## January 29, 2009

### The Third Time is the Charm

#### Posted by John Baez

A puzzle:

Name as many instances as you can of mathematicians or physicists who introduced a logical sequence of three concepts, each depending on the previous one, where the third was crucial for the applications they had in mind.

Posted at January 29, 2009 6:14 AM UTC

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### Re: The Third Time is the Charm

For Alexander Grothendieck the number would be more like ten. At least.

Posted by: estraven on January 29, 2009 8:02 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Eilenberg and Mac Lane introduced the notion of natural transformation in their 1945 category theory paper with applications to algebraic topology in mind.
To define natural transformations they introduced functors.
To define functors they introduced categories.

Posted by: Dmitri Pavlov on January 29, 2009 8:04 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Yes, that’s one of the examples I had in mind! Great!

But there’s another, even more familiar. I only noticed the resemblance recently.

Posted by: John Baez on January 29, 2009 6:53 PM | Permalink | Reply to this

### 2 out of 3 ain’t bad; Re: The Third Time is the Charm

Albert Einstein.

(1) Special Relativity;
(2) General Relativity;
(3) The Unified Field Theory of 1957 which, after Godel’s time travel solutions to the field equations, was used to extend his life and win his second Nobel Prize.

Whoops, (3) is that not on the branch of the multiverse where we reside…

Posted by: Jonathan Vos Post on January 29, 2009 4:27 PM | Permalink | Reply to this

### classical mechanics; Re: The Third Time is the Charm

It was a community, not a single person, who established this sequence of three formulations of classical mechanics:
(1) Newtonian,
(2) Lagrangian, and
(3) Hamiltonian.

Posted by: Jonathan Vos Post on January 29, 2009 4:32 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

For a long time I’ve enjoyed this example:

• categories,
• functors,
• natural transformations.

Saunders Mac Lane is said to have remarked, “I didn’t invent categories to study functors; I invented them to study natural transformations.” The reason is that mathematicians were comparing different cohomology theories, and needed a precise concept of when two such theories were ‘naturally isomorphic’.

One reason this example is noteworthy is that it means Eilenberg and Mac Lane were really inventing the 2-category of categories. This illustrates the ‘upwards pull’ of categorification: to really understand $n$-categories, you need the $(n+1)$-category of all $n$-categories, which means you need to understand $(n+1)$-categories.

But yesterday, while teaching an undergraduate class, I noted another trio of concepts in which the ‘highest’ and least obvious of the three is really the one needed in the most glorious application of the theory.

What is it?

I like ‘special relativity, general relativity, unified field theory’ — but that’s not it, and unfortunately Einstein never really found the third item.

Posted by: John Baez on January 29, 2009 4:48 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Hmm, you’re teaching differential geometry aren’t you? A rich area for towers of concepts.

I remember being shocked at the definition of a tangent vector as an equivalence class of paths.

Is that the particular example you are after? Or is it curvature or geodesic?

You need the three concepts to be devised by the same person? Gauss or Riemann then.

Posted by: David Corfield on January 29, 2009 5:10 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I was thinking of something along the lines of Jonathan’s suggestion except I would go

- Newtonian
- Hamiltonian (symplectic)
- 2-symplectic

Or something…

Posted by: Eric on January 29, 2009 5:36 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

You guys are going to be disgusted when I finally reveal my answer. You’re thinking about stuff that’s way too fancy. I’m thinking about a trio that rolls off the tongue as easily as “category, functor, natural transformation” — but one that many more people know about.

Posted by: John Baez on January 29, 2009 5:42 PM | Permalink | Reply to this

### Newton again; Re: The Third Time is the Charm

Newton again, finally with application to rocketry:

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

II. The relationship between an object’s mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.

III. For every action there is an equal and opposite reaction.

Posted by: Jonathan Vos Post on January 29, 2009 5:44 PM | Permalink | Reply to this

### Re: Newton again; Re: The Third Time is the Charm

That’s interesting, but Newton’s three laws are not “a logical sequence of three concepts, each depending on the previous one, where the third was crucial for the applications he had in mind.”

Neither are the three laws of thermodynamics.

Posted by: John Baez on January 29, 2009 6:29 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Give us a hint — what field?

Posted by: Jamie Vicary on January 29, 2009 5:53 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

You know the answer to this question, so you don’t need a hint.

Posted by: John Baez on January 29, 2009 6:28 PM | Permalink | Reply to this

### 3 Math theories of Communications; Re: The Third Time is the Charm

Again, creation by a community:

Krippendorff, Klaus. “Information” Paper presented at the annual meeting of the International Communication Association, TBA, San Francisco, CA, . 2008-12-11

Abstract:

Much contemporary communication research stands on the shoulders of three mathematical theories: the logical, statistical, and algorithmic theories of information. Logical theory concerns distinctions and selections among logical possibilities. As such, it addresses a semantic space of possible realities. Statistical theory concerns observed probabilities, and is familiar from Claude Shannon’s theory of communication. Algorithmic theory concerns the information that is required to compute a function or problem (i.e., some software requires more memory than others). Each of these theories center on the distinctions (possibilities, probabilities, and length of instructions) which, from the human perspective, give rise to meaning. The translation of distinguishable signals into informative messages depends on at least three aspects of their context: (i) temporality (e.g., a comprehensible sequence); (ii) rules (e.g., situational constraints); (iii) a history of relationships between senders and receivers; and (iv) the known intentions of senders and needs of receivers.

Posted by: Jonathan Vos Post on January 29, 2009 5:31 PM | Permalink | Reply to this

### Re: 3 Math theories of Communications; Re: The Third Time is the Charm

I think Krippendorf is using ‘at least three’ to mean ‘four or more’. I want a sequence of three concepts, in clear logical succession.

As the Second Brother put it:

Three shalt be the number thou shalt count, and the number of the counting shall be three. Four shalt thou not count, nor either count thou two, excepting that thou then proceed to three. Five is right out.

Posted by: John Baez on January 29, 2009 6:44 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

From mathematical finance:

1.) Black-Scholes (Stock options from “point” prices)
2.) LIBOR Market Model (Bond options from “yield curve” prices)
3.) Not established yet, but I would like to :)

Posted by: Eric on January 29, 2009 5:40 PM | Permalink | Reply to this

### Kurt Gödel again; Re: The Third Time is the Charm

Since I mentioned Kurt Gödel [28 April 1906 - 14 Jan 1978]:

(1) doctoral dissertation under Hahn’s supervision in 1929 submitting a thesis proving the completeness of the first order functional calculus;

(2) Gödel’s Incompleteness Theorems”. In 1931 he published these results in Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme. He proved fundamental results about axiomatic systems, showing in any axiomatic mathematical system there are propositions that cannot be proved or disproved within the axioms of the system. In particular the consistency of the axioms cannot be proved;

(3) His masterpiece Consistency of the axiom of choice and of the generalized continuum-hypothesis with the axioms of set theory (1940) is a classic of modern mathematics. In this he proved that if an axiomatic system of set theory of the type proposed by Russell and Whitehead in Principia Mathematica is consistent, then it will remain so when the axiom of choice and the generalized continuum-hypothesis are added to the system. This did not prove that these axioms were independent of the other axioms of set theory, but when this was finally established by Cohen in 1963 he built on these ideas of Gödel.

Reference for above is Kurt Gödel, by J J O’Connor and E F Robertson

Posted by: Jonathan Vos Post on January 29, 2009 5:41 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

1) Heavens, 2) Earth, 3) Light.

Posted by: Urs Schreiber on January 29, 2009 5:50 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

In the beginning God created the heaven and the earth.

And the earth was without form, and void; and darkness was upon the face of the deep. And the Spirit of God moved upon the face of the waters.

And God said, Let there be light: and there was light.

However, it’s not the one I had in mind. I was definitely thinking about human-made mathematics or physics.

Posted by: John Baez on January 29, 2009 6:38 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Given that John is teaching Differential Geometry:

(1) Vector bundle
(2) Vector bundle with connection
(3) The Riemannian connection on the tangent bundle

I know that I found Diff. Geom. got much easier when I started classifying results according to which of these three concepts they were really about. It doesn’t exactly roll off the tongue, though.

Posted by: DavidS on January 29, 2009 5:53 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Sorry, I don’t think first anybody invented vector bundles, and then vector bundles with connection, and then the Riemannian connection on the tangent bundle… with his goal being to study the third one. These ideas may seem like a logical sequence now, but in fact the last of them was probably invented first.

Also, I don’t see them as a natural progression with same sort of inevitability as ‘category, functor, natural transformation’.

Posted by: John Baez on January 29, 2009 6:34 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Hmm…
pointed homotopy of pointed paths, homotopy *class* of a pointed path, universal covering space?

Posted by: some guy on the street on January 29, 2009 6:28 PM | Permalink | Reply to this

### Hamilton again; Re: The Third Time is the Charm

Okay, since I mentioned Hamilton:

(1) Reals;
(2) Complex;
(3) Quaternion
(4) bonus, by Cayley et al.: Octonions

“On a new Species of Imaginary Quantities connected with a theory of Quaternions”

By Sir William R. Hamilton

[Proceedings of the Royal Irish Academy, Nov. 13, 1843, vol. 2, 424-434]

It is known to all students of algebra that an imaginary equation of the form i^2 = -1 has been employed so as to conduct to very varied and important results. Sir Wm. Hamilton proposes to consider some of the consequences which result from the following system of imaginary equations, or equations between a system of three different imaginary quantities:

i^2 = j^2 = k^2 = -1.

ij = k, jk = i, ki = j.

ji = -k, kj = -i, ik = -j.

no linear relation between i, j, k being supposed to exist…

Sir W. Hamilton calls an expression of the form Q a quaternion; and the four real quantities w, x, y, z he calls the constituents thereof. Quaternions are added or subtracted by adding or subtracting their constituents….

Posted by: Jonathan Vos Post on January 29, 2009 6:40 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Leray invented (1) spectral sequences to compute (2) sheaf cohomology and for that he had to invent (3) sheafs.

Posted by: Gonçalo Marques on January 29, 2009 7:25 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Another possibility is
1)Manifolds
2)Differential forms
3)Stokes theorem

Posted by: Gonçalo Marques on January 29, 2009 7:31 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

The first two don’t really go in order here, but maybe

1. differentiation
2. integration
3. the fundamental theorem of calculus
Posted by: Mike Shulman on January 29, 2009 8:15 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I feel a strange blend of amusement and despair.

When you see the trio I’m thinking of, you’ll recognize it as forming an inevitable logical sequence, as fundamental as “1,2,3” or “category, functor, natural transformation” — and just as important in mathematics and physics. And as with “category, functor, natural transformation”, the third member of this sequence is its crowning glory.

Posted by: John Baez on January 29, 2009 8:27 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

i)Vectors
ii)Vector spaces
iii)Linear morphisms

Posted by: Gonçalo Marques on January 29, 2009 8:45 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Perhaps a slightly better variant on this might be (I don’t actually know the history well enough):
1. Limits
2. Differentiation
3. Integration

Certainly most of the big applications of calculus use integration, but you can’t compute them without derivatives, and those require limits. (Though I’m not sure if Newton thought about it that way.)

Posted by: Noah Snyder on January 30, 2009 5:52 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

A dark horse candidate, and almost certainly not what you had in mind… but it does roll of the tongue.

Claude Shannon

1) Information
2) Compression
3) Capacity

Posted by: Steve Flammia on January 29, 2009 8:20 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

When someone finally guesses the trio I have in mind, everyone else will probably get angry. They’ll probably argue that the first — and maybe even the second — concept in this trio was not really invented by the person who put the third to such astoundingly effective use.

And in a sense they’ll be right. But surely, the whole formalism that makes this trio into a tightly connected sequence was first clearly understood by the person in question. And surely everyone here is familiar with what I’m referring to. So I don’t feel at all guilty for posing this puzzle.

Posted by: John Baez on January 29, 2009 8:47 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Integers, sets, groups; Galois

Posted by: Charlie C on January 29, 2009 9:06 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Nice, but much more erudite than the one I have in mind… and I don’t see these three concepts as forming an inevitable progression, where each one leads to the next in a systematic way.

You’re going to be really annoyed when someone comes up with the answer I’m thinking of.

Posted by: John Baez on January 29, 2009 9:17 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Is it: 1) Tangent, 2) Normal, and 3) Frenet frame?

