# DeSitter group SO(4,1) intro

Gold Member
Dearly Missed
probably some of us need to get more familiar with the DeSitter group SO(4,1)

the best online introduction I know is a few short paragraphs towards the end of Baez TWF 235
http://www.math.ucr.edu/home/baez/week235.html

you have to scroll down about 5/8 of the way. on my printer it is arond page 5 out of 8 pages total.

If anyone here wants to teach us about SO(4,1), explain stuff, discuss, ask questions about it, that would be constructive.

the point is that most everything in physics is built on MINKOWSKI spacetime which is FLATTEST possible spacetime. It is so flat that it doesnt even expand!
It is like what spacetime would be if there were no matter in the universe at all and the gravitational field were zero everywhere.

DeSitter space is sort of next-of-kin to Minkowski space in the sense that it is FLATTEST POSSIBLE SPACETIME THAT EXPANDS A LITTLE.
It is the flattest, most uniform, most symmetric, most empty, most totally vanilla that spacetime can possibly be, if it is required to have a little bit of cosmological constant Lambda in it making it expand.

you just add a tiny bit of dark energy, very evenly distributed, so it doesnt introduce any structure or disturb the symmetry any more than absolutely necessary.

I never liked the name Minkowski, it reminds me of a stout lady in a fur coat. I used to live in Westchester, and sometimes in New York. I actually would be very glad if we could get off of Minkowski spacetime and move on to something more interesting.

But you always have to have an idea of vanilla.

Anyway, SO(4,1) is the symmetries of DeSitter spacetime, just like Lorentz group or Poincaré is the symmetries of Minkowski spacetime.

Gold Member
Dearly Missed
the way to study DeSitter spacetime is to set up a 5D euclidean space and take a HYPERBOLOID subset of it
and that 4D hyperboloid is the spacetime we want

{(w,x,y,z,t) : w2 + x2 + y2 + z2 - t2 = k2}

the Lie group SO(4,1) is just the group of transformations that STAYS ON THE HYPERBOLOID
because it doesnt change the k.
it doesnt change this "measure" of the size of a point:
w2 + x2 + y2 + z2 - t2
whatever that is before the transformation, it is the same after
so the Lie group is just the matrices that transform normal 5D space in such a way that RESPECTS the slicing up into hyperboloids and doesnt mix up one k with some different k "sheet".

that might not be how the actual WORLD is, but mathematically it is convenient and easy to imagine (relatively any way).

basically we are all just greeks drawing flat planes and parabolas on the sand or on papyrus, and these are idealizations of the world. they are not the world. but they are incredibly useful idealizations. so this hyperboloid is JUST ANOTHER PLANE, just another diagram of the world such as euclid might have been studying. that is my opinion.

so now it's time to move attention to DeSitter space and ITS symmetries, because maybe that is the new "flat vanilla" local approximation of the world, the new "tangent plane" to reality. maybe.

=================
motivation
Baez 235 gives some motivation, also some other threads here at PF "Beyond" forum are talking about DeSitter group stuff
=================

BTW correct me if I am wrong. no way am I an authority, I just think we need to take this introductory step in what we are able to discuss here.
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Was reading TWF 235 some more, it is really good. Garrett, thanks for the extra savvy in the next post!
I am for looking at every clue right now.
Let's put TWF 235 (which this thread is partly about) together with what JB just said today in the "Baratin Freidel" thread. I will quote in full and we can see if we can pick up and highlight something from it:
john baez said:
So I guess the question is, what do you think? Do you also think that the promised paper
"Background Independent Perturbation Theory for Gravity Coupled to Particles: Classical Analysis" is actually the new one we have in hand called "Particles as Wilson Lines of Gravitational Field" but renamed?

I don't know. I remember Laurent saying they weren't even sure how many papers they were writing on this subject: two or three. They've done a lot of work, obviously, and for a big project like this one needs to keep rethinking the best way to slice the work into papers.

The paper they wrote doesn't actually do any "perturbation theory", apart from writing the MacDowell-Mansouri Lagrangian as the BF Lagrangian plus two extra terms, and analysing what this means... which they'd already done in a previous paper. The big new thing is to introduce particle worldlines as "defects" - curves removed from spacetime - much as had already been done in 3d gravity. So, it makes sense for their title to emphasize this.

In fact, their title is a bit more dramatic than what I might have chosen, because they don't really study these particle worldlines in the context of MacDowell-Mansouri gravity, except for one equation right near the end. Mostly they study these particles in the context of plain old 4d BF theory.

