}, s_{1}] → Y ⊂ T^{∗}M

where Y = {p_{0} = −H(q, p)} for our Hamiltonian H : T^{∗}X → ℝ. We also need a ‘canonical p-form’, and moreover, on what? (So that we can integrate it over our p-dimensional membrane.) We make the following table:

Particles (p = 1)

Membranes (p ≥ 0)

(1) Particles have a one-dimensional worldline γ : [s_{0}, s_{1}] → M.

(1) Membranes have a p-dimensional worldvolume γ : Σ → M. (Called a worldsheet for p = 2, or a particle that exists only for an instant if p = 0... These latter (for p = 0) are called instantons.)

(2) The extended phase space is T^{∗}M (which keeps track of momentum too). To get this, look at the tangent to our curve γ, and take the dual space, so we get (T M)^{∗}.

(2) What is the extended phase space here? Here, the tangent “sheet” consists of pwedges (wedges for orientation reasons) of tangent vectors, and dualizing gives (⋀^{p}T M)^{∗}.

(3) We start studying γ̃ : [s_{0}, s_{1}] → T^{∗}M, because T^{∗}M has a canonical 1-form α̃ on it.

(3) We start studying γ̃ : Σ → ⋀^{p}(T^{∗}M), because it has a canonical p-form on it. (We’ll check this later.)

8.1. More derivations. Recall that our worldline has a tangent vector γ′(s) ∈ T_{γ(s)}M, called its velocity. (Also, we need this velocity - or even the analogue for general p - to vary continuously/smoothly, like a section.) To generalise this to general p, one option is that γ : Σ → M gives dγ : T Σ → T M (or dγ(x) : T_{x}Σ → T_{γ(x)}M).

Figure 10. tangent vector

But over in this situation, we actually want our answer to lie in ⋀^{p}T M. Thus, let us (also) call the analogue γ′(x) ∈ ⋀^{p}T M, called its multi-velocity or p-velocity, which says how fast, and in which direction (of coordinates on Σ), γ moves. For example, for p = 2, the picture looks like the following “parallelogram” or area therein, inside ⋀^{2} T_{γ(x)}M:

Figure 11. shaded parallelogram = determinant

Here’s one possible definition of the multi-velocity for general p: let’s define γ′ using some choice of local coordinates s_{1}, …, s_{p} on Σ:

γ′(x) = \frac{∂γ(x)}{∂s_{1}} ∧ ⋯ ∧ \frac{∂γ(x)}{∂s_{p}} ∈ ⋀^{p}T_{γ(x)}M

Remark 8.1.

(1) Another choice of local coordinates s_{i} should give a “rescaled” section.

(2) This definition is a section, as we will eventually see. For now, we look at it only at a point γ(x) ∈ M.

(3) This vanishes if the \frac{∂γ(x)}{∂s_{i}}’s are linearly dependent (i.e. γ is not an immersion). For instance, if a cylinder is shrunk to a line (or in general, dim M < p etc.).

(4) The challenge here, is to invent a coordinate-free definition of γ′.

8.2. Generalizing the Lagrangian. Earlier, we had L : T X → ℝ. This is not good, since we want eventually to forget about X, and only use M = X × ℝ. Thus, we generalize this notion, and consider general L : T M = T (X × ℝ) → ℝ. (The earlier notion of Lagrangian is, now, merely independent of time.)

This is a good thing to do because now we have (γ(s), γ′(s)) ∈ T M, i.e. velocity is really in T M.

To generalize to p-membranes, we now want to consider the Lagrangian to be a function L : ⋀^{p}T M → ℝ, since now, (γ(x), γ′(x)) ∈ ⋀^{p}T M.

Question. Why is space-time still X × ℝ = M here, instead of X × Σ One possible reason: Σ is not “all-time-coordinates”, so to speak; it’s just “one-time-and-(p − 1)-space” coordinates!

9. Nov 28, 2006: More on particles “vs.” membranes

Recall from last time, that we were trying to generalize our framework from p = 1 to general p. This resulted in our changing the base parameter space [s_{0}, s_{1}] to a p-dimensional manifold Σ, possibly with boundary (and corners).