Posted by: Nick on January 29, 2009 9:28 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I feel sure this isn’t right, but I’ll throw it out there:

• Vector field
• Connection
• Curvature

All three are important in both math and physics, with the third the most important, the crowning glory.

Posted by: Todd Trimble on January 29, 2009 9:43 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Position, velocity, acceleration; Newton?

Posted by: Todd Trimble on January 29, 2009 9:48 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Dangit, I just had that exact same thought.

Posted by: Mike Shulman on January 29, 2009 10:59 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I know these are not what you have in mind. I am not even sure of the accuracy as I am not sure what name to put to them. How about?

1) vector space
2) dual vector space
3) cohomology

or

1) Hilbert space
3) eigenstate

Posted by: Bruce Westbury on January 29, 2009 11:16 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

You’re not thinking of something as simple as:
are you?

(Even if you’re not, this suggestion might get people to aim a bit lower on the sophistication scale!)

Posted by: Stuart on January 29, 2009 11:37 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

After posing the puzzle I did indeed think about addition, multiplication and exponentiation as a famous trio of concepts in logical progression. It’s very nice, and you’re right — it’s closer to the level of sophistication I was looking for. But it doesn’t quite match my puzzle: “Name as many instances as you can of mathematicians or physicists who introduced a logical sequence of three concepts, each depending on the previous one, where the third was crucial for the applications they had in mind.”

Posted by: John Baez on January 30, 2009 1:40 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Several examples come to mind:
(1) Integers, rational, reals.
(2) Chart, atlas, manifold
(3) Subset, open set, continuous map.

Based on David’s detective work, I think JB had (2) in mind.

Posted by: Scott Carter on January 29, 2009 11:47 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Todd wrote:

Position, velocity, acceleration; Newton?

YES!!!

Each one is clearly obtained from the previous one in a systematic way, and Newton invented this way — differentiation!

But, he needed differentiation not for first derivatives, but for second derivatives. His grand discovery was this: the laws of motion become clear not when we try to explain an object’s position, nor when we try to explain its velocity, but only when we try to explain its acceleration!

So, just as Eilenberg and Mac Lane needed categories to define functors, and needed functors to define natural transformations, so they could make precise sense of cohomology theories being ‘naturally isomorphic’…

… Newton needed position to define velocity, and velocity to define acceleration, so he could make precise sense of

$F = m a .$

There’s even a nice analogy between these two trios. An category is a place for an object to sit. A functor is a way of changing an object. A natural transformation changes a way of changing an object.

I haven’t quite figured out how to exploit this analogy. It’s interesting to note that an $n$-category internal to $Vect$ is automatically a chain complex and is thus related to second derivatives by the formula

$d^2 = 0 .$

But somehow I should get natural transformations (chain homotopies) into the game.

Anyway, it might be a sign of genius for somebody to study a trio of concepts, each systematically derived from the previous one, where the third one is the charm.

Posted by: John Baez on January 30, 2009 12:24 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

This is one of the few times where my not actually knowing any mathematics would have helped me answer a question on the n-Category Café! Unfortunately for me, I didn’t notice the thread until the question had already been answered.

Posted by: Blake Stacey on January 30, 2009 2:33 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I haven’t quite figured out how to exploit this analogy.

This sounds like another challenge :)

Is there some way to imagine a derivative as somehow being internal to something?

Posted by: Eric on January 30, 2009 3:37 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I’m obviously thinking out loud where “thinking” is probably even a stretch, but going the other way:

1.) acceleration
2.) velocity
3.) position

makes me think of “integration without integration” for some reason…

Posted by: Eric on January 30, 2009 6:00 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Hi John,

Maybe there is another kind of generalization for newton’s law that uses a kind of of 3 charm:

“BF-theory is a gauge theory of n-gerbes with connection over n+3-dimensional spacetime S (often, and for larger n always, assumed to be trivial in the existing literature). The gerbe connection provides us, locally, on patches U i⊂S, with an n+1-form”

http://golem.ph.utexas.edu/string/archives/000777.html

See equation 4. Isn’t like taking the equations of movement from that action like getting a kind of generalized newtons law for D-dimensional particles with exotic statistics in a D+3 space? No wonder that is called a motion group. And isn’t the reason that there is a 3 dimensional step related to the fact that the surface in which the motion happens follows d^2=0?

Posted by: Daniel de França MTd2 on January 30, 2009 4:29 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Are there infinitesimal versions of functors and natural transformations that make the analogy better? I can think of one situation: This is for fiber functors on Tannakian categories. Then the invertible natural transformations form a group scheme, so that one can look at its Lie algebra.

But are there some general notions?

Posted by: Minhyong Kim on January 30, 2009 11:52 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I haven’t quite figured out how to exploit this analogy.

Actually you have! There was some article we once wrote:

let $G$ be a Lie group – and let’s concentrate here on $G = \mathbb{R}$ to get closest to the Newtonian example regarded as a 1-object groupoid $\mathbf{B}\mathbb{R}$.

Functors from $\mathbf{B}\mathbb{R}$ to itself and natural transformations between these arrange themselves into the smooth 2-groupoid $\mathbf{B}\mathbf{E}(\mathbb{R})$.

To find the infinitesimal version of this triple gadget (object = position, morphism = functor, 2-morphism = natural transformation) we proceed the way one differentiates any Lie group: we map smooth 1- and 2-parameter paths into this:

So let $P_2(\mathbb{R}^2)$ be the strict 2-groupoid of smooth paths in the plane and consider smooth 2-functors

$F : P_2(\mathbb{R}^2) \to \mathbf{B}\mathbf{E}(\mathbb{R}) \,.$

These 2-functors are in bijection with 1-forms $A$ on $\mathbb{R}^2$.

The way the equivalence works is precisely the Newtonian example:

Regarding the value of the functor $F$ on a path $\gamma$ in $\mathbb{R}^2$ as a trajectory in $\mathbb{R}$, $A$ gives the velocity of this trajectory with respect to “parameter time”. For instance if you think of the standard speed parameterized path $\gamma : t \mapsto (t,0)$ in $\mathbb{R}^2$ and write $A = A_t d t$, then $A_t = v$ is precisely the velocity of the trajectory.

And the value of the 2-functor $F$ on 2-morphisms then, i.e. its value in natural transformations of functors on $\mathbf{B}\mathbb{R}$, is indeed given by the (“exterior”) acceleration $d A$.

This example, incidentally, when we pass from $\mathbb{R}$ to more general groups and include linear representations, then also subsumes the first Trimblean triple as a “special case” of “category-functor-natural transformation”.

Posted by: Urs Schreiber on January 30, 2009 1:17 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

It is funny how you can write something that looks like english and makes us feel as if it should have been obvious.

I WISH I understood what you just said because it sounds neat, but I really have no clue :) The failing is entirely mine, but reading this gave me a smile this morning (which is rare these days)

Posted by: Eric on January 30, 2009 2:42 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

It is funny how you can write something that looks like english and makes us feel as if it should have been obvious.

What is also a bit funny is that I have been talking about this here a lot, with the increasing impression of having pretty much decoupled from my suroundings thereby in an unfortunate way.

This analogy

$\array{ category & functor & natural transformation \\ bundle & connection & curvature }$

has occupied me quite a bit in my attempts to find the right abstract way of looking at curvature in nonabelian differential cohomology. It led me to those “tangent categories” (which yield the $\mathbf{E}\mathbb{R}$ appearing above) and “arrow theoretic differentials” etc, which you may remember, as for instance from page 10 here. Cleaned-up bits of that are in my articles with Konrad Waldorf.

By now I have a more, say, high-brow way of thinking about curvature things, which comes back to language that more people may be able to connect to at least vaguely, but it is less close to the above analogy.

Posted by: Urs Schreiber on January 30, 2009 3:10 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

If $A$ is a 1-form on $\mathbb{R}^2$ given by

$A = v dt,$

then

$dA = v_x dx\wedge dt.$

Should we throw a Hodge star in there so that

$\star A = v dx$

and

$d\star A = a dt\wedge dx,$

where $a$ is acceleration?

Posted by: Eric on January 30, 2009 3:04 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

To improve the analogy

$\array{ position & velocity & acceleration \\ fiber & connection & curvature }$

it is good to think of movement of relativistic particles moving at speed of light: for them all acceleration is always transversal and given by a bivector.

Indeed, for a charged relativistic particle the above analogy becomes an honest duality: the connection (1-form) of the background field pairs with the velocity (1-vector) of the particle, and the curvature (2-form) of the background field with the acceleration (bivector).

In the “Geometric Algebra” corner they are good at emphasizing this point that acceleration is really a bivector (in relativistc mechanics) and that the Lorentz force law (which is Newton’s law for the charged particle!) just equates the acceleration bivector with the curvature of the background field – let’s see if I find a good reference to point to… ah, maybe slide 6 here.

Posted by: Urs Schreiber on January 30, 2009 3:32 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

it is good to think of movement of relativistic particles moving at speed of light: for them all acceleration is always transversal and given by a bivector.

I happen to think that this situation applies to all particles in nature, i.e. instantaneous velocity is always the speed of light, but average velocity depends on the frequency of changes in direction (which depends on mass via ideas similar to the Feynman checkerboard). BUT that is another story :)

Posted by: Eric on January 30, 2009 4:00 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I happen to think that this situation applies to all particles in nature, i.e. instantaneous velocity is always the speed of light, but average velocity depends on the frequency of changes in direction

Sure. Fundamentally, all particles are massless.

Only because some of them get kicked around by the Higgs background field a), b), c) don’t they quite manage to go straight at the speed of light and effectively seem to go straight at a lower speed f).

So any question concerning physics at its fundamental level (and I suppose that’s what we are interested in here) should concentrate on massless particles.

Posted by: Urs Schreiber on January 30, 2009 4:18 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Neat. Although it probably should have been obvious if I were paying attention, I never thought about Higgs this way before. My formal field theory education ended with Sakurai in 1992 :)

Posted by: Eric on January 30, 2009 5:06 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Hi people,

Doesn’t this 3 charm look more natural in the discussions of the Motion Group in terms of BF-Theory?

This post at string coffe table seems to put BF theory in relation to this 3 charm in a more motivational and interesting position, that is, leading to more richer and genercial (anyon or nonabelion) structures. It seems that BF is “critical” in relation to 3 charm.

I am sorry, but I can’t see the point in discussing so much classical theories with such nice ideas and formalisma. It is like trying to kill an ant with a nuclear bomb. Wouldn’t it a much better use of them for more “advanced” theories, like branes, topological strings and LQG?
I say that, because in that Urs’s post, the Motion Group is related to this things.

Well, maybe I am ignorant, and for all of you, when you talk about classical theories, everything that is more advanced is obvious for you guys.

So, just as a curiosity, how, or would it be possible, that anything that is non classic could fit in there? Hall Effects and branes, for example…

Posted by: Daniel de França MTd2 on January 31, 2009 11:29 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Daniel wrote something like:

I am sorry, but I can’t see the point in discussing so much classical theories with such nice ideas and formalisms. It is like trying to kill an ant with a nuclear bomb. Wouldn’t it a much better use of them for more “advanced” theories, like branes, topological strings and LQG?

I disagree.

I want to get to the bottom of things. So, if a mathematical idea offers new viewpoints on physics, I want to start by thinking about the simplest situations where these new viewpoints apply. If the idea only applies to fancy “advanced” theories, then that’s what I’ll think about. But if the idea already sheds new light on something simpler, like classical mechanics, then that’s where I want to start.

Posted by: John Baez on February 1, 2009 5:12 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

The professor for the only “physics” class I took in “college” (there’s an unimportant meaning to each set of scare-quotes) once told me that the first question about any new model is “where is the hydrogen atom?” I’d wager that by now “where is the simple harmonic oscillator?” should come first, if not “where is the free particle?”