This nicely complements my own study, with http://arxiv.org/abs/gr-qc/0603085" [Broken].

Unfortunately, Crans, Wise, Perez and I studied strings coupled to 4d BF theory for a general gauge group but didn't work out the details for the gauge group Freidel uses, namely SO(4,1). We focused on SO(3,1). It should be easy to do the SO(4,1) case now, though since Freidel & Company have worked out a lot of the necessary stuff.

After a talk I gave, Freidel guessed that the strings may be related to gravitons... or replace them, somehow. It's a big mystery: a nice structure is emerging, but it's not clear what it means! This is what makes physics fun.

BTW notice that Baez is throwing handfuls of good research problems to the crowd these days. PhD-size problems are coming down like confetti. In TWF 235 at one point he tosses off a PhD problem and then says
"Oh gee, I should have checked that long ago. How careless of me!" and then he actually says "Ahem", when you thought it could not be made more obvious. What a time this is!

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garrett
Gold Member
The best way to understand the DeSitter group, SO(4,1), is to first understand the DeSitter algebra, so(4,1). This is the algebra of rotations for a five dimensional space with signature (4,1) -- that's four directions with positive norm and one direction with negative norm. (The same way Minkowski space has signature (1,3) or (3,1).) The DeSitter algebra has 10 generators which have commutation relations (structure coefficients) between them under the antisymmetric product (bracket).

But there's another, equivalent, way of understanding the DeSitter algebra that I think is better, physically. A rotation algebra can be constructed explicitly as the algebra of Clifford algebra bivectors under the antisymmetric product -- each bivector corresponding to an algebraic generator, and producing the same structure coefficients and norms between them. So, in terms of Clifford algebra, the DeSitter algebra is just the algebra of bivectors of the corresponding Clifford algebra, $Cl^2_{4,1}$. Now, an interesting fact about Clifford algebras is that the subalgebra of Clifford bivectors is equivalent to the subalgebra of Clifford bivectors AND vectors of the Clifford algebra of one less dimension. So, it works out that the DeSitter algebra of five dimensional rotations is nothing but the Clifford subalgebra of vectors and bivectors of four dimensional Euclidean space, $Cl^{1,2}_4$. This is really neat. And it's easy to build explicitly.

Start with four Clifford basis vectors with positive norm, $\gamma_\alpha$. Their product is 1 if they're the same, or a bivector if they're different, and different basis vectors anticommute. This means their symmetric product gives 1 or 0:
$$\gamma_\alpha \cdot \gamma_\beta = \frac{1}{2} ( \gamma_\alpha \gamma_\beta + \gamma_\beta \gamma_\alpha ) = \delta_{\alpha,\beta}$$
and their antisymmetic product gives the six bivectors, such as
$$\gamma_1 \times \gamma_2 = \frac{1}{2} ( \gamma_1 \gamma_2 - \gamma_2 \gamma_1 ) = \gamma_{12}$$
$$= \gamma_1 \gamma_2 = - \gamma_2 \gamma_1$$
The algebra of these 10 vectors and bivectors under the antisymmetric product, such as
$$\gamma_1 \times \gamma_{12} = \gamma_2$$
is the DeSitter algebra.

This is very simple, powerful, and cool. And things really get hopping when you deal with fermions, which are naturally expressed Clifford elements. (The reason I use the same notation for Clifford vectors that you may have seen for Dirac matrices is that they are the same objects -- the Dirac matrices are simply a matrix representation.) I'm not sure why John Baez, Freidel, etc., don't take advantage of this Clifford stuff yet in their work, explicitly, but I am quite sure that they will. :)

john baez
Gold Member
visualizing deSitter spacetime

garrett said:
The best way to understand the DeSitter group, SO(4,1), is to first understand the DeSitter algebra, so(4,1).

For computations Lie algebras are better, but for visualization and intuition I think Lie groups are better. Elements of Lie groups describe symmetries in various kinds of geometry. That stuff is nice to visualize! One can figure out lots of stuff without resorting to pencil and paper. That's appealing to us lazy sorts.

If anyone has trouble with SO(4,1) they should start with SO(2,1). SO(2,1) is the 3d Lorentz group: in other words, linear transformations of R^3 that preserve the Minkowski metric

t^2 - x^2 - y^2

But can also think of this group as the symmetries of the hyperboloid

t^2 - x^2 - y^z = 1 (t > 0)

This hyperboloid gets a distance function on it from the Minkowski metric, and then it's called the hyperbolic plane.

