# Fiber bundles

1. Jan 2, 2015

### Fredrik

Staff Emeritus
I was asked what a bundle structure is in a PM. I'm posting my reply here, to give others a chance to add comments.

I didn't get to see the exact sentence in which the term "bundle structure" was used. I would guess that it was used informally to say that some smooth manifold is the total space of a fiber bundle, so the term that really needs to be explained is "fiber bundle".

The basic idea is this: Consider a function $\pi:E\to B$ that's surjective onto $B$. The sets $\pi^{-1}(b)$ with $b\in B$ are mutually disjoint, and their union is $E$. It's convenient to call each $\pi^{-1}(b)$ the "fiber over $b$" and to think of $E$ as a "bundle" of fibers. $E$ is also called the "total space". $\pi$ is called "the projection". If there's a set $F$ such that each $\pi^{-1}(b)$ can be mapped bijectively onto $F$, then $F$ is called "the fiber".

This concept is however pretty useless when $E$ and $B$ are just sets. The term "fiber bundle" is usually only defined when $E,B,F$ are structures of the same type, e.g. when they're all topological spaces or all smooth manifolds. Then we make additional requirements on the relationship between $F$ and the fibers $\pi^{-1}(b)$ that's appropriate for the type of structure we're dealing with. If we're dealing with topological spaces, then we require that $F$ is homeomorphic to each of the $\pi^{-1}(b)$. If we're dealing with smooth manifolds, then we require that $F$ is diffeomorphic to each of the $\pi^{-1}(b)$.

The 4-tuple $(E,B,F,\pi)$ is then called a "fiber bundle". A nice example is when E is a cylinder, B is a circle and F is a line segment. Another nice example is when E is a Möbius strip, B is a circle and F is a line segment.

Note that if $\{U_i|i\in I\}$ is a collection of open subsets of $B$ that covers $B$, then $\{\pi^{-1}(U_i)|i\in I\}$ is a collection of subsets of $E$ that covers $E$. The total space $E$ can be thought of as consisting of the overlapping pieces $\pi^{-1}(U_i)$ that are glued together. In the case when B is a circle and F is a line segment, we can pick two open sets $U,V$ such that $B=U\cup V$. The sets $\pi^{-1}(U)$ and $\pi^{-1}(V)$ will be like two rectangular strips of paper that are glued together to form the space $E$, and as we all know two rectangular strips of paper can be glued together to form either a circle or a Möbius strip.

The most useful fiber bundle is the tangent bundle $TM$ of a smooth manifold $M$. There are a few slightly different ways to define it. One is to simply take $TM$ to be the union of all the tangent spaces, i.e. $TM=\bigcup_{p\in M} T_pM$. The base space is the smooth manifold $M$. The projection is the map $\pi:TM\to M$ defined by saying that for each $v\in TM$, $\pi(v)$ is the unique $p\in M$ such that $v\in T_pM$.

A section of the tangent bundle $TM$ is a function $X:U\to TM$ such that $U\subseteq B$ and $X(p)\in T_pM$ for all $p\in U$. A section of $TM$ is also called a "vector field". (Edit: I fixed a mistake in this paragraph after it was pointed out by Orodruin below).

Last edited: Jan 2, 2015
2. Jan 2, 2015

### Orodruin

Staff Emeritus
Very nice and concise up to this point. The section should be mapping points in the base space to the bundle (i.e., not to $\mathbb R$) such that $\pi(X(p)) = p$.

3. Jan 2, 2015

### Fredrik

Staff Emeritus
Thank you. I guess I was in too eager to finish up the post when I was adding that last paragraph. The codomain should definitely be $TM$, as you said. I have edited it into my post now. My condition "$X(p)\in T_pM$ for all $p\in M$" should have ended with "...all $p\in U$", but is otherwise OK, as long as we're dealing specifically with the tangent bundle. If we're going to define a section of an arbitrary bundle, we need to use your condition instead of mine.

4. Jan 3, 2015

### lavinia

Hi Fredrik

A couple of things.

- Usually a fiber bundle is locally trivial. This is certainly true of vector bundles and their relatives such as sphere bundles and principal bundles.
Without local triviality I am not sure what you get. Maybe a fibration.

- Also, fiber bundles usually have a structure group. This is a Lie group that acts by homeomorphisms on the fiber. The transition functions of the bundle lie in this structure group. In Steenrod's old book The Topology of Fiber Bundles, a group is part of the definiton of a fiber bundle.

