Thread: Fibre Bundles

A strange enough name for constructions that have been long known to mathematicians, and relatively recently found applications in physics - by relatively I mean quite a long time ago.

I start with a simple, and possibly familiar example.

Suppose a manifold - if it helps you can think of the sphere, a 2-manifold - but it is better to think more generally.

Now, at any point I may define a vector space denoted as to signify the fact that these define a set of vectors tangent to at each and every point. A problem arises, however.......

Taking the naive view of a tangent, we must assume that our tangent vectors "protrude" into some "surrounding" space in which our manifold is embedded. We cannot allow this - the inventors of differential geometry originally called it "intrinsic geometry". So we need a different definition of a tangent vector. Let us nonetheless be a little naive, and say a vector has direction and magnitude (not the best definition).

So one defines the gadget as being a tangent vector, which we can very loosely think of as the "tendency" of our point to move in the direction on . It's a definition! Magnitude comes from applying a real scalar to this gadget, and, assuming for simplicity , the real 2-sphere, we write where

Obviously this extends to any n-manifold.

So the set-theoretic union (actually it's strictly a disjoint union) of all for all is called the "tangent bundle over ". One writes somewhat confusingly, for this bundle

So a section of this bundle - in the usual usage of the word section - defines a vector field on , that is it "selects" a single vector and assigns it to each and every point . It does this as follows........

First one defines a projection , so that for any point , which is quite obviously a vector, is vector at . Like all projections known to man, beast or bacterium, this mapping is highly surjective - given a point , the pre-image set at that very point.

For reasons that escape me, this is called the "fibre at ". We are of course entitled to regard all such fibres at all points in our manifold as equivalent, so hence the term - the disjoint union of all such fibres at all such points is called a "fibre bundle".

Let's return to our section of this bundle, and write it as . Then quite obviously if any of the above is to make any sense - and it may not! - we must have, if our section is to be field, that .

This may seem a very complicated way of explaining something rather simple, and indeed it is, But I have my reasons. There are also some complications around what are called local vs. global triviality, which, if you want to follow the direction I am travelling, I need to explain.

But for now, I am out of puff, and you, no doubt, are out of patience

looks cool especially the klein bottles. I really like constructions wish i was better at them myself
some of the things i have seen constructed are really cool.
how would one go about constructing this obviously 3d space prob CAD because paper here would be a pain. Any idea of a good app for it??

So. where was I? Oh yes, we have our tangent bundle and our vector field defined as a section of this bundle over our manifold.

This is well and good except for one annoying fact - manifolds do not in general enjoy the luxury of a global coordinate system/ Those that do - say for any are pretty boring to be frank, so let's forget about these.

Now remember that a vector field assigns to each and every a single tangent vector of the form , the set being the coordinates at the point .

It follows that, if there can no single coordinate set that covers our manifold, there can be no global section of our bundle resulting in a coherent (i.e. with no redundancy) vector field. But as always we have the solution - our bundle is itself a manifold, so we need, as usual in the case of manifolds, to work "locally", and define the open set such that, recalling a fibre over the point is defined as the pre-image of our projection, I can write as a "local fibre bundle".

So we will say that our tangent bundle is globally trivial if and only if it inherits the product topology as . Otherwise one says it is locally trivial if the local bundle inherits the product topology .

Once again, I am using excessively fancy language to describe something really rather simple. But wait while I draw breath - the principles in these 2 posts can be used to describe something that is far from simple, namely the concepts of gauge theory

Originally Posted by Guitarist

Now, at any point I may define a vector space denoted as to signify the fact that these define a set of vectors tangent to at each and every point.

Are there (possibly) an infinite number of tangent vectors at any point?

Originally Posted by Strange

Are there (possibly) an infinite number of tangent vectors at any point?

Yes, this is true in spite of the fact that the cardinality of the set of basis vectors may be finite, or even 1, Think of it like this - if the cardinality of the scalar field that acts multiplicatively on a finite set of of basis vectors is infinite (and it almost always is), then the number of possible vectors in a vector space must be infinite, no matter the cardinality of the set of basis vectors.

This applies to any vector space, not only to tangent vector spaces

Could you give a quick definition of the term "topology" as opposed to "geometry" before we progress further ? I never really got the distinction between the two.

That is a good question, Markus. The reason being that I am using the word "topology" in several different ways. My excuse, and it's no excuse at all, is that everybody does it.

So we suppose a set . We define a topology on as a subset of the powerset of , id est (recall the powerset of a set is the set of all its subsets. Confused? You should be!).

One calls the indivisible pair a topological space, it being understood that the subset consists of the open sets in our topological space

There's one usage. Here's another......

The study of topological spaces is the subject known as "topology", which comes in 2 flavours - point-set top. and algebraic top., the second of which I am very weak on, and never discuss.

Returning to the point-set flavour, we know we can "turn" this into a manifold by insisting on certain reasonably exacting properties - essentially a local continuous mapping from our top. space to a subset of for some that has a continuous inverse. We say out manifold locally "looks like" . Clearly this manifold has a topology, but this does NOT mean it has a "shape", or as one might put it a "geometry". More on that in a moment....

