Fiber Bundle Projection Map


by Matterwave
Tags: bundle, fiber, projection
Matterwave
Matterwave is offline
#1
Feb3-12, 09:40 PM
P: 2,078
Hey guys,

I've often seen in the definition of a Fiber bundle a projection map [itex]\pi: E\rightarrow B[/itex] where E is the fiber bundle and B is the base manifold. This projection is used to project each individual fiber to its base point on the base manifold.

I then see a lot of references to the inverse of this projection map. It seems to me, that in general, this map should be many to one, since it should project a whole fiber to its base point. In this case, how can one define an inverse to this map? It seems odd to me that there can be an inverse to a many to one mapping...or have I missed something basic?
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
morphism
morphism is offline
#2
Feb3-12, 09:44 PM
Sci Advisor
HW Helper
P: 2,020
It is many-to-one in general (and that's really one of the points of this whole construction!) so it makes no sense to talk about its inverse map. Are you perhaps confusing the notation for the preimage [itex]\pi^{-1}(U) = \{ e \in E \colon \pi(e) \in U \}[/itex] of a subset U of B under [itex]\pi[/itex] with the identical notation for the image of U under the inverse of [itex]\pi[/itex] (a function that doesn't necessarily exist)?

P.S. A standard abuse of notation is to use [itex]\pi^{-1}(b)[/itex] to denote the preimage of the singleton {b}, i.e. the fiber lying over b.
Matterwave
Matterwave is offline
#3
Feb3-12, 09:51 PM
P: 2,078
That's probably it, but I haven't seen it termed "pre-image" in any of my books, maybe they are just getting a little sloppy here.

The first mention of this, my book simply says: "...which means that the bundle over any set [itex]U_j[/itex], which is just [itex]\pi^{-1}(U_j)[/itex], has a homeomorphism..."

I guess since the book didn't call it an inverse nor a preimage, I simply assumed it meant inverse and got confused.

For example, the wikipedia article: http://en.wikipedia.org/wiki/Fiber_bundle also simply just uses this notation without clarification.

morphism
morphism is offline
#4
Feb3-12, 09:56 PM
Sci Advisor
HW Helper
P: 2,020

Fiber Bundle Projection Map


Yes, that's the inverse image, and it's pretty standard notation in topology. The reason you haven't seen it explicitly defined in your books is (probably) due to the fact that books dealing with bundles tend to assume basic topology (and all the accompanying notation and terminology) as a prereq.
Matterwave
Matterwave is offline
#5
Feb4-12, 12:44 AM
P: 2,078
So it's just overloaded notation huh...that seems inconvenient...
Matterwave
Matterwave is offline
#6
Feb4-12, 02:11 AM
P: 2,078
A follow up question. My book discusses the structure group of fiber bundles and seems to imply that a trivial structure group (one where the structure group can be reduced to simply {1}) means a trivial fiber bundle (one where the fiber bundle is isomorphic to a product space).

The book isn't very explicit about this, however. I was just wondering if this condition is "necessary and sufficient"? I.e. Is a fiber bundle trivial if and only if its structure group is trivial? Or perhaps this condition is only necessary but not sufficient, or sufficient but not necessary?

Also, my book makes mention of a "frame bundle". I can't see the difference between a frame bundle and a tangent bundle other than that the frame bundle doesn't include the base element in its fibers. What's the difference? The wikipedia article is not really accessible to me.
morphism
morphism is offline
#7
Feb4-12, 12:39 PM
Sci Advisor
HW Helper
P: 2,020
Yes, a bundle is trivial iff its structure group can be reduced to {1}.