Everyone should learn a bit of http://en.wikipedia.org/wiki/Hyperbolic_geometry" [Broken], because the sphere and hyperbolic plane are the simplest examples of non-Euclidean geometry - and that's where general relativity came from. All the deSitter and anti-deSitter spacetimes are just generalizations of the sphere and the hyperbolic plane with various numbers of space and time dimensions!

There are various handy http://www.geom.uiuc.edu/docs/forum/hype/model.html" [Broken]. In the Klein model you draw the hyperbolic plane as a disk, and in one coordinate system geodesics are just straight lines in this disk. In the Poincare model you also draw a disk, but now lines are portions of circles that hit the edge of the disk at right angles.

SO(2,1) has 1 dimension worth of "rotations" and 2 dimensions worth of "Lorentz boosts". If we see how these act on the hyperbolic plane we see the rotations are just rotations, while the boosts act like "translations". But, unlike translations in Euclidean geometry, they don't commute!

SO(2,1) is also the group of symmetries of the hyperboloid

t^2 - x^2 - y^2 = -1

This hyperboloid also gets a metric on it from the Minkowski metric, but it's a 2d spacetime instead of a 2d space. In fact, it's the 2d version of deSitter spacetime! Remember that the 4d version is given by

t^2 - w^2 - x^2 - y^z - z^2 = -1

So, if you want to visualize 4d deSitter spacetime and its symmetry group SO(4,1), you'd better start with the 2d version and its symmetry group SO(2,1). And the cool thing is, 2d deSitter spacetime is just like the hyperbolic plane, but with one dimension of space replaced by one dimension of time! - with all that entails.

The geodesics in 2d deSitter spacetime are obtained by taking planes through the origin of R^3 and intersecting them with our hyperboloid

t^2 - x^2 - y^2 = -1

Can you see how, just as in the hyperbolic plane, the parallel postulate is drastically violated by geodesics in deSitter spacetime? Do you see what it means? Two people moving relative two each other may never crash into each other - unlike in 2d Minkowski spacetime. The reason is, the universe is expanding!

The group SO(2,1) acts in an interesting way. We have, as with the hyperbolic plane, 1 dimension of "rotations" that just rotate the x and y axes around. But, we also have the boost in the xt plane and the boost in the yt plane. As with the hyperbolic plane, boosts act vaguely like "translations" on deSitter spacetime, since they can move any point on this spacetime to any other point. But as before, they don't commute.

To dig deeper, you want to see how to slice deSitter spacetime with spacelike slices so that it looks like an exponentially expanding universe. There are various ways to slice deSitter spacetime, which caused some interesting discussions not just http://groups.google.com/group/sci....physics.*+author:baez&hl=en#9e562ab0d38e18fc" is going to be so lonely if the current exponential expansion keeps up.

I'm not sure why John Baez, Freidel, etc., don't take advantage of this Clifford stuff yet in their work, explicitly, but I am quite sure that they will. :)

We want to leave something for you to do! Like add the Standard Model and win a Nobel prize! :tongue2:

But seriously, we'll use Clifford algebras whenever we feel it helps - like when fermions get into the game. You can already find Clifford algebras in http://xxx.lanl.gov/abs/gr-qc/0607014" - see especially the appendix.

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garrett
Gold Member

john baez said:
But seriously, we'll use Clifford algebras whenever we feel it helps - like when fermions get into the game. You can already find Clifford algebras in http://xxx.lanl.gov/abs/gr-qc/0607014" - see especially the appendix.

Yes, I suppose I hadn't fully appreciated that was there. Kind of turns my cocky prediction into a weak postdiction. But it's certainly good to see. I know your familiarity with Clifford algebra predates mine by a good ways. And, from that appendix, it looks like Freidel et al are starting to take it seriously -- dipping their TOEs in the water beneath Hamilton's bridge before jumping in. This makes me especially happy, since it's converging with what I've been doing. I've been mostly working on pedantic algebraic model building, while you guys are coming from the quantum gravity side. When the two meet, I'm hoping it will work out very well.

For computations Lie algebras are better, but for visualization and intuition I think Lie groups are better.

