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

[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
<br /> g \simeq 1 + \theta^A T_A<br />
The "rolling without slipping constraint" relates the velocity of the table contact point to the angular velocity:
<br /> v^1 = R \frac{d \theta^2}{d t}<br />
<br /> v^2 = - R \frac{d \theta^1}{d t}<br />
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
<br /> \frac{d}{d t} g = \vec{v} \underrightarrow{\omega} g<br />
which they are iff our connection is
<br /> \underrightarrow{\omega} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1<br />
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
<br /> \underrightarrow{\underrightarrow{F}}=<br /> \underrightarrow{\partial} \underrightarrow{\omega} + \underrightarrow{\omega} \underrightarrow{\omega} = \underrightarrow{dx^1} \underrightarrow{dx^2} \frac{1}{R^2} T_3<br />
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:
<br /> e^A = 1 + A + \frac{1}{2!} A A + ...<br />
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:
<br /> \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<br />
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
<br /> \underrightarrow{\omega} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1 + \underrightarrow{d\theta^3} T_3<br />
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:

<br /> \omega = dx^i (e_i)^\alpha T&#039;_\alpha + dx^i A_i{}^A T_A<br />

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:
<br /> e^A = 1 + A + \frac{1}{2!} A A + ...<br />
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

<br /> e^{A + B} \ne e^A e^B <br />

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?[/color]

 

[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?[/color]

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?[/color]

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&#039;}

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!
[/color]

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
[/color]

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
===============
Hurkyl beat me to it! Thanks for answering Hurkyl.

 

[garrett]

OK, spiffy. I'm liking your rolling sphere example more and more, and looking forward (OK, maybe impatiently) to seeing it come together. We can describe the rolling of the sphere locally with a connection. The "without slipping or twisting" restriction is what GR aficionados think of as "parallel transport." If a point on the sphere is labeled r, and written in good old cartesian 3-space, relative to the sphere's center, as a column vector, then any rotation of the sphere moves this to the new point,
<br /> r&#039; = g r<br />
with g an element of the rotation group, G=SO(3), represented as a matrix. (phew, I got the O right that time) Of course, we could also do a rotation by writing r as a row vector, and operating with g from the right, or by writing r as a quaternion and operating with g as g r g^- (but I don't think it will matter). Adopting the left acting representation, the connection is a 1-form over the surface, M, valued in the Lie algebra of G,
<br /> \underrightarrow{w} \in \underrightarrow{Lie(G)} = \underrightarrow{so(3)}<br />
Now, there are two ways clear to me to build this into a fiber bundle picture. The first is to use the sphere, S^2 = SO(3)/SO(2), as the typical fiber for a bundle that is locally M \times S^2. The "covariant derivative" of a "S^2 valued field" is
<br /> \underrightarrow{\nabla} r(x) = \underrightarrow{\partial} r + \underrightarrow{w} r<br />
and a sphere position is parallel transported (rolled without slipping or twisting) along a path with velocity \vec{v} over the surface iff
<br /> 0 = \vec{v}\underrightarrow{\nabla} r = \frac{d}{d t} r + \vec{v} \underrightarrow{w} r <br />
The second way to choose a fiber bundle is to see that we can choose some arbitrary starting position on the sphere, and write any position using a SO(3) group element,
<br /> r(x) = g(x) r_0<br />
This way, parallel transport along a path happens iff
<br /> \frac{d}{d t} g(x(t)) = - \vec{v} \underrightarrow{w} g <br />
with the SAME connection, \underrightarrow{w}. This is called the principle bundle associated to M \times S^2, locally Q = M \times SO(3) -- "associated" because it has the same connection (technically, the same structure group, G), and "principle" because the structure group and fiber are the same.

I went ahead and worked out what this connection should be for the sphere rolling over a flat 2D surface:
<br /> \underrightarrow{w} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1<br />
And computed its curvature.

Now, John Baez is promising to somehow relate this to a Cartan connection, \underrightarrow{\omega}, which is a 1-form over the ENTIRE space of a different bundle, locally P = M \times SO(2), and still valued in the so(3) Lie algebra. In terms of the various basis, this will be able to be written as:
<br /> \underrightarrow{\omega} = \underrightarrow{dx^1} \omega_1{}^A T_A + \underrightarrow{dx^2} \omega_2{}^A T_A + \underrightarrow{d\theta} \omega_3{}^A T_A<br />
And have to satisfy some restrictions.