For instance, a twisted torus is a trivial circle bundle over the circle if SO(2) is its structure group but is not trivial if its structure group is Z/2Z..

Sphere bundles may have various structure groups. If the group is the orthogonal group then the sphere bundle comes from a vector bundle. But there are sphere bundles whose structure group is the group of diffeomorphisms of the sphere and these may not come from a vector bundle.

- If you want to say that the tangent bundle is the most useful bundle, you should explain why this is. Perhaps you meant most useful in the study of smooth manifolds, or most useful in Physics.

Last edited: Jan 3, 2015
5. Jan 3, 2015

### Fredrik

Staff Emeritus
Hi Lavinia. Thank you for the input.

1. Yes, the local triviality condition is included in all textbook definitions I've seen. The fact that I didn't mention it reflects both a personal preference of mine (I like to define "fiber bundle" without that condition and then define "locally trivial") and the fact that I had temporarily forgotten that texbook definitions are different from what I would want them to be. I'll add a comment about local triviality later.

2. Sounds like you're thinking about principal bundles, a slightly different concept. I will need to refresh my memory about the details before I can comment further.

3. Good point. I just meant that for most people, and especially physics students, it's the first fiber bundle they encounter that's actually described as a fiber bundle (in order to define vector fields as sections), and then they will encounter it more often than any other fiber bundle that's actually described as a fiber bundle. ($\mathbb R^3$ can be thought of as an $\mathbb R^2$ bundle over $\mathbb R$, but this is rarely useful).

6. Jan 3, 2015

### weirdoguy

No, structure groups appear also in some definitions of fibre bundles.

7. Jan 4, 2015

### Fredrik

Staff Emeritus
As mentioned above, I need to add a comment about local triviality. I like to do things a bit differently from most textbooks. I'll do it my way first, and then explain the difference.

A 4-tuple $(E,B,F,\pi)$ such that $E,B,F$ are topological spaces, is said to be a fiber bundle if

(a) $\pi:E\to B$ is surjective onto $B$ and continuous.
(b) For all $b\in B$, $\pi^{-1}(b)$ is homeomorphic to $F$.

A fiber bundle $(E,B,F,\pi)$ is said to be trivial if $E=B\times F$, and is said to be locally trivial if

(c) There exists sets $I$, $\{U_i|i\in I\}$, $\{\phi_i|i\in I\}$ such that

(i) For all $i\in I$, $U_i$ is an open subset of $B$.
(ii) $\bigcup_{i\in I}U_i=B$.
(iii) For all $i\in I$, $\phi_i$ is a homeomorphism from $\pi^{-1}(U_i)$ to $U_i\times F$ such that $\operatorname{Proj}_1\circ\phi_i =\pi|_{\pi^{-1}(U_i)}$.​

Here $\operatorname{Proj}_1$ is the projection onto the first variable, i.e. $\operatorname{Proj}_1 (x,y)=x$.

The difference between my approach and the textbook approach is that they include this local triviality condition in the definition of "fiber bundle". This looks ugly to me, for several reasons. One reason is that I feel that axioms (a)-(b) are the reason why these things are called fiber bundles. The local triviality condition isn't part of it.

Another reason is that their approach makes the term "locally trivial" kind of useless. What sort of thing can be locally trivial? All fiber bundles are locally trivial by definition, and every other thing in mathematics is neither locally trivial nor not locally trivial, since the definition doesn't apply.

A third reason is that when the local triviality is included in the definition, then it seems much more natural to start the definition with "A 7-tuple $(E,B,F,\pi,I,\{U_i|i\in I\},\{\phi_i|i\in I\})$..." instead of "A 4-tuple $(E,B,F,\pi)$..." This makes the definition much uglier to me.

On the other hand, if we include local triviality in the definition, axiom (b) is no longer needed, as it's implied by the local triviality condition.

Last edited: Jan 4, 2015
8. Jan 4, 2015

### lavinia

This reply is to suggest why local triviality is central to the whole idea of bundles and also to motivate why structure groups are needed.

One can think of fiber bundles as generalizations of Cartesian products. From this point of view local triviality generalizes global triviality i.e. it generalizes the Cartesian product.

One would then want to understand how local triviality fails to be global and this is done by describing the way fibers are glued together on intersections of locally trivial neighborhoods..

This naturally leads to the idea of a structure group as the glue.