But does the global Lie group geometry matter in this BF theory context? Maybe the answer to a quick question will clear this up for me:

The Lie groups SO(4,1) and Spin(4,1) have the same Lie algebra, $so(4,1) = Cl^{1,2}_4$. In the paper you sited above (and for MacDowell-Mansouri model in general?) would all the results be the same if we used Spin(4,1) instead of SO(4,1)?

If the answer is "yes," then I'm not sure why one would think of the Lie group as better to work with, and visualize with, than the Lie algebra, which describes the local Lie group geometry -- since using different groups with the same algebra would produce the same theory.

Now, I do think that the global Lie group geometry will matter when we start looking at fermions and charge assignments, but that's a different kettle of fish.

Oh, and, umm, Thanks!

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john baez
Gold Member
garrett said:

Thanks! It's fun here.

But does the global Lie group geometry matter in this BF theory context? Maybe the answer to a quick question will clear this up for me:

The Lie groups SO(4,1) and Spin(4,1) have the same Lie algebra, $so(4,1) = Cl^{1,2}_4$. In the paper you cited above (and for MacDowell-Mansouri model in general?) would all the results be the same if we used Spin(4,1) instead of SO(4,1)?

The paper I cited may not go far enough to hit this issue, but when one really tries to understand fermions coupled to MacDowell-Mansouri gravity, one needs to think of the connection as a Spin(4,1) connection rather than an SO(4,1) connection.

And don't forget, Freidel is a physicist, so he's allowed to say SO(4,1) and mean Spin(4,1). :tongue:

If the answer is "yes," then I'm not sure why one would think of the Lie group as better to work with, and visualize with, than the Lie algebra...

I would never want to use Lie groups without the freedom to use Lie algebras, nor Lie algebras without Lie groups. Why tie one hand behind my back? If you look at my work with Derek and Alissa on string-like excitations in BF theory, you'll see that particle and string types are described by conjugacy classes in the Lie group. Only in the limit where the Planck mass goes to infinity do these look like adjoint orbits in the Lie algebra. The global Lie group topology matters a lot here. But it's the interplay between Lie algebras and Lie groups that makes this fun - multiplication of group-valued momenta reduces to ordinary addition of momenta in the Lie algebra as the Planck mass goes to infinity.

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garrett
Gold Member
I would tie one hand behind your back because if I didn't you would make up too much math! ;)

My real point here is that I don't think BF gauge theory, by itself, involves DeSitter space -- it only involves DeSitter algebra. Any space like DeSitter space locally will do just fine. So it confuses me when people start talking about DeSitter space in the context of BF gauge theory. I guess the point is kind of subtle.

I do think it's an interesting observation that gauge fields (connections) don't care about Lie group (fiber) topology -- they only care about the Lie algebra (geometry) -- but that fermions do care about topology.

The reason I'm being difficult on this point is I think it's actually bad to use DeSitter space in discussing BF gauge theory. It's just confusing when all you really need is the DeSitter algebra. I'm probably the kind of physicist that mathematicians hate, because if the universe doesn't clearly care about the math being used, I don't usually care either. Nature seems beautiful enough as she is, without making extra dresses for her that don't fit.

For my part, I promise not to confuse SO(4,1) with Spin(4,1), even as a dumb physicist. But I think if they're interchangeable in a certain physical context it's really so(4,1) that matters, and talking about DeSitter space qualifies as going off on a tangent.

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john baez
Gold Member
garrett said:
My real point here is that I don't think BF gauge theory, by itself, involves DeSitter space -- it only involves DeSitter algebra. Any space like DeSitter space locally will do just fine. So it confuses me when people start talking about DeSitter space in the context of BF gauge theory. I guess the point is kind of subtle.

Well, I can only understand the geometrical meaning of MacDowell-Mansouri gravity in terms of Cartan geometry, where you try to understand fairly general spacetimes using special, highly symmetrical spacetimes. In ordinary Riemannian geometry we think of each point in a curved space as having a "tangent plane" which looks like flat Euclidean space. In Cartan geometry, we can also set things up so each point has a "tangent sphere" - which is nice when your space is close to a round sphere, perhaps with some bumps and wiggles here and there - or a "tangent hyperbolic space" - which is nice when your space is negatively curved.

In the usual approach to general relativity we think of each spacetime point as having a "tangent plane" which looks like Minkowski spacetime. In Cartan geometry we can also assume a "tangent deSitter spacetime" or a "tangent anti-deSitter spacetime". And, MacDowell-Mansouri gravity is really - secretly - based on Cartan geometry.