My pedantic self will be especially happy when I see exactly how \underrightarrow{\omega} relates to \underrightarrow{w}.

And I can see another reason why this Cartan connection construction is especially pretty. When one plays around with the geometry of a Lie group, G, one of the best things to build and work with is Cartan's 1-form,
<br /> \underrightarrow{W} = g^- \underrightarrow{\partial} g \in \underrightarrow{Lie(G)}<br />
which is a 1-form over the entire group manifold. From the definition, its curvature vanishes. But what if this 1-form over the group manifold were a little different, and some of its curvature wasn't necessarily zero? Say, if the part of the curvature valued in the Lie algebra of some subgroup, H, could be nonzero, while the rest of the curvature was zero. I think this would then be a Cartan connection. And this point of view may tie together how \underrightarrow{\omega} relates to \underrightarrow{w}. But, I'm not sure.

I really am eager to see how John fits all this together! Hmm, maybe I should promise to pay him twice as much... Wait, of course, I can get him to work for me if I can figure out his True Name.

 

[garrett]

It looks like maybe
<br /> Q = M \times G = (M \times H) \times G/H = P \times G/H<br />
would be consistent with \underrightarrow{\omega} being the Cartan connection over P. But I'm not sure exactly how to pull that off.

 

[Hurkyl]

I hope nobody minds -- I've deleted two of my previous posts so I can say the same thing in a better and more cohesive way!

Since when do natural transformations get to masquerade as functors?

This is actually something that bothered me when I was first learning this stuff, but had forgotten about until marcus's answer. In fact, it made me want to think of natural transformations as some sort of functor!


But in some sence, a natural transformation is really just a pair of functors (source and target), with some extra information thrown in! Composition of functors is just a special case of horizontal composition of natural transformations, which makes it even easier for natural transformations to masquerade as functors.

In our case, we have a natural transformation from the identity, which makes the masquerade complete.


So, for our particular example of a gauge transformation, a natural transformation T:1 ==> F can act on any connection by horizontal composition! The connection C gets turned into the connection CF, and we even have the natural transformation CT:C ==> CF which tells us how it's done!

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.

But this isn't quite true in the general case. If g is something in the center of our group (that's not the identity), then translating all of the fibers by g gives us a nontrivial natural transformation T:1 ==> 1.

But I guess in the case of interest, the center is trivial, so we don't have this problem. There's probably an easy proof that T is unique in this case, but I don't see it yet.

 

[john baez]

got 'em all

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
P has dimension 3, because P = M x SO(2), and 3 = 2 + 1
Q has dimension 5, because Q = M x SO(3), and 5 = 2 + 3

The only other interesting connection I see is that P 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!
[/color]

Yes, you got 'em all! - though you mixed up P and Q, so I fixed that in the quoted text above. You got to mind your P's and Q's...

Particularly important are the relations that hold in general, whenever we're doing Cartan geometry. There are basically four of them:

dim(G/H) = dim(G) - dim(H)

is just a fact about the quotient of a Lie group by a subgroup. (In our example G = SO(3), H = SO(2) and G/H = S^2.)

dim(Q) = dim(M) + dim(G)

follows because Q is a G-bundle over M, and similarly

dim(P) = dim(M) + dim(H)

because P is an H-bundle over M. The special thing about Cartan geometry is

dim(M) = dim(G/H)

since we want M to have "tangent G/H's" - in our example, our surface M has "tangent spheres".

Given the first three equations, the last one is equivalent to

dim(P) = dim(G)

which is the way you expressed it.

By the way, folks - tomorrow morning I'm going to http://www.chinavista.com/suzhou/home.html" for three days, to see the gardens and canals... so I won't be posting until I get back. Don't worry, I'm eager to return to this thread.

 

[john baez]

OK, spiffy. I'm liking your rolling sphere example more and more, and looking forward (OK, maybe impatiently) to seeing it come together.

Me too! As I'm explaining this, I'm noticing various things I didn't understand as well as I thought... but this is why I became a math teacher in the first place: so I'd have an excuse to learn the darn stuff! Anyway, I think it will all fit together quite nicely pretty soon.