For instance, for vector bundles, the glue must be an element of the general linear group for otherwise the vector space structure on the fibers would not be well defined. This is the topological way to describe a change of coordinates.

Once this setting is established, one can move on to showing how new bundles can be constructed from old ones.

So for instance, from vector bundles one gets Whitney sums, tensor product bundles, dual bundles and so forth. For each, the new structure group is derived from the old. Otherwise put, if one knows how to glue the fibers of the vector bundles, one knows how to glue the fibers of the derived bundles.

Last edited: Jan 4, 2015
9. Jan 4, 2015

### DarthMatter

Thank you for this great post. My background in differential geometry is rather weak, so if you don't mind I will try to specify the cylinder example later and hopefully get some feedback.

10. Jan 10, 2015

### Fredrik

Staff Emeritus
I'm convinced about the local triviality, but I still don't understand why the structure group should be included. It's possible that my understanding of some of the basics is flawed. Here's what I'm thinking. Feel free to point out any mistakes.

Any surjection $\pi:E\to B$ partitions its domain into sets of the form $\pi^{-1}(b)$. If there's an $F$ such that each $\pi^{-1}(b)$ is isomorphic to $F$, then $E$ can be viewed as consisting of one copy of $F$ attached to each point of $B$. This is part of what we want a fiber bundle to be, so we will build up the definition around the 4-tuple $(E,B,F,\pi)$, and we will need to make sure that the definition either states explicitly or implies that each $\pi^{-1}(b)$ is isomorphic to $F$. But we also want to be able to view a fiber bundle as consisting of cartesian products $U_i\times F$ that are glued together in the overlap regions $U_i\cap U_j \times F$, in such a way that for each $b\in U_i\cap U_j$, the copy of $\pi^{-1}(b)$ from $U_i\times F$ is glued to the copy of $\pi^{-1}(b)$ from $U_j\times F$.

The local triviality condition is included in the definition because it's strong enough to grant both of these wishes. Unless there's something else that we really want a fiber bundle to be, the following definition will be appropriate.

A 6-tuple $(E,B,F,\pi,I,\{(U_i,\phi_i)|i\in I\})$ such that $E,B,F$ are topological spaces is said to be a fiber bundle if

(a) $\pi:E\to B$ is continuous and surjective onto $B$.
(b) For all $i\in I$, $U_i$ is an open subset of $B$.
(c) $\bigcup_{i\in I}U_i =B$.
(d) For all $i\in B$, $\phi_i$ is a homeomorphism from $\pi^{-1}(U_i)$ to $U_i\times F$ such that for all $b\in U_i$ and all $e\in\pi^{-1}(b)$, we have $\pi(\phi_i(e))=b$.

These conditions imply that for all $i,j$, the map
$$\phi_i\circ\phi_j^{-1}:U_i\cap U_j\times F\to U_i\cap U_j\times F$$ is such that
$$\phi_i\circ\phi_j^{-1}(b,f)=(b',f')\ \Rightarrow\ b'=b.$$ This implies that for each $b\in U_i\cap U_j$, we can define a function $t_{ij}(b):F\to F$ by $\phi_i\circ\phi_j^{-1}(b,f)=(b,t_{ij}(b)f)$ for all $f\in F$. (This notation seems to be common). This also defines functions $t_{ij}:U_i\times U_j\to\operatorname{Aut}(F)$.

The $t_{ij}(b)$ functions are our gluing instructions. I'm guessing that the structure group would be a subgroup of $\operatorname{Aut}(F)$ that all the $t_{ij}(b)$ would belong to, but I'm not even sure I got that right. I don't see how such a group enters the picture naturally, or what "wish" it grants about what a fiber bundle is supposed to be. It seems to me that the structure group is irrelevant to the wishes I wrote down before the definition above, but maybe I just left out something important.

I would also be interested in some thoughts about how to state the definition in such a general way that it works for all categories, including sets, topological spaces, topological vector spaces, smooth manifolds, etc. Is that possible? Can we prove, in a category-independent way, e.g. that if $\pi:E\to B$ is a surjective homomorphism, then each $\pi^{-1}(b)$ is a structure of the same type as $E$?

Last edited: Jan 11, 2015
11. Jan 11, 2015

### lavinia

I think you are correct that a fiber bundle does not require a structure group. But structure groups are key to the majority of bundles commonly encountered.

In the few examples from physics that I have some across, the bundles have structure group,s for instance the principal Lie group bundles, or the tangent bundle of Space-time.
The group structure is key to the physics.

