## November 8, 2006

### Quantization and Cohomology (Week 3)

#### Posted by John Baez Sorry for the long pause! Here are the notes for the October 17th class on Quantization and Cohomology:

• Week 3 (Oct. 17) - From Lagrangian to Hamiltonian dynamics. Momentum as a cotangent vector. The Legendre transform. The Hamiltonian. Hamilton’s equations.

Last week’s notes are here; next week’s notes are here.

Posted at November 8, 2006 5:20 AM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1024

Read the post Quantization and Cohomology (Week 2)
Weblog: The n-Category Café
Excerpt: The Lagrangian approach to classical mechanics
Tracked: November 8, 2006 5:32 AM
Read the post Quantization and Cohomology (Week 4)
Weblog: The n-Category Café
Excerpt: Hamiltonian dynamics and symplectic geometry.
Tracked: November 8, 2006 5:58 AM

### Re: Quantization and Cohomology (Week 3)

The Legendre transform cropped up in my last post too. You don’t happen to have one of those lovely little phrases which can help me see what all uses of the transform have in common, and further what the Legendre transform has in common with the Laplace transform so that they’re only a change of rig away.

Posted by: David Corfield on November 8, 2006 8:47 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

I think if you take the definition of Laplace transform and take the temperature $\to 0$ limit, you’ll see it becomes the Legendre transform.

More precisely: according to Gibbs and Boltzmann, for any temperature $T > 0$ there’s a map from energies to relative probabilities given by

$E \mapsto exp(-E/T)$

This lets us take the rig of relative probabilities

$([0,\infty), + , 0, \times, 1)$

and pull back the entire rig structure to the set of energies

$\mathbb{R} \cup \{+\infty\}$

We get a rig $\mathbb{R}^T$ which is isomorphic to

$([0,\infty), + , 0, \times, 1)$

When we take the $T \to 0$ limit, this rig $\mathbb{R}^T$ reduces to the rig

$(\mathbb{R} \cup \{+\infty\}, min, +\infty, +, 0)$

Now: take the definition of Laplace transform and write in terms of the rig $\mathbb{R}^T$, and then take the $T \to 0$ limit. You’ll get one of the standard definitions of the Legendre transform!

Some hints:

1. I think the exponential in the Laplace transform disappears when we describe it using the rig $\mathbb{R}^T$ instead of the isomorphic rig $([0,\infty), + , 0, \times, 1)$ The exponential is in the isomorphism, as mentioned above.
2. The integral in the definition of Laplace transform becomes the minimum in the definition of Legendre transform - or maximum, depending on your conventions. I like minimizing energy and action, but these guys like maximizing. It’s not a big deal.
3. I don’t mention the definition of Legendre transform using minimization in my course notes, but you can find it on Arnold’s book on classical mechanics, and you’ll see it’s basically equivalent. I’m secretly minimizing something by setting its derivative to zero.
4. Innocent bystanders who don’t know about any of this stuff should read this, and then look at some books on “tropical mathematics” or “idempotent analysis”.
Posted by: John Baez on November 8, 2006 8:55 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

Thanks. Litvonov resorts to group characters (p. 11 of this) to explain the link. But I don’t feel I’ve got to the essence of these things yet. If the Legendre transform passes between tangent and contangent spaces, what spaces does the Laplace transform pass between?

Posted by: David Corfield on November 9, 2006 10:19 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

But I don’t feel I’ve got to the essence of these things yet.

Yeah, sorry. I’ve been really busy getting ready for some talks, so I didn’t have time to concoct “one of those lovely little phrases” that suddenly clarifies everything; I just sketched how you can do the calculation that shows the Laplace transform has the Legendre transform as a low-temperature limit.

If the Legendre transform passes between tangent and cotangent spaces, what spaces does the Laplace transform pass between?

I’m sorry, I’m still too busy right now, so I can only toss you some more clues. Forget tangent and cotangent spaces - that’s a red herring. Forget everything I said about the Legendre transform in my course notes - that’s also a red herring, for what you’re interested in. (Eventually it all fits together, but only eventually.)

Here’s how to think of it: both the Legendre transform and the Laplace transform take a function on a vector space $V$ and give a function on the dual vector space $V^*$. This is evident from the formulas you’ll see if you click the links I just provided. Don’t be fooled by the the fact that these formulas consider the special case $V = V^* = \mathbb{R}$, or by the fact that they take the integral in the Laplace transform and restrict it to the positive half of $\mathbb{R}$. That’s just designed to throw you off the scent. Look at the formulas and see how one is using the rig $\mathbb{R}^T$ and the other is using the $T \to 0$ limit of this rig!

Posted by: John Baez on November 9, 2006 5:02 PM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

I think maybe it’s the Laplace transform I see less clearly. Should we expect that wherever the Legendre transform is used there’s a non-zero temperature/quantum equivalent which uses the Laplace/Fourier transform?

The Legendre transform works on functions on Banach spaces, where I’m seeing it in statistical learning theory. The Laplace transform also works with Banach spaces, I see.

In the Wikipedia article, the Laplace transform is represented as generally operating between a time and frequency representation. How then can we tell ourselves a story about temperature and its zero limit?

Here’s another interpretation:

we interpet the Laplace transform of probability density functions as the probability that the corresponding random variable “wins a race” against an exponentially distributed catastrophe.

I see that to add to the fun there’s also a Laplace-Stieltjes transform, where the Laplace-Stieltjes transform of $g$ is the Laplace transform of the derivative of $g$.

Posted by: David Corfield on November 10, 2006 8:58 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

Alex Hoffnung’s answer to the Hamiltonian vector field homework can be found here.