Let me drag out that old definition of "Cartan geometry" and see where we are:

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 w on P satisfying:

1. w is a linear isomorphism from the tangent space of P to Lie(G)

2. R(h)^* w = Ad(h^{-1}) w for all h in H

3. w(X+) = X for all X in Lie(H).

where 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.

Naturally the scary-looking equations 2 and 3 will be the very last things we'll try to understand. As I tell my calculus students, it's always good to start with the easiest part of a math problem and slowly work your way to the hard part - because if you're lucky, a disaster might strike, killing you before you ever get to the hard part. We'll follow that important principle here.

But, the stuff on the top should make lots of sense now:

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, ...

In our favorite example, M is our surface in 3-space, G is the 3d rotation group SO(3), H is the subgroup SO(2), G/H is the sphere, and P is the space of all ways of placing the sphere on our surface so its south pole touches the surface!

The general definition just generalizes the heck out of this.

Clearly P is the trickiest ingredient so far. We'll see why it's important in just a second.

So, now we're ready for the star of the show: the Lie(G)-valued 1-form on P, which I'm calling w! :!)

For those who aren't up on their Lie(G)-valued 1-forms, this may require some preparation.

A 1-form on a manifold is a gadget that eats tangent vectors and spits out numbers in a linear way. A vector-valued 1-form eats tangent vectors and spits out some other sort of vectors in a linear way. In our example, Lie(G) is the vector space of infinitesimal rotations - folks normally just call it R^3, but since G = SO(3) we should call it so(3), the Lie algebra of SO(3).

So, w eats tangent vectors on P and spits out infinitesimal rotations.

What's the point? If you know how to roll a ball around on a surface, I claim you actually know a linear map that eats tangent vectors to P and spits out infinitesimal rotations.

In other words, I claim that you can guess how a tiny motion of a point on P gives a tiny rotation. Can anyone here guess it?

(I like the word "tiny" because it's less scary than "infinitesimal". It suggests a cute little quantity that's getting ready to approach zero, so we can take a limit, as in calculus. Why in the world did people pick a scary, enormously long word like infinitesimal to mean "tiny"? That was just bad PR. )

Of course to do this you have to really remember what P is!

To see if you're on the right track, it's worth looking at clause 1:

1. w is a linear isomorphism from the tangent space of P to Lie(G).

In other words, w is one-to-one and onto, so it has an inverse!

So, if you guess w correctly, you'll see you also know how a tiny rotation yields a tiny motion of a point on P!

I think I'll stop here, eat dinner, go to bed and go to Suzhou next morning. By the time I'm back, I bet everyone will understand this w guy, and we'll be ready to see why it satisfies clauses 2 and 3!

 

[Dcase]

w=1 may be a solution.
One arc-second of rotation should yield one arc-second of movement.
One arc-nano-second of rotation should yield one arc-nano-second of movement.

The concept "infinitesimal" probably better correlates as the inverse of the Georg Cantor demonstration that there exist more than one degree of infinity.
http://www.gap-system.org/~history/Biographies/Cantor.html

 

[marcus]

In our favorite example, M is our surface in 3-space, G is the 3d rotation group SO(3), H is the subgroup SO(2), G/H is the sphere, and P is the space of all ways of placing the sphere on our surface so its south pole touches the surface!

...
So, now we're ready for the star of the show: the Lie(G)-valued 1-form on P, which I'm calling w! :!)
...
In our example, Lie(G) is the vector space of infinitesimal rotations - folks normally just call it R^3, but since G = SO(3) we should call it so(3), the Lie algebra of SO(3).

So, w eats tangent vectors on P and spits out infinitesimal rotations.

What's the point? If you know how to roll a ball around on a surface, I claim you actually know a linear map that eats tangent vectors to P and spits out infinitesimal rotations.

In other words, I claim that you can guess how a tiny motion of a point on P gives a tiny rotation. Can anyone here guess it?

To see if you're on the right track, it's worth looking at clause 1:
...In other words, w is one-to-one and onto, so it has an inverse!