9.1. (Functorial) construction of the multivelocity alternating tensor. Here’s how to define the multivelocity γ′ of a map γ : Σ → M, i.e. of a p-dimensional membrane in the space-time M:

Figure 12. [s_{0}, s_{1}] maps to a tangent at a point on a path γ; Σ maps to a shaded parallelogram at a point on a surface M

To define γ′ in a coordinate-free way, take γ : Σ → M and apply various functors to get γ′ (if we want to be really precise!).

(1) The tangent functor: γ : Σ → M goes to its differential dγ : T Σ → T M, or to be very precise, the tangent functor T γ : T Σ → T M. Here, T is the tangent functor that takes manifolds to vector bundles over them (more precisely, T : M ↦ T M), and morphisms between manifolds, to their differentials, which are vector bundle maps between the corresponding tangent bundles.

(2) The (top-)wedge functor: We now apply the functor ⋀^{p} that takes finite-rank vector bundles on M to finite-rank vector bundles on M. Thus, if (E → M) has fiber E_{x} of dimension m over x ∈ M, then it is mapped to ⋀^{p}(E → M) = (⋀^{p}E) → M, with fiber (⋀^{p}E)_{x} = ⋀^{p}E_{x} of dimension \binom{m}{p} for all x. In particular, if m = p, then we get a line bundle on M, also called the determinant line bundle.

How about morphisms? Given a vector bundle map f : E → E′ (both E, E′ are over M), we get ⋀^{p}f : ⋀^{p}E → ⋀^{p}E′, given by

(⋀^{p}f)_{x}(e_{1} ∧ ⋯ ∧ e_{p}) := f(e_{1}) ∧ … f(e_{p}), ∀e_{i} ∈ E_{x}

9.2. Volume forms. Since we are working over Σ, hence we thus get, given γ : Σ → M, the map

⋀^{p}dγ : ⋀^{p}T_{x}Σ → ⋀^{p}T_{γ(x)}M

But the multivelocity was just one vector inside it, not a function! Actually, the domain above is just one-dimensional, being the top exterior power, but this means that there still is the choice of a scalar involved. Moreover, this scalar must be nonzero for any x (so that our domain vector is nonzero), which means that we are looking for a nonvanishing section of the top exterior power.

Definition 9.1. A volume form is a nonvanishing section of the top exterior power.

For example, ∫ (−)dx ∧ dy ∧ dz takes in functions and spits out numbers.

So, assume for now, that Σ comes equipped with a volume form vol ∈ Ω^{p}(Σ), so that 0 ≠ vol_{x} ∈ ⋀^{p}T_{x}^{∗}Σ for all x ∈ Σ. Then the top (pth) exterior powers of the tangent and cotangent bundles on Σ are both of rank one, hence the evaluation map on their fibers gives an isomorphism between them. More precisely, if V_{x} = ⋀^{p}T_{x}Σ, then dim V_{x} = dim V_{x}^{∗} = 1, where V_{x}^{∗} = ⋀^{p}T_{x}^{∗}Σ. One then defines φ_{x} : V_{x}^{∗} →^{∼} V_{x}, by φ_{x}(0) = 0, and

⟨φ_{x}(ω), ω⟩ = 1 ∀0 ≠ ω ∈ V_{x}^{∗}

Warning. Note that the map φ_{x} is not linear. It is “inverse-linear”.

We can now define the multi-velocity.

Definition 9.2. Given a volume form vol ∈ Ω^{p}(Σ) and γ : Σ → M, define the multi-velocity γ′ at γ(x) ∈ M, to be

γ′(x) = (⋀^{p}dγ)(φ^{−1}(vol_{x})) ∈ ⋀^{p}T_{γ(x)}M

Remark 9.3. The crucial fact is that this definition is not independent of vol! However, what we will see later, is that the notion of the volume (that generalizes the length of a string and area of a membrane) is now independent of choice of vol (i.e. choice of normalization).

We once again develop the two pictures side-by-side, for particles and for general membranes.

Particles (p = 1)

Membranes (p ≥ 0)

(1) A particle comes equipped with a worldline γ : [s_{0}, s_{1}] → M = space-time.

(1) A membrane has a world-volume γ : Σ → M.