I could say much more, but I won't, because this will all appear in Derek Wise's PhD thesis, or possibly a paper that'll appear on the arXiv before he finishes his thesis.

I do think it's an interesting observation that gauge fields (connections) don't care about Lie group (fiber) topology -- they only care about the Lie algebra (geometry) -- but that fermions do care about topology.

I don't quite believe that. There's more information in a Spin(4,1) connection than an SO(4,1) connection, and even if you've just got gauge fields around, the path integral in a Spin(4,1) gauge theory should often give different answers than in the corresponding SO(4,1) gauge theory. This effect is much easier to see in simpler theories, like SU(2) versus SO(3) Yang-Mills theory on a 2d spacetime, where you can compute everything explicitly and see how things work out differently.

(Witten was the first to exactly solve 2d Yang-Mills theory on a Riemann surface; having things exactly soluble is just nice for making sure one isn't getting fooled by divergences or some other tricky issue here or there.)

The reason I'm being difficult on this point is I think it's actually bad to use DeSitter space in discussing BF gauge theory.

Okay, don't do it then. I think it's actually good, for the several reasons I've explained and plenty more, including the fact that the quantum version of SO(4,1) BF theory has a very simple beautiful state representing the "quantum version of deSitter spacetime" - an exponentially expanding universe not unlike our own, including quantum effects. The background-free perturbation theory that Freidel and Starodubtsev are trying to develop will probably involve perturbation around this state. And I don't see how I'm going to understand this state without first understanding the classical deSitter solution!

It's just confusing when all you really need is the DeSitter algebra.

Maybe that's all you need; it's not all I need.

I'm probably the kind of physicist that mathematicians hate, because if the universe doesn't clearly care about the math being used, I don't usually care either.

You may know what "the universe doesn't clearly care about". I don't. So, if I'm studying a theory that involves some piece of math, and that piece is integrally connected to some other pieces, I think about those other pieces too, and see what I can do with them.

I don't think further discussion of this general difference in our philosophies will be profitable or even fun - at least, not for me.

garrett
Gold Member
OK. I'll reconsider my close minded position. On the one hand, I could look at these points you've raised as issues that are "something extra" outside the context of the classical BF / MacDowell-Mansouri gauge gravity construction. But you're right that some of them may turn out to be physically relevant in a larger context, and I think I even said that in the beginning of this discussion. My only point was that I thought starting out with a description of DeSitter space instead of starting with DeSitter algebra was confusing in this MacDowell-Mansouri context. But it's possible I just don't understand this stuff well enough yet.

I'm particularly curious about the Cartan geometry approach, as an alternative to the standard fiber bundle construction, especially now that you've mentioned it's the basis for your understanding. I know a little about it, but I never really groked it. I don't want to take the wind out of Darek's thesis, but are you sure we can't discuss that a bit, maybe just on the introductory level? (Phyiscs Forums is neat because you can do LaTex here.)

As far as I know, a Cartan geometry is sort of like a fiber bundle. It involves starting with a large Lie group, G, with a closed subgroup, H. One builds the "base" space as the coset, G/H, and the "fiber" as H. But there's this interesting connection-like thing which has an extra piece in it:
$$\omega = dx^i (e_i)^\alpha T'_\alpha + dx^i A_i{}^A T_A$$
in which $T_A$ are the generators for H, $T'_\alpha$ are the generators for G that aren't for H, and the $x^i$ coordinates are for the coset space, G/H. Is that right, and where does one go from there?

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john baez
Gold Member
rolling without slipping or twisting

garrett said:
OK. I'll reconsider my close minded position.

You don't need to! Just don't try to get me to reconsider mine. :tongue2:

My only point was that I thought starting out with a description of DeSitter space instead of starting with DeSitter algebra was confusing in this MacDowell-Mansouri context. But it's possible I just don't understand this stuff well enough yet.

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.

I know a little about it, but I never really groked it. I don't want to take the wind out of Derek's thesis, but are you sure we can't discuss that a bit, maybe just on the introductory level?

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!

A couple of questions:

What (if any) are the (theoretically) observable phenomena with regards to the expansion of coordinate time in the DeSitter space?

How is the mapping of non-gravitational forces envisioned with regards to the expansion. In other words, does it follow the stretching or do we have to adjust for it?

Hope my questions make any sense. :shy:

garrett
Gold Member
Cartan geometry

Ah, OK, I think I had it backwards. Can you tell me if this is the correct way to describe a Cartan geometry as a fiber bundle:

The fiber is the homogeneous space, G/H.
The base manifold needs to have the same dimension as G/H.
The structure group, acting on G/H, is G.