So, if you guess w correctly, you'll see you also know how a tiny rotation yields a tiny motion of a point on P!
...

thanks for leaving us something to think about. Hope you all enjoyed Suzhou. I've was fascinated by Marco Polo's description of Hongzhou (you may also visit?)---he was impressed that if you had some free time you could rent a boat and take a picnic lunch and a party of friends out on the lake or the canals---and if you didnt happen to have a picnic lunch and ladies with you then you could rent a boat that was ALREADY PROVISIONED with both. I think Suzhou is known as China's "Venice" but IIRC Hongzhou was Polo's favorite. Unless I'm dreaming he also said that those 13th century Hongzhou merchants were so rich that they burned ornamentally carved logs in the fireplace.

this P is a PRINCIPAL BUNDLE with the circle group, so I picture the fibers as circles over points in the base manifold M, and one could take a local trivialization and look at the tangent space. A tangent vector would consist of a tiny bit of motion on the floor (M) and a tiny bit of turning.

Naively it wouldn't matter for such small things which you did first so you could imagine it as a tiny turn followed by a tiny budge in a certain direction

this has to be mapped linear isometrically to so(3) the infinitesimal rotations, but I picture them as the same thing: a tiny turn around the axis where you are, and a tiny roll in a certain direction.

so at the moment this w thing, the linear map between tangentspace P and so(3), looks to me remarkably like a natural identification. but I could be wrong and would feel much better if someone else confirms this-----maybe Dcase already has (I didnt study and figure out his notation with complete certainty) or maybe Hurkyl or selfAdjoint will kindly step in.

 

[derekwise]

a paper on this subject

Hi!

Interesting conversation you all have been having here. This is a subject very near to my heart (i.e. my thesis):!) , and I'd love to join in! But I can't really, because I'm too busy trying to finish up a paper on this stuff before my advisor gives away all of my secrets about the connection (pardon the pun) between MacDowell-Mansouri gravity and Cartan geometry! This paper has been in the works for some time, and might already be out by now if I hadn't gotten tied up finishing a little project on http://math.ucr.edu/~derek/pform/pform_cqg.pdf first. Hopefully very soon, though, the paper will appear on the arXiv, and there will be a more formal place to read about the geometrical foundations of MacDowell-Mansouri gravity. Those of you who have been following this thread will recognize a bunch of the stuff I talk about in the paper. I even talk about what has become the favorite example here as well---the sphere rolling on a surface:
http://math.ucr.edu/~derek/spheres.gif (I tried embedding this picture, but the IMG commands seem to be turned off.)

It's nice to see Garrett is still thinking about this stuff... and that John is getting him to come around to the geometric side. Garrett and I were trying to talk about MacDowell-Mansouri about a year ago or so, but we didn't seem to communicate very well about it at that point---probably because I like Lie groups and geometry, while he likes Clifford algebras and writing funky little arrows over and under everything in sight:
<br /> \underrightarrow{W} = g^- \underrightarrow{\partial} g \in \underrightarrow{Lie(G)}<br />
We'll have to talk more sometime, Garrett.

this P is a PRINCIPAL BUNDLE with the circle group, so I picture the fibers as circles over points in the base manifold M, and one could take a local trivialization and look at the tangent space. A tangent vector would consist of a tiny bit of motion on the floor (M) and a tiny bit of turning.

Naively it wouldn't matter for such small things which you did first so you could imagine it as a tiny turn followed by a tiny budge in a certain direction

this has to be mapped linear isometrically to so(3) the infinitesimal rotations, but I picture them as the same thing: a tiny turn around the axis where you are, and a tiny roll in a certain direction.

Sounds to me like Marcus is on just the right track here. I don't want to give too much away before my paper comes out, but let me say that another interesting example to think about from this perspective is an ISO(2)/SO(2) connection. I'll let you think about the details here... I need to get back to writing!

-DeReK

 

[marcus]

Yay MacDowell-Mansouri! Come on Cartan!

looking forward to seeing that thesis!

thanks for posting news and the gif picture.

 

[garrett]

Hi Derek, good to see you on physics forums!

I'd love to join in! But I can't really, because I'm too busy trying to finish up a paper on this stuff before my advisor gives away all of my secrets

Well, you probably shouldn't follow my example, but I've always thought scientific "secrets" are best when shared as much as possible (without annoying people). It is important to get papers out, so you have a technical reference to point to and say "I did it here" -- but the thing is... not many people read those unless you're either already famous or have been talking with others about your ideas. I think openness generates interest, and the widest readership for your work. It kind of saddens me when physicists clam up about what they're working on. Seems like not as much fun, too. I do understand the desire for "priority" -- but with the net changing the world as it has, I'm not sure getting ink out on dead tree establishes priority more effectively than describing your work on an open forum.