Posted by: John Armstrong on February 1, 2009 6:29 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Here is a the way a kindergartener might think about it…

We want $A$ to be a velocity 1-form, but a velocity should probably come from a position 0-form $p$. An obvious thing to do given a position 0-form is to take its exterior derivative

$dp = p_x dx + p_y dy.$

A particle will follow a path

$\gamma:\mathbb{R}\to\mathbb{R}^2$

so that pulling back we get

$\gamma^* dp = v dt.$

I don’t think we want to set $A = dp$ though because the curvature $F = dA$ would be zero. So instead, maybe we want

$A = \star dp.$

If that is the case, then the curvature

$F = dA = d\star d p = \star \nabla p$

is the (Hodge star of the) Laplace-Beltrami operator applied to the position one form. I’m in no position to check, but I would guess that

$F = dA = a dx\wedge dy,$

where $a$ is the acceleration.

Then if I were to further guess, I’d guess that

$\gamma^* (i_v F) = a dt,$

which would mean

$i_v F = \mathcal{L}_v A.$

I warned that this was kindergarten math, but it’d help to understand it in the more traditional language and then see how to relate that to the groupoid version.

Posted by: Eric on January 30, 2009 4:56 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I like this example very much and think it could help illustrate a lot of neat concepts that you guys are working on.

If we were to work through the relativistic charged particle in an electomagnetic field in detail, we would rather need to think of paths in a directed space.

Posted by: Eric on January 30, 2009 5:24 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

If we were to work through the relativistic charged particle in an electomagnetic field in detail, we would rather need to think of paths in a directed space.

Now that you say this in the present context, it made me have an idea:

Let $X$ be a globally hyperbolic Lorentzian manifold.

Suppose we have an electromagnatic background field $\nabla$ on $X$, given by a smooth parallel transport functor on all paths $P_1(X)$ in $X$.

Then for a charged particle zipping through $X$, what equals its acceleration bivector $\dot v v = \dot v \wedge v$ is not the full curvature bivector $F_\nabla$ of $\nabla$, but just its electric part, which is its projction onto the timelike direction.

(Still see the equation on slide 6 here).

So, how can we understand this functorially? Well, let’s follow your suggestion and restrict our parallel transport functor to (future-)directed paths, i.e. to those paths which correspond to morphisms in the poset structure on $X$.

This gives us a category $P_1(X)^{directed}$ of future-directed paths (being a subcategory of $P_1(X)$). So we can restrict our background field

$tra : P_1(X) \to U(1)Tor$ along the inclusion $i : P_1^{directed}(X) \hookrightarrow P_1(X)$ and consider the curvature of tis $i^* \mathrm{tra}$. I’d need to think about this more in quiet, but it seems that the curvature of $i^* tra$ would be precisely the 2-form/bivector which equals the acceleratoin bivector in the Newton-Lorentz force law.

But let me think about this. First of all I haven’t before defined curvature of a functor which is not a functor on a groupoid of paths. But roughly the above looks like it might be a useful way of looking at things here.

Posted by: Urs Schreiber on January 31, 2009 2:20 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Yeah. This example, although already pretty simple, may not even be the simplest.

An even “simpler” example (for undergrads) might be to consider a Galilean spacetime $\mathbb{R}^3\times\mathbb{R}$ as a directed space. Seeing how Newton’s laws fall out of groupoids would be very neat. Seeing acceleration as curvature would also make a nice segue to general relativity. In fact, I suspect that acceleration can be seen as curvature in all three frameworks

* Netwonian

* Special relativistic

* General relativistic

That would be a great way to teach this stuff.

Note: I think the Lorentzian version is actually “simpler” than the Galilean version, but undergrads may not be ready for that (depending on background).

Posted by: Eric on January 31, 2009 5:02 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

It’s been a while since I wrote down the Laplace-Beltrami operator, but since you’re using a connection $\nabla$, I should probably use $\Delta$ instead so

$F = dA = \star \Delta p = a vol,$

where $a$ is acceleration and $vol$ is the volume form on $\mathbb{R}^2$.

Posted by: Eric on January 31, 2009 5:41 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

John wrote:

I haven’t quite figured out how to exploit this analogy.

Urs wrote:

Actually you have! There was some article we once wrote…

Hmm. Maybe you’re right. At least you’re getting close, but I still don’t feel like I see a perfectly beautiful example of a situation where:

• position is given by a category (or maybe an object in a category),
• velocity is given by a functor (or maybe the value of a functor applied to an object in a category),
• acceleration is given by a natural transformation (or maybe the value of a natural transformation applied to an object in a category).

Working out this analogy in the most beautiful way in the simplest example would be a very good thing. It might help some people understand category theory and other people understand physics.

To improve the analogy

$position:velocity:acceleration::fiber: connection:curvature$

it is good to think of movement of relativistic particles moving at speed of light: for them all acceleration is always transversal and given by a bivector.

I’ll have to think about that more…

What is also a bit funny is that I have been talking about this here a lot, with the increasing impression of having pretty much decoupled from my suroundings thereby in an unfortunate way.

You spend a lot more time thinking about math and physics than explaining it (and thinking about how to explain it), and you’re smart, so you go faster along your own particular line of research than anyone can keep up with you.

I spend more time explaining math and physics than actually thinking about it, so I don’t have this particular problem.

With luck, someday you’ll be famous enough that a bunch of people will spend their full time trying to understand you and develop your ideas further. This seems to be a key element of Witten’s and Connes’ success… though of course they also do spend a lot of time giving talks and writing books and trying in other ways to make their ideas easy to follow.

The way the $n$-Category Café works, you have a lot of people reading what you write… but it’s a blog, so if I’m a typical example, a lot of these people start skimming whenever it gets too hard to understand.

Posted by: John Baez on January 30, 2009 6:29 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Okay, let me try to find a setup in which this becomes very obvious.

First of all, the more obvious part of the analogy is

$\array{ category & functor & natural transformation \\ fiber & connection & curvature }$

Here a simple situation is that where

- each fiber looks like a group $G$, regarded as a one-object groupoid (hence a category), which I write as $\mathbf{B}G$;

- all functors go from $\mathbf{B}G$ to itself;

- and the natural transformations in question are the natural transformations between these functors.

And we want to assume that $G$ is a Lie group, so that we can proceed with the analogy which clearly presupposes a smooth setup.

In this case, the main subtlety in making sense of the above analogy comes from the existence of outer automorphisms on $G$. So, in the spirit of making everything as clear as possible, let me assume a situation where these vanish. For instance assume $G = E_8$ (i.e. the compact real form of $E_8$).

Then (I am saying this to be self contained for the record and for other readers) the relation between the two lines in the above analogy is established by the notioin of smooth 2-functors

$tra : P_2(X) \to Cat$

from the 2-groupoid $P_2(X)$ whose

- objects are points in a manifold $X$;

- morphisms are classes of smooth paths in $X$;

- 2-morphisms are classes of smooth surfaces in $X$.

Assume this 2-functor $tra$ assigns

1) to each point the fiber of a $G$-principal bundle $P$ over $X$, which we can locally identify with $G$ and hence with the category $\mathbf{B}G$ $\array{ \mathbf{B}G \\ category \\ fiber \\ P_x }$

2) to each path $\gamma : x \to y$ a functor from the fiber over $x$ to the fiber over $y$.

Theorem: such smooth assignments of functors to paths are in bijection with connections $\nabla$ on the $G$-principal bundle: upon identifying the fibers with $\mathbf{B}G$ the functor is given by an automorphism of $G$ hence, by assumption, by an element of $G$, and this element is the parallel transport $P \exp(\int_\gamma \nabla)$.

$\array{ \mathbf{B}G & \mathbf{B}G \stackrel{tra(\gamma)}{\to} \mathbf{B}G \\ category & functor \\ fiber & connection \\ P_x & \nabla }$

3) to each surface $\Sigma : \gamma \to \gamma' : x \to y$ a natural transformation of functors $tra(\gamma) \rightarrow tra(\gamma')$. Since $\mathbf{B}G$ has a single object, such natural transformations are given by elements of $G$ once we identify the fibers with $\mathbf{B}G$.

Theorem: this element of $G$ is the group element $P_\nabla \exp(\int_\Sigma F_\nabla)$ where $F_\nabla$ is the curvature 2-form of $\nabla$ and $P_\nabla \epx(\int_\cdot \cdot)$ denotes the surface ordered exponential integral of a 2-form with respect to $\nabla$ (as appearing in the nonabelian Stokes theorem).

$\array{ \mathbf{B}G & \mathbf{B}G \stackrel{tra(\gamma)}{\to} \mathbf{B}G & tra(\gamma) \stackrel{\exists !}{\Rightarrow} tra(\gamma') \\ category & functor & natural transformation \\ fiber & connection & curvature \\ P_x & \nabla & F_\nabla } \,.$

So that’s how this works. Before going into the further analogy

$\array{ category & functor & natural transformation \\ fiber & connection & curvature \\ position & velocity & acceleration }$ let’s massage the above kind of observation into a form that we are all happy with.

Posted by: Urs Schreiber on January 31, 2009 1:43 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Using the symbol BG in that way is confusing to us of an earlier generation.

Posted by: jim stasheff on January 31, 2009 2:14 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Using the symbol BG in that way is confusing to us of an earlier generation.

It’s actually meant to be maximally non-confusing and adapted to the needs of the earlier generation, because everything you do with the groupoid $\mathbf{B}G$ remains true after geometric realizatoin for the space $B G$.

Please see our long discussion about this at the end of $n$Lab: category algebra.

Posted by: Urs Schreiber on January 31, 2009 2:24 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I think I can sympathize with what Jim is saying, because it’s taken me a while to get used to the notation $B G$ as you are using it here. I was brought up thinking that $B G$ is a space, not a 1-object category. And old habits die hard.

The philosophy, if I understand it, seems to be: we should be thinking of the one-object category $B G$, or the space $B G$, as incarnations (in different categories) of the same basic concept, and so one is within rights to use the same notation for each.

But, this philosophy is in some conflict with the philosophy that a concept (e.g., the concept of the real number system) is never in isolation, but is always relative to some specified categorical context in which the concept is regarded as living (e.g., real numbers as topological group, real numbers as discrete field, and so on). In other words, the trouble with $B G$ is that most mathematicians will probably reflexively think, “$B G$ lives in $Top$.” Followed by, “no wait, that doesn’t make sense here.”

I don’t recall whether Urs uses different notation for the topological sense of $B G$, but all I’m saying is that the topological meaning is probably the default meaning of $B G$ for most mathematicians. I think I’m at the point where I don’t mind Urs’s way of using $B G$ for the one-object groupoid, but for me it took some getting used to, some additional mental training, and that may be the added difficulty that Jim is referring to.

Posted by: Todd Trimble on January 31, 2009 3:32 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I have to agree with Jim and Todd, here. To a topologist, $B$ is a functor from categories to spaces, and $G$ means either the group or the one object category.

Posted by: Dan Christensen on January 31, 2009 9:17 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Because of my short attention span, I haven’t checked if someone has already pointed this out, but there is a stack-theoretic reason for using Urs’ convention.

One can define the stack BG in various reasonable settings by taking its sections BG(X) on some object X to be the groupoid of principal G-bundles on X. The notation here seems to be perfectly consistent with the occurrence of BG in homotopy theory.

But then, it’s straightforward to check (say for usual topological spaces, or schemes with the etale toplogy) that BG as a stack is represented by the groupoid BG under discussion, that is, with one object and automorphism group G, thought of as the point mod G.’

Posted by: Minhyong Kim on February 1, 2009 10:44 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

taking its sections BG(X) on some object X to be the groupoid of principal G-bundles on X.

so in the top context BG(X) = [X,BG]
where [ , ] = homotopy classes of maps?

would that be a definition or a theorem?
an important to me distinction

Posted by: jim stasheff on February 1, 2009 2:37 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Sorry about the confusing formulation. Let’s define the stack $P$ by taking its sections $P(X)$ on $X$ to be the groupoid of principal $G-$bundles on $X$. So the objects are principal $G$-bundles and the morphisms are isomorphisms. On the other hand, consider the groupoid $BG$ with one object having automorphism group $G$. Then it is a (small) theorem that $BG$ represents $P$.

To spell out a bit what this means, recall that any groupoid, say in topological spaces,

$A_1 \rightarrow A_0$

(sorry, there should be two arrows, but I’ve forgotten how to do that in latex)

determines a pre-stack, whose sections on a space $X$ are the groupoid whose objects are $A_0(X)$ and whose morphisms are $A_1(X)$. There is a rather obvious notion of stackification’ of a pre-stack, which I’ll refrain from formulating for now. Then the theorem is that the stackfication of $BG$ is exactly $P$. Of course if you think about it a bit, this theorem is essentially the definition of a principal bundle. So maybe the original assertion is a definition after all.

