What is the connection between DeSitter group SO(4,1) and Minkowski spacetime?

[selfAdjoint]

This is new, probably by some other guy named Ward.

Just one more in the OT bunch. FYI.
This Ward is a phenomenologist. Apparently he went to Baylor in Waco TX back when the SST was still a live future, and has stayed there. His big thing is this YFL resumming, which he has updated and adapted to nonabelian gauge theory. He has a long series of papers with various coauthors applying it to different areas of big collider physics.

This venture into Feynman's graviton theory is a new thing for him.

 

[john baez]

Back to SO(4,1)?

The new opening is that Freidel and Starodubtsev's background-free approach is based on MacDowell-Mansouri gravity, which in turn is based on geometry where every point in spacetime has a tangent de Sitter space.

Or "osculating" de Sitter space, if you prefer.

This is a theme that I hope can be developed in this thread.

It is a major reason why I (and other people might) want to learn about the deSitter group SO(4,1).

Sounds great. So, maybe someone should say something about SO(4,1), or ask a question about it!

Does everyone grok how SO(4,1) reduces to the Poincare group when the cosmological constant goes to zero, for example?

Again, for this, it's good to start with lower-dimensional versions, like how the symmetry group of the sphere (SO(3)) reduces to the Euclidean group of the plane when the radius of the sphere approaches infinity.

This is closely related to the Cartan geometry thread. That was going nicely for a while, but now Garrett is gone - probably spending his FQXi money on a wild spree.

 

[marcus]

Sounds great. So, maybe someone should say something about SO(4,1), or ask a question about it!

Does everyone grok how SO(4,1) reduces to the Poincare group when the cosmological constant goes to zero, for example?

...

No, as a matter of fact, now that you mention it, I DONT grok that and it has been bothering me.

I understand that the cosmological constant is 1/k
where k defines a "surface"
$$w^2 + x^2 + y^2 + z^2 - t^2 = k^2$$

so making CC go to zero is the same as making k go to infinity.

what I don't then see is how that surface "flattens out" to minkowski space

but wait, maybe I do! you suggest it is like an ordinary sphere becoming flat if you increase the radius to infinity

OK

that was nice. it was easier than I thought
============

now to tie up the loose ends. SO(4,1) is the symmetry group of that surface, by definition.
so now we let k -> oo
and the "surface" defined earlier becomes approximable by Minkowskispace
Since SO(4,1) is the symmetries of the "surface", while Poincaré is the symmetries of Minkowski
there is an intuitive sense in which the two groups are getting close

but we need some FORMALITY.
let's try dividing all the coordinates by k
$$(w/k)^2 + (x/k)^2 + (y/k)^2 + (z/k)^2 - (t/k)^2 = 1$$

I am thinking out loud. perhaps should...
well the answer to your question is no: I don't grasp how the GROUP "reduces to" the group.
I only can see how the "surface" approximates Minkowski space.

 

[marcus]

symmetry group of the sphere (SO(3)) reduces to the Euclidean group of the plane when the radius of the sphere approaches infinity.

osculating picture. also reminds me of the Lie algebra of SO(3).
vector addition replacing "rolling the ball" multiplication.
OK this is a helpful idea.

---------
re Garrett
or else, instead of a spree it had kind of the opposite effect.
getting the award made him suddenly hyperserious. he doesn't have time to play any more.
either way (surfboard or chalkboard) we arent likely to see as much of him here

 

[garrett]

Yes, I'm much more serious now. Here's proof:
http://sifter.org/~aglisi/stuff/tux.jpg
I spent FQXi's money on a new formal wardrobe.

Actually, I was the flower girl at my sister's wedding in Santa Cruz. How's that for serious? But now I'm back.

And I've been goofing off over in this other thread:
https://www.physicsforums.com/showthread.php?t=124233
But I'll have a look at what you boys have been rolling around here.

 

[marcus]

hi Garrett!
I'm having difficulty understanding how SO(4,1) "reduces to" Poincaré

The trouble is not exactly with the intuition. Intuitively I feel sure it has to.
Maybe what I need is just the right formal element---a bit of algebra.

I think of SO(4,1) as certain 5x5 matrices
how can a 5X5 matrix reduce to a 4X4 (Lorentz) matrix and a translation?

OK yeah you look sharp in a tux.

 

[john baez]

The Matrix Reloaded

Wow - the next time I see Garrett I expect him to wear that tux.

I'm having difficulty understanding how SO(4,1) "reduces to" Poincaré

Good - I asked that question to start the ball rolling. Rolling, without slipping or twisting.

The trouble is not exactly with the intuition. Intuitively I feel sure it has to.
Maybe what I need is just the right formal element---a bit of algebra.

I think of SO(4,1) as certain 5x5 matrices
how can a 5X5 matrix reduce to a 4X4 (Lorentz) matrix and a translation?

I think the trick for this is to describe the Poincare group as a group of 5x5 matrices.

Do you happen to know how, or can you figure out how, to describe the group of Euclidean transformations of the plane as 3x3 matrices? If you can do that, you can probably use the same idea here.

This is a nice puzzle... personally I prefer a more intuitive geometric approach, but matrices are nice too.

 

[garrett]

OK, let me see if I can grok the rolling sphere, then go from there to Cartan geometry.
If we have a two dimensional sphere, its orientation is described by an element, g, of the three dimensional rotation group, SO(3). If we keep this sphere touching a two dimensional surface, and roll it around without slipping or twisting, its contact point with the surface will be a path specified locally by coordinates, $x^i(t)$, and having velocity
$$v^i(t) = \frac{d x^i}{d t}$$
Now, the sphere will have a changing orientation, $g(t)$, as it rolls along the path, with the specific change dependent on the shape of the surface. I think the equation governing this is
$$\frac{d}{d t} g = \vec{v} \underrightarrow{\omega} g = v^i(t) \omega_i{}^B(x(t)) T_B g$$
with the so(3) valued 1-form, $\underrightarrow{\omega}$, the surface position dependent connection. This equation should integrate to give the orientation of the sphere at any time along the path (the holonomy):
$$g(t) = e^{\int \underrightarrow{dt} v^i \omega_i{}^B T_B} = e^{\int \underrightarrow{\omega}}$$
And I imagine the curvature of this connection corresponds to the curvature (lumpiness) of the surface.

But what's to say the sphere doesn't twist (rotate around the axis determined by the contact point and sphere center) as it rolls? We can refer to this twist as $\theta$ and its twist velocity along a path as
$$\alpha(t) = \frac{d \theta}{d t}$$
So what we've done is enlarged the configuration space for our problem -- we have the two coordinates for the contact point, and the twist, $x^1,x^2,\theta$. Now in order to find the sphere's orientation along some path through this configuration space, we need to solve
$$\frac{d}{d t} g = \vec{v'} \underrightarrow{\omega'} g = ( v^i \omega_i{}^B T_B + \alpha \omega^B T_B ) g$$
and we're going to need to know the connection (the Cartan connection?) over this larger configuration space:
$$\underrightarrow{\omega'} = \underrightarrow{dx^i} \omega_i{}^B (x,\theta) T_B + \underrightarrow{d\theta} \omega^B (x,\theta) T_B$$
Then we can again calculate the holonomy by integrating over the path in this larger configuration space.