You should, of course, heed the advice of your advisor -- but if he says it's OK, I'd love to hear your thinking on Cartan geometry and MM gravity. Especially since he seems pretty darn busy, and you might have fun with the rapid exchange possible here on PF.

This paper has been in the works for some time, [...] Hopefully very soon, though, the paper will appear on the arXiv, and there will be a more formal place to read about the geometrical foundations of MacDowell-Mansouri gravity.

I certainly look forward to reading it.

It's nice to see Garrett is still thinking about this stuff...

What's that expression? "Like a dog with a bone"? Once I think something's interesting and physically relevant, I don't let go until I've tracked down every implication and possibility. Plus, well, I'm slow. ;)

and that John is getting him to come around to the geometric side.

Sheesh, you guys are like playground kids egging on a fight. I love geometry! As far as I can tell, there's total agreement that one needs to understand MM from the perspective of a global geometric construction and with a local algebraic framework. I guess the perceived conflict arose from me expressing my opinion that learning things starting from the algebraic side is easier.

Garrett and I were trying to talk about MacDowell-Mansouri about a year ago or so, but we didn't seem to communicate very well about it at that point

I remember I enjoyed our brief exchange, but I don't remember that we talked much. I would have like to talk more. Especially about what you were doing. I remember I sent you a copy of the paper I was working on and did get some good feedback. And that was appreciated. But yah, our correspondence didn't take off. Perhaps it was the lack of TeX in email that killed it? Something they've got working here. :)

---probably because I like Lie groups and geometry, while he likes Clifford algebras and writing funky little arrows over and under everything in sight:
<br /> \underrightarrow{W} = g^- \underrightarrow{\partial} g \in \underrightarrow{Lie(G)}<br />

Ooh, how it warms my heart to see little arrows, even sarcastically. I guess as a bit of an eccentric I feel compelled to use my own notation. On the practical side, once one gets rid of indices it's too easy to loose track of which bundles a geometric object is valued in -- thus the arrows, to remind me.

I'd go so far as to say I love Lie groups. :) I just don't see how one can fully understand them without Lie algebra. And a Clifford algebra is just a Lie algebra (having an antisymmetric product (the Lie bracket)) that also has a symmetric product, and is great for describing rotations. Since you like Lie groups, maybe you would enjoy the fun we've been having over in this introductory thread:
https://www.physicsforums.com/showthread.php?t=124233
So I guess it's just that you don't like my arrows. ;) If I had to write "\underrightarrow" for each one, instead of "\f" like I do on my wiki, I wouldn't like them either.

We'll have to talk more sometime, Garrett.

Would love to! If you'd like to talk in person, I'll be in Newport Beach the last two weeks of September -- just a short drive for me up to Riverside. I'm actually here now, but I'm heading to Burning Man for a week and then hopping around the bay area.

But talking here on PF is good too.

but let me say that another interesting example to think about from this perspective is an ISO(2)/SO(2) connection.

Ahh, your thesis appears to have come from a SPR post:
http://www.lepp.cornell.edu/spr/2000-01/msg0021182.html

I'll let you think about the details here... I need to get back to writing!

OK, don't want to hinder you -- but I'm certainly happy to talk whenever you feel like it. This stuff is supposed to be fun and somewhat social you know. :)

This post seems light on physics -- so, maybe jumping the gun, I'll put out my biggest question on MM:
"Why is the MM action what it is? (Besides the anthropish reason that it's what it needs to be to reproduce GR.)"

That's a question on the far horizon... for the mean time, I'd love to hear more from you or JB on Cartan geometry. It's something I got the feeling I liked from skimming through Sharpe's book, but I didn't fully understand it, and would like to.

 

[john baez]

Hi! I'm back from the http://math.ucr.edu/home/baez/diary/suzhou/suzhou-master-of-nets-1.jpg" , and back from a near-death experience with my laptop! I'll post more soon...

It kind of saddens me when physicists clam up about what they're working on. Seems like not as much fun, too. I do understand the desire for "priority" -- but with the net changing the world as it has, I'm not sure getting ink out on dead tree establishes priority more effectively than describing your work on an open forum.

Surely Derek doesn't think getting his work published on paper is relevant. These days in math and physics, it's putting stuff on the arXiv that proves to everyone that you've done it, and makes everyone able to read it. Publishing in a journal is also important, but only for getting jobs and promotions - it's a way of proving that your ideas are accepted by your peers.