Posted by: Minhyong Kim on February 1, 2009 4:08 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I’m happier with represents’ rather than is’

Posted by: jim stasheff on February 1, 2009 9:18 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

But, this philosophy is in some conflict with the philosophy that a concept (e.g., the concept of the real number system) is never in isolation, but is always relative to some specified categorical context in which the concept is regarded as living.

Of course, as you probably know, the argument is that Gpd should be considered as a full sub-$(\infty,1)$-category of Top, so that $B G$ as a groupoid and $B G$ as a space really are the same object. Geometric realization from groupoids to spaces (as opposed to from categories to spaces) is, homotopically speaking, a no-op.

I also want to point out that $B$ actually has two different meanings in classical topology:

1. If $G$ is a topological monoid (or an $A_\infty$-space), then $B G$ is its classifying space, aka “delooping.”
2. If $C$ is a category, then $B C$ is the realization of its nerve.

I suppose the intuition behind this conflation of notation is that if you consider a group as a one-object groupoid and take the realization of its nerve, you get the same thing as if you consider it as a discrete topological group and take its classifying space.

But the point, for me, is that these two meanings are actually in conflict. For example, if $G$ is a topological group then $\pi_n(B G) = \pi_{n-1}(G)$. But if $C$ is a groupoid, then $\pi_0(C)$ should clearly mean its set of isomorphism classes of objects, and $\pi_1(C,x)$ should clearly mean the automorphism group of $x\in C$, but these are equal to $\pi_0$ and $\pi_1$ of the realization of the nerve of $C$. Likewise, the inverse of the delooping $B$ is the loop space, $G\simeq \Omega B G$, but the realization of the nerve has no inverse for categories, and for groupoids its inverse is the fundamental groupoid $\Pi_1$, again a no-op on 1-types.

Or suppose that $C$ is a monoidal groupoid. I can take the nerve of its classifying space to get a topological monoid (or $A_\infty$-space if it is not strict monoidal), and then I can deloop that monoid. In classical notation I would have to write the result as $B(B C)$, where the two $B$s denote completely different operations!

I do realize that it is difficult for people to get used to changes in notation. But I think that using $B$ to mean both (a) the delooping of a topological group and (b) the one-object groupoid associated to a group is less confusing than using it for the two classical meanings, because (a) and (b) are much more closely related (in fact, essentially the same, modulo the equivalence of groupoids with 1-types). It took me quite a while to get over my confusion about the two classical uses of $B$, which is one reason I am on such a crusade to expunge its use for the realization of nerves.

I also feel that the identification of groups with one-object groupoids does, overall, more harm than good—more harm than the identification of groupoids with 1-types! (Groups can instead be identified with pointed connected groupoids.) So I think it is really important to have a notation for passage from groups to one-object groupoids, to remind us that there really is something going on.

I think that someone, John maybe, has used ordinary $B$ for the delooping of a topological group and script $\mathcal{B}$ for the one-object groupoid associated to a group. If that makes you less confused, I would be happy to use that.

Posted by: Mike Shulman on January 31, 2009 10:46 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Mike Shulman wrote:
Of course, as you probably know,

Not me - as I say, consider your audience

the argument is that Gpd should be considered as a full sub-(infty,1)-category of Top, so that BG as a groupoid and BG as a space really are the same object. Geometric realization from groupoids to spaces (as opposed to from categories to spaces) is, homotopically speaking, a no-op.

What’s a no-op?

I also want to point out that B actually has two different meanings in classical topology:

1. If G is a topological monoid (or an A_infty-space), then BG is its classifying space, aka “delooping”.

2. If C is a category, then BC is the realization of its nerve.

I suppose the intuition behind this conflation of notation is that if you consider a group as a one-object groupoid and take the realization of its nerve, you get the same thing as if you consider it as a discrete topological group and take its classifying space.

YES!

but the realization of the nerve has no inverse for categories,

perhaps only up to infty-homotopy

Or suppose that C is a monoidal groupoid. I can take the nerve of its classifying space to get a topological monoid (or A_infty-space if it is not strict monoidal), and then I can deloop that monoid. In classical notation I would have to write the result as B(BC), where the two Bs denote completely different operations!

am I the only one confused by that - the nerve of a space? and completely different?? in what categories?

So I think it is really important to have a notation for passage from groups to one-object groupoids, to remind us that there really is something going on.

That I’d agree with.
I think that someone, John maybe, has used ordinary B for the delooping of a topological group and script B for the one-object groupoid associated to a group. If that makes you less confused, I would be happy to use that.

Something like that but don’t depend on fonts.

Posted by: jim stasheff on February 1, 2009 2:32 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Of course, as you probably know,

Not me - as I say, consider your audience

Sorry, “you” in that sentence was addressed to Toby, not Jim, since it was his specific comment that I was responding to. I wouldn’t expect a classical topologist to know “that Gpd should be considered as a full sub-(infty,1)-category of Top”!

What’s a no-op?

Means “does nothing;” from computer science. I refer to passing across an equivalence of categories, or an equivalence of homotopy theories, as a no-op because no information is lost or gained; we only change our point of view on some “real” underlying object. In this case, the relevant fact is that the homotopy theory of groupoids is equivalent to the homotopy theory of 1-types.

but the realization of the nerve has no inverse for categories,

perhaps only up to infty-homotopy

No, a category cannot be recovered up to equivalence from the singular complex of its nerve. There is a (zigzag of) functors relating a category to the category of simplices of the singular complex of the realization of its nerve, which induces a weak equivalence of nerves, but I don’t think it is reasonable to view that as recovering the category itself. Equivalence of nerves is not the fundamental notion of “sameness” for categories.

Or suppose that C is a monoidal groupoid. I can take the nerve of its classifying space to get a topological monoid (or A_infty-space if it is not strict monoidal), and then I can deloop that monoid. In classical notation I would have to write the result as B(BC), where the two Bs denote completely different operations!

am I the only one confused by that - the nerve of a space? and completely different?? in what categories?

Sorry, where I said “nerve of its classifying space” in that paragraph I should have said “realization of its nerve.”

Posted by: Mike Shulman on February 1, 2009 10:31 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Sorry, “you” in that sentence was addressed to Toby, not Jim, since it was his specific comment that I was responding to.

You mean Todd, not Toby. You already convinced me back on the Lab!

Posted by: Toby Bartels on February 6, 2009 8:04 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Sorry! I knew who I meant… but apparently my fingers didn’t.

Posted by: Mike Shulman on February 6, 2009 8:43 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

1) actually I am typing

“mathbf{B}G”

to retain an indication that this is closely related to but different from the topological $B G$. You can see this in my LaTeX notes. But for some reason, lately the boldface math seems not to be displayed here anymore (??)

2) You may recall that we had this discussion about a year ago already. Back then I used to write $\Sigma G$ and “you” (people here) convinced me that it would be better to use $\mathbf{B}G$.

3) and I was happy with this: as Todd and Minhyong mentioned: when writing $\mathbf{B}G$ most statement you make just literally remain true whether you interpret them

- in groupoids

- in simplicial sets

- in stacks

- in topological spaces.

- etc.

4) It seems there is no globally consistent notation that fits every single convntion.

5) finally: I claim we need some suggestive notation to distinguish the discrete monoidal groupoid $G$ from its one-object groupoid incarnation.

6) of course we can always fall back to writing $pt//G$.

Posted by: Urs Schreiber on February 2, 2009 12:59 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

It’s unfortunate that time and pixels were wasted over a bug in the itex2mml code.

It seems that “\mathbf{B}” does not work because it is expecting more than one character (?). For example, “\mathbf{BE}” does work. I’m trying to find a hack to work around it, but hopefully someone somewhere can fix the bug.

Posted by: Eric on February 2, 2009 4:37 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

For more on this problem with \mathbf{}, see my comment over at TeXnical Issues. We should discuss such problems there, not here.

Posted by: John Baez on February 3, 2009 1:44 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Since we’re all expressing our feelings about the notion $B G$ for ‘the group $G$ regarded as a one-object groupoid’, I can’t resist joining in.

While I’ve occasionally teased Urs about making a big deal about the distinction between groups and one-object groupoids, there are certainly times when this distinction must be made clear. I think the notation $B$ is good for this — although it needs to be explained every time one starts talking to a new person!

In fact, Jim Dolan and I advocated this notation back in 1998, on page 13 of our paper on categorification. We listed a bunch of ways to hop around the periodic table, and this was one:

Delooping: given a $k$-tuply monoidal $n$-category, there should be a $(k-1)$-tuply monoidal $(n+1)$-category $B C$ with one object obtained by reindexing, the $j$-morphisms of $B C$ being the $(j+1)$-morphisms of $C$. We use the notation ‘$B$’ and call $B C$ the delooping of $C$ because of its relation to the classifying space construction in topology.

Right now we’re talking about the special case $k = 1$, $n = 0$ — and the special case of that where our monoid is actually a group.

Anyone puzzled about the relation between this concept of ‘$B$’ and another construction with the same name — the classifying space of a topological group — should check out page 22 of the same paper.

I agree with Mike that the nerve of a category deserves a different name. Not being very creative, I use $N$.

Posted by: John Baez on February 2, 2009 6:38 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

John, Mike,

thanks for the comments on $B G$! I am glad we are finally discussing this properly and converging to a joint attitude.

Here is one aspect which one might add to the discussion (but please check critically what I say now):

I am thinking that we have the operations of looping and delooping $B \leftrightarrow \Omega$ whenever we are in the context of a category with interval object ($\to$ $n$Lan) $I$.

Given any pointed object $X$ in such a category, we always have a notion of its loop object $\Omega X$, defined as the pullback $\array{ \Omega X &\to& [I,X] \\ \downarrow && \downarrow^{d_0 \times d_1} \\ pt &\stackrel{pt_X \times pt_X}{\to}& X \times X }$ which always inherits an $A_\infty$ monoidal structure. So we can then ask for an object $\mathbf{B}(\Omega X)$ with an essentially unique point such that $\Omega \mathbf{B}(\Omega X) \simeq \Omega X$.

Posted by: Urs Schreiber on February 3, 2009 10:56 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Maybe we should write an nLab page called “on the uses of B” or something.

Your notion of loop object in a category with interval object sounds reasonable. Another general notion of loop object is the following homotopy pullback: $\array{\Omega X & \to & 1\\ \downarrow && \downarrow \\ 1 & \to & X}$ As pointed out by Lurie, from this point of view the existence of deloopings is a type of $(\infty,1)$-categorical exactness. The loop object $\Omega X$, or more accurately the groupoid $\Omega X \;\rightrightarrows\; 1$, is the kernel of the point $1\to X$. Then if you have any internal group(oid) $G\;\rightrightarrows\; 1$, exactness says that it is a kernel, i.e. that $G\simeq \Omega(B G)$ for some $B G$.

Posted by: Mike Shulman on February 3, 2009 2:07 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

again equiv as?

also true for a more general class of monoids but not all

Posted by: jim stasheff on February 3, 2009 2:44 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Well, classically, as I’m sure that you (Jim) know (since you helped invent the subject), but other readers might not, one has the equivalence for any “grouplike $A_\infty$-space.” In an exact $(\infty,1)$-category one has the equivalence for any “internal group(oid) object;” this is made precise in Lurie’s work but amounts to about the same thing: a space that has a multiplication which is associative, unital, and has inverses up to all higher homotopies. (Formally, Lurie’s notion is more akin to a Segal space than to the operadic approach.) In a grouplike $A_\infty$-space, the inverses are only stipulated to exist up to homotopy rather than all higher homotopies, but one of the special things about inverses, like adjoints, is that they can always be “boosted up” to become coherent.

Posted by: Mike Shulman on February 3, 2009 3:45 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Another general notion of loop object is the following homotopy pullback

Good that you mention this, because it gives me an excuse to raise again an issue which I tried to discuss recently in the Ben-Zvi thread:

first, notice that the $n$Lab already essentially mentions this example (but I’ll make it more explcit):

at generalized universal bundle it says that the fiber of the universal $X$-bundle $\mathbf{E}_{pt} X$ (for pointed $X$ as in your comment above) is $\Omega X$. This is computed by the pullback

$\mathbf{E}_{pt}X \to X \leftarrow pt$

and this happens to be a fibrant replacement of the diagram

$pt \to X \leftarrow pt$

and hence computes the homotopy limit (as described there in detail) if $X$ is an un-directed object (i.e. a groupoidal object), i.e. if $[I,X]$ is a path object: in this case we have $\mathbf{E}_{pt} X \simeq pt$ and $\Omega X$ is an ($A_\infty$-)group.