Now, I think this is OK, with the fiber, H=SO(2), corresponding to the twist, and the enlarged configuration space corresponding to P. But I think I need some help understanding what further restrictions need to be placed on $\underrightarrow{\omega'}$ for it to be a Cartan connection. If these restrictions could be explained within this simple context and notation I've been working with, that would be great. So... help?

 

[marcus]

Do you happen to know how, or can you figure out how, to describe the group of Euclidean transformations of the plane as 3x3 matrices? ...

unless I'm mistaken, the way to do that is to add a point at infinity to the plane and take SO(3) as the group of 3x3 matrices

but wait, there is the Cartan picture!

the way you can describe the Euclideans of the plane is just to PUT A BIG BALL ON THE PLANE and work the appropriate 3x3 matrices on that ball!*

and the bigger the radius of the ball is (like the bigger "k" is) the less you are bothered by periodicity

(this could be erroneous, but at least it is an idea---OK so that is my guess as to how to do what JB says---if anybody sees a dumbness with it please let me know so I can correct it before JB gets back )

=============
* footnote to make the idea clearer. rotating the ball on the contact point gives you the rotations
and rolling the ball to a new contact point gives you the translations. and mixing gives you the whole works

now JB's suggestion is that if you can see how the 3x3 rotations "reduce" to the Euclidean group on the plane (and I do think I see that) then you can FIGURE OUT how SO(4,1) "reduces to" the Poincaré
in the magic intuitionworld where things always work out, it would seem that one would take a HUGE PLASTIC DUCK (which is the deSitter space that SO(4,1) is symmetries of) and ROLL IT ON MINKOWSKISPACE!

the reason I say it is a huge yellow plastic duck is to remind everybody that it is not a ball, like in the earlier simple example---so there could well be problems with rolling it that I haven't considered.

however letting k -> oo is still going to do good things for us, the bigger k is the better it works to roll the deSitter duck on top of the Minkowski table. I expect this is not a solution, just the beginning of a guess----but the suggestion was "if you can understand this then you might be able to figure out that" and I had to give it a try.

 

[john baez]

flattening out

Does everyone grok how SO(4,1) reduces to the Poincare group when the cosmological constant goes to zero, for example?

No, as a matter of fact, now that you mention it, I DONT grok that and it has been bothering me.

I understand that the cosmological constant is 1/k
where k defines a "surface"
$$w^2 + x^2 + y^2 + z^2 - t^2 = k^2$$

so making CC go to zero is the same as making k go to infinity.

Almost; not quite. I think I got something a bit wrong in the first edition of http://math.ucr.edu/home/baez/week235.html" , so you can blame me. Here's the story:

k is the "radius" of the deSitter spacetime defined by the equation you just wrote, where I use "radius" to keep in mind the analogy to a sphere. Since curvature involves second derivatives, it has units of $$1/length^2$$, so the curvature is proportional to $$1/k^2$$. So, up to some silly factor of 2 or 6 or something, this number $$1/k^2$$ is the cosmological constant.

what I don't then see is how that surface "flattens out" to minkowski space

but wait, maybe I do! you suggest it is like an ordinary sphere becoming flat if you increase the radius to infinity

Right, exactly!

In the case of

$$x^2 + y^2 + z^2 = k^2$$

it's easy to see we get a sphere, and it's easy to see that this sphere gets bigger and biggers as k goes to zero, and that the curvature of this sphere is proportional to 1/k.

Now imagine this sphere is a balloon, and someone starts blowing it up, so $$k \to \infty$$. Imagine we're little guys sitting on this balloon. From our point of view, the balloon will look like it's flattening out to a plane. When it's really big, we'll think we're on a plane, and we'll think the symmetries of the balloon are symmetries of this plane: rotations and translations of the plane.

Do you grok how what we little guys think are rotations of the plane, are actually rotations of the sphere? More importantly: do you grok how what we little guys think are translations of the plane, are also rotations of the sphere?

Can anyone set up some coordinates on the sphere and on our imaginary plane (the tangent plane), so we can relate these symmetries of our plane:

rotations around the origin
translations in the x direction
translations in the y direction

to these symmetries of the sphere:

rotations around the x axis
rotations around the y axis
rotations around the z axis ?

If you can do this, you're well on the way to grokking how the symmetries of deSitter spacetime reduce to symmetries of Minkowski spacetime as $$k \to \infty$$. The only difference is that we need to add one space dimension and one time dimension to our story.

SO(4,1) is the symmetry group of that surface, by definition. So now we let $$k \to \infty$$ and the "surface" defined earlier becomes approximable by Minkowski space. Since SO(4,1) is the symmetries of the "surface", while Poincaré is the symmetries of Minkowski, there is an intuitive sense in which the two groups are getting close.

Exactly! It takes a bit of work to make precise the sense in which one group "approaches" or "reduces to" another. This was first done by Inonu and Wigner, so it's called "Inonu-Wigner contraction", or contraction for short.

but we need some FORMALITY.

Really? Okay. Garrett - put on your tux!

Well the answer to your question is no: I don't grasp how the GROUP "reduces to" the group. I only can see how the "surface" approximates Minkowski space.

Well, to understand how the group SO(4,1) "contracts" to the Poincare group as $$k \to \infty$$, we can write them both down using 5x5 matrices and see how one gets close to the other. Or, we can think about the puzzles I just outlined, to grok how the rotations of a sphere "get close" to symmetries of the plane as the sphere flattens out.

It's best to do both and see how the two viewpoints coincide. For SO(3) contracting to the Euclidean group we just need 3x3 matrices, and it's easy to visualize everything. Then later, if we're feeling masochistic, we can make our matrices 5x5 and do the full-fledged case. Or even better, let's not do that but act like we did.

 

[marcus]

... sense in which one group "approaches" or "reduces to" another. This was first done by Inonu and Wigner, so it's called "Inonu-Wigner contraction", or contraction for short.

BTW found something about this in SPR archive by Arnold Neumaier with comments by J. Dolan and Oz
http://www.lepp.cornell.edu/spr/2004-09/msg0063888.html
Seems at least as good though to just think about the example we have here and chew it over. only mention the SPR posts in passing, don't especially recommend

... do the full-fledged case. Or even better, let's not do that but act like we did.

excellent idea!
===================
another BTW: I found an interesting historical note by Inonu
www.physics.umd.edu/robot/wigner/inonu.pdf[/URL]
describing how he came to Princeton as postdoc for six months and worked with Wigner, and how they
came upon the idea of group contractions. Wigner gave him a more elementary problem and after he'd solved that they
kept going with it.

 

[john baez]

rolling, twisting, but no slipping

OK, let me see if I can grok the rolling sphere, then go from there to Cartan geometry.