(2) Then comes a Lagrangian and an action. For instance, in general relativity, M is a Lorentzian manifold, and the action for a particle with mass m and charge e, is

S(γ) = mℓ(γ) + e ∫_{γ} A

where A ∈ Ω^{1}(M) is the “electromagnetic vector potential”, and ℓ is the length, obtained using the metric on M: ℓ(γ) = ∫_{s_{0}}^{s_{1}} |γ′(s)| ds (2) The usual Nambu-Goto action for a membrane is S(γ) = m Vol(γ) + e ∫_{γ} A

where e is the charge, m is the membrane tension now, and A is the p-form potential. Finally, Vol stands for the volume of γ, also obtained using the metric on M:

Vol(γ) = ∫_{Σ} |γ′(x)| vol_{x}

(3) (Marginal cases.) If A = 0, a particle extremizing the action traces out a geodesic.

(3) (Marginal cases.) If A = 0, a membrane extremizing the action traces out a minimal surface.

(4) On the other hand, if m or the metric vanishes, i.e. we consider only the electromagnetic part of the action, namely e ∫_{γ} A, then this gives a change of phase exp(ie ∫_{γ} A) ∈ U(1) = S^{1}, when we move a quantum particle along a path γ. In fact, if γ is a loop homologous to the trivial loop, and it “bounds a surface” S, then by Stokes’ Theorem, ∫_{γ} A = ∫_{S} F, where ∂S = γ, and dA = F is the electromagnetic field.

(4) If we do the same thing here, i.e. consider only the p-form potential, then we get a change of phase exp(ie ∫_{γ} A) ∈ U(1), when we move a quantum membrane along the surface γ. In fact, if γ is a surface homologous to the trivial one, and it bounds a higherdimensional manifold S, then by Stokes’ Theorem, ∫_{γ} A = ∫_{S} F, where ∂S = γ and F = dA.

Remark 9.4.

(1) In the Nambu-Goto action above, A is the p-form generalization in Maxwell’s equations, hence is also called p-form electromagnetism.

(2) Also, |γ′(x)| uses the metric on ⋀^{p}T M, which in turn is obtained from the inner product/metric on T M. Moreover, here we use our chosen volume form vol on Σ.

(3) Then the formula for the volume is independent of choice of vol! This is because any scaling of γ′(x) involves a scaling inside φ (by the definition above), which was “inverse-linear”, and the corresponding choice of vol_{x} is rescaled by exactly the reciprocal of the earlier scaling. This makes the entire expression independent of choice of normalization.

(4) Also note that the integral had better converge. Thus, we need to have some restriction on Σ, or else at least on γ. For example, the image of γ may need to be compact in M.

(5) In either of these two cases, e ∫_{γ} A is independent of parametrization as well. Therefore the entire expression in the formula above, is also independent of parametrization, as we wanted.

(6) Finally, we get a map C_{γ} : Ω^{p}(M) → U(1) = S^{1} for each such γ, namely: C_{γ}(A) = exp(i ∫_{γ} A). (Here, we’re arranging the units so that ℏ = 1.) This brings us closer to cohomology, which we will in fact encounter next time.

9.3. The canonical p-form. We now do the computations similar to the p = 1 case above. Let us denote the canonical p-form on ⋀^{p}T^{∗}M by α_{p}. Since we will use (q^{i}, p_{i}) for coordinates on the cotangent bundle, let us compute instead, the canonical n-form α_{n} on ⋀^{n} T^{∗}M.

Using coordinates: An element of ⋀^{n} T^{∗}M, at a point q ∈ M, has a “Plücker” basis p_{I}, where I = {i_{1} < i_{2} < ⋯ < i_{n}} ⊂ {1, 2, …, dim M}. Define corresponding differentials dq^{I} := dq^{i_{1}} ∧ ⋯ ∧ dq^{i_{n}}, and the n-form by

α_{n} := p_{I}dq^{I}

where we still use Einstein summation for index sets - or for n-tuples of indices.

