Fiber Bundle Basics: An Overview of Penrose's Twisted Bundles

[Rasalhague]

In The Road to Reality, § 15.2, Roger Penrose introduces the concept of a fibre bundle. Here I'll modify his notation to that of Wikipedia and other sources, so that B stands for base space (rather than the bundle), F for fibre, and E for the total space, which I think Penrose calls the (fibre) bundle itself.

His first example is the real line bundle over S¹. This is the product space S¹ x R¹, where S¹ is a circle (any circle?) and R¹ is the 1-dimensional vector space of real numbers over the reals with the natural definition of scalar multiplication. A product space is called a trivial bundle of F over B. So far so good. He identifies this structure with a (the?) cylinder.

I'm having trouble with his next example, which he calls a twisted bundle, which he identifies with a (the?) Möbius strip. There are pictures of Möbius strips. I can see what twisted means when it refers to a strip of paper, but what does twisted mean when it refers to a fibre bundle?

I think I follow what a (cross-)section means, at least when the bundle is trivial: a function s : B --> E such that the projection $p \; \circ \; s(x) = x$. (This according to Wikipedia's more formal intro; Penrose's cross-section is only the image of this map, or can be "thought of as" the image; he's a little vague here.) At least, I think can see how this works for a product B x F.

But how is it different in the twisted case? Penrose, having just described the case for a trivial bundle B x F: "This is like the ordinary idea of the graph of a function [...] More generally, for a twisted bundle B, any cross-section of B defines a notion of a 'twisted function' that is more general than the ordinary idea of a function." In Fig. 15.7, he shows a curve drawn in a loop around a cylinder, and one around a Möbius strip. I think the idea is that the loop can't go all the way around the Möbius strip and join up with itself (and thus can't be continuous) unless it crosses the central line around the strip (which line Penrose identifies with the zero section).

I suppose what's puzzling me is that these copies of F are disjoint, so how can one copy of, say, the vector space (R¹, R¹, regular multiplication) be said to have something analogous to an orientation or direction compared to another? Whereabouts in the definition of the Möbius strip bundle does this idea come in?

 

[Ben Niehoff]

I suppose what's puzzling me is that these copies of F are disjoint, so how can one copy of, say, the vector space (R¹, R¹, regular multiplication) be said to have something analogous to an orientation or direction compared to another? Whereabouts in the definition of the Möbius strip bundle does this idea come in?

The missing piece here is the connection. The connection describes how the fibers are connected to each other; i.e., it defines a notion of parallel transport. Or alternatively, it splits the tangent space into vertical and horizontal subspaces; the verticals moving within a fiber, and the horizontals moving you to the adjacent fiber.

Also remember that a fiber bundle is a manifold, so it can be covered by charts. Within a chart, it is easy to define an orientation. One can propagate this orientation to nearby charts by insisting on orientation-preserving transition functions (i.e., with positive Jacobians). Thus one can build up the notion of global orientability by demanding that it be possible to assign a single orientation to the entire atlas.

Put a few charts on a Mobius band and you'll see that a global orientation is not possible.

 

[arkajad]

I can see what twisted means when it refers to a strip of paper, but what does twisted mean when it refers to a fibre bundle?

It means that there is no (global) bundle isomorphism with the trivial (product) bundle. Bundle isomorphism means homeomorphism that maps fibers onto fibers (thus, in particular, it induces a homeomorphism of the base spaces).

 

[quasar987]

In The Road to Reality, § 15.2, Roger Penrose introduces the concept of a fibre bundle. Here I'll modify his notation to that of Wikipedia and other sources, so that B stands for base space (rather than the bundle), F for fibre, and E for the total space, which I think Penrose calls the (fibre) bundle itself.

His first example is the real line bundle over S¹. This is the product space S¹ x R¹, where S¹ is a circle (any circle?) and R¹ is the 1-dimensional vector space of real numbers over the reals with the natural definition of scalar multiplication. A product space is called a trivial bundle of F over B. So far so good. He identifies this structure with a (the?) cylinder.

I'm having trouble with his next example, which he calls a twisted bundle, which he identifies with a (the?) Möbius strip. There are pictures of Möbius strips. I can see what twisted means when it refers to a strip of paper, but what does twisted mean when it refers to a fibre bundle?

I think I follow what a (cross-)section means, at least when the bundle is trivial: a function s : B --> E such that the projection $p \; \circ \; s(x) = x$. (This according to Wikipedia's more formal intro; Penrose's cross-section is only the image of this map, or can be "thought of as" the image; he's a little vague here.) At least, I think can see how this works for a product B x F.

Note that this is just an abstract way of saying: a cross section is a map from B to E such that a point b in B is sent to an element of F_b, the fiber of E over b. This is a generalisation of the notion of vector field on a manifold. In that case, B=M, the base manifold, E=TM, the tangent bundle, and a cross section is just a vector field on M.

But how is it different in the twisted case? Penrose, having just described the case for a trivial bundle B x F: "This is like the ordinary idea of the graph of a function [...] More generally, for a twisted bundle B, any cross-section of B defines a notion of a 'twisted function' that is more general than the ordinary idea of a function." In Fig. 15.7, he shows a curve drawn in a loop around a cylinder, and one around a Möbius strip. I think the idea is that the loop can't go all the way around the Möbius strip and join up with itself (and thus can't be continuous) unless it crosses the central line around the strip (which line Penrose identifies with the zero section).

I suppose what's puzzling me is that these copies of F are disjoint, so how can one copy of, say, the vector space (R¹, R¹, regular multiplication) be said to have something analogous to an orientation or direction compared to another? Whereabouts in the definition of the Möbius strip bundle does this idea come in?

It makes sense to compare elements of different fibers, because the fibers are "glued together" by the topology of E. Or in other words, the topology on E imposes a notion of nearness on the fibers.

And note that a fiber bundle is not topologized in just any arbitrary way. No, Penrose says "A twisted bundle resembles a product bundle locally." What he means if that around each point b of B, there is an open set U and an homeomorphism $h:p^{-1}(U) \rightarrow U\times \mathbb{R}^1$ called a local trivialization of E over U.. That is, locally, E is topologized as a product space, and so to check that a cross section of E is continuous at a point e of E over b, just look at what the cross section looks like in a local trivialization U x R^1, in which it is easy to check for continuity.

Also, the local trivializations are assumed to be linear isomorphisms from fiber to fiