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Frame bundle

by Matterwave
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Matterwave
#1
Mar1-12, 04:45 PM
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Ok, so I don't have much of an intuition for frame bundles, so I have some basic questions.

A frame bundle over a manifold M is a principle bundle who's fibers are the sets of ordered bases for the vector fields on M right.

1) This means that any point in the fiber (say, over a point m in M) is literally a set of ordered bases right?

2) Since the frame bundle is a principle fiber bundle, each fiber has to be isomorphic to its structure group, which I gather is GL(n,R) right. So, a frame bundle over a 4-d manifold is 16 dimensional? Why so many dimensions?

3) What do these dimensions mean? Going "in a different direction" in this fiber corresponds to doing what to my ordered bases?

This stuff seems really confusing to me...
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quasar987
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Mar1-12, 06:15 PM
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I will try to clear things up a little for you, if that is possible.

Quote Quote by Matterwave View Post
Ok, so I don't have much of an intuition for frame bundles, so I have some basic questions.

A frame bundle over a manifold M is a principle bundle who's fibers are the sets of ordered bases for the vector fields on M right.
I wouldn't put it that way, because it sounds like you're saying that fibers are made up of basis for the vector space of the vector fields on M, which is false of course. More precise would be that the frame bundle over a manifold M is a principle GL(n,R)-bundle who's fibers are the sets of ordered bases for the tangent space of M at p. (where n=dim(M))

Quote Quote by Matterwave View Post
1) This means that any point in the fiber (say, over a point m in M) is literally a set of ordered bases right?
Yes, it is literally an n-tuple of vectors spanning TmM.

Quote Quote by Matterwave View Post
2) Since the frame bundle is a principle fiber bundle, each fiber has to be isomorphic to its structure group, which I gather is GL(n,R) right. So, a frame bundle over a 4-d manifold is 16 dimensional? Why so many dimensions?
You have to be careful here because a fiber bundle (such as a principal bundle) is itself a manifold. So you have to distinguish the dimension of the bundle as a manifold and the dimension of its fibers. Here, 16=4 is the dimension of the fiber. The dimension of the frame bundle as a manifold is 16+4=20. What do you mean "why so many dimensions"? I would think that you know the answer to that since you computed correctly that GL(4,R) has dimension 16!

Quote Quote by Matterwave View Post
3) What do these dimensions mean? Going "in a different direction" in this fiber corresponds to doing what to my ordered bases?
Ah, perhaps you do not understand what the connection is between the fact on the one hand that the fiber over m are the frames of TmM and the claim that this space is just GL(4,R). This is simply because in a vector space V of dimension n, once you fix a basis (vi), you can then express any vector w in V as a set of n real numbers; namely the so-called coordinates of w wrt to (vi):
[tex]w=\sum_i a_iv_i[/tex]
In particular, if (wj) is another basis of V, then this corresponds to n numbers (n numbers for each wj), which you can arrange in a matrix by declaring that the first row is to be made up of the n coordinates of w1 and so on. Then the fact that these (wj) are linearly independant is equivalent to saying that the matrix thus constructed has nonvanishing determinant. That is, it is a matrix in GL(n,R). So you see that there is a (non canonical) bijective correspondance between the frames F(V) of V and GL(n,R). Use that to transfer the smooth structure of GL to F(V). You can verify that this will be independant of the choice of bijection and so puts a well-defined canonical smooth structure on F(V) such that given any choice of basis in V resulting as above in a bijection F(V)<-->GL, this bijection is a diffeomorphism. So that is how each fiber of the frame bundle is (non canonically) diffeomorphic to GL(n,R).
quasar987
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Mar2-12, 07:37 AM
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In particular, now we see "why" there are 16 dimensions to F(TmM) when dim(M)=4. Namely, once a basis (vi) for TmM has been fixed, there is one degree of freedom/dimension corresponding to each of the 4=16 ways to choose each coordinate aji of each vector wj of another basis (wj=Ʃajivi).