Is that right?

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john baez
Gold Member
Cartan geometries as bundles

garrett said:
Ah, OK, I think I had it backwards. Can you tell me if this is the correct way to describe a Cartan geometry as a fiber bundle:

The fiber is the homogeneous space, G/H.
The base manifold needs to have the same dimension as G/H.
The structure group, acting on G/H, is G.

Is that right?

Yes, this is exactly right. Of course a "Cartan geometry" has more information than this, but it does have all this.

The http://en.wikipedia.org/wiki/Cartan_connection" [Broken] of a Cartan geometry is short and to the point, and maybe it will make sense now.

Hmm, but there's a little mistake in it, so let me post a corrected version here:

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).

Here X+ is their goofy notation for the vector field on P corresponding to the Lie algebra element X in Lie(H); you get such a vector field because the group H acts on P. Similarly, R(h) is their goofy notation for how an element h of H acts on P on the right, and the little extra superscript * says it's acting on 1-forms on P via pullback.

All this stuff makes more sense if you talk about it a lot before hitting someone with the formulas....

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john baez
Gold Member
exponentially expanding universe

MeJennifer said:
A couple of questions:

What (if any) are the (theoretically) observable phenomena with regards to the expansion of coordinate time in the DeSitter space?

It's not "coordinate time" that expands in DeSitter space; space expands as coordinate time passes.

Since our universe is pretty well approximated by DeSitter space nowadays, the observable phenomena have actually been observed and are pretty well-known. For example, distant galaxies are moving away from ours, so we see their light red-shifted, and more careful studies show that this expansion is speeding up!

How is the mapping of non-gravitational forces envisioned with regards to the expansion. In other words, does it follow the stretching or do we have to adjust for it?

It sounds like you're raising a famous old puzzle: "If the whole universe is expanding, how do we know? Why don't we just expand along with it?" You can read an answer in this physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/GR/expanding_universe.html" [Broken]

The quick answer to both questions is no. The fun part is understanding why the answer is no.

For the long-term effects of an exponentially expanding universe, try my cheery page on http://math.ucr.edu/home/baez/end.html" [Broken].

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garrett
Gold Member
One has to appreciate the fact that I just tried to give a Lie group oriented definition, while JB gave a Lie algebra oriented one. Heh.

john baez
Gold Member
garrett said:
One has to appreciate the fact that I just tried to give a Lie group oriented definition, while JB gave a Lie algebra oriented one. Heh.

The definition I gave includes everything you said, and more. It crucially uses both Lie groups and Lie algebras - as usual, they're inseparable.

I find it far, far less interesting to argue the merits of Lie groups versus Lie algebras than to actually talk about physics and geometry. It's like arguing about the relative merits of odd versus even numbers when one could be seriously discussing number theory. The interesting question is not what percentage of the definition of "Cartan geometry" uses Lie groups and what percentage uses Lie algebras. The interesting question is whether one understands the definition and can do things with it.

garrett
Gold Member
Agreed. I just can't resist pointing out irony, even when I'm its source.

So, can we get back to helping me understand Cartan geometry? In mathematics you can often define something in different but equivalent ways. I took a swing at defining it as:
The fiber is the homogeneous space, G/H.
The base manifold needs to have the same dimension as G/H.
The structure group, acting on G/H, is G.
And you said
Yes, this is exactly right. Of course a "Cartan geometry" has more information than this, but it does have all this.
and went on to give the precise definition. I'm wondering what more the definition I gave needs (the "more information") in order to match the precise definition you quoted?

From what I can tell, the connection, w, that the wiki definition uses is the same as the http://en.wikipedia.org/wiki/Ehresmann_connection" [Broken]
that would naturally be defined over the total space of the fiber bundle as I described it.

If I were to guess: a piece of information I think might be missing could be that:
The base manifold needs to have the same topology as G/H.
Is that right, and is there more?

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john baez
Gold Member
Cartan connections

garrett said:
So, can we get back to helping me understand Cartan geometry?

Yeah, that's more interesting.

As we've seen, a Cartan geometry involves a Lie group G with a subgroup H, a base manifold X, a principal H-bundle P over X, and the start of the show: the Cartan connection w.

It's really helpful to keep in mind our example of a sphere rolling over a lumpy bumpy surface in Euclidean space. In this example G is the 3d rotation group, H is the subgroup of rotations that preserve a given point on the sphere, and G/H is the sphere. X is our lumpy bumpy surface. P is space of all ways we can set a sphere tangent to our lumpy bumpy surface.

The star of the show is the Cartan connection, w. w says which direction our sphere starts rolling when we set it on our surface X and start rotating it. So, it gives an isomorphism between the tangent space of P and the Lie algebra of G.

In mathematics you can often define something in different but equivalent ways. I took a swing at defining it as:

The fiber is the homogeneous space, G/H.
The base manifold needs to have the same dimension as G/H.
The structure group, acting on G/H, is G.

And I said yes, that's great - but there's more information in a Cartan geometry, namely the Cartan connection:

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.

I'm wondering what more the definition I gave needs (the "more information") in order to match the precise definition you quoted?

Well, as you can see, it's the Cartan connection w. Note that item 1 guarantees P has the same dimension as G:

dim(P) = dim(G)

Since P is a principal H-bundle over X we have

dim(P) = dim(H) + dim(X)

and of course

dim(G) = dim(H) + dim(G/H)

so we get

dim(G/H) = dim(X)

as you stated.

The principal H-bundle P has fibers that look like H, so it's not the bundle with fibers G/H that you were talking about. However, we can get a bundle with fibers G/H from P by the usual "associated bundle" trick:

$$P \times_{H} G/H$$

and this is the bundle you were talking about.

So, we get everything you said, but more.

I have the feeling that I gave the precise definition a bit too soon; it makes perfect sense to me, but probably no sense to anyone who hasn't thought about this stuff for a long time. It's best to take the example I gave and work through in detail what each item of the definition does!!!

From what I can tell, the connection, w, that the wiki definition uses is the same as the http://en.wikipedia.org/wiki/Ehresmann_connection" [Broken] that would naturally be defined over the total space of the fiber bundle as I described it.

The relation between Cartan connections and Ehresmann connections is a bit sneaky, so it's just as likely to be confusing as enlightening until one understands both concepts quite well.

In an Ehresmann connection we have a principal H-bundle P, and a map w giving an isomorphism between Lie(H) and the "vertical" part of the tangent space of P.

In a Cartan connection we have a principal H-bundle P but H sits in a bigger group G. Item 1 says that w gives an isomorphism between Lie(G) and the whole tangent space of P. Item 3 implies that w gives an isomorphism between Lie(H) and the "vertical" part of the tangent space of P.

For both kinds of connection w needs to be covariant with respect to the action of H on the bundle P, and that's item 2 item 2 in the above definition.

Again, it's probably more distracting than helpful for me to say all this stuff at this stage of the game... but you asked for it.

If I were to guess: a piece of information I think might be missing could be that:
The base manifold needs to have the same topology as G/H.
Is that right, and is there more?

No, that's not right - remember our key example, where we get a Cartan connection by rolling a sphere on a lumpy bumpy surface. The lumpy bumpy surface is the base manifold, G/H is the sphere, and they don't need to have the same topology. Each point on the lumpy bumpy surface has a "tangent sphere".

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Gold Member
Dearly Missed
very brief interruption. back in post #13 there was this definition of a Cartan geometry involving a group G and subgroup H and the quotient G/H was to serve as "tangent space" to the base manifold.
===quote Baez===

Hmm, but there's a little mistake in it [the Wiki definition], so let me post a corrected version here:

===quote Wiki (corrected)===
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.
===endquote===

All this stuff makes more sense if you talk about it a lot before hitting someone with the formulas....
=====endquote====

just want to mention this reminds me that DeSitter space is SO(1,4)/SO(1,3)
that is, DeSitter space is the quotient of a group by a subgroup---an homogeneous space (like the rolling ball)---
so I am not sure if this is relevant but it evokes the idea that one could have a manifold with a cartan geometry where the tangent space at each point of the manifold is DeSitter space, which would be taking the place of the ball which JB talks about rolling around on a bumpy surface. There may be some obvious reason not to think along these lines, but I don't immediately see it.

a possible primer on DeSitter group with many elementary facts is http://arxiv.org/abs/hep-th/0411154

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john baez
Gold Member
marcus said:
just want to mention this reminds me that DeSitter space is SO(4,1)/SO(3,1)
that is, DeSitter space is the quotient of a group by a subgroup---an homogeneous space (like the rolling ball)---
so I am not sure if this is relevant but it evokes the idea that one could have a manifold with a cartan geometry where the tangent space at each point of the manifold is DeSitter space, which would be taking the place of the ball which JB talks about rolling around on a bumpy surface.