You should, of course, heed the advice of your advisor -- but if he says it's OK, I'd love to hear your thinking on Cartan geometry and MM gravity.

Derek's advisor is an old guy who feels no qualms about spilling the beans - he has tenure, he has no higher job ambitions, he just likes talking about physics. He also knows from experience that the hard part is not keeping your ideas secret or establishing priority - it's getting people to pay attention to them!

So, Derek's advisor leaves it up to Derek how much he wants to reveal his ideas online before getting that paper on the arXiv.

But Derek and his advisor both know that as a young grad student aiming eventually for a tenured job, Derek needs to spend a lot of time publishing his butt off. Not chatting with us.

 

[john baez]

thanks for leaving us something to think about. Hope you all enjoyed Suzhou.

Yes, thank you! We enjoyed it immensely - the many gardens, the network of canals, and especially the http://math.ucr.edu/home/baez/diary/suzhou/suzhou-ping-tan.jpg" - you may need to scroll down a bit.

I've was fascinated by Marco Polo's description of Hangzhou (you may also visit?)

No, we didn't visit the nearby city of Hangzhou on this trip, and my plans to visit Zhenghan Wang there have fallen through, but we will visit there and see the West Lake, and we're sort of dreaming of taking a boat from there to Suzhou, so we can spend a night or two in an old http://www.the-silk-road.com/hotel/pingjianghotel/" that we chanced upon during our last visit.

He was impressed that if you had some free time you could rent a boat and take a picnic lunch and a party of friends out on the lake or the canals---and if you didn't happen to have a picnic lunch and ladies with you then you could rent a boat that was ALREADY PROVISIONED with both.

I like that idea, but Lisa will be there, so we'll probably just take the boat and maybe a picnic lunch, and skip the ladies.

I think Suzhou is known as China's "Venice" but IIRC Hangzhou was Polo's favorite. Unless I'm dreaming he also said that those 13th century Hongzhou merchants were so rich that they burned ornamentally carved logs in the fireplace.

Wow. Hangzhou was the capital of China during the Southern Sung, after the Mongols invaded the northern capital in 1127, and before they invaded Hangzhou in 1279. It's been a beautiful place ever since, I guess. Suzhou is a bit less important - I don't actually know its history well at all - but it's famous for gardens, canals and silk.

Anyway, let's talk about Cartan geometry a bit. I'm afraid my output here will slow down now that I have my own blog - or rather, a blog I'm sharing with David Corfield and Urs Schreiber, called http://golem.ph.utexas.edu/category/" .

But, we should at least get this Cartan connection business straight!

This P is a PRINCIPAL BUNDLE with the circle group, so I picture the fibers as circles over points in the base manifold M, and one could take a local trivialization and look at the tangent space. A tangent vector would consist of a tiny bit of motion on the floor (M) and a tiny bit of turning.

That sounds about right - there are some tricky things here, but I think it's right.

Naively it wouldn't matter for such small things which you did first so you could imagine it as a tiny turn followed by a tiny budge in a certain direction.

To first order, yes.

This has to be mapped linear isometrically to so(3) the infinitesimal rotations, but I picture them as the same thing: a tiny turn around the axis where you are, and a tiny roll in a certain direction.

Nothing about "isometrically" in the definition of Cartan connection, but if you just said "isomorphically" you'd be right - and more importantly, your picture is the right one.

(Isomorphic mean 1-1 and onto (in this context), while isometric means distance-preserving.)

So at the moment this w thing, the linear map between tangentspace P and so(3), looks to me remarkably like a natural identification. But I could be wrong and would feel much better if someone else confirms this...

It's not a natural identification unless we trivialize P, writing it as a product of M and the circle. Trivializing a bundle gives it a connection in the usual Ehresmann sense, and I guess that's true for Cartan connections too. But, we don't want to trivialize our bundles; we want to keep things '"floppy" enough so there's a choice of connection.

But anyway: yes - if we know how to roll our surface on our surface M, we know a linear isomorphism from any tangent space of P to the Lie algebra so(3). And this is our Cartan connection.

 

[john baez]

Last call

I'll be glad to explain the rest of the clauses in this definition if anyone wants...