Yes, everyone should at some point try to understand the math of a sphere rolling on a surface in 3d space. The fun part is figuring out what it means for this sphere to "slip" or "twist". It reminds me of some killer classical mechanics problems I had to do in college. Even with nothing sneaky like special relativity thrown in, they could still be quite mindbending and frustrating!

Back in classical mechanics class, they called "not slipping or twisting" an anholonomic constraint. The reason is that you can take a ball resting on a plane, roll it around a small loop without slipping or twisting, and it'll come back resting slightly rotated. TRY IT!

So, rolling the ball around a loop gives a rotaton, called the holonomy around this loop. Because the constraint allows this holonomy, it's called "anholonomic", meaning... umm... "no holonomy".

I guess they just wanted to make the terminology as confusing as possible.

Now, however, we see that "rolling without twisting or slipping" defines an SO(3)/SO(2) Cartan connection on the plane, and this connection has curvature!

Why? Well, we say a connection has curvature when it has holonomy around some small loops.

If we have a two dimensional sphere, its orientation is described by an element, g, of the three dimensional rotation group, SO(3). If we keep this sphere touching a two dimensional surface, and roll it around without slipping or twisting, its contact point with the surface will be a path specified locally by coordinates, $x^i(t)$, and having velocity
$$v^i(t) = \frac{d x^i}{d t}$$
Now, the sphere will have a changing orientation, $g(t)$, as it rolls along the path, with the specific change dependent on the shape of the surface. I think the equation governing this is
$$\frac{d}{d t} g = \vec{v} \underrightarrow{\omega} g$$
with the su(3) valued 1-form, $\underrightarrow{\omega}$, the surface position dependent connection.

su(3)? This is a rolling ball, not a rolling quark!

Of course Garrett means so(3); he must have the strong force on his mind.

(I've deleted all expressions that have too many superscripts and subscripts for my feeble brain to process; luckily you also give the simplified versions.)

This equation should integrate to give the orientation of the sphere at any time along the path (the holonomy):
$$g(t) = P e^{\int \underrightarrow{\omega}}$$

I stuck in a "P" to remind us that this is a path-ordered exponential.

And I imagine the curvature of this connection corresponds to the curvature (lumpiness) of the surface.

Sort of...

But, the curvature of the plane is nonzero in this approach! Why? Because if you roll a ball around a small loop on the plane, it comes back rotated. Nontrivial holonomy, hence curvature. TRY IT!

This illustrates a cool feature of Cartan geometry. Since we're rolling a TANGENT SPHERE around to define our Cartan geometry, we find that the PLANE is curved! It's curved "compared to this sphere". Similarly, if we rolled a TANGENT PLANE around to define our Cartan geometry, we'd find that the SPHERE is curved!

See? Everything is relative, man.

Puzzle: what surface would turn out to have no curvature at all, if we studied it by rolling a certain sized sphere on it?

But what's to say the sphere doesn't twist (rotate around the axis determined by the contact point and sphere center) as it rolls?

Ah, now that's getting a bit trickier - and maybe a bit ahead of ourselves, since perhaps you hadn't even realized that the plane isn't flat anymore, when we study it using a rolling sphere.

I'll say one thing, though: if we allow our tangent sphere (or tangent plane!) to twist in a specified way as we roll it, we still get a Cartan connection - but this connection has torsion. Torsion means "twisting", so this time the terminology actually makes sense.

But I think I need some help understanding what further restrictions need to be placed on $\underrightarrow{\omega'}$ for it to be a Cartan connection. If these restrictions could be explained within this simple context and notation I've been working with, that would be great. So... help?

This is very worthwhile, so I suggest that we revive the official definition of "Cartan connection", look at it, and see what it says. I'm too lazy to do it right now, so I'll just sketch the idea.

We're working locally, so we have a trivial bundle, so it simplifies a lot. Our Cartan connection will be an so(3)-valued 1-form $$\underrightarrow{\omega}$$ on our surface, which says how our ball rotates (infinitesimally) as we roll it in any direction (infinitesimally). You already got that far.

I believe the definition of "Cartan connection" will put a condition on this guy $$\underrightarrow{\omega}$$ which says the ball doesn't slip. I think it's allowed to twist! But, let's see what the definition actually says, and work out its consequences. Next time.

 

[marcus]

Puzzle: what surface would turn out to have no curvature at all, if we studied it by rolling a certain sized sphere on it?
...

a sphere of the same size?

yes I just did the experiment with a pair of colorfully patterned juggling balls (a family-member likes circus-arts)

Many thanks to Garrett for holding up the other half of the dialog so ably. Garrett we need to keep you around so that it stays this interesting!

This illustrates a cool feature of Cartan geometry. Since we're rolling a TANGENT SPHERE around to define our Cartan geometry, we find that the PLANE is curved! It's curved "compared to this sphere". Similarly, if we rolled a TANGENT PLANE around to define our Cartan geometry, we'd find that the SPHERE is curved!
...

this is really neat! one could speculate Glast might see energydependent speed of gammaray photons, or something else surprising because momenta live in a "tangent" space that is is secretly bent--as I think has been done.
If stuff can be curved that we previously didnt realize was curved because earlier we rolled the wrong thing on it, well that could be positively delightful. sorry about the excited noises! please keep on the Garrett-topic and don't let this distract.

 

[Hurkyl]

a sphere of the same size?

yes I did the experiment with two colorfully patterned juggling balls

Wow, I didn't even think of that; I put my tangent sphere on the inside... in which case "rolling" the tangent sphere amounts to no motion at all.

 

[marcus]

Hurkyl this stuff is magic!
Glad to see you.
Is it obvious to you how SO(4,1)/SO(3,1) is deSitter space?
I suspect you of having an organized mind and keeping track of things like this.
Sometimes I understand things and later can't remember how I did.
I forget if it's easy to see about deSitter space being the space of cosets?

 

[garrett]

Yes, everyone should at some point try to understand the math of a sphere rolling on a surface in 3d space. The fun part is figuring out what it means for this sphere to "slip" or "twist". It reminds me of some killer classical mechanics problems I had to do in college. Even with nothing sneaky like special relativity thrown in, they could still be quite mindbending and frustrating!

Ahh, the good old days... When I saw the Lagrangian formulation worked out for the first time I thought it was the most beautiful thing I'd ever seen. Other high points have been GR and dynamical chaos. But you know, the Lagrangian formulation may still have all others beat for aesthetics. I mean, extremize an integral and get equations of motion out -- how do you beat that?

Back in classical mechanics class, they called "not slipping or twisting" an anholonomic constraint. The reason is that you can take a ball resting on a plane, roll it around a small loop without slipping or twisting, and it'll come back resting slightly rotated. TRY IT!
So, rolling the ball around a loop gives a rotaton, called the holonomy around this loop. Because the constraint allows this holonomy, it's called "anholonomic", meaning... umm... "no holonomy".
I guess they just wanted to make the terminology as confusing as possible.