Posted by: John Baez on November 8, 2006 9:03 PM | Permalink | Reply to this
Weblog: The n-Category Café
Excerpt: I'm in one of those phases where everywhere I look I see the same thing. It's Fourier duality and its cousins, a family which crops up here with amazing regularity. Back in August, John wrote: So, amazingly enough, Fourier duality...
Tracked: January 11, 2007 2:17 PM

### Re: Quantization and Cohomology (Week 3)

Questions for qc.pdf - week 3
=================================

Section 4.3, page 15: “From now on, assume that L is strongly regular” - You are also assuming that L is not an explicit function of time, right? But if L is L(q,\dot{q},t) (for instance, with a time varying potential)? I think it would be interesting to qualitatively mention in brief the difficulties involved. I’m thinking about this paper:

Also, I’m thinking whether there is another restriction for the Lagrangian so that you can define a Legendre transform, that is, would L have to be a convex function (or is it q that must be so)? Also, I’m interested in learning a little bit more on how can it be that a Legendre transform does the trick on the duality \dot{q} and p.

(I can see there are previous comments on this, I’ll read them. What I have learned from Legendre transforms comes from Arnold’s book).

Still on page 15: “In what follows, not that” - typo: “note that”

Page 16: Please define what the arrow with a tilde means.

Thanks,
Christine

Posted by: Christine Dantas on August 1, 2007 5:52 PM | Permalink | Reply to this

### The Problem of Time

Christine writes:

Section 4.3, page 15: “From now on, assume that L is strongly regular” - You are also assuming that L is not an explicit function of time, right?

Be careful! In this situation, ‘time’ can mean two drastically different things — the parameter describing a path $\gamma: [t_0,t_1] \to M,$ or a coordinate on the manifold $M$!

In this class, the Lagrangian is always a function

$L: T M \to \mathbb{R}$

where $M$ is some manifold called the ‘configuration space’. A point in $T M$ is a pair $(q, \dot q)$, so the Lagrangian can depend on $q$ and $\dot q$, but not on the parameter $t$ parametrizing a path in $M$.

But note: in applications to the physics of a single particle, $M$ can be space or spacetime! When we’re doing general relativity and studying a particle moving around on a curved spacetime, we should always take $M$ to be spacetime. In fact, it’s even good to do this in special relativity!

In this situation we get a Lagrangian of the form $L(q,\dot q)$. There’s no explicit dependence on the parameter $t$. So, Noether’s theorem applies and we get a conserved quantity from time translation invariance. But, this quantity is zero!

This is the simplest place to start understanding ‘the problem of time’.

To see how this works in special relativity, see section 3.4.3 in my classical mechanics book. To see how it works in general relativity, see section 3.8.

But if $L$ is $L(q,\dot{q},t)$ (for instance, with a time varying potential)? I think it would be interesting to qualitatively mention in brief the difficulties involved.

If the Lagrangian depends on the parameter $t$ that parametrizes the path in $M$, everything becomes difficult and mathematically ugly. So, this sort of time-dependent Lagrangian is something I’d rather not discuss here.

It might be good to talk about parameter-dependent Lagrangians in my general-purpose classical mechanics course — and I do intend to keep expanding the book for that course. But, it’s not something I want to discuss when I’m trying to find the most elegant way to do everything… and then categorify it all.

Posted by: John Baez on August 2, 2007 11:26 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

Also, I’m thinking whether there is another restriction for the Lagrangian so that you can define a Legendre transform, that is, would $L$ have to be a convex function (or is it $q$ that must be so)?

We’re assuming our Lagrangian is strongly regular. Often a convenient strategy to prove a Lagrangian is strongly regular begins by showing $L(q,\dot q)$ is a convex function of $\dot q$ for each $q$. But, convexity is neither necessary nor sufficient.

‘Recall’ from section 4.1 of my classical mechanics book that a Lagrangian is regular if the map sending velocity to momentum:

$\array{ \lambda: T M & \to & T^* M \\ (q,\dot q) & \mapsto & (q, \frac{d L}{d\dot q}) }$

gives a diffeomorphism between $T M$ and some open set $U \subseteq T^* M$. This means we can use position and momentum as our basic variables in physics, since we can recover the velocity from these.

The Lagrangian is strongly regular if $U$ is all of $T^* M$. Then any momentum is possible!

(See section 4.2.4 of the book for examples that are regular but not strongly regular. In these situations, not any momentum is possible!)

For our Lagrangian to be regular, we certainly need

$(q,\dot q) \mapsto (q, \frac{d L}{d\dot q})$

to be one-to-one. This will be true if $L$ is strictly convex in the $\dot q$ directions. But, it will also be true if $L$ is strictly concave! There may be other possibilities too — I don’t know offhand.

Of course the most familiar example has $L$ being quadratic in $\dot q$, namely

$L(q,\dot q) = m|\dot q|^2/2 - V(q)$

This is convex and strongly regular if $m > 0$.

Please define what the arrow with a tilde means.

It means ‘isomorphism’: a map with an inverse.

Thanks for the comments, corrections, questions and so on! I’m glad you’re reading the notes carefully. Don’t read them too fast, though — it takes a long time to write these replies. Posted by: John Baez on August 2, 2007 11:56 AM | Permalink | Reply to this

### Re: Quantization and Cohomology (Week 3)

Thank you! I appreciate very much your careful and detailed responses. Things are getting much clear now.

Best,
Christine

Posted by: Christine Dantas on August 2, 2007 12:17 PM | Permalink | Reply to this