But for many applications it is important to speak of $\Omega_{pt} X$ in the caser that all this does not hold, but where $\Omega_{pt} X$ still exists as an ($A_\infty-$)monoid.

For instance for $X = Vect_k$ with ground field $k$ as the point we do not have $\mathbf{E}_{pt} X \simeq pt$ and $\Omega_{pt} X$ (using my definition above) is not a group but is the monoid $k$. Similarly for $X = 2Vect$ with $Vect$ as a point, in which case $\Omega_{pt} X$ is $Vect$, etc.

In all these situations the definition above still works, but the one in terms of homotopy limits does not.

It is these kind of examples (where we are talking about $X$-bundles with $X$ not groupoidal) which make me hesitate to pass, for instance, to the $(\infty,1)$-category associated with my underlying homotopical category: this forgets the directed homotopies which are important in some applications.

Posted by: Urs Schreiber on February 3, 2009 4:44 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Well, I would argue that when there are “directed homotopies” in sight, often you aren’t really in an $(\infty,1)$-category but some sort of $n$-category for $n\gt 1$. In that case the homotopy pullback should be replaced by a comma object. I just wrote a bit about that in my own web at exactness hypothesis.

Posted by: Mike Shulman on February 3, 2009 5:09 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Well, I would argue that when there are “directed homotopies” in sight, often you aren’t really in an $(\infty,1)$-category […]

Yes, that’s exactly my point. Well, I guess I shouldn’t be arguing here in the context of a discussion that I had with somebody else in some other thread.

But I wanted to amplify this because it felt to be of some relevance in the context of the $B G$ discussion here. It makes me grow more fond of the concept of a closed monoidal homotopical category with interval object as a good context for higher abstract nonsense.

I just wrote a bit about that in my own web at exactness hypothesis.

Your comment about the comma object there provokes an obvious question: can we find the open entry in the analogy

$\array{ pseudo-limit & homotopy limit \\ comma object & ?? }$ ?

Well, I suppose this is asking for directed homotopy theory

One last comment, concerning your remark about $\mathbf{B} M$ as a codescent object:

by the general nonsense of codescent, I think it should be true that the realization of $\mathbf{B}M$ as a strict $\infty$-category is the coend

$\mathbf{B}M = \int^{[n] \in \Delta} M^{\times n} \cdot O(\Delta^n)$

(with $O(\Delta^n)$ the $n$th oriental, as you know).

Incidentally, this is (even if not in this language) the motivating example that Lafont, Métayer et al. give at the beginning of a coupel of their articles on strict $\infty$-categories for motivating the use of (strict) higher categories in computer science.

Posted by: Urs Schreiber on February 3, 2009 6:56 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

can we find the open entry in the analogy pseudo-limit : homotopy limit :: comma object : ??

I would personally say “weighted homotopy limit”. The best way I know of to use homotopy theory to get at non-invertible higher morphisms is to use enriched homotopy theory: the homotopy theory builds in the homotopies and the enrichment gives you the directionality.

by the general nonsense of codescent, I think it should be true that the realization of $B M$ as a strict $\infty$-category is the coend

Yes, certainly!

Posted by: Mike Shulman on February 4, 2009 4:15 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I take it [I,X] means Map(I,X)
and not homotopy classes as elsewhere?

That last equiv is as ???

and BOmegaX equiv X? as ???

Note also that if that bottom arrow were the diag X –> X x X
then the pull back is the free loop space more beloved of physicists.

Posted by: jim stasheff on February 3, 2009 2:42 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Note also that if that bottom arrow were the diag $X \to X \times X$ then the pull back is the free loop space more beloved of physicists.

Sure, this is discussed at

Posted by: Urs Schreiber on February 3, 2009 5:03 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I take it $[I,X]$ means $Map(I,X)$

Essentially, yes. The thing is that when talking about this object $I$ we are (or at least I am in the above discussion) placing ourselves in the context of a category $V$ with interval object, where $I$ is that interval object, which means in particular that our category $V$ is closed monoidal homotopical and hence in particular closed monoidal.

With this context in mind I am following standard convention in enriched category theory to denote the internal hom from $I$ to $X$ by $[I,X]$.

So in particular if $V =$ nice topological spaces then, yes, $[I,X] = Maps(I,X)$.

But if for instance $V = Groupoids$ (the other case occuring frequently in our discussion here) then $[I,X] = Funct(I,X)$ is the functor category.

Posted by: Urs Schreiber on February 3, 2009 7:16 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I didn’t mean Maps to denote topology
Perhaps I should have said Mor

Posted by: jim stasheff on February 4, 2009 1:47 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Urs wrote a nice explanation, ending with:

$\array{ \mathbf{B}G & \mathbf{B}G \stackrel{tra(\gamma)}{\to} \mathbf{B}G & tra(\gamma) \stackrel{\exists !}{\Rightarrow} tra(\gamma') \\ category & functor & natural transformation \\ fiber & connection & curvature \\ P_x & \nabla & F_\nabla } \,.$

So that’s how this works. Before going into the further analogy

$\array{ category & functor & natural transformation \\ fiber & connection & curvature \\ position & velocity & acceleration }$ let’s massage the above kind of observation into a form that we are all happy with.

I’m very happy with how you described this analogy:

$\array{ category & functor & natural transformation \\ fiber & connection & curvature }$

Maybe other people find the path 2-groupoid $P_2(X)$ scary, or smooth 2-functors

$tra: P_2(X) \to [B G, smooth endofunctors on B G, smooth natural isomorphisms]$

but I’m comfy with these concepts, and eager to explain them to anyone who has questions.

So maybe we can focus on the next stage, which is understanding the analogy

$\array{ fiber & connection & curvature \\ position & velocity & acceleration }$

I don’t understand this very well! It’s embarrassing, since I’ve spent years thinking about each row.

Posted by: John Baez on January 31, 2009 7:31 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

This thought keep crossing my mind and I’m not sure if it is relevant, but thought I would say it out loud and see what happens…

Given a smooth curve

$\gamma:I\to M$

in a smooth manifold $M$, what if we thought of $M$ as a fiber over $I$ so that a curve in $M$ is a section of the bundle $\pi:M\to I$?

Then “position” is like a tangent vector and “velocity” is like a connection telling you how to parallel transport a tangent vector, i.e. position, from $p_t\in M_t$ to $p_{t+\epsilon}\in M_{t+\epsilon}$, where

$p_t = \gamma(t)$

and

$M_t = \pi^{-1}(t).$

But then I get lost on acceleration and curvature…

Treating $M$ as a directed space might help because then curvature/acceleration might come about as a type of Aharanov-Bohm effect, which has a certain poetic aspect to it.

I’m starting to see diamonds, so I better stop their because I’ve already rambled too much.

Posted by: Eric on January 31, 2009 11:36 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Ok. When I said I would stop I lied :)

I just started a draft here:

—-

Consider two paths $\gamma_+,\gamma_-$ on a directed graph $G$ that both begin at the point $(x_i,t_i)$ and end at the point $(x_f,t_f)$ in the figure below

$\array{ {} &{} & {} & (x_f,t_f) & {} & {} \\ {} &{} & {} & \bullet & {} & {} \\ {} &{} & {}^{v_{-,f}}\nearrow & {} & \nwarrow^{v_{+,f}} & {} \\ (x_-,t_-)&\bullet & {} & a & {} & \bullet & (x_+,t_+) \\ {} &{} & {}_{v_{i,-}}\nwarrow & {} & \nearrow_{v_{i,+}} & {} \\ {} &{} & {} & \bullet & {} & {} \\ {} &{} & {} & (x_i,t_i) & {} & {} }$

The edges represent velocities and the area enclosed represents the acceleration.

If the acceleration is zero, then the area collapses and any two paths that connect the same two points must correspond to the same straight line, i.e. velocity does not change.

Posted by: Eric on February 1, 2009 12:13 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Ok. If I’m going to ramble…

Combining three ideas: acceleration as area, Feynman checkerboard, and Higgs, then it is tempting relate the area to mass. If the edges are lightlike, then mass = 0 corresponds to acceleration = 0, so particles would travel at the speed of light.

Massive particles have a nonzero acceleration 2-form.

Posted by: Eric on February 1, 2009 12:19 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I’ll state the crux even though at this point it is obvious.

Fiber = Position
Connection = Velocity
Curvature = Acceleration

Posted by: Eric on February 1, 2009 12:24 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Coupling this to electromagnetic theory should be trivial too. The minimal principle probably leads to the area 2-form

$F_{area} = d A_{velocity} = d\star dp$

being replaced with

$F_{area} + F_{em},$

where $F_{em}$ is the electromagnetic curvature

$F_{em} = dA_{em}.$

Then “zero acceleration” means

$F_{area} = -F_{em}.$

There may be a Hodge star or sign missing, but I’m pretty sure the basic idea is correct.

Posted by: Eric on February 1, 2009 12:42 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Like the portrayal of Salieri in Amadeus, I feel teased by seeing a flash of something beautiful, but cursed because I do not have the gifts needed to write it down.

I’ve been doodling trying to make some sense of this in simplistic terms, but with little luck. I think I’ve settled on a picture that even a high school student could understand, so I hope it fits into the picture that I’m anxiously waiting to hear about from Urs. Anyway, here is the picture:

———-

Galilean Spacetime

Consider two paths that begin at $x_0$ and end at $x_{2\Delta t}$ enclosing the spacetime region depicted below

$\array{ {} & {} & {} & x_{2\Delta t} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v^-_{\Delta t}}\nearrow & {} & \nwarrow^{v_{\Delta t}^+} & {} & {} \\ x^-_{\Delta t} = x_0 - v^-_0 \Delta t & \bullet & {} & \stackrel{a}{\Rightarrow} & {} & \bullet & x^+_{\Delta t} = x_0 + v^+_0 \Delta t \\ {} & {} & {}_{v_0^-}\nwarrow & {} & \nearrow_{v_0^+} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }$

Since both paths end up at the same point, the velocities must satisfy the relation

$v^-_{\Delta t} - v^-_0 = v^+_0 - v^+_{\Delta t},$

which implies that we can write

$v^-_{\Delta t} = v^-_0 + (+a)\Delta t$

and

$v^+_{\Delta t} = v^+_0 + (-a) \Delta t$

for some constant $a$.

———-

In haste, I wrote some nonsensical equations earlier, but from the picture above, if you think of the velocity as a “holonomy along an edge”, then the expression

$v^-_{\Delta t} - v^-_0 = v^+_0 - v^+_{\Delta t},$

says there is no curvature, i.e.

$A = dp\quad\quad\text{("velocity connection")}$

and

$F = dA = 0.$

Nonetheless, I remain convinced that acceleration is somehow related to a 2-form defined over the 2-dimensional Galilean spacetime region, i.e. 2-diamond, above. This 2-form tells you how to (2-)transport the left path across region to the right path.

It actually seems to fall out nicely in the discrete world, so it should have an acceptable continuum formulation as well.

Another indication, which is also simplistic but meaningful in my opinion, is that

* It takes one tick of a clock (zero intervals) to measure position.

* It takes two ticks of a clock (one interval) to measure velocity.

* It take three ticks of a clock (two intervals) to measure acceleration.

—-

In a directed graph:

* It takes one tick to traverse a node (or 0-diamond)

* It takes two ticks to traverse an edge (or 1-diamond)

* It takes three ticks to traverse a 2-diamond

——

Putting these together:

* Position lives on 0-diamonds and is a 0-form

* Velocity lives on 1-diamonds and is a 1-form

* Acceleration lives on 2-diamonds and is a 2-form

That is about as far as I’ve been able to get so far…

Posted by: Eric on February 2, 2009 6:30 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Hi John,

So maybe we can focus on the next stage […]

Okay. In my comment which started this discussion I essentially pointed out that:

the bijection/equivalence

(smooth 1-functors from $paths in X \to endomorphisms of G$) $\simeq$ (1-forms)

is – manifestly the way one proves it – controlled by the concept of velocity:

if you think of $X$ as 1-dimensional, then the 1-form on $X$ with values in $Lie(G)$ is nothing but the gadget which reads in a bit of “coordinate time” for a trajectory on $G$ and spits out the velocity tangent vector of that trajectory with respect that coordinate time.