And re: frame bundle. The fiber at a point x in the tangent bundle is the tangent space TxM at x; the fiber at a point x in the frame bundle is the set of all ordered bases (i.e. frames) for TxM. So we're really dealing with completely different objects: the tangent bundle is a vector bundle, i.e. its fibers are vector spaces, whereas the frame bundle isn't.
Matterwave
Matterwave is offline
#8
Feb4-12, 05:16 PM
P: 2,078
Can you explain "the set of all ordered bases (i.e. frames) for TxM"? When I hear that I'm thinking of basis vectors, of which it's pretty arbitrary.
mathwonk
mathwonk is offline
#9
Feb4-12, 06:56 PM
Sci Advisor
HW Helper
mathwonk's Avatar
P: 9,428
every surjection has a right inverse, and these are called sections.
Matterwave
Matterwave is offline
#10
Feb4-12, 07:05 PM
P: 2,078
How is that defined? Without much context, I'm not sure which part of my post you are referring to, or what point you wanted to make...
morphism
morphism is offline
#11
Feb5-12, 12:20 PM
Sci Advisor
HW Helper
P: 2,020
Quote Quote by Matterwave View Post
Can you explain "the set of all ordered bases (i.e. frames) for TxM"? When I hear that I'm thinking of basis vectors, of which it's pretty arbitrary.
What's a fiber bundle over M? It's literally a bundle of fibers over M! I.e. you glue a set F (the fiber) to each point x in X. And then you require some 'niceness' conditions of this gluing.

Now, what's the frame bundle of M? It's a fiber bundle over M where the fiber glued to x consists of ordered bases for the tangent space at x. Literally a point in the frame bundle above x is an ordered basis [v_1,...,v_n] for TxM.
Matterwave
Matterwave is offline
#12
Feb5-12, 02:02 PM
P: 2,078
Hmmm...so, that seems pretty arbitrary in my choice of bases. If I can choose "all ordered bases", don't I get back T[sub]x[\sub]M? I'm thinking "all ordered bases" would include pretty much every single vector other than the 0 vector because certainly every vector is in some arbitrary choice of bases...
Matterwave
Matterwave is offline
#13
Feb6-12, 04:11 PM
P: 2,078
After rereading your last post, i think I get it. Thanks =D
arkajad
arkajad is offline
#14
Feb7-12, 03:34 AM
P: 1,412
Quote Quote by Matterwave View Post
Can you explain "the set of all ordered bases (i.e. frames) for TxM"? When I hear that I'm thinking of basis vectors, of which it's pretty arbitrary.
It is arbitrary indeed - up to an arbitrary transformation by an element of the group GL(n). So, the structure group of the frame bundle (a principal bundle) is GL(n).
Bacle2
Bacle2 is offline
#15
Feb7-12, 09:56 AM
Sci Advisor
P: 1,168
Quote Quote by arkajad View Post
It is arbitrary indeed - up to an arbitrary transformation by an element of the group GL(n). So, the structure group of the frame bundle (a principal bundle) is GL(n).
Unless you specify that you want some additional condition, to , e.g., preserve orientation, in which case you use SL(n).
arkajad
arkajad is offline
#16
Feb7-12, 10:03 AM
P: 1,412
Quote Quote by Bacle2 View Post
Unless you specify that you want some additional condition, to , e.g., preserve orientation, in which case you use SL(n).
Indeed, but, in such a case, we are dealing not with the whole "frame bundle", but with "a reduced frame bundle".
Bacle2
Bacle2 is offline
#17
Feb7-12, 12:25 PM
Sci Advisor
P: 1,168
Quote Quote by arkajad View Post
Indeed, but, in such a case, we are dealing not with the whole "frame bundle", but with "a reduced frame bundle".
Right, I missed that.
Matterwave
Matterwave is offline
#18
Feb7-12, 12:43 PM
P: 2,078
I need to go back and read this part again lol...


Register to reply

Related Discussions
On transition functions of fiber bundle Differential Geometry 4
Fiber bundle basics Differential Geometry 35
Is an associated bundle isomorphic to the principal bundle? Differential Geometry 0
Foliation, fibration, fiber bundle? Differential Geometry 1