Yes, that was exactly my point. To drive it home even further:

The Cartan geometry where you let a sphere roll on a surface involves the group SO(2) sitting in SO(3), with the quotient space SO(3)/SO(2) being the sphere. This is easy to visualize, and there's a lot one can learn from studying it in detail.

If we then add one dimension of space and one dimension of time, we get the Cartan geometry where you let a deSitter spacetime roll on spacetime. This involves the group SO(3,1) sitting in SO(4,1), with the quotient space SO(4,1)/SO(3,1) being deSitter space. This is the sort of Cartan geometry you need to understand for MacDowell-Mansouri gravity.

Luckily, many of the lessons from the easy example generalize to the harder one, since this sort of trick doesn't really care much about the number of space and/or time dimensions.

Gold Member
Dearly Missed
It's nice that the tangent space to a manifold doesn't have to be flat.
This shows that nature has imagination, or rather would show that, if it should turn out the tangent "plane" to actual spacetime is bent.
That would be so much fun it almost has to be true!
(just to keep in practice I will use the "big grin" emoticon----appropriate in this case)

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Staff Emeritus
Gold Member
Dearly Missed
marcus said:
It's nice that the tangent space to a manifold doesn't have to be flat.
This shows that nature has imagination, or rather would show that, if it should turn out the tangent "plane" to actual spacetime is bent.
That would be so much fun it almost has to be true!
(just to keep in practice I will use the "big grin" emoticon----appropriate in this case)

Well, of course "osculationg" curves and surfaces are old, but the beautiful thing is that this particular osculation has all the sweet properties it does.

Makes me wonder; Ward has developed Feymann's original idea of perturbing GR spacetime around the tangent Minkowski space. See the "Can this be true?" thread. What would be the result of peturbing around the osculant de Sitter space?

john baez
Gold Member
whoops

john baez said:
Yeah, that's more interesting. As we've seen, a Cartan geometry involves a Lie group G with a subgroup H, a base manifold X, a principal H-bundle P over X, and the Cartan connection w.

Sorry, I accidentally posted two nearly identical articles on this thread - numbers 18 and 19. The second one is better. I can't see how to delete the first one. Could one of the local demigods delete the first one, dated 07-22-2006 07:11 PM? And also this post?

Or, if that's impossible, people should just know that the second post is a bit more detailed.

Staff Emeritus
Gold Member
Dearly Missed
I deleted it for you. If you go into EDIT, do you see a button for DELETE?

marcus said:
the way to study DeSitter spacetime is to set up a 5D euclidean space and take a HYPERBOLOID subset of it
and that 4D hyperboloid is the spacetime we want

{(w,x,y,z,t) : w2 + x2 + y2 + z2 - t2 = k2}

the Lie group SO(4,1) is just the group of transformations that STAYS ON THE HYPERBOLOID
because it doesnt change the k.
it doesnt change this "measure" of the size of a point:
w2 + x2 + y2 + z2 - t2
whatever that is before the transformation, it is the same after
so the Lie group is just the matrices that transform normal 5D space in such a way that RESPECTS the slicing up into hyperboloids and doesnt mix up one k with some different k "sheet".
...

Cool topic Marcus

After seeing the 4D hyperboloid

{(w,x,y,z,t) : w2 + x2 + y2 + z2 - t2 = k2}

which is a quadric surface, I felt compelled to express it in terms of its corresponding symmetric matrix of coefficients. Here's what I got:

$$f=v^{T}Av= \left(\begin{array}{ccccc}w & x & y & z & t\end{array}\right) \left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{array}\right) \left(\begin{array}{c}w \\ x \\ y \\ z \\ t \end{array}\right).$$

Now our various values of k2 give level curves of this quadric surface via:

$$f=v^{T}Av=k^{2}.$$
Some example level curves are:
$$f_1=v^{T}Av=k_1^{2}=3$$
$$f_2=v^{T}Av=k_2^{2}=1$$
$$f_3=v^{T}Av=k_3^{2}=-1$$

So in this way, the Lie group SO(4,1) can be viewed as the group preserving the symmetric matrix A. The matrix A would then be a kind of metric tensor $$g_{\mu\nu}$$ in 4+1 dimensional DeSitter spacetime with SO(4,1) acting as an isometry group.

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