This is really what “parallel transport” of 1-forms means: find the trajectory such that its velocity is at every point prescribed by the given 1-form.

Think in the case $X = \mathbb{R}$ of the functor as describing a particle which propagates on a group manifold. The fact that the functor is smooth says that we have a smooth trajectory. The fact that it is functorial means that this trajectory is obtained by intergrating a velocity 1-form.

Now, in my comment which started this discussion I next said that on 2-cells our parallel transport computes the “exterior acceleration”: given a 1-parameter family of trajectories, the 2-form that controls the value of a smooth 2-functor on it is the exterior differential of the “velocity 1-form”.

At first it is admittedly a bit of a stretch as far as the Newtonmian concept of acceleration goes to call this curvature 2-form the “acceleration” here, though it does consist of second derivatives of the original trajectories.

But then notice the Newton-force law for charged relativitic particles: the Lorentz-force law. it says that the accelreation bivector $\dot v v$ of the relativistic particle equals the (electric component of) the curvature 2-form. So we have

$\array{ fiber & inner degrees of freedom at given position \\ connection & action functional in terms of velocity \\ curvature & force law in terms of acceleration } \,.$

There must be better ways to say this. But maybe this helps to indicate what I am thinking of.

Posted by: Urs Schreiber on February 2, 2009 1:26 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I was playing around with this a little last night…

Let $V$ denote the 4-velocity of a charged particle and $v$ denote the corresponding dual 1-form, the acceleration 1-form may be written as

$a = \frac{q}{m} i_V F$

where $F$ is the curvature 2-form and the “acceleration 2-form” (corresponding to the acceleration bivector) is

$v\wedge a = \frac{q}{m} v\wedge(i_V F).$

I guess this is equal to $F$. To see this, note that

$i_V(v\wedge F) = (i_V v)F - v\wedge(i_V F)$

We have

$i_V v = v(V) = |V|^2 = 1$

and

$i_V(v\wedge F) = 0$

so that

$F = v\wedge(i_V F) = \frac{m}{q} v\wedge a.$

I believe this is the clue that Urs mentioned, but made a bit more explicit. On the left is curvature and the right is the acceleration 2-form.

In my Galilean spacetime example, I think it is the “acceleration 2-form” $v\wedge a$ that specifies the 2-transport even in the absence of an electromagentic field.

Posted by: Eric on February 2, 2009 9:52 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Eric wrote:

$i_V v= v(V)= |V|^2 = 1$

Hmm. Before you guys were saying something really nice happens when we assume our particle is massless. Then its velocity vector is lightlike, which gives

$|V|^2 = 0$

Your calculation seems to work better for a massive particle, where we can use the metric to turn its velocity vector $v$ into a 1-form $V$ with $v(V) = |V|^2 = 1$. Of course we can always find some 1-form with $v(V) = 1$, but probably not in a canonical way (using the metric) if $v$ is lightlike.

I need to think about this more… there should indeed be something fun going on here, and Urs has probably explained it already, but I need to unpack it and ponder it.

In my Galilean spacetime example, I think it is the “acceleration 2-form” $v \wedge a$ that specifies the 2-transport even in the absence of an electromagentic field.

If you take some light whose color is magenta, is that an example of an electromagentic field?

(Sorry, I’m just a goofball.)

Posted by: John Baez on February 3, 2009 1:10 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

You can rest assured knowing that when I start writing down equations, a good 10-15% of them will not be correct. That is a small price to pay if we can push the idea along in my opinion :)

I don’t think

$i_V(v\wedge F) = 0,$

but it was a convenient mistake to illustrate a partial relationship between curvature and the acceleration 2-form.

As far as massive vs massless particles, one of the points (I think) is that for massless particles, the acceleration is zero so they will always propagate in a straight line. Bringing Higgs in would be fun, but then I think the definition of “mass” becomes tricky.

If there is any truth to anything I’m saying, it is only because I’m trying to parse what Urs already said in a way I might understand. There is a good chance I am just spewing a bunch of nonsense though (it wouldn’t be the first time!).

I hope the spacetime diagram ends up being useful because it is very pretty.

Posted by: Eric on February 3, 2009 3:34 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Before you guys were saying something really nice happens when we assume our particle is massless.

I may have to take this back, sorry. What I really argued so far is that the nice 1-2-3-sequence we are looking for is obtained when thinking of acceleration for relativistic particles, which is naturally a bivector that naturally pairs with the curvature 2-form of the background field which exerts the force.

Above I thouhgt lightlike particles would make things even nicer, but possibly I was being silly. Let me think about it a bit more…

Posted by: Urs Schreiber on February 3, 2009 4:53 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Oh.

When there is no electric field, then $F = B$ and the acceleration, velocity, and magnetic field are mutually orthogonal. In this case, the geometric product

$va = v\wedge a + v\cdot a = v\wedge a$

is the same as the wedge product and you can “invert” the interior product so (I think)

$v\wedge(i_V B) = B.$

In this case, you have

$v\wedge a = \frac{q}{m} B$

and the relation between curvature and the acceleration 2-form is explicit.

I think the math is correct this time. To check

$i_V(v\wedge a) = a - i_V(a) = a$

since $i_V(v) = 1$ and $i_V(a) = v\cdot a = 0$. Also

$i_V\left(\frac{q}{m} B\right) = a$

which is the Lorentz force law.

Is that what you’ve been saying all along? :)

Posted by: Eric on February 3, 2009 5:55 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

I’m getting closer. Promise. I think what I did above is a special case of what you’ve been saying.

Since

$a = \frac{q}{m} i_V F,$

it is generally true that $v\cdot a = 0$ since $i_V$ is nilpotent, i.e.

$v\cdot a = i_V a = \frac{q}{m} i_V i_V F = \frac{q}{m} F(V,V) = 0$

since $F$ is antisymmetric. Therefore, the acceleration 2-form is always the same as the geometric product of velocity and acceleration

$va = v\wedge a + v\cdot a = v\wedge a.$

At least when the acceleration comes from a curvature.

SOOO…

When the acceleration comes from some curvature, then the acceleration 2-form $v\wedge a$ is “special”. I think it is special because it specifies the 2-transport between particle world lines.

Posted by: Eric on February 3, 2009 6:29 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Help me Mozart!

When the acceleration comes from some curvature, then the acceleration 2-form $v\wedge a$ is “special”. I think it is special because it specifies the 2-transport between particle world lines.

Does that paragraph make anyone else think about general relativity?

I wish I was smarter!

Posted by: Eric on February 3, 2009 6:33 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

With that last comment in mind and going back to the 2-diamond example:

$\array{ {} & {} & {} & x_{2\Delta t} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v^-_{\Delta t}}\nearrow & {} & \nwarrow^{v_{\Delta t}^+} & {} & {} \\ x^-_{\Delta t} & \bullet & {} & \stackrel{va}{\Rightarrow} & {} & \bullet & x^+_{\Delta t} \\ {} & {} & {}_{v_0^-}\nwarrow & {} & \nearrow_{v_0^+} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }$

If we switch to a relativistic perspective and think of the edges as being lightlike (geodesics), then this depicts a discrete general relativistic picture where the two paths are actually geodesics implying the spacetime surface contains curvature and the acceleration 2-form specifies the 2-transport from the left world line to the right world line.

Posted by: Eric on February 3, 2009 6:50 PM | Permalink | Reply to this

### Geometric Algebra

In the spacetime diagram

$\array{ {} & {} & {} & x_{2\Delta t} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v^-_{\Delta t}}\nearrow & {} & \nwarrow^{v_{\Delta t}^+} & {} & {} \\ x^-_{\Delta t} & \bullet & {} & \stackrel{va}{\Rightarrow} & {} & \bullet & x^+_{\Delta t} \\ {} & {} & {}_{v_0^-}\nwarrow & {} & \nearrow_{v_0^+} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }$

I was trying to figure out exactly what should go on the edges and what should go on the surface. I knew the edges had to do with velocity and the surface had to do with the acceleration bivector.

Then I got it (by following the link the Urs gave to the geometric algebra stuff!). I had the 2-arrow pointed in the wrong direction. Instead of 2-transporting the left worldline to the right worldine, it should actually be a “boost” along the time direction, i.e. the 2-arrow should be “up”!

$\array{ {} & {} & {} & x_{2\Delta t} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{R(v^-_{\Delta t})}\nearrow & {} & \nwarrow^{R(v_{\Delta t}^+)} & {} & {} \\ x^-_{\Delta t} & \bullet & {} & \stackrel{R(va)}{\Uparrow} & {} & \bullet & x^+_{\Delta t} \\ {} & {} & {}_{R(v_0^-)}\nwarrow & {} & \nearrow_{R(v_0^+)} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }$

But now what gets assigned the edges are “rotors” from geometric algebra. The edge rotors represent a Lorentzian “boost” along the edge direction. The surface rotor represents a rotation of the initial velocity edge into the final velocity edge.

If we think of this this as an infinitessimal region, then globbing these things together amounts to intergration. The velocity and acceleration rotors are constant on the infinitessimal 2-diamond.

That is neat!

Posted by: Eric on February 5, 2009 4:05 AM | Permalink | Reply to this

### Re: Geometric Algebra

A little more explicitly…

$\array{ x^+_{\Delta t} &= R(v^+_0) x_0 \tilde{R}(v^+_0) \\ x^-_{\Delta t} &= R(v^-_0) x_0 \tilde{R}(v^-_0) \\ v^+_{\Delta t} &= R(v^+_0 a) v^+_0 \tilde{R}(v^+_0 a) \\ v^-_{\Delta t} &= R(v^-_0 a) v^-_0 \tilde{R}(v^-_0 a) }$

Almost there.

Posted by: Eric on February 5, 2009 4:42 AM | Permalink | Reply to this

### Re: Geometric Algebra

By the way, I modified the spacetime diagram to make it look more like a strict 2-category at the nLab:

$\array{ {} & {} & {} & x_{2\tau} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v^-_{\tau}}\nearrow & {} & \nwarrow^{v_{\tau}^+} & {} & {} \\ x^-_{\tau} & \bullet & \stackrel{F^-}{\Uparrow} & \stackrel{v_0}{\uparr} & \stackrel{F^+}{\Uparrow} & \bullet & x^+_{\tau} \\ {} & {} & {}_{v_0^-}\nwarrow & {} & \nearrow_{v_0^+} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {} & x_0 & {} & {} & {} }$

• Objects $\{x_0,x^+_\tau,x^-_\tau,x_{2\tau}\}$
• Morphisms $\{x_0\stackrel{v_0}{\to}x_{2\tau},x_0\stackrel{v^+_0}{\to}x^+_\tau,x_0\stackrel{v^-_0}{\to}x^-_\tau,x^+_\tau\stackrel{v^+_\tau}{\to}x^+_{2\tau},x^-_\tau\stackrel{v^-_\tau}{\to}x^-_{2\tau}\}$
• 2-Morphisms $\{v^+_0\stackrel{F^+}{\Rightarrow}v^+_\tau,v^-_0\stackrel{F^-}{\Rightarrow}v^-_\tau\}$

Now, I’m hoping to do something along the lines of explicitly constructing a 2-transport functor

$tra: P^2(\mathbb{D}^2)\to\mathcal{M}^4$

somehow, where $\mathcal{M}^4$ is the spacetime algebra.

If things work out, it looks like we might be able to express “surface ordered surface integrals” via geometric algebra.

Posted by: Eric on February 6, 2009 4:26 PM | Permalink | Reply to this

### Re: Geometric Algebra

where $\mathcal{M}^4$ is the spacetime algebra.

This “spacetime algebra” is just the Clifford algebra $CL_{3,1}$. This is not a 2-category, so your proposed to functor does not quite make sense as stated, as its target should be some 2-category.