Matterwave
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Mar2-12, 03:17 PM
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Frame bundle

Sorry, I got a little busy. Gimme a little bit of time to digest your post and then I'll respond with more questions. =D
quasar987
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Mar2-12, 03:42 PM
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Quote Quote by quasar987 View Post
More precise would be that the frame bundle over a manifold M is a principle GL(n,R)-bundle who's fibers are the sets of ordered bases for the tangent space of M at p. (where n=dim(M))
Here I mean to say

"More precise would be that the frame bundle over a manifold M is a principle GL(n,R)-bundle who's fiber over p is the set of ordered bases for the tangent space of M at p. (where n=dim(M))"
Matterwave
#6
Mar3-12, 01:24 AM
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Ok, I had some time to look at your post and digest, I get it now thanks. I had a wrong conception that it would only take 4 numbers to specify a basis in 4-D because you only need 4 vectors, but I forgot that each vector actually consists of 4 numbers.

Thanks for the help. =]
quasar987
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Mar3-12, 07:11 AM
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Nice, yw!
Matterwave
#8
Mar8-12, 04:18 PM
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As a follow up, frame bundles, and principle bundles in general, are not vector bundles right? Since GL(n,R) doesn't seem to form a vector space. I'm a little confused because this book I'm reading is discussing the "distribution on P of vertical vectors...", where P is a principle bundle, but it seems to me that a principle bundle is not a vector bundle so where are the vectors?
quasar987
#9
Mar8-12, 05:07 PM
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in the tangent to P!

Let [itex]\pi[/itex]:P-->B be a principal G-bundle. Then consider the derivative [itex]\pi_*[/itex]:TP-->TB and for each point p in P, form the subspace V_p:=Ker([itex]\pi_{*,p}[/itex])[itex]\subset T_pP[/itex]. Since p is a submersion, this has dimension dim(G) and corresponds to all the directions (tangent vectors) along the fiber through p. According to the standard picture of fiber bundles as a bunch of G's lined up vertically against a horizontally drawn base space, V_p is called the vertical space at p since it is tangent to the fibers. The collection of all the Vp's form a subbundle (aka a tangent distribution!) of TP called the vertical subbundle V.

By the way... a connexion in a principal G-bundle is, by definition, a choice of a subbundle H complementary to V which we call a horizontal subbundle. Such an H is isomorphic to TM via [itex]\pi_*[/itex] and so we think of it as giving the direction "along M" in P. This allows one to define a notion of parallel transport: given a point p in P and a path c(t) in M, to parallelly transport p along c, first lift the velocity vector field v(t) of c up to H via [itex]\pi_*[/itex]. Then use this to lift c(t) up to the unique path c'(t) in P such that
a) c'(0)=p
b) c' is an integral curve of the lifted velocity field
Then by definition, c'(t) is the parallell translate of p along c. Hence, the name "connexion" is justified. And of course, when the bundle is a vector bundle, it can be shown that this definition of connecxon is equivalent to the more common one in terms of specifying an operator on sections [itex]\nabla[/itex].
Matterwave
#10
Mar8-12, 06:06 PM
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Wow, you explain it a lot better than this book I got haha.

Can you explain a little more your last sentence regarding "when the bundle is a vector bundle..."? Can vector bundles be principle bundles? This seems odd to me because a vector space needs the 0 vector, but the structure group which is usually GL(n,R) (or some subset thereof) for vector bundles cannot contain the 0 matrix.

Or does the above construction of a connexion work even for vector bundles?
quasar987
#11
Mar8-12, 07:47 PM
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Quote Quote by Matterwave View Post
Can you explain a little more your last sentence regarding "when the bundle is a vector bundle..."? Can vector bundles be principle bundles?
Oh, sorry. All along I wrote about principal bundles because that is what you seem to be interested in primarily, but actually this definition of connection holds for a general fiber bundle. And while I agree that a vector bundle cannot be a principal bundle, a vector bundle is always a fiber bundle, so this definition of connection in terms of horizontal subbundle applies to it. This is what I had in mind when I wrote that awkward last comment.
Matterwave
#12
Mar8-12, 11:54 PM
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Interesting, because I assumed that the connection on the tangent bundles we are used to would arise out of some kind of induced connexion on the principle bundle it is associated with. hmm.

I'm only "interested" in principle bundles in so far as that's what the book I'm reading seems to be concerned with. I had assumed that that was so because it "leads" to the regular connection due to some kind of inducement.