Coordinate-free definition: As earlier, keep in mind that α_{n} likes to eat n-wedges of tangent vectors to T^{∗}M, i.e.

v_{1} ∧ ⋯ ∧ v_{n}, v_{i} ∈ T_{(q,p)}(T_{q}^{∗}M)

and spit out numbers. But we have the projection π_{1} from T^{∗}M, taking (q, p) to q. Thus, dπ_{1}(v_{i}) : T_{(q,p)}(T^{∗}M) → T M. Moreover, π′_{2}(v_{i}) ∈ T^{∗}M (as earlier), so we now define

α_{n}(∧_{i}v_{i}) = (∧_{i}π′_{2}(v_{i}))(∧_{i}dπ_{1}(v_{i}))

As above, we have Theorem 9.5. If q^{i} are any local coordinates on M, and (q^{i}, p_{i}) the corresponding local coordinates on T^{∗}M, then α_{n} = p_{I}dq^{I}.

Proof. (This is a (somewhat tortuous!) exercise in keeping track of multiindices in wedging, so feel free to skip it!) Choose any v_{i} ∈ T_{(q,p)}T_{q}^{∗}M for 1 ≤ i ≤ n. Then, as earlier,

∧_{i}v_{i} = ∑_{K,J} a_{J}^{K} ∧_{k∈K} \frac{∂}{∂q^{k}} ⋀ ∧_{j∈J} \frac{∂}{∂p_{j}}

for some choice of scalars a_{J}^{K}, and where K, J vary over disjoint subsets of \{1, 2, …, dim M\}, so that |K| + |J| = n. In fact, if v_{i} = ∑_{l} a^{l,i} \frac{∂}{∂q^{l}} + a_{l,i} \frac{∂}{∂p_{l}}, then

a_{J}^{K} = ∑_{i_{k},i′_{j}} ± ∏_{k∈K} a^{k,i_{k}} · ∏_{j∈J} a_{j,i′_{j}}

where we sum over all tuples i_{k} and all i′_{j}, so that they are all distinct, and the ± is the product of the signs of the two permutations, that rearrange the sets {i_{k} : k ∈ K} and {i′_{j} : j ∈ J} in increasing order.

As above, we also condense our notation by writing

\frac{∂}{∂q^{K}} := ∧_{k∈K} \frac{∂}{∂q^{k}} , \frac{∂}{∂p_{J}} := ∧_{j∈J} \frac{∂}{∂p_{j}}

where all wedges are taken with indices in increasing order (just as in dq^{I}, or even in p_{I}). Now by our “random facts” on wedges, we know that

dq^{I}(\frac{∂}{∂q^{K}} ∧ \frac{∂}{∂p_{J}}) = δ_{K}^{I}δ_{∅}^{J}

since |I| = n = |K| + |J|. In other words, it vanishes if J is nonempty, or if J = ∅ (so |I| = |K|) but I ≠ K.

Method 1: We therefore start our computations. Firstly, evaluating ∧_{i}v_{i} at p_{I}dq^{I}, we have

p_{I}dq^{I}(∧_{i}v_{i}) = ∑_{J,K} p_{I}a_{J}^{K}δ_{K}^{I}δ_{∅}^{J} = p_{I}a_{∅}^{I}

where most terms in the sum vanish because of the Kronecker deltas.

Method 2: We now want to reconcile this to the coordinate-free definition in these coordinates. In this case, we note that if

v = (q, p, ∑_{i} a^{i} \frac{∂}{∂q^{i}} + a_{i} \frac{∂}{∂p_{i}} )

then dπ_{1} : T (T^{∗}X) → T X sends it to (q, ∑_{i} a^{i} \frac{∂}{∂q^{i}}), where q^{i}’s on the righthand side are local coordinates in a different space (T X) than the q^{i}’s on the left-hand side. In other words, dπ_{1}(v_{i}) removes the vertical components \frac{∂}{∂p_{j}} ’s from each v_{i}, and takes the \frac{∂}{∂q^{i}} ’s to themselves in the tangent space to the vector space T_{q}^{∗}X.

Therefore, ∧_{i}dπ_{1}(v_{i}) precisely equals a_{∅}^{K} \frac{∂}{∂q^{K}}, since there are no vertical components remaining!

