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

[john baez]

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

I assume everyone here knows the sexy original meaning of "osculating".

Makes me wonder; Ward has developed Feymann's original idea of perturbing GR spacetime around the tangent Minkowski space.

Whoops! You just got 5 points from item 8 on the http://math.ucr.edu/home/baez/crackpot.html". It's spelled "Feynman".

More seriously, studying classical GR by perturbing around Minkowski spacetime (not the tangent space) is a wonderful thing. https://www.amazon.com/gp/product/0691087776/?tag=pfamazon01-20 wrote an enormous book in which they proved that in GR, Minkowski spacetime is stable under small perturbations - a surprisingly difficult result!

Studying quantum gravity by perturbing GR around Minkowski spacetime was initiated by Feynman and then studied intensively by many people. People have concluded that it's problematic, since it's not renormalizable. More precisely, the http://math.ucr.edu/home/baez/week195.html" seems to be:

• In 4 dimensions, pure gravity without matter is renormalizable to 1 loop, but not 2.
  
• In 4 dimensions, pure gravity with non-supersymmetric matter is generically not renormalizable even to 1 loop.
  
• In 4 dimensions, supergravity theories are renormalizable up to 2 loops. It is believed that most of these theories are not renormalizable to 3 loops, since a candidate divergent term is known. However, http://www.livingreviews.org/Articles/Volume5/2002-5bern/index.html" that "no explicit calculations have as yet been performed to directly verify the existence of the three-loop supergravity divergences."
  
• Maximally supersymmetric supergravity theories behave better than people had expected. In 4 dimensions, it seems that so-called "N = 8 supergravity" is renormalizable up to 4 loops, but not 5. However, neither of these have been proved, and this theory could even be renormalizable to all orders: see pages 33-35 in http://online.kitp.ucsb.edu/online/strings05/bern/" on the subject
  
• 11-dimensional supergravity is renormalizable to 1 loop but not 2.

All these results are based on the usual background-dependent perturbation theory. Now Freidel and Starodubtsev are trying to develop a background-free perturbation theory, and this could change things.

What would be the result of perturbing around the osculant de Sitter space?

I don't know careful work on this subject, but I kind of doubt things will work better here in the usual background-dependent perturbation theory.

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

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

 

[selfAdjoint]

Whoops! You just got 5 points from item 8 on the crackpot index. It's spelled "Feynman".

The cracks are not in my pot, but in my keyboard, where my fingers spend a lot of time. I am a 3-finger-look-at-the-keyboard-and-not-what-you're-typing sort. Ususally I go back and clean stuff up, but this one I missed.

More seriously, studying classical GR by perturbing around Minkowski spacetime (not the tangent space) is a wonderful thing. Christodoulou and Klainerman wrote an enormous book in which they proved that in GR, Minkowski spacetime is stable under small perturbations - a surprisingly difficult result!

Studying quantum gravity by perturbing GR around Minkowski spacetime was initiated by Feynman and then studied intensively by many people. People have concluded that it's problematic, since it's not renormalizable.

B.F.L. Ward, in
hep-ph/0607198
Exact Quantum Loop Results in the Theory of General Relativity
Authors: B.F.L. Ward (1) ((1) Dept. of Physics, Baylor University, Waco, TX, USA)
"We present a new approach to quantum general relativity based on the idea of Feynman to treat the graviton in Einstein's theory as a point particle field subject to quantum fluctuations just as any such field is in the well-known Standard Model of the electroweak and strong interactions. We show that by using resummation techniques based on the extension of the methods of Yennie, Frautschi and Suura to Feynman's formulation of Einstein's theory, we get calculable loop corrections that are even free of UV divergences. One further by-product of our analysis is that we can apply it to a large class of interacting field theories, both renormalizable and non-renormalizable, to render their UV divergences finite as well. We illustrate our results with applications of some phenomenological interest. "
Does cite Feynman as you see, and critically uses 't Hooft and Veltmann's results to evaluate his loops. His contribution is to apply his own development of YFS resummation of the IR infinities leading to finite one-loop results.

Ward's references to these authors:

[4] G. ’t Hooft and M. Veltman, Nucl. Phys. B44,189 (1972) and B50, 318 (1972); G. ’t
Hooft, ibid. B35, 167 (1971); M. Veltman, ibid. B7, 637 (1968).

[11] R. P. Feynman, Acta Phys. Pol. 24 (1963) 697.
[12] R. P. Feynman, Feynman Lectures on Gravitation, eds. F.B. Moringo and W.G.
Wagner (Caltech, Pasadena, 1971)

[18] D. R. Yennie, S. C. Frautschi, and H. Suura, Ann. Phys. 13 (1961) 379; see also
K. T. Mahanthappa, Phys. Rev. 126 (1962) 329, for a related analysis.


And finally he references Reuter's work on asymptotic safety of GR to evade the nonrenormalizability issue. See for example

hep-th/0511260
Asymptotic Safety in Quantum Einstein Gravity: nonperturbative renormalizability and fractal spacetime structure
Authors: O. Lauscher, M. Reuter

What all this is worth, I can't tell.

 

[Careful]

Ward's references to these authors:

[4] G. ’t Hooft and M. Veltman, Nucl. Phys. B44,189 (1972) and B50, 318 (1972); G. ’t
Hooft, ibid. B35, 167 (1971); M. Veltman, ibid. B7, 637 (1968).

[11] R. P. Feynman, Acta Phys. Pol. 24 (1963) 697.
[12] R. P. Feynman, Feynman Lectures on Gravitation, eds. F.B. Moringo and W.G.
Wagner (Caltech, Pasadena, 1971)

[18] D. R. Yennie, S. C. Frautschi, and H. Suura, Ann. Phys. 13 (1961) 379; see also
K. T. Mahanthappa, Phys. Rev. 126 (1962) 329, for a related analysis.