"rolling on the floor laughing" indeed. OK, let me put some math where your mouth is. If we're going to treat the simplest case of a sphere on a flat table... We can line up the x1 and x2 axis through the center of the sphere with the x1 and x2 axis on the table. For small counter clockwise rotations around the x1, x2, and x3 axis the SU(3) group element is going to be approximately
$$g \simeq 1 + \theta^A T_A$$
The "rolling without slipping constraint" relates the velocity of the table contact point to the angular velocity:
$$v^1 = R \frac{d \theta^2}{d t}$$
$$v^2 = - R \frac{d \theta^1}{d t}$$
I seem to recall that holonomic constraints can be imposed as a relationship between configuration variables, but these anholonomic constraints are imposed between the velocities. And these equations, using a connection, should be equivalent to
$$\frac{d}{d t} g = \vec{v} \underrightarrow{\omega} g$$
which they are iff our connection is
$$\underrightarrow{\omega} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1$$
That should supply a hint as to what a Cartan connection should look like in general. I hope. But this is an awfully simple case.

Now, however, we see that "rolling without twisting or slipping" defines an SO(3)/SO(2) Cartan connection on the plane, and this connection has curvature!
Why? Well, we say a connection has curvature when it has holonomy around some small loops.

Sure enough, the curvature of our connection is
$$\underrightarrow{\underrightarrow{F}}= \underrightarrow{\partial} \underrightarrow{\omega} + \underrightarrow{\omega} \underrightarrow{\omega} = \underrightarrow{dx^1} \underrightarrow{dx^2} \frac{1}{R^2} T_3$$
So when we roll our ball around in small loops, it should spin around the (vertical) x3 axis. Hmm, I'm not much of a ball sports guy... the closest thing to a ball around right now is my head. But yah, if I use my head, I can see it spinning around that axis. Neat.

su(3)? This is a rolling ball, not a rolling quark!
Of course Garrett means so(3); he must have the strong force on his mind.

Clearly I would never hack it on the professional poker circuit.

(I've deleted all expressions that have too many superscripts and subscripts for my feeble brain to process; luckily you also give the simplified versions.)

Yes, I also like to put arrows on my symbols to help my feeble brain keep track of form order. Only a mathematician could call some object a "Lie algebra valued Grassmann 2-form" and label it "$\Omega$" with no decoration. ;)

I stuck in a "P" to remind us that this is a path-ordered exponential.

Hey, I had a mini question about that. Since presumably the exponential of any algebraic element can be defined via:
$$e^A = 1 + A + \frac{1}{2!} A A + ...$$
Why would that "P" be necessary?

This illustrates a cool feature of Cartan geometry. Since we're rolling a TANGENT SPHERE around to define our Cartan geometry, we find that the PLANE is curved! It's curved "compared to this sphere". Similarly, if we rolled a TANGENT PLANE around to define our Cartan geometry, we'd find that the SPHERE is curved!

OK. I get it. That's funky.

I'll say one thing, though: if we allow our tangent sphere (or tangent plane!) to twist in a specified way as we roll it, we still get a Cartan connection - but this connection has torsion. Torsion means "twisting", so this time the terminology actually makes sense.

Ooh, neat. So I could have put a T_3 piece in our connection:
$$\underrightarrow{\omega} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1 - \underrightarrow{dx^2} x^1 \frac{1}{R^2} T_3$$
And that would be a connection with torsion. Hmm, and maybe I could have made it so that new contorsion piece in the connection cancels out the curvature?

This is very worthwhile, so I suggest that we revive the official definition of "Cartan connection", look at it, and see what it says. I'm too lazy to do it right now, so I'll just sketch the idea.
We're working locally, so we have a trivial bundle, so it simplifies a lot. Our Cartan connection will be an so(3)-valued 1-form $$\underrightarrow{\omega}$$ on our surface, which says how our ball rotates (infinitesimally) as we roll it in any direction (infinitesimally). You already got that far.
I believe the definition of "Cartan connection" will put a condition on this guy $$\underrightarrow{\omega}$$ which says the ball doesn't slip. I think it's allowed to twist! But, let's see what the definition actually says, and work out its consequences. Next time.

Great! I'm all ears. And I'm not particularly lazy, but I only have an hour or two in the morning and evenings the next few days, since I'm helping my parents move. But here's the definition you gave for a Cartan connection, so we can go from there:

A Cartan geometry consists of the following. A smooth manifold M, a Lie group H having Lie algebra Lie(H), a principal H-bundle P on M, a Lie group G with Lie algebra Lie(G) containing H as a subgroup, and a Cartan connection. A Cartan connection is a Lie(G)-valued 1-form $\underrightarrow{\omega}$ on P satisfying
1. $\underrightarrow{\omega}$ is a linear isomorphism from the tangent space of P to Lie(G).
2. $R(h)^* \underrightarrow{\omega} = h^- \underrightarrow{\omega} h$ for all h in H.
3. $\vec{\xi_X} \underrightarrow{\omega} = X$ for all X in Lie(H).
where $\vec{\xi_X}$ is the vector field on P corresponding to the Lie algebra element X in Lie(H), and R(h)* says how an element h of H acts on 1-forms on P.

I think in order to satisfy 1 and 3 our simple ball on a table connection needs to be
$$\underrightarrow{\omega} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1 + \underrightarrow{d\theta^3} T_3$$
And, if that's right, I need clarification on what 2 means.

 

[George Jones]

Is it obvious to you how SO(4,1)/SO(3,1)

The idea is that SO(3,1) is (isomorphic to) a little (or isotropy) group.

de Siitter space is the set of all (w, x, y, z, t) such that w^2 + x^2 +y^2 + z^2 - t^2 = k^2.

Consider q = (k, 0, 0, 0, 0) as an element of R^5, and let O_q = {gq | g in SO(4,1)} be the orbit of p under the action of SO(4,1) on R^5. Then, de Sitter space is the subset O_q of R^5. Let L_q be the little group of q, i.e., the subgroup of SO(4,1) that leaves q invariant, so q = h q for every h in L_q \subset SO(4,1). L_q is isomorphic to SO(3,1).

Now let g be an arbitrary element of SO(4,1), and set p = g q. Clearly, p is in O_q, and a little work shows that the map p -> g L_q gives a bijection of sets between O_q and the coset space SO(4,1)/L_q, i.e., between de Sitter space and SO(4,1)/SO(3,1).

 

[john baez]

rolling a sphere on a sphere of the same size

Puzzle: what surface would turn out to have no curvature at all, if we studied it by rolling a certain sized sphere on it?

a sphere of the same size?

Excellent! Yes!

Yes I just did the experiment with a pair of colorfully patterned juggling balls (a family-member likes circus-arts).

Good - I'm glad you actually checked! Quantum gravity experiments are few and far between; we need all the experiments we can get.

Indeed: if we roll one ball on the surface of another ball of the exact same size, without slipping or twisting, it comes back unrotated after we roll it around in a loop! If the balls differ in size, it will typically come back rotated (relative to its original orientation).

We had an http://groups.google.com/group/sci....2933fd73f918a2f?tvc=1&hl=en#e2933fd73f918a2f" about this on sci.physics.research, back when I was a moderator there! It makes fun reading, I think. Of course I'm biased.

Why does it work like this?

The answer is obvious if you imagine a mirror placed between the two balls: each ball is then a mirror image of the other. The problem then reduces to the problem of rolling a ball on a mirror.

Clearly a ball does not come back rotated relative to its mirror image, when we roll it around a loop! That would be like looking in the mirror, doing some somesaults and dancing around a bit, looking in the mirror again and finding that your mirror image was now tilted compared to you!

However, there are other ways of thinking about this problem, that make it delightfully difficult.

In fact, you can even get confused about this mirror image stuff if you think about it carefully.

It's also fun to imagine one ball rolling around, not outside another ball, but inside it. What happens when one ball is much smaller than the other? What happens in the limit where they become the same size?

Anyway, the main point is that "rolling a sphere of radius r, without slipping or twisting" defines an SO(3)/SO(2) Cartan connection on any surface with a Riemannian metric on it. And, this Cartan connection will be flat when the surface is a sphere of the same radius!

Similarly, "rolling a deSitter spacetime of cosmological constant $$1/k^2$$, without slipping or twisting" defines an SO(4,1)/SO(3,1) Cartan connection on any 4d spacetime with a Lorentzian metric on it. And, this Cartan connection will be flat when our spacetime is a deSitter spacetime with the exact same cosmological constant!

 

[john baez]

clever, clever

Wow, I didn't even think of that; I put my tangent sphere on the inside... in which case "rolling" the tangent sphere amounts to no motion at all.

Ah, so you already answered a puzzle I asked later. Clever, clever!

Note that mapping a tangent sphere on the outside to a tangent sphere on the inside is precisely what a mirror would do. That's how your clever trick is related to the mirror image trick I mentioned.

 

[duke_nemmerle]

Either I'm not as pedestrian as I used to be or this explanation is just really good because it's been a great read for me. Not to the point that I could go and explain it to someone but simply that I'm actually following what's going on, which says a lot about you folks seeing as I'm about as far from differential geometry and group theory as one can be while still being mathematically inclined. Keep it up, it's a solid motivator for those of us still very early in our studies

 

[duke_nemmerle]

You don't need to! Just don't try to get me to reconsider mine.



Okay, so the discussion isn't quite done yet... I'm in a slightly better mood today, so I'll say a bit more.

I'm unable to understand a Lie algebra without also understanding the corresponding Lie group. And, I'm unable to understand a Lie group without knowing a bunch of things it's the symmetry group of. They're all connected in my mind.

For example, suppose you ask me why there's no nonzero element of the Lie algebra so(4,1) whose bracket with all other elements is zero. I could explain this in various ways:

1) laboriously take a 5x5 matrix in so(4,1), calculate its brackets with a second matrix of the same form, and check that if we always get zero, the first matrix must have been zero.

2) cite a theorem that Lie algebras of the form so(p,q) are semisimple, and semisimple elements have vanishing center: they have no nonzero elements whose bracket with all other elements vanishes.

3) note that if there were such a nonzero Lie algebra element, we could exponentiate it and get a nontrivial element of SO(4,1) which commutes with all other elements - since brackets come from commutators. This would be a symmetry operation on deSitter spacetime which we could define independent of our reference frame And such a thing obviously does not exist, if you know what deSitter spacetime looks like.

Actually it's best to know all 3 approaches.

Approach number 1 is the best if you want to minimize the prerequisites - you only need to know how to multiply matrices, and how to tell when a matrix is in so(4,1). The downside is, it's boring and not terribly illuminating.

Approach number 2 is the best if you want to show off. Seriously, it's the best if you want to know how to answer, not just this one question, but a huge swathe of similar questions. The downside is, it relies on general theorems that take a fair amount of work to learn - especially if you want to really understand in an intuitive way while they're true.

Approach number 3 is the most fun - for me, anyway. It definitely has prerequisites, but it's pleasantly geometrical: at the end, you can just stare off into space, imagine the situation, and say sure, I see why this is true! We have converted a Lie algebra question into a question about Lie groups, and answered it by seeing these Lie group elements as symmetries of a space we can visualize.

Different people proceed different ways; some people might be satisfied by having one approach to this question, but I feel safest when I have all three at my disposal, and can check that the answers agree. Admittedly, I would only use approach 1 as a last resort, because I'm lazy. I've explicitly checked by hand, many times, that there's no nonzero element of so(3) whose bracket with all others vanishes - so I'm willing to believe that the calculation will go the same way for so(4,1), especially since approach 2 says it should. It's often reassuring to know that you could check something by explicit calculation, even if you're too lazy to actually do so. When I get stuck in some situation that seems like a paradox, I will break down and do calculations to see what the hell is going on - to discover which of my assumptions is screwed up.


About Cartan geometry...



Sure - until I started thinking about with Derek, Cartan geometry just seemed like a bunch of symbols to me:

$$\omega = dx^i (e_i)^\alpha T'_\alpha + dx^i A_i{}^A T_A$$

I didn't really see the geometry behind it. But now I do, and Derek is supposed to explain this in his thesis!

I'll just give the intuitive idea. Instead of thinking of your spacetime as having tangent planes, you think of it as having tangent spheres, or tangent deSitter spaces... or whatever sort of nice symmetrical "homogeneous space" you like. (A homogeneous space is one of the form G/H where G is a Lie group and H is a subgroup.)

Let's do the case with tangent spheres, since everyone can imagine a sphere. Say we have a lumpy bumpy surface, and a path from P to Q
along the surface. We can set our sphere so it's tangent to P, and then
roll it along our path - rolling without slipping or twisting - until it's tangent to Q. When we're done, our sphere has rotated a certain amount from its initial position. So, we get a certain element of the rotation group SO(3) from our path on our surface!

This is the holonomy of the Cartan connection along the path.

In other words, the usual geometry of Euclidean space puts a natural Cartan connection on any surface in space - a Cartan connection with

G = SO(3)

and

H = SO(2)

The homogeneous space

G/H = SO(3)/SO(2)

is our sphere. Each point in our surface has a copy of G/H tangent to it. And, parallel translation along a path in our surface gives an element of G.

Now just replace G = SO(3) by G = SO(4,1), replace H = SO(2) by H = SO(3,1), and G/H becomes deSitter space, and we're ready to get a Cartan connection on any lumpy bumpy 4d spacetime by rolling a copy of deSitter spacetime over it!

This could be a really stupid thing to ask and my ignorance would certainly allow for that just fine, but ever since watching John's Higher-dimensional Algebra lecture I've been thinking category-theoretically as much as possible. I was just wondering if this parallel transition discussion could possibly be framed in those terms? I'm seeing as the "objects" these actual points P and Q on our space and the groups of symmetries or possibly the connections as the "morphisms" on these objects.