Moreover, as in the earlier proof (for p = 1), we have that π_{2}(v) = p_{i}dq^{i} in the cotangent space at q. Wedging this over all v_{i}’s gives us all possible terms of type p_{I}dq^{I}. Thus, we can now compute α_{n} using the coordinate-free definition, to get

α_{n}(∧^{i}v_{i}) = p_{I}a_{∅}^{K}dq^{I}(\frac{∂}{∂q^{K}}) = δ_{K}^{I}p_{I}a_{∅}^{K} = p_{I}a_{∅}^{I}

and since the two definitions agree at all ∧_{i}v_{i}, hence the forms themselves are equal, as desired. □

10. Dec 05, 2006: Phases and connections on bundles

Today we shall return to our extended phase space, so to speak. From now on, we forget about X, and only talk about our phase space M. Thus, the paths and 1-forms are henceforth denoted by γ, α etc. (and not γ̃, α̃ etc.).

A glimpse of what is to come: Having reviewed classical particle mechanics, and its stringy and brany generalizations, next we will quantize these. We’ll mainly talk about geometric quantization.

10.1. When connections come in. In particle mechanics, a phase space is often a symplectic manifold (M, ω) (but we’ll see an example today, that is not of this form). The role of the 2-form ω is to (help) compute the action of a path γ ∈ M : S(γ) = ∫_{γ} α, where dα = ω (upto a sign, which we shall henceforth ignore, since we do not care about reconciling this to physicists’ notation anymore!).

What about ω itself? If γ = ∂Σ (i.e. γ is homologous to the boundary of a surface Σ - always assumed to be compact here, in order to integrate on it), then S(γ) = ∫_{Σ} ω by Stokes’ Theorem.

Figure 13. γ bounds Σ that might have handles on it

This raises two questions:

(1) What if γ is a path not cohomologous to zero, i.e. γ ≠ ∂Σ?

(2) What if ω ≠ ∂α for any α? (For cotangent bundles, this problem does not arise, but in general symplectic manifolds, ω is merely closed, not exact.)

(One obvious solution is: if γ is cohomologous to zero but ω is not exact, we use S(γ) = ∫_{Σ} ω - and if γ is not cohomologous to zero but ω is exact, then ω = dα, and we use S(γ) = ∫_{γ} α. Thus, we only get in trouble if both these problems arise simultaneously.)

But we look here, at a generalization of a 1-form that works, called a connection on a U(1)-bundle.

10.2. Phases and relative phases. In quantum mechanics, what matters is not the action S(γ) ∈ ℝ, but the phase e^{iS(γ)} ∈ U(1). Note that this already loses information, being periodic:

0 → 2πℤ → ℝ ⟶^{s↦e^{is}} U(1) → 0

Actually, if one has (in quantum mechanics) a bunch of paths between points m_{0}, m_{1} in M, i.e.

℘ ⊂ P_{m_{0}→m_{1}}M := {γ : [s_{0}, s_{1}] → M : γ(s_{i}) = m_{i}}

(depending on what we need, γ might need to be differentiable, smooth, etc.), then the amplitude to get from m_{0} to m_{1} (via ℘) is

A_{℘} := ∫_{γ∈℘} e^{iS(γ)} Dγ ∈ ℂ

(modulo convergence issues, of course). Such an expression is called a path integral (it is one of the two major types of Feynman integrals).

The amplitude now gives a probability |A_{℘}|^{2} ∈ [0, 1], upto a suitable renormalization.

So, for physics, changing e^{iS(γ)} ↦ ce^{iS(γ)} for some c ∈ U(1), does not change any probabilities.

However, a connection on a U(1)-bundle lets us calculate a definite phase (i.e. an actual, concrete value) for any loop γ, and thus a ratio of phases for any pair of paths γ_{1}, γ_{2} from m_{0} to m_{1}, as follows:

Figure 14. two paths γ_{1}, γ_{2} : m_{0} → m_{1}; maps to the loop
γ_{1} ∘ γ_{2}^{−1}

We define \frac{Phase(γ_{1})}{Phase(γ_{2})} = Phase(γ), where γ is the loop γ_{1} ∘ γ_{2}^{−1} (based at

m_{0}, so note the order of composing). These relative phases are what we can actually measure: they are unchanged by multiplying Phase(γ_{1}), Phase(γ_{2}) by any c ∈ U(1).

Next, any U(1)-connection has a curvature 2-form ω ∈ Ω^{2}(M), which should be our symplectic structure in applications to geoemtric quantization. If γ is a loop with ∂Σ = γ for some surface Σ, then

Phase(γ) = e^{i ∫_{Σ} ω}

This is just like e^{i ∫_{γ} A} = e^{i ∫_{Σ} ω}, but now Phase(γ) is not coming from a 1-form A anymore.