Now $Cl_{3,1}$ itself is not a 2-category. But it can naturally be regarded as being an object in a 2-category: namely the 2-category $Vect-Mod$!

if you conmsider that, then the situation you indicate starts to look very much like the situation discussed on page 25 of AQFT from $n$-functorial QFT (#) (the corrected version to appear in print I’ll upload in a few days.)

Posted by: Urs Schreiber on February 6, 2009 5:17 PM | Permalink | Reply to this

### Re: Geometric Algebra

Thanks Urs. I know you are traveling and very busy (as usual), so I’m slogging along trying to make some progress. When things return to normal levels of insanity, I’m hoping I make enough progress that you can help clean up any remaining issues with just a few brush strokes.

I’ll try to understand the “more correct” object I think I’m after:

$tra: P^2(\mathbb{D}^2)\to 2Vect,$

where $P^2(\mathbb{D}^2)$ is the strict 2-category representation of the diamond spacetime region $\mathbb{D}^2$. That sounds pretty.

I’m guessing that this is just some extremely simple situation of the generally picture you’ve already written down, but making it explicit at this elementary level would be helpful I think.

Posted by: Eric on February 6, 2009 5:40 PM | Permalink | Reply to this

### Re: Geometric Algebra

I just noticed I read that wrong. I need to learn about $Vect-Mod$.

Posted by: Eric on February 8, 2009 9:07 PM | Permalink | Reply to this

### Complex Numbers and Vect-Mod

I know it is probably extreme overkill, but is there a way to think of complex numbers $\mathbb{C}$ as sitting inside $Vect-Mod$ somehow?

Posted by: Eric on February 9, 2009 4:19 PM | Permalink | Reply to this

### Re: Geometric Algebra

Eric wrote:

I know it is probably extreme overkill, but is there a way to think of complex numbers $\mathbb{C}$ as sitting inside $Vect-Mod$ somehow?

Sure! $\mathbb{C}$ is the unit object of the unit object of $Vect-Mod$.

Here’s what I mean:

$Vect-Mod$ is a monoidal bicategory, i.e. there’s a tensor product of $Vect$-modules.

The unit object for this tensor product is $Vect$. In other words, you can tensor any $Vect$-module $A$ with $Vect$ and get a $Vect$-module equivalent to $A$.

$Vect$ is a monoidal category: i.e. there’s a tensor product of vector spaces (which are just $\mathbb{C}$-modules).

The unit object for this tensor product is $\mathbb{C}$. In other words, you can tenso rany $\mathbb{C}$-module $A$ with $\mathbb{C}$ and get a $\mathbb{C}$-module isomorphic to $A$.

I hope you see, from the singsong repetitive quality of this comment, that this is just the beginning of a pattern that continues forever.

By the way, speaking of patterns that continue forever: when you keep commenting on your own comments to your own comments, it creates a highly nested thread that’s not much fun to read. All those vertical lines on the left get a bit dizzying! To avoid them, it’s better to keep commenting on the topmost relevant comment.

Posted by: John Baez on February 9, 2009 8:57 PM | Permalink | Reply to this

### Re: Geometric Algebra

is there a way to think of complex numbers $\mathbb{C}$ as sitting inside $Vect−Mod$ somehow?

John replied:

Sure! $\mathbb{C}$ is the unit object of the unit object of $Vect−Mod$.

Maybe Eric is coming from my statement above that the Cliffford algebra $Cl_{3,1}$ can be regarded as an object of $Vect-Mod$. In that case, maybe he is asking if the complex numbers $\mathbb{C}$ regarded as an algebra are also an object of $Vect-Mod$. (?)

In this case the answer is Yes, but this is another way we can think of the symbol $\mathbb{C}$ as naturally denoting something “inside” $Vect-Mod$:

every algebra $A$ defines an object of $Vect-Mod$. We can think of these algebras

- as being the objects of $Vect-Mod$ (if we adopt the point of view of $Vect$-enriched category theory in which case the objects of $Vect-Mod$ are $Vect$-enriched categories which are algebras if they have a single object)

- or as being placeholders for the true objects of $Vect-Mod$ (namely if we regard this as the bicategory of $Vect$-module categories), in which case the true object is the category $Mod_A$ of modules over the algebra $A$ (vector spaces on which the algebra is represented!).

The complex numbers $\mathbb{C}$ regarded as an algebra are an object of $Vect-Mod$ in the first sense. As such they may be regarded as representing the category $Mod_{\mathbb{C}}$ of representations (= modules) of the algebra of complex numbers. But every complex vector space is tautologically such a representation. So $Mod_{\mathbb{C}} = Vect_{\mathbb{C}}$. This leads to the perspective that John described as we can find $\mathbb{C}$ sitting inside its own category of representations

$\mathbb{C} \in Mod_{\mathbb{C}} = Vect_{\mathbb{C}} \,.$

Posted by: Urs Schreiber on February 9, 2009 11:27 PM | Permalink | Reply to this

### Re: Geometric Algebra

I wonder if it is a coincidence that the functor

$End: (E\stackrel{e^{itH}}{\to}E)\mapsto (End(E)\stackrel{e^{itH} o(-) e^{-itH}}{\to}End(E))$

on Page 4 looks like a rotation $a\mapsto e^{\frac{ItW}{2}} a e^{-\frac{ItW}{2}}$ in geometric algebra for any multivector $a$?

Posted by: Eric on February 7, 2009 5:53 AM | Permalink | Reply to this

### Re: Geometric Algebra

I wonder if it is a coincidence…

I don’t know if one should call it a coincidence or not, but in any case it is not supposed to be a deep. You should think of the Heisenberg picture of quantum mechanics, where time evolution is given by an inner automorphism, i.e. by conjugation with a unitary operator. Also rotations in quantum mechanics are represented by such conjugations on other operators. Whether we have a spin representation or not.

Posted by: Urs Schreiber on February 9, 2009 11:10 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

John wrote:

The way the n-Category Café works, you have a lot of people reading what you write; but it’s a blog, so if I’m a typical example, a lot of these people start skimming whenever it gets too hard to understand.

I would say:
The way the n-Category Café works, you have NOT a lot of people reading what you write; IN COMPARISON TO YOUR POTENTIAL AUDIENCE. YOU DO GIVE TALKS BUT PUBLICATION MIGHT HELP. ;-) but it’s a blog, so if I’m a typical example, a lot of these people CAN’T KEEP UP.

It’s like those physics talks in which the slides flash by at the speed of light.

Posted by: jim stasheff on January 31, 2009 2:12 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Jim wrote:

The way the n-Category Café works, you have NOT a lot of people reading what you write, IN COMPARISON TO YOUR POTENTIAL AUDIENCE.

It’s hard to know how many people are really paying attention to this blog, but it’s a lot more than the number of people who post comments. It’s not a substitute for putting papers on the arXiv and then publishing them, but it’s a good supplement.

54,902 people have visited the main page of the n-Café since September 25th, 2008. And they come from all over:

United States (US) 18,782

United Kingdom (GB) 6,268

Germany (DE) 4,493

France (FR) 2,271

Australia (AU) 1,471

Netherlands (NL) 1,015

Japan (JP) 1,006

Switzerland (CH) 894

Sweden (SE) 806

Spain (ES) 774

Brazil (BR) 724

Belgium (BE) 649

Portugal (PT) 640

Italy (IT) 498

Poland (PL) 443

Finland (FI) 426

Norway (NO) 421

Austria (AT) 363

China (CN) 352

New Zealand (NZ) 337

Russian Federation (RU) 320

Denmark (DK) 293

India (IN) 284

Greece (GR) 277

Czech Republic (CZ) 222

Argentina (AR) 210

Israel (IL) 205

Bulgaria (BG) 176

Korea, Republic of (KR) 158

Ireland (IE) 155

Croatia (HR) 149

Luxembourg (LU) 146

Thailand (TH) 140

Taiwan (TW) 132

Mexico (MX) 128

Colombia (CO) 122

Singapore (SG) 99

Malaysia (MY) 93

Hungary (HU) 82

Hong Kong (HK) 77

Turkey (TR) 62

Estonia (EE) 62

Chile (CL) 60

Romania (RO) 58

Europe (EU) 48

Philippines (PH) 48

Slovenia (SI) 47

Vietnam (VN) 43

South Africa (ZA) 42

Algeria (DZ) 31

Egypt (EG) 26

Iran, Islamic Republic of (IR) 25

Saudi Arabia (SA) 25

Ukraine (UA) 24

United Arab Emirates (AE) 24

Iceland (IS) 22

Indonesia (ID) 21

Venezuela (VE) 15

Lithuania (LT) 13

Uruguay (UY) 7

Costa Rica (CR) 7

Latvia (LV) 7

Asia/Pacific Region (AP) 7

Serbia (RS) 6

Slovakia (SK) 6

Jamaica (JM) 6

Oman (OM) 5

Nepal (NP) 5

Peru (PE) 5

Jordan (JO) 4

Pakistan (PK) 4

Qatar (QA) 3

Belarus (BY) 3

Moldova, Republic of (MD) 3

Libyan Arab Jamahiriya (LY) 2

Kuwait (KW) 2

Cyprus (CY) 2

Palestinian Territory (PS) 2

Bahrain (BH) 2

Tunisia (TN) 2

Bolivia (BO) 1

Rwanda (RW) 1

Paraguay (PY) 1

Ethiopia (ET) 1

Guam (GU) 1

Nicaragua (NI) 1

Guatemala (GT) 1

Cuba (CU) 1

Antigua and Barbuda (AG) 1

Dominican Republic (DO) 1

Iraq (IQ) 1

Syrian Arab Republic (SY) 1

Georgia (GE) 1

Bosnia and Herzegovina (BA) 1

Azerbaijan (AZ) 1

Yemen (YE) 1

Malta (MT) 1

Brunei Darussalam (BN) 1

Posted by: John Baez on January 31, 2009 6:55 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Impressive
but are those people or hits?

Posted by: jim stasheff on February 1, 2009 2:18 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

The precise way those numbers are computed can be seen here.

Briefly: these numbers count ‘visits’. Each day you go to this blog’s main page counts as a ‘visit’. If you go to main page more than once in a day, that still counts as a single ‘visit’. And if you don’t go to the main page, that counts as no visits at all.

Posted by: John Baez on February 3, 2009 1:27 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

So those of us who never go to the main page, but click through directly from the RSS feed to an individual blog entry, aren’t counted at all?

Posted by: Mike Shulman on February 3, 2009 2:55 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

I believe you’re not counted. I’m getting my data from that ClustrMaps thing on the main page of this blog, and I think it only counts people who visit that specific page.

So, if lots of people use RSS feeds to view individual comments, the above figures may seriously underestimate the number of visits.

Anyway, it’s clear we have masses of followers. Now it’s time to take over the world.

Posted by: John Baez on February 3, 2009 4:27 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

What if we change the picture like that:

• Position is given by 2-paths in 2-graph
• Velocity is given by paths
• Acceleration is given by vertices

Then we can use following boundary-equations:

$dom(dom) = dom(cod)$ $cod(dom) = cod(cod)$

Now if we code boundaries of given n-path $\alpha$ as $\delta \alpha = cod(\alpha)-dom(\alpha)$ and assume linearity then we have

$\delta^2 \alpha = 0$

from boundary-equation. The case with forces encodes 2-paths with some “hanging” edges, calibration encodes the way of “binding” these hanging edges (but I don’t know how to explain this in categorical way :)).

Is it ugly analogy? I think it seems like some “duality”.

Posted by: osman on October 1, 2009 1:27 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Zeh, in writing about time, mentions
initial conditions, which may be
analogous to position:

“It has indeed proven appropriate to
divide the formal dynamical description
of Nature into laws and initial
conditions. Wigner (1972), in his Nobel
Prize lecture, called this conceptual
distinction Newton’s greatest discovery,
since it demonstrates that the laws by
themselves are far from determining
Nature.”

Posted by: Stephen Harris on January 30, 2009 8:22 AM | Permalink | Reply to this

### Roger Penrose?

I’ll take the risk of suggesting something I not only don’t understand in any depth but also something that may be too specialised to count as concepts rather than specific gadgets, but how about Roger Penrose with:

1. Spinors networks.

2. Twistors.

3. Twistor sheaf cohomology.

Posted by: bane on January 30, 2009 10:09 AM | Permalink | Reply to this

### Re: The Third Time is the Charm

Not to be deterred…

How about: charge, current, magnetic flux: Maxwell?
And is there a functorial relationship between those and position, velocity, acceleration?