This is actually very trivial, it's just how I've been looking at things since watching this lecture and is thus the first curiosity when reading this. Can these groups of objects/actions be framed in this language at all? If so, which are the objects, which are the morphisms, etc. Very trivial, but still interesting to me.

 

[Hurkyl]

Is it obvious to you how SO(4,1)/SO(3,1) is deSitter space?
I suspect you of having an organized mind and keeping track of things like this.
Sometimes I understand things and later can't remember how I did.
I forget if it's easy to see about deSitter space being the space of cosets?

George explained it one way... I'll explain it another!

Actually, I usually don't like to think in terms of cosets. And I'll talk about spheres instead of deSitter space, since it's easier to visualize.


The elements of SO(3) act as rotations of the sphere. In fact, SO(3) acts transitively on the sphere. That's just a fancy way of saying that if you have two points P and Q, there's something in SO(3) that moves P to Q.

The problem is that there are lots of things in SO(3) that move P to Q. If you picture it (or play with juggling balls), you'll see that once you move P to Q, you're free to spin the sphere around the axis containing Q and you haven't changed the fact you moved P to Q.

To say that differently, we have precisely an SO(2) amount of freedom in selecting how to move P to Q.

In group-theory land, the standard way to get rid of extra information is by taking the quotient group. There is a unique element of SO(3)/SO(2) that corresponds to moving P to Q. (note that we cannot think of SO(3)/SO(2) as actual rotations of the sphere, though maybe we can think of it as a "partially specified" rotation)

In fact, for any point R, there is a unique element of SO(3)/SO(2) that corresponds to moving P to R. So, there is a bijection between SO(3)/SO(2) and points of the sphere.

 

[garrett]

Hmm, was hoping to hear back from JB on clarifying what restrictions need to be placed on a Cartan connection. Hope he didn't get a bad xiaolong bao.

 

[john baez]

connections are functors

... ever since watching John's Higher-dimensional Algebra lecture I've been thinking category-theoretically as much as possible.

Excellent!

I was just wondering if this parallel transition discussion could possibly be framed in those terms?

Sure! A connection on a bundle

$$P \to B$$

is a functor from the "path groupoid" of the base space B to the "transport groupoid" of the bundle.

Very roughly speaking, it goes like this:

The path groupoid has points of B as objects and paths in B as morphisms.

The transport groupoid has fibers of P as objects and maps betwen fibers as morphisms.

Here's how we get a functor. First send each point x in B to the fiber $$P_x$$ over that point - that's what our functor does to objects. Then send each path from x to y in B to the "parallel transport" map from $$P_y$$ to $$P_y$$ - this is what our functor does to morphisms.

I'm omitting a lot of details. All this is explained more precisely in http://math.ucr.edu/home/baez/barrett/" . But, we go a lot farther: we think of "2-connections" as 2-functors between 2-categories. This let's us do parallel transport not just along curves, but along surfaces.

You can read a lot more about "connections as functors" in my http://math.ucr.edu/home/baez/qg-spring2005/" . This may be a little less stressful than picking what you want out of a more fancy discussion.

Here's a little puzzle: if we think of a connection as a functor, what's a gauge transformation?

 

[john baez]

No bad xiaolong bao, Garrett, thanks. That place is a little hole in the wall, but it makes damn good xiaolong bao, and I've never gotten sick here in Shanghai. I'm just suffering from a lot of distractions!

OK, let me put some math where your mouth is. If we're going to treat the simplest case of a sphere on a flat table... We can line up the x1 and x2 axis through the center of the sphere with the x1 and x2 axis on the table. For small counter clockwise rotations around the x1, x2, and x3 axis the SU(3) group element is going to be...

SU(3) again? Why must you make everything so complex?

Since presumably the exponential of any algebraic element can be defined via:
$$e^A = 1 + A + \frac{1}{2!} A A + ...$$
Why would that "P" be necessary?

When we compute a holonomy, we're not exponentiating a Lie algebra element. We're not first doing an integral to get a Lie algebra element and then exponentiating it. That would be wrong, except when our Lie algebra is abelian. In the nonabelian case

$$e^{A + B} \ne e^A e^B$$

so we have to be careful.

To compute a holonomy, we take our path, chop it up into lots of tiny pieces, compute a Lie algebra element as an integral for each one, exponentiate them all, and then multiply them in the right order... and take the limit where the pieces get really tiny. That's what a path-ordered exponential does!

This is different from taking our path, chopping it up into lots of tiny pieces, computing a Lie algebra element as an integral for each one, adding them all up, and then exponentiating them... and then taking the limit where the pieces get really tiny. This would give non-gauge-invariant nonsense - except when our Lie algebra is abelian.

In general, we need path-ordered exponentials whenever we solve a differential equation like

$${d \over dt} v(t) = A(t) v(t)$$

where A(t) is a matrix-valued function of t, and v(t) is a vector-valued
function. That's what we're doing when we're computing a holonomy!

To answer the rest of your questions, I need to do it my own way. We talk different languages, so you can translate what I say into your language. I just want to explain Cartan connections using this example of a ball rolling on a surface, with a minimum of extra stuff going on. I'll only do a little bit today, 'cause it's time for dinner.

Anyway, imagine we have a little piece of surface sitting in Euclidean 3-space. Call it M. Let P be the space of all ways we can place a sphere on top of M. Since we can make the sphere touch any point of M, and we can also rotate the sphere arbitrarily, we have

$$P = M \times SO(3)$$

This gadget P is an example of a "principal bundle", but I think I'll call it the space of placements, where a placement is a way of placing a sphere on top of M.

If our surface M is topologically tricky, like a sphere, we may need a "nontrivial" principal bundle, which only looks locally like what I've written down. It's up to us how much we want to get into that - the general definition of Cartan connection handles this issue.

Our Cartan connection will tell us precisely how the sphere rolls as we move it around on the surface. If I move the sphere a little bit on the surface, it rolls a bit and its placement changes slightly.

We need to formalize this...

 

[duke_nemmerle]

Excellent!



Sure! A connection on a bundle

$$P \to B$$

is a functor from the "path groupoid" of the base space B to the "transport groupoid" of the bundle.

Very roughly speaking, it goes like this:

The path groupoid has points of B as objects and paths in B as morphisms.

The transport groupoid has fibers of P as objects and maps betwen fibers as morphisms.

Here's how we get a functor. First send each point x in B to the fiber $$P_x$$ over that point - that's what our functor does to objects. Then send each path from x to y in B to the "parallel transport" map from $$P_y$$ to $$P_y$$ - this is what our functor does to morphisms.

I'm omitting a lot of details. All this is explained more precisely in http://math.ucr.edu/home/baez/barrett/" . But, we go a lot farther: we think of "2-connections" as 2-functors between 2-categories. This let's us do parallel transport not just along curves, but along surfaces.

You can read a lot more about "connections as functors" in my http://math.ucr.edu/home/baez/qg-spring2005/" . This may be a little less stressful than picking what you want out of a more fancy discussion.