Warning. Before we move on to a concrete example, let us remark that not every loop is the boundary of a (compact) surface Σ. For example, consider the unit circle in the plane punctured at the center of the circle. Even as topological spaces, the boundary of the punctured disk is the union of a circle and a point, not just the circle. And as manifolds with boundaries, the point is not even of codimension one.

10.3. Example: Rigid rotor. Let M = S^{2} ⊂ ℝ^{3} be the unit sphere, with the 2-form ω = sin^{2} ϕ dθ dϕ, where θ, ϕ denote the longitude, and the latitude from the north pole (not equator) respectively.

[figure: sphere with equatorial plane and angle θ; prime meridian and angle ϕ with north pole]

Then ∫_{Σ} ω = area(Σ).

Figure 15.

What’s our U(1) connection here? Given a loop γ, pick any unit tangent vector v at a point x, and parallel transport it around γ, to get a new tangent vector v′.

Thus, v′ is obtained by rotating v in the tangent plane at x, and this gives Phase(γ) ∈ U(1). Moreover, in this case we really do have

Phase(γ) = e^{i ∫_{Σ} ω} = area(ω)

when ∂Σ = γ. We say that ω is the curvature of our U(1)-connection.

Figure 16. shows the tangent plane at a point on a sphere, the loop γ, and the vectors v, v′

M is then the phase space (not configuration space) of a rigid rotor i.e. a rigid spinning ball with fixed rate of rotation (angular speed), but unknown axis and “direction of spinning”.

Remark 10.1.

(1) Note that the ambiguity of antipodal points giving the same axis of rotation is removed, if we decree that all rotations are specified by the north pole of the axis, and the right-hand-rule to determine the direction of rotation.

(2) M is also the Riemann sphere, and the phase space of a cubit. The quantization of this is the phase space of an electron.

(3) Also note that M is the phase space of a rigid rotor, because both position and velocity are mentioned here.

Note that M is 2-dimensional, so the 2-form ω is obviously closed. However, we claim that M is not the cotangent bundle of any manifold. In fact,

Proposition 10.2. ω is not exact.

Proof. Suppose not, i.e. say ω = dα for some 1-form α. Then if ∂Σ = γ for some (connected) compact region Σ on S^{2}, denote Σ′ to be the closure of the complement S^{2} \ Σ. This too has boundary γ, but perhaps with the opposite orientation. In either case, we now use Stokes’ Theorem to get

area(Σ) = ∫_{Σ} ω = ∫_{γ} α = ± ∫_{Σ′} ω = ±(area(S^{2}) − area(Σ))

for all Σ as above. This is a contradiction (e.g. take Σ = S^{2}, Σ′ = ∅). □

10.4. Integral cohomology and Max Planck. More generally, say (M, ω) is any symplectic manifold equipped with a U(1)-connection whose curvature is ω, and say Σ is a surface inside M. Then Phase(γ) = e^{i ∫_{Σ} ω} when ∂Σ = γ.

Now consider two surfaces Σ_{1}, Σ_{2} with common boundary γ - or more precisely, ∂Σ_{1} = γ, ∂Σ_{2} = γ^{−1}. Let Σ = Σ_{1} ∪ Σ_{2}.

Figure 17. a surface (say with genus) obtained by joining two half-surfaces

Then we have

Phase(γ) = e^{i ∫_{Σ_{1}} ω} = e^{i−∫_{Σ_{2}} ω}

so that if Σ is as above, then

e^{i ∫_{Σ} ω} = e^{i ∫_{Σ_{1}} ω+∫_{Σ_{2}} ω} = 1

Therefore ∫_{Σ} ω ∈ 2πℤ!

This argument works for any Σ ⊂ M, with ∂Σ = ∅, since we can always write Σ = Σ_{1} ∪Σ_{2} as in the picture above. So, ω describes an integral second cohomology class.

Remark 10.3.

(1) This is one origin of the word “quantization”, as Max Planck and Bohr-Sommerfeld essentially realised. The energy states have to occur with some integer-conditions...

(2) We’ll see later (next quarter!) that this generalizes to (p + 1)-forms for all p!