Of course, anything along these lines was already properly formulated and detailed long ago, but I’m just getting started, so I don’t know anything yet.

Posted by: Charlie C on January 30, 2009 2:28 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

How about: charge, current, magnetic flux: Maxwell? And is there a functorial relationship between those and position, velocity, acceleration?

One can formulate all things “charge, current, magnetic flux” along the lines of smooth $n$-functors as in the above example. I talked about that in Charges and twisted $n$-bundles, I.

Doing so unifies phenomena. Though I think it will also show that the triple you mention in not the kind of 1-2-3-sequence analogoues to position-velocity-acceleration, bundle-connection-curvature and category-functor-transformation.

For instance charge and electric current are really two aspects of the same entity, one integral, one local. And magnetic flux is dual.

The best discussion of the subtleties of this after Dirac himself is in the introductory section of Freed’s Dirac Charge Quantization and Generalized Differential Cohomology

Posted by: Urs Schreiber on January 30, 2009 2:50 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Thank you for the comments, Urs. Your Charges and Twisted n-Bundles, I post provides exactly the kind road map needed to guide the thinking and motivate the tools required to understand your approach. This kind of high-level overview is indispensable as a guide to my studies as I assemble a minimal set tools and techniques necessary to understand the mathematics of modern physics. I can not hope to contribute; I can not hope to solve problems; but I can hope to understand how the pieces fit together and basically how each piece works. My dream is someday to see a guide for the intelligent, motivated, undergraduate-level non-specialist that leads one step-by-step, starting with the most basic concepts, through all the relevant fields of math (no proofs required, but no shirking the tough topics either), to the solution of a simple but uniquely contemporary physical problem. John and you and others have already done much work in this direction. Maybe a corner of the n-Lab would be a good place to start gluing all the necessary pieces together. I know you are too busy creating to teach kindergarten classes. But maybe someone out there can help.

Posted by: Charlie C on January 31, 2009 10:48 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Fourier series
Convergence
Sets

Posted by: anon on January 30, 2009 7:36 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

Someone’s been reading Everything and More. But no, even though considering one may lead one to have to consider the next, the original goal was to explain Fourier series, not to explain sets.

Posted by: John Armstrong on January 30, 2009 10:31 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

My first thoughts were:
1. Paths and loops
2. Homotopies
3. Fundamental groups
(Poincare?)

Posted by: Jan de Wit on February 2, 2009 3:26 PM | Permalink | Reply to this

### All about B and Omega

Just posted and quite relevant

arXiv:0902.0177
Date: Mon, 2 Feb 2009 00:45:24 GMT (49kb)

Title: Operadic bar constructions, cylinder objects, and homotopy morphisms of
Authors: Benoit Fresse
Categories: math.AT
MSC-class: 18D50; 18G55
\
The purpose of this paper is twofold. First, we review applications of the
bar duality of operads to the construction of explicit cofibrant replacements
in categories of algebras over an operad. In view toward applications, we check
that the constructions of the bar duality work properly for algebras over
In a second part, we use the operadic cobar construction to define explicit
cyclinder objects in the category of operads. Then we apply this construction
to prove that certain homotopy morphisms of algebras over operads are
equivalent to left homotopies in the model category of operads.
\ ( http://arxiv.org/abs/0902.0177 , 49kb)

Posted by: jim stasheff on February 3, 2009 2:52 PM | Permalink | Reply to this

### Re: The Third Time is the Charm

How about examples of questions where the third step is where things go wrong?

For example, in ordinary differential equations? In one dimension everything is “trivial”: everything boils down to monotone regions; in two dimensions, again the asymptotics are generically straight-forward (because separatrices actually separate different dynamics). But in three dimensions, we typically get chaos!

Or how about simplicial cycles? In an n-cycle, the interior of an n-cell is a good manifold, and so is the interior of an (n-1)-cell and even (!) the interior of an (n-2)-cell; but in codimension 3, we can get singular loci. Or perhaps you might consider that something going wrong at the *fourth* step, I don’t know…

Has anyone any less-geometric examples?

Posted by: some guy on the street on February 6, 2009 5:52 AM | Permalink | Reply to this

### Explicit Construction of a Transport 2-Functor via Clifford Algebra

Thanks John and thanks Urs.

My questions about $Vect-Mod$ come from my scribblings here:

==================

Formulation

First, due to the symmetry of the diagram above, we can consider the simplified region

$\array{ {} & {} & {} & x_{\tau} & {} & {} & {} \\ {} & {} & {} & \bullet & {} & {} & {} \\ {} & {} & {}^{v_0}\nearrow & {}_F\Rightarrow & \searrow^{v_{\tau}} & {} & {} \\ x_0 & \bullet & {} & \stackrel{v}{\longrightarrow} & {} & \bullet & x_{2\tau} }$

This diagram represents a very simple strict 2-category we will denote by $P^2(\mathbb{D}^2)$ with

• Objects $\{x_0,x_\tau,x_{2\tau}\}$
• Morphisms $\{x_0\stackrel{v_0}{\to}x_{\tau},x_\tau\stackrel{v_\tau}{\to}x_{2\tau},x_0\stackrel{v}{\to}x_{2\tau}\}$
• 2-Morphism $\{v_0\stackrel{F}{\Rightarrow}v_\tau\}$

We will explicitly construct a 2-transport functor

$tra: P^2(\mathbb{D}^2)\to Vec-Mod,$

where

$x\mapsto tra(x)\in\mathcal{M}^2$

===============

I was initially going to write $\mathcal{M}^4$, but thought the simple example could be described by $\mathcal{M}^2$, but then thought you might describe $\mathcal{M}^2$ as $CL_{1,1}$, but then recognized that as $\mathbb{C}$.

As you can tell, I am floundering about, but not giving up. Thanks again for your help.

Posted by: Eric on February 10, 2009 12:42 AM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

Eric,

as I said, a 2-functor has to send objects of a 2-category to objects of a 2-category, so $tra(x)$ has to be an object of a 2-category. But $\mathcal{M}^2$ is supposed to denote an algebra, not a 2-category, as far as I understand. This algebra, however, may be regarded itself as being an object of a 2-category. So writing $tra(x) = \mathcal{M}^2$ would parse. $tra(x) \in \mathcal{M}^2$ does not quite parse.

$Cl_{1,1}$, but then recognized that as $\mathbb{C}$.

Let’s see, maybe ou are reading the notation different from its intention: $Cl_{n,m}$ is supposed to denote the Clifford algebra of the vector space $\mathbb{R}^{n + m}$ equipped with the canonical bilinear form which has $n$ entries 1 and $m$ entries -1 on its diagonal.

So $\mathbb{C}$ can be identified with $Cl_1$, not $Cl_{1,1}$. $Cl_1$ is the real algebra generated by a unit and a single new generator which squares to -1. This is $\mathbb{C}$.

But $Cl_{1,1}$ is generated from a unit, a generator $\gamma_0$ that squares to +1 and yet another generator $\gamma_1$ which squares to -1 such that $\gamma_0$ and $\gamma_1$ anticommute.

Posted by: Urs Schreiber on February 10, 2009 7:26 AM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

Hi Urs,

I am stuck on this idea of making “elements” of $\mathcal{M}^4$ “objects” of a 2-category.

Since $\mathcal{M}^4$ is an algebra, then if we declare “objects” of our 2-category to be elements of $\mathcal{M}^4$, then “morphisms” are also elements of $\mathcal{M}^4$.

I haven’t worked it out yet, obviously, but this is kind of what I’m thinking…

I want objects of my 2-category to be vectors in $\mathcal{M}^4$ intended to represent “position”.

I want morphisms of my 2-category (to be)/(come from) vectors in $\mathcal{M}^4$ intended to represent “velocity”. I haven’t settled on how a velocity vector becomes a morphism. I see two possibilities:

$v:x\mapsto x + v\tau$

or

$v:x\mapsto R(v\tau) x \tilde{R}(v\tau)$

for some rotor $R$ derived from $v$.

I started out thinking about the first option, but then have been coming around to the second. I’m still not sure…

Finally, I want my 2-morphisms to come from acceleration, but I think what I really want to work with is the acceleration bivector $va$. Here again, I’m not sure I want to deal directly with $va$ or some rotor $R(va)$ derived from $va$.

The slogan I’m after is

Velocity translates position.

Acceleration rotates velocity.

This should fit neatly into the analogy that started me down this path

$\array{ category & functor & natural transformation \\ fiber & connection & curvature \\ position & velocity & acceleration }$

I don’t think I’m trying to do anything other than what you already said quickly in passing, but there is no way I can understand what you said initially due primarily to language barriers. Everything here is my feeble attempt to translate what you said.

Posted by: Eric on February 10, 2009 3:14 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

By the way, I borrowed the idea of vectors as objects from Step 3 in

So maybe what I have in mind is an action 2-groupoid?

Posted by: Eric on February 10, 2009 7:21 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

Eric wrote:

Since $\mathcal{M}^4$ is an algebra, then if we declare “objects” of our 2-category to be elements of $\mathcal{M}^4$, then “morphisms” are also elements of $\mathcal{M}^4$.

Why? Is this something you want to be true? You’re making it sound like something that must be true, by the sheer force of pure mathematics. But I only know two ways to make algebras into the objects of a 2-category, and for neither of these are the morphisms given by algebra elements!

Way 1. There’s a strict 2-category where:

• objects are algebras $A,B,C, \dots$
• morphisms are algebra homomorphisms $f : A \to B$.
• a 2-morphism from the algebra homomorphism $f : A \to B$ to the algebra homomorphism $g : A \to B$ is an element $b \in B$ such that $g(a)b = b f(a).$

Since an algebra is a $Vect$-enriched category with one object, we can think of the above more elegantly as a full and 2-full sub-2-category of the 2-category of $Vect$-enriched categories, $Vect$-enriched functors and $Vect$-enriched natural transformations. The funny-looking equation above is really the definition of natural transformation in disguise.

Way 2.

There’s a weak 2-category where:

• objects are algebras.
• morphisms are bimodules.
• 2-morphisms are bimodule homomorphisms.

I discussed this second construction in detail in week209. In week211 I explained how it generalized from algebras to superalgebras… and here, interestingly, Clifford algebras play an incredibly important role!

But, I’m not claiming this will be helpful to you. I don’t see how either of these 2-categories is relevant to your current quest.

Posted by: John Baez on February 10, 2009 8:24 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

Thanks! I’ll have to think about the “Two Ways” you outlined. I think I had a different way in mind, which means my way is likely wrong.

Looking back at the Exercise in Groupoidification, I was reminded of something I called

Exploding a Category

Since an algebra $A$ is a monoid in Vect, then

$Explode(A)$

is a category with one object for each vector in the underlying vector space of $A$. Each morphism of $A$ “explodes” into a huge spaghetti mess of morphisms in $Explode(A)$, i.e. for each morphism

$a: A\to A$

in $A$, we get a ton of morphisms

$a(v): v\to a(v)$

in $Explode(A)$. One for each vector in $v\in A$.

I was thinking, and could very well be wrong, that if we begin with an algebra and “explode” it so that each vector is an object, then any other object can be thought of as a morphism, i.e. given $a\in A$, then given any other $b\in A$, then

$b:a\to ba.$

So maybe I am thinking of an action groupoid $A//A$?

Posted by: Eric on February 10, 2009 9:21 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

By the way, if that made any sense (I won’t hold my breath) and elements of $A$ are both objects AND morphisms, then I thought they must also be 2-morphisms, and 3-morphisms, etc, etc.

In other words, I thought maybe that exploding an algebra gave you an $\infty$-category.

Posted by: Eric on February 10, 2009 9:27 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

Oh oh! I think what I had in mind is related to Way 1, but maybe $Explode($Way 1$)$.

Exploding a 2-category gives another 2-category in an obvious way. If we consider Way 1, but with only one object $A$, then exploding this gives what I had in mind I think.

Posted by: Eric on February 10, 2009 9:44 PM | Permalink | Reply to this

### Re: Explicit Construction of a Transport 2-Functor via Clifford Algebra

If you only have one object, is Way 1 closely related to Way 2?

Posted by: Eric on February 10, 2009 10:21 PM | Permalink | Reply to this

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