EDIT: The book I'm using (Bishop and Crittenden) seems to always suggest that V and H are in P not TP, it'd be pretty weird if they were just blatantly wrong, is there some subtleties there? Or are they just wrong and it should be that V and H live in TP?
quasar987
#13
Mar9-12, 08:12 AM
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Well, on page 74, they say V is a distribution on P. By definition, this means it is a vector subbundle of TP. Also, note that they are using the notation Pp to mean TpP. Also, they use the notation [itex]d\pi[/itex] while I use [itex]\pi_*[/itex]. Other than that, their definition of V is exactly the same as mine. They put

[tex]V_p:=\{t\in T_pP | \pi_*(t)=0\}=\mathrm{Ker}(\pi_{*,p})[/tex]
Matterwave
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Mar9-12, 02:29 PM
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Oh, so it's just notation...ok. @_@

You've been of great help thanks. =]
Matterwave
#15
Mar9-12, 05:23 PM
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Some follow up questions. This formulation of parallel transport is pretty foreign to me.

Let's restrict ourselves to vector bundles since that is ultimately my interest as a physicist.

You said that with this construction, I can get a unique curve c'(t) which is the horizontal lift of c(t) given c'(0)=p and c'(t) being the integral curve of the horizontally lifted tangent vectors to c(t).

Does this uniqueness define a unique way for me to define parallel transport? The way that I've always learned about parallel transport and connections is that there is some degree of arbitrariness to them. If I am not given a metric structure on my manifold, then I can have connections with torsion or without torsion, for example, and there was no way for me to a priori define which matrices in GL(n,R) to act on my tangent spaces to transform my vectors as they move along a path.

This seems to clash with the uniqueness of the horizontal lift. Does this uniqueness imply that given a vector bundle E over a manifold M, I have only 1 unique way to define parallel transport of a vector in E along a curve in M?
Matterwave
#16
Mar9-12, 05:50 PM
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Another follow up question. This book is now talking about how the connexion in the principle bundle will "induce" in some special way a connexion in its associated fiber bundle. Is this construction the same as one would get if one simply redid the construction, as you said was possible, on this associated fiber bundle without first doing the construction on the principle fiber bundle?

For example, if I have a connexion on my frame bundle to M, it should induce a connexion on my tangent bundle TM.

If I had, instead, just started with TM as a vector bundle without any regard to the fact that it was associated with the frame bundle and did the construction of vertical distributions, etc., would I have arrived at the same result?
quasar987
#17
Mar9-12, 06:09 PM
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Quote Quote by Matterwave View Post
Some follow up questions. This formulation of parallel transport is pretty foreign to me.

Let's restrict ourselves to vector bundles since that is ultimately my interest as a physicist.

You said that with this construction, I can get a unique curve c'(t) which is the horizontal lift of c(t) given c'(0)=p and c'(t) being the integral curve of the horizontally lifted tangent vectors to c(t).

Does this uniqueness define a unique way for me to define parallel transport? The way that I've always learned about parallel transport and connections is that there is some degree of arbitrariness to them. If I am not given a metric structure on my manifold, then I can have connections with torsion or without torsion, for example, and there was no way for me to a priori define which matrices in GL(n,R) to act on my tangent spaces to transform my vectors as they move along a path.

This seems to clash with the uniqueness of the horizontal lift. Does this uniqueness imply that given a vector bundle E over a manifold M, I have only 1 unique way to define parallel transport of a vector in E along a curve in M?
The freedom is in the choice of the horizontal bunble H. Given V, there is an infinite number of ways to choose a complementary bundle to it.
Matterwave
#18
Mar9-12, 06:27 PM
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Hmmm, can you elaborate a bit on that?

I can see that if H is arbitrary, then my definition of what is a horizontal lift is aribtrary, and that would mean arbitrariness in parallel transport. But I can't see how H would be arbitrary, doesn't it have to encompass every "direction" that V doesn't in order to be a compliment to it?

I mean dim(V)=dim(G) and dim(H)=dim(M) and dim(P)=dim(G)+dim(M) right, so I'm not seeing how it can be arbitrary.

Perhaps a simple example would help? Or perhaps one could try to explain, for example, if my manifold is now endowed with a metric, how that limits my choice of H to be compatible somehow.


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