Here's a little puzzle: if we think of a connection as a functor, what's a gauge transformation?

Thanks for framing it that way, I find this way of thinking to be quite seductive, I'll certainly have a look at those references and think about that puzzle.

As I said in another post of mine, there couldn't be anyone more novice than me, but this all seems pretty accessible the way it's being described, bravo :)

 

[marcus]

hint? spoiler warning. disclaimer: apt to be wrong.
a functor from the transport groupoid to itself? one that happens to be the identity on objects?

 

[duke_nemmerle]

hint? spoiler warning. disclaimer: apt to be wrong.
a functor from the transport groupoid to itself? one that happens to be the identity on objects?

I almost came to that conclusion myself, maybe it's just the change of coordinates from the path groupoid to the transport groupoid, but that's what the connection functor itself seems to be. Still thinking :)

EDIT: Oh and John you underestimate your expository gift, the Higher Gauge Theory transparencies are quite accessable and lovely, I can't wait to print them off and kill time at work checking them out :)

You know what, I think you're right Marcus, reading here in this link http://math.ucr.edu/home/baez/qg-spring2005/ it looks like a gauge transformation is an isomorphism from principal bundle P to itself.

It goes on to show how this morphism maps the Px fiber to itself. Essentially showing (I think, correct me if I'm wrong)what someone was saying earlier, that from a path P to Q irrespective of "symmetrical rotations" one still arrives at Q. I've got to look into this more as it's seems just above my head, but it sure is fascinatingly close to understable to me that I will keep at it.

So to make long-windedness short, I think the gauge transformations may be the rotational symmetries along the path discussed earlier?

 

[john baez]

I will keep on talking about Cartan geometry in a quite leisurely way, always using our example of a sphere rolling around on a surface, since this is easy to visualize.

I'll gradually fumble my way to the definition of "Cartan connection" in this special case. But then, at the end, we'll be able to instantly generalize everything we did, simply by replacing the sphere and surface by more general spaces, the rotation group by a more general group, and so on. So, while Garrett will be bored silly and bristling with impatience for quite a while, eager for me to bring on the fancy formulas, at the end everything will be clear (I hope).

Anyway, imagine we have a little piece of surface sitting in Euclidean 3-space. Call it M. Let P be the space of all ways we can place a sphere on top of M.

I should not have called this P, since it doesn't match the thing called "P" in the general definition of "Cartan connection" that I gave quite a while back on this thread. So, let's call it Q. Let me restate some stuff I said, with this new notation:

Let Q be the space of all ways we can place a sphere on top of M.

Since we can make the sphere touch any point of M, and we can also rotate the sphere arbitrarily, we have

$$Q = M \times SO(3)$$

This gadget Q is an example of a "principal bundle", but I think I'll call it the space of placements, where a placement is a way of placing a sphere on top of M.

(If our surface M is topologically tricky, we may need a "nontrivial" principal bundle, which only looks locally like what I've written down.)

Now, sitting inside the 3d rotation group SO(3) is the 2d rotation group SO(2). Think of this as consisting of all rotations of our sphere that leave the south pole fixed - imagine the Earth spinning around on its axis.

The sphere, the group SO(3), and the group SO(2) have a special relationship - if you don't understand this you can't possibly understand Cartan geometry. This relation is:

$$S^2 = SO(3)/SO(2)$$

Here $$S^2$$ is our name for the sphere, since its surface is 2-dimensional.

What does this mean? The group SO(3) acts as rotations of the sphere, and it acts transitively: we can carry any point on the sphere to any other point using some rotation. So, we can get to any point on the sphere by taking the south pole and applying some rotation. But, lots of different rotations carry the south pole to the same point! Indeed lots of different rotations carry the south pole to itself, and these form the group SO(2).

So, we can get all points on the sphere by rotations in SO(3) but this description of points on the sphere is redundant due to SO(2). This is what we mean by saying

$$S^2 = SO(3)/SO(2)$$

One can explain this more precisely, and I think somr people on this thread already did that when discussing a fancier example, namely

$$deSitter spacetime = SO(4,1)/SO(3,1)$$

But, right now I'm having fun trying to keep everything as jargon-free as possible. So, I'll just emphasize that this business is very general. If we have any space X, we can look for a group G that acts transitively on it. If we find one, this means we can describe any point of X by taking our favorite point (the "south pole") and applying some transformation in G. But, this description of points in X will usually be redundant: there will some transformations in G that carry the south pole to itself, and these will form a group H sitting inside G. So, we will have

$$X = G/H$$

So, we're developing Cartan geometry in the special case

$$X = S^2, G = SO(3), H = SO(2)$$

but everything will generalize painlessly. Note that in our special case the surface M has the same dimension as X. We always want that in Cartan geometry!

Okay, let's see where's a good place to wrap up before everyone's eyes glaze over. (Well, not everyone - Garrett will be very frustrated at my slow pace, but he's not paying me enough to write an exposition crafted personally for him!)

I guess I should point out this. We had a "space of placements"

$$Q = M \times SO(3)$$

whose points were all ways we could place a sphere on top of the
surface M. But, we also have a smaller space

$$P = M \times SO(2)$$

and this is really the star of the show in Cartan geometry.

What is this smaller space? It's the space whose points are
ways we can place a sphere on top of M, such that the south pole of the sphere touches M!

Do you see why? A point in $$M \times SO(2)$$ is a point in M, which says where your sphere touches the surface M, and a rotation in SO(2), which says how the sphere is rotated - but making sure the south pole touches the surface.

Now, it's sort of surprising that this P guy is the star of the show in Cartan geometry, because we want to talk about a rolling ball, and P is about a ball whose south pole touches the surface M - it can't really roll!

That's actually why I screwed up and brought in that other guy, Q. But, introducing P is part of Cartan's cleverness.

Just so you have something to do, in case you're sitting there bored browing the web, here's a little puzzle. What are the dimensions of these various space (in our example):

$$M, SO(3), SO(2), S^2, P, Q$$

and how are these dimensions related? There are a bunch of simple relationships that are not coincidences - relationships that hold in any Cartan geometry.

 

[Hurkyl]

Here's a little puzzle: if we think of a connection as a functor, what's a gauge transformation?

hint? spoiler warning. disclaimer: apt to be wrong.
a functor from the transport groupoid to itself? one that happens to be the identity on objects?

My first instinct is that it's obviously a natural transformation! But then another part of me says marcus is obviously right!

As I think more about it, I suspect that we may both be right! We need a functor to change the morphisms, but we need a natural transformation to remember what we did to each fibers.

So my proposed answer is that, when looking at the transport groupoid, a gauge transformation amounts to two pieces of information:

(1) A functor F from the transport groupoid to itself
(2) A natural transformation from the identity functor to F.

(Well, technically I don't even need to specify the functor, since it can be recovered from the natural transformation. So maybe I was right after all! )

You know what, I think you're right Marcus, reading here in this link http://math.ucr.edu/home/baez/qg-spring2005/ it looks like a gauge transformation is an isomorphism from principal bundle P to itself.