All of the differential geometry that I know is defined on bundles with some Lie group as the structure group.

Some bundles must have a structure group, for instance, vector bundles.

12. Jan 11, 2015

### Terandol

I’m not sure entirely what you mean by a structure of the same type as E but if you are asking whether given any morphism in a category, there exists a “preimage” of every “point” which is again an object in the category then the answer is no in general. I put preimage and point in quotations because these do not even make sense in an arbitrary category. The categorical way of writing the analogue of a “point” $b\in B$ in a category is a morphism $b: 1\to B$ (so we have already restricted to the special case where categories have terminal objects. ) Then the preimage $\pi^{-1}(b)$ can be defined as the pullback of the morphism $\pi$ along the morphism b so we need to further restrict to categories in which pullbacks exist. So you would need to prove you are in a category with a terminal object and enough limits (all pullbacks of the required form) to say that a preimage of a point exists in the category. Most of the categories that come up in differential geometry do have these so it probably isn't much of a practical limitation.

The requirement that the pre image of every point is isomorphic to F then becomes the condition that given an epimorphism $\pi: E\to B$, the pullback of $\pi$ along any two morphisms $b,c:1\to B$ is isomorphic to F. There is probably a way to phrase the local triviality condition in this categorical setting but I’m not sure what it is off the top of my head.

—Edit: The nlab article defines local triviality in terms of slice categories here: http://ncatlab.org/nlab/show/fiber bundle . The nlab is one of the first places I always go if I want to find a categorical generalization of anything.

If you have the background in category theory (ie. things like fibered categories, Grothendieck topologies, stacks etc. ) you may be interested to read up on descent data. It is a very general framework within which it makes sense to glue things together. You can even do things like take “fiber bundles” whose fiber is itself a category. I think the standard reference for descent in fibered categories is still a chapter in Volume I of SGA.

13. Jan 12, 2015

### Fredrik

Staff Emeritus
Thanks for the post and the link, Terandol. That web page seems to sum it up nicely. The category-theory definitions are very elegant, but still a little more complicated than I had hoped. (I was wondering if they would be so simple and elegant that even someone who doesn't already know category theory might prefer them).

I know very little about categories. It didn't even occur to me that an object in a category may not have elements, and I had even forgotten the term "object". That's how bad I am at categories. So I certainly don't know anything about those things you listed at the end.

14. Jan 12, 2015

### lavinia

That is correct. The $t_{ij}(b)$ are called the transition functions of the bundle and they are always in the group of homeomorphisms of the fiber.

15. Jan 13, 2015

### Terandol

Yeah, all this categorified differential geometry stuff seems somewhat esoteric still even among mathematicians (of course algebraic geometers have been doing this since Grothendieck so it's nothing new in that field) but the group of people associated with the nLab view higher categories as the natural mathematical setting in which to do the type of gauge theory required for quantum gravity.

For example, both string theory and LQG agree that the fundamental objects are higher dimensional rather than points so for example if your string follows some path (ie. you have a surface) and you want to parallel transport say velocity vectors above the string along this path then you need to define the holonomy of a 2 dimensional surface which, for nonabelian groups leads either to gerbes or categorified principal bundles depending on your perspective (the difference is essentially whether you are thinking in terms of sheaves or of bundles.) Perhaps this viewpoint will eventually catch on with physicists also.

I know it isn't particularly relevant to this thread but this is actually fairly misleading so I wanted to correct it anyway. The statement that the natural categories in differential geometry have pullbacks is obviously not true. As I said in that post, given $f:M\to N$ the preimage of a point $f^{-1}(n)$ is a pullback and this of course is not a manifold in general so clearly the category of smooth manifolds does not have the required pullbacks.

There is a way to fix this though and take the category of diffeological spaces or Chen smooth spaces etc. (I seem to recall reading somewhere that these are all special cases of sheaves on a site so maybe that is the best way to describe these more general smooth spaces) to get a well-behaved category. Of course doing this creates much worse behaved objects in the category so there is a tradeoff between getting well-behaved smooth spaces or getting a well-behaved category.

16. Feb 5, 2015

### ChrisVer

Can there be an illustration of Fibre Bundles and Gauge groups, on a known object in particle physics?
Like a $U(1)$ local or global example?

nice thread :) well done

17. Feb 8, 2015

### Calabi

Hello evryone nice thread thanks you very much Frederik.

Good afternoon.

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