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 w on P satisfying:

1. w is a linear isomorphism from the tangent space of P to Lie(G)

2. for all h in H R(h)^* w = Ad(h^{-1})w

3. w(X+) = X for all X in Lie(H).

where X+ is the vector field on P corresponding to the Lie algebra element X in Lie(H), and R(h^{-1})^* says how an element h of H acts on 1-forms on P.

... or maybe this thread has done its job.

 

[marcus]

I'll be glad to explain the rest of the clauses in this definition if anyone wants...

Nobody in his right mind turns down such an offer.
careless of me to say metric when I meant morphic, sorry.

BTW I'm beginning to crave the next Baratin-Freidel paper. Wonder if it will be coming out in the next month or two. (or if anyone who is not actually Baratin or Freidel would have any way of telling)

 

[Dcase]

Hi John Baez and Marcus:

I know that I would like to learn more if I can.

The concept is more of a one to one relationship than an identity.

I probably am taking this concept too far, but if this concept could be genelaized to roll, yaw, pitch with an eigen axis then there may be correlation with NCG-space: X+, Y+, Z+ as vector fields, all X,Y,Z in Lie(H).

Could this concept also be expanded to cycldes? - or generalized to non-distance-preserving, but ratio-perseving situations such as gauge or scale transformations [2-1, 3-1 ,,,]?
http://mathworld.wolfram.com/Cyclide.html

 

[john baez]

Nobody in his right mind turns down such an offer.
careless of me to say metric when I meant morphic, sorry.

No prob. But actually it seems my attention has drifted elsewhere, along with everyone else's... I'm too absorbed posting stuff about logic on my blog! So, let's leave it at this for now.

 

[garrett]

last call still answerable?

I'll be glad to explain the rest of the clauses in this definition if anyone wants...

Yes, please!

I've been away for a couple of weeks at the Burning Man festival, where I harassed (in a friendly way) some string theorists by making a public bet with them regarding the (non)appearance of superparticles at the LHC. I also hung out and talked with an interesting fellow, Michael Edwards, in Santa Cruz about BRST geometry. I was hoping JB would continue this thread on Cartan geometry here, but it looks like he's been sidetracked :( , though no doubt productively. Is there any hope of bringing him back for a little clarification on this definition? I think I mostly understand this:

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{w} on P satisfying
1. \underrightarrow{w} is a linear isomorphism from the tangent space of P to Lie(G).
2. R(h)^* \underrightarrow{w} = h^- \underrightarrow{w} h for all h in H.
3. \vec{\xi_X} \underrightarrow{w} = 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.

Here's my take on it:

(1) says the star of the show is a Lie(G) valued 1-form, which we can write on local coordinate patches of P as
<br /> \underrightarrow{w} = \underrightarrow{dx^i} w_i{}^B T_B + \underrightarrow{dy^a} w_a{}^B T_B<br />
where group elements, h, of H are described by some coordinates, y^a, and the x^i are coordinates over M. So p=(x,y) are coordinates over patches of P, which can also be labeled by the M coordinate and H element, p=(x,h), since H elements are specified by y coordinate, h(y). This 1-form allows us to take any path on P and map its tangent vector at each point to a Lie(G) element, which we can then integrate along the path to get an element of G. In the rolling sphere example, this G element tells us how the sphere rotates.

(3) directly relates motion along the "vertical" H part of P to the corresponding Lie(H) generators which are a subalgebra of Lie(G). Specifically, a Lie(H) generator gives a flow of the points of P by (I presume?) acting on the vertical parts from the right: p T_a = (x, h T_a). This flow is associated with a vector field,
<br /> \vec{\xi_a} \underrightarrow{\partial} p = \xi_a{}^b \partial_b p = p T_a<br />
and this vector field, \xi_a{}^b, can be calculated explicitly, depending on how P (and H) is coordinatized by x and y. Now, (3) says the Cartan connection must satisfy
<br /> \vec{\xi_a} \underrightarrow{w} = T_a<br />
which, since \xi is an invertible square matrix, completely determines the "vertical acting" part of w,
<br /> w_a{}^B = \xi^-_a{}^B<br />
Looked at another way, this says the vertical part of the Cartan connection is the same as the Maurer-Cartan connection for the H. The other part of the Cartan connection must describe the bumpiness of the surface.