I don't know if we're supposed to, but I would like to think of the principal G-bundle P-->B as a groupoid. It's objects are elements of B, and for each b in B, Hom(b, b) is a copy of G.

If we think this way, then the gauge transformation is a natural transformation from the identity functor to itself!



Er, ack. If I like to think of a principal G-bundle as a groupoid, does that mean I should like to think, not of a transport groupoid, but of a transport... um... bigroupoid? Eep!

Edit: nevermind, that's a silly thing, I think.

 

[john baez]

cool!

Here's a little puzzle: if we think of a connection as a functor, what's a gauge transformation?

hint? spoiler warning. disclaimer: apt to be wrong.
a functor from the transport groupoid to itself? one that happens to be the identity on objects?

Hmm! That may be right! And the interesting thing is, I had a quite different answer in mind, which I know is right... but I like yours better!

I didn't intend for anyone to actually think to answer my question; you were just supposed to use the tao of mathematics, and say to yourself

connections : gauge transformations :: functors : ?

and make an obvious wild guess, even kind of a pun:

connections : gauge transformations :: functors : natural transformations

This is actually right: if we have a gauge transformation carrying a connection A to a connection A', and we think of A and A' as giving functors as I explained, there will be a natural transformation from the first functor to the second.

(I explained this here, especially in the "sophisticated digression" on page 6, though you'll have to read the stuff before to understand what I'm talking about.)

But, this viewpoint neglects a basic fact, which is that we can speak of a single gauge transformation acting on all possible connections to give new connections. We can't in general think of a natural transformation acting on all functors to give new functors; it only goes from one specific functor to another.

So, what gives?

I think your viewpoint solves this puzzle: if a connection is a functor from the path groupoid to the transport groupoid, and a gauge transformation is a functor from transport groupoid to itself, we can compose these to get a new connection. And, a given gauge transformation can be composed with all possible connections to give new connections. I'll have to check to see if this is really what's going on.

But, if it works, it immediately leads to a new puzzle: how come we can think of a gauge transformation in these two ways - as a natural transformation but also as a functor? Since when do natural transformations get to masquerade as functors? What's the proper level of generality for understanding this phenomenon? There's got to be some general fact about categories that explains it - or at least groupoids. And, it can't be all that complicated, because this stuff really isn't all that complicated, despite all the multi-syllable words I'm throwing around.

So, thanks! I've got something to think about tonight.

You know what, I think you're right Marcus, reading here in this link http://math.ucr.edu/home/baez/qg-spring2005/" it looks like a gauge transformation is an isomorphism from principal bundle P to itself.

That's right: it's a smooth invertible map from P to itself that preserves all the extra structure (the projection down to the base manifold and the action of the gauge group). Since the transport groupoid Trans(P) is defined using just P and this extra structure, any gauge transformation of P will give a functor from Trans(P) to itself. And yes, as Marcus said, it will be one that's the identity on objects. So, he nailed it.

Isn't "Nemmerle" something from A Wizard of Earthsea? That was one of my favorite books as a kid... but it made me really frustrated, because I couldn't do magic.

Now I can.

 

[john baez]

My first instinct is that it's obviously a natural transformation! But then another part of me says marcus is obviously right!

As I think more about it, I suspect that we may both be right!

I think so. Man, we've got some real category-theoretic physicists on this forum!

So my proposed answer is that, when looking at the transport groupoid, a gauge transformation amounts to two pieces of information:

(1) A functor F from the transport groupoid to itself
(2) A natural transformation from the identity functor to F.

(Well, technically I don't even need to specify the functor, since it can be recovered from the natural transformation. So maybe I was right after all! )

Of course a natural transformation T: 1 => F determines F, by definition - hence your wink.

The funny thing is that in this situation, F uniquely determines T. I still haven't figured out the principle at work here. I could even be seriously mixed up.

I don't know if we're supposed to, but I would like to think of the principal G-bundle P-->B as a groupoid. It's objects are elements of B, and for each b in B, Hom(b, b) is a copy of G.

That's fine. This sits inside the transport groupoid I was discussing:

The transport groupoid has elements of B as objects, and for b,b' in B, Hom(b,b') consists of all "transporters" from b to b'. A transporter is a map

$$f \colon P_b \to P_{b'}$$

that commuts with the right action of G on P:

$$f(pg) = f(p)g$$

In other words, it's a possible way to do parallel transport from b to b'.

I believe your groupoid is just the "diagonal" in this one, where we throw out all morphisms except from an object to itself.

Anyway, Trans(P) is a smooth groupoid in an obvious sense - it has a smooth manifold of objects, a smooth manifold of morphisms, and all the groupoid operations are smooth. http://math.ucr.edu/home/baez/2conn.pdf" that smooth functors from the path groupoid of B to Trans(P) are in 1-1 correspondence with connections on P.

 

[john baez]

Just so you have something to do, in case you're sitting there bored browing the web, here's a little puzzle. What are the dimensions of these various space (in our example):

$$M, SO(3), SO(2), S^2, P, Q$$

and how are these dimensions related? There are a bunch of simple relationships that are not coincidences - relationships that hold in any Cartan geometry.

I ain't going to explain more of this stuff until some folks take a crack at these puzzles. It'll be more fun if people actually think about this stuff a bit.

Digression: I'm happy because I finally uploaded my http://math.ucr.edu/home/baez/cohomology.pdf" .

 

[Hurkyl]

Spoiler!

M has dimension 2, because it's a surface in 3-space.
SO(2) has dimension 1, because it's the circle.
S^2 has dimension 2, because it's a sphere.
SO(3) has dimension 3, because S^2 = SO(3) / SO(2), and 2 = 3 - 1
Q has dimension 3, because Q = M x SO(2), and 3 = 2 + 1
P has dimension 5, because P = M x SO(3), and 5 = 2 + 3

The only other interesting connection I see is that Q and SO(3) have the same dimension. And that will be true in the general case, because we always want the thing we're rolling around on M to have the same dimension as M!


edit: changed text to white, so as not to be a spoiler!

 

[marcus]

...Just so you have something to do, in case you're sitting there bored browing the web, here's a little puzzle. What are the dimensions of these various space (in our example):

$$M, SO(3), SO(2), S^2, P, Q$$

and how are these dimensions related? There are a bunch of simple relationships that are not coincidences - relationships that hold in any Cartan geometry.

spoiler warning
$$SO(2)$$ is the rotations in the plane, so it looks like the unit circle----it is 1D

$$SO(3)$$ is the rotations in euclidean 3-space to it is a 2D choice of axis and a 1D rotation around that axis---so it is 3D

$$S^2$$ is the 2-sphere---so it is 2D and we expect that because
dimension of SO(3)/SO(2) = 3-1 = 2


everybody here knows what I just said, but I am not sure about M, P, Q.
I suspect you told us that M was 2D, which would make Q 3D and P 5D
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Hurkyl beat me to it! Thanks for answering Hurkyl.