And finally he references Reuter's work on asymptotic safety of GR to evade the nonrenormalizability issue. See for example

hep-th/0511260
Asymptotic Safety in Quantum Einstein Gravity: nonperturbative renormalizability and fractal spacetime structure
Authors: O. Lauscher, M. Reuter

What all this is worth, I can't tell.

Understanding classical non-linear theories is extremely hard, perturbation around solutions of a (sub) theory (or linearisation thereof) which we can solve exactly is the only thing we can do. One must make sure however that the number of physical degrees of freedom in the subtheory is large enough so that Hilbertspace methods on the latter are sufficient to approximate the solution of the full theory. Understanding perturbation theory is something which still requires lots and lots of future work.

 

[marcus]

... kind of doubt things will work better here in the usual background-dependent perturbation theory.

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

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

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

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

we seem to be contemplating the possibility of a
BACKGROUND-FREE PERTURBATION THEORY

that sounds paradoxical because one is used to thinking of perturbing around a fixed metric like Minkowski.

but the Freidel-Starodubtsev approach is AFAIK a PERTURBATION AROUND BEEF----that is there is NO METRIC ON THE MANIFOLD.
It is a perturbation theory where there is no background metric
required to start the process of approximation.

that is the new opening that I think JB is talking about, in what I quoted
and that is actually why the TOPIC OF THE THREAD is to give an introduction to SO(4,1)!

 

[john baez]

oh, something new!

The cracks are not in my pot, but in my keyboard, where my fingers spend a lot of time. I am a 3-finger-look-at-the-keyboard-and-not-what-you're-typing sort. Ususally I go back and clean stuff up, but this one I missed.

Heh, okay. Btw, the 5 points only ate up your -5 point starting credit, so your crackpot index is still 0, and if I subtracted points each time someone said something sensible, you'd be way negative.

B.F.L. Ward, in
http://www.arxiv.org/abs/hep-ph/0607198"
Exact Quantum Loop Results in the Theory of General Relativity

Oh, interesting. When you said "Ward", I thought you were talking about something old, because the "Ward identities" have been important in quantum field theory for ages. This is new, probably by some other guy named Ward.

And finally he references Reuter's work on asymptotic safety of GR to evade the nonrenormalizability issue. See for example

http://www.arxiv.org/abs/hep-th/0511260"
Asymptotic Safety in Quantum Einstein Gravity: nonperturbative renormalizability and fractal spacetime structure
Authors: O. Lauscher, M. Reuter

What all this is worth, I can't tell.

Lauscher and Reuter's results are suggestive but far from definitive, since they make some drastic approximations - so it is completely sensible for skeptics like Distler to http://golem.ph.utexas.edu/~distler/blog/archives/000648.html" . Ward's work will have to stand or fall on its own merits. I haven't read it yet.

 

[selfAdjoint]

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

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

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

 

[john baez]

Back to SO(4,1)?

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

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

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

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

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

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

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

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

 

[marcus]

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

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

...

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

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

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

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

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

OK

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

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

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

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

 

[marcus]

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

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

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

 

[garrett]

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

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

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

 

[marcus]

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

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

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

OK yeah you look sharp in a tux.

 

[john baez]

The Matrix Reloaded

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

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

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

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

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

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

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

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

 

[garrett]

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

But what's to say the sphere doesn't twist (rotate around the axis determined by the contact point and sphere center) as it rolls? We can refer to this twist as \theta and its twist velocity along a path as
<br /> \alpha(t) = \frac{d \theta}{d t}<br />
So what we've done is enlarged the configuration space for our problem -- we have the two coordinates for the contact point, and the twist, x^1,x^2,\theta. Now in order to find the sphere's orientation along some path through this configuration space, we need to solve
<br /> \frac{d}{d t} g = \vec{v&#039;} \underrightarrow{\omega&#039;} g = ( v^i \omega_i{}^B T_B + \alpha \omega^B T_B ) g<br />
and we're going to need to know the connection (the Cartan connection?) over this larger configuration space:
<br /> \underrightarrow{\omega&#039;} = \underrightarrow{dx^i} \omega_i{}^B (x,\theta) T_B + \underrightarrow{d\theta} \omega^B (x,\theta) T_B<br />
Then we can again calculate the holonomy by integrating over the path in this larger configuration space.

Now, I think this is OK, with the fiber, H=SO(2), corresponding to the twist, and the enlarged configuration space corresponding to P. But I think I need some help understanding what further restrictions need to be placed on \underrightarrow{\omega&#039;} for it to be a Cartan connection. If these restrictions could be explained within this simple context and notation I've been working with, that would be great. So... help?

 

[marcus]

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

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

but wait, there is the Cartan picture!

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

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

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

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

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

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

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

 

[john baez]

flattening out

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

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

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

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

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

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

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

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

Right, exactly!

In the case of

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

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

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

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

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

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

to these symmetries of the sphere:

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

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

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

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

but we need some FORMALITY.

Really? Okay. Garrett - put on your tux!

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

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

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

 

[marcus]

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

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

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

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

 

[john baez]

rolling, twisting, but no slipping

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Sort of...

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

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

See? Everything is relative, man.

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

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

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

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

But I think I need some help understanding what further restrictions need to be placed on \underrightarrow{\omega&#039;} for it to be a Cartan connection. If these restrictions could be explained within this simple context and notation I've been working with, that would be great. So... help?

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

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

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

 

[marcus]

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

a sphere of the same size?

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

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

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

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

 

[Hurkyl]

a sphere of the same size?

yes I did the experiment with two colorfully patterned juggling balls

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

 

[marcus]

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

 

[garrett]

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

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

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

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