Now (2) I'm not so clear on, and could use some help. It sort of looks like the way you'd want to describe a gauge transformation of a connection, but seems a little strange to me. Maybe John Baez can chime in and let me know if my expanded descriptions of (1) and (3) are OK, and help me out by clarifying the meaning of (2)?

 

[garrett]

Hmm, crickets... ah well, I'm used to talking to myself, so I can have a go at explaining (2) to myself. I think the key is that R(h) is an automorphism of P -- mapping points of P to other points of P via right action by h:
<br /> p_2 = R(h) p_1 = R(h) (x,h_1) = (x,h_1 h) = (x,h_2)<br />
So R(h)^* \underrightarrow{w} is the pullback of the \underrightarrow{w} 1-form at p_2 to p_1, where it is restricted to equal h^- \underrightarrow{w} h. This restriction implies that if we know what \underrightarrow{w} is along any section in P, (2) tells us what \underrightarrow{w} is over all of P -- since we can get to the other points by acting with R(h) for any h. Cool. And, once again, this same restriction is satisfied by the Maurer-Cartan connection over group manifolds.

Subsequently, (2) implies that we can describe the connection via a Lie(G) valued 1-form over patches of M, with some local trivializing section presumed. I think this use came up earlier.

The next interesting things to consider are covariant derivatives, and the curvature of this Cartan connection. Hmm, I think if the curvature vanishes, \underrightarrow{w} is the Maurer-Cartan connection, with P taken as the base space for G.

Maybe someone will correct me. Or maybe not.

 

[derekwise]

MacDowell-Mansouri gravity and Cartan Geometry

Hi!

For anyone still interested in how Cartan geomtry is about "rolling a homogeneous space on a manifold", I explain a bunch of this in my new paper:

MacDowell-Mansouri gravity and Cartan Geometry
http://arxiv.org/abs/gr-qc/0611154

-DeReK

 

[marcus]

Hi!

For anyone still interested in how Cartan geomtry is about "rolling a homogeneous space on a manifold", I explain a bunch of this in my new paper:

MacDowell-Mansouri gravity and Cartan Geometry
http://arxiv.org/abs/gr-qc/0611154

-DeReK

thanks. looks like a very helpful paper!

 

[garrett]

Yep, Derek's written an excellent expository paper that fills in a lot of geometric details and background behind BF gravity. It's also quite readable.

Perhaps now that he's let his cat out of its bag he'll come talk to us again on this thread, or on his.

 

[marcus]

this thread has a lot of conversation between Garrett Lisi and John Baez about things that might be relevant to Garrett's recent paper, so i thought I'd bring it back from the limbo of Forgotten Threads

 

[marcus]

This thread needs updating. Quite a lot of research has come out in the wake of what John Baez and Garrett Lisi were discussing here. I will try to catch it up a little. Others' help woud be very welcome.

In fact, JB and Garrett were talking about some work by Derek Wise and as it happens Derek published a paper just last month (August 2009) which continues the line of research they were discussing.

http://arxiv.org/abs/0904.1738
Symmetric Space Cartan Connections and Gravity in Three and Four Dimensions
Derek K. Wise
Article prepared for special journal issue dedicated to Elie Cartan
SIGMA 5 (2009), 080, 18 pages
(Submitted on 10 Apr 2009, revised for publication 3 Aug 2009)
"Einstein gravity in both 3 and 4 dimensions, as well as some interesting generalizations, can be written as gauge theories in which the connection is a Cartan connection for geometry modeled on a symmetric space. The relevant models in 3 dimensions include Einstein gravity in Chern-Simons form, as well as a new formulation of topologically massive gravity, with arbitrary cosmological constant, as a single constrained Chern-Simons action. In 4 dimensions the main model of interest is MacDowell-Mansouri gravity, generalized to include the Immirzi parameter in a natural way. I formulate these theories in Cartan geometric language, emphasizing also the role played by the symmetric space structure of the model. I also explain how, from the perspective of these Cartan-geometric formulations, both the topological mass in 3d and the Immirzi parameter in 4d are the result of non-simplicity of the Lorentz Lie algebra so(3,1) and its relatives. Finally, I suggest how the language of Cartan geometry provides a guiding principle for elegantly reformulating any 'gauge theory of geometry'."

Derek will be giving a talk next week in Corfu, on Saturday 19 September. John Baez is giving a series of 5 lectures at the Corfu QG School.