Another usage of the word topology in common usage is to say - here is an object that has the topology of, say, the 2-sphere

Now we know that the 2-sphere is a manifold. The statement above means there is a continuously invertible mapping between them as top. spaces, called a homeomorphism.

So every manifold is a top. space. If our manifold has a metric, we can consider it as having a "shape". Specifically we know what we mean by length (equivalently distance) and angle, and it is the study of these one calls "geometry"

Confusingly, perhaps, if the metric in use is the Euclidean metric, one says it satisfies the axioms of, or has, Euclidean geometry. Likewise for manifolds with the Riemann or semi-Riemann metric. There are other metrics with which I am less familiar.

I hope I haven't muddied the waters too much, but I fear I might have.........

Ok, thanks Guitarist, I think I get the general idea, if not all the fine details.
Let's proceed

OK, Markus, but I hope you are not the only one trying to follow this stuff.

So quickly, as it is my turn to cook dinner.....

We have our tangent bundle , and a section I called a vector field. Recall this assigns to each and every point a vector. The following is true.....

It is NOT the case that every point in has a unique vector, in the sense that the same vector may be assigned to different points (this is over-simplistic, but will do for now).

Neither is it the case that two different sections - different fields - may not assign the same vector to different points. I am sure this is garbled. Math-speak is better....

First consider a section for any . Then I do not insist that for that . \edit in words - any two or more points may "share" an element in our field

Now consider another section, say, and call our 2 sections - vector fields recall - as .

Then I do not insist that, for a fixed point that . \edit - in words any 2 (or more) different fields may assign the same vector to the same point

Finally, for a laugh....

The set of all sections - all vector fields over a given manifold - is itself vector space. Worse, or better, depending on your taste, it is a real Lie algebra, though the proof of this is a bit difficult

OK, Markus, but I hope you are not the only one trying to follow this stuff.

No, I think there are plenty of silent followers, there always are.
But do continue on.

EDIT : I very interesting topic, and I am looking forward to seeing how you connect this to gauge theories, specifically in the light of our recent PM conversation. I have a certain idea here, but won't blurt it out until your presentation is finished.

<lurk>

<same>

OK, let's now get to the meat of it.

I define a manifold in which is taken as a particular Lie Group and as the base manifold, just as in our tangent bundle. is a manifold, pretty much because both and are.

Then the conditions I placed on the tangent bundle - local triviality, existence of a projection , and the fact that ones calls the preimage of this projection, for any as a fibre at the point - carry over. But of course our so-called fibre is not now a vector space but a Lie Group.

Some complications arise thereby, which I will now try to explain....

First let's assume a non-flat manifold. Them it is easy to show that concept of parallelism is not well-defined. For, transport a vector at the North Pole of the 2-sphere down a meridian to the South Pole in the direction our vector is "pointing" and we will find these two vectors are anti-parallel. But transport the same vector at the North Pole to the South Pole, but this time in a direction perpendicular to the direction our original vector is "pointing", and we will find they are indeed parallel. Likewise, other other transport of our vector results in another vector that is neither parallel nor anti-parallel, Thus parallelism is not well-defined.

(Interestingly, Euclid himself had reservations about parallelism in the 2-plane - he doubted his own 5-th postulate!)

Anyway, on a Riemann manifold the Levi-Civita connection "corrects" this - it preserves parallelism or, what amounts to the same thing, preserves the inner product under this sort of transport of vectors from any point to any another. So the connection on the tangent bundle - points in which are vectors, recall - over a Riemann manifold may be canonically chosen as the Levi-Civita connection.

Not so for our principal bundle. We need a new kind of connection, one that maps points in one fibre - an element in a Lie Group - to points on another fibre.

That will have to wait for now - I am supposed to be working, albeit from home, but working nonetheless.

Yeah well, I fear I have bitten off more than you guys can chew (to muddle a metaphor). Lemme try to make a frantically hand-wavey restution....

But recall first we are trying to find a connection on our principal bundle . Just to spoil the show (and to drag the physicists in here) let me pre-empt myself by saying that when , the 1-dimensional; rotation group in the complex plane, and is spacetime, then the connection is the 1-form , otherwise known as the vector potential in standard Electromagnetism.

We need all the following (sorry!)

1. Given a group , then the groups axioms specify that, for any that where the "dot" denotes an arbitrary group operation, which may be arithmetic multiplication, arithmetic addition, matrix multiplication, permutation ....blah blah. Accordingly it is customary to omit the centre dot and write

2. There is an element in our group such that for any called the "identity"

3. A Lie group, as the name implies is a group satisfying these axioms. It is also a manifold.

4. Then it follows I can "roam" over this manifold by repeated application of the group operation

5. As a manifold, it is entitled, at every point to a tangent vector space - call it

6. The tangent space at the identity is referred to as the Lie algebra associated to this group - this association is unique in the sense that no group may have more than 1 associated algebra. The converse is false - more than 1 group may share an algebra. For this algebra one writes

7. From elementary linear algebra, 2 (or more) vector spaces are taken to be isomorphic if their sets of basis vectors have the same cardinality. So that, for any we may write . This is a simple consequence of our ability to "roam" freely over our manifold/group.

Phew. Out of puff for now