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?
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.
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.
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.
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.
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 T_{x}M at x; the fiber at a point x in the frame bundle is the set of all ordered bases (i.e. frames) for T_{x}M. 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.
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.
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...
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 T_{x}M.
Hmmm...so, that seems pretty arbitrary in my choice of bases. If I can choose "all ordered bases", don't I get back T_{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...}
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).
Indeed, but, in such a case, we are dealing not with the whole "frame bundle", but with "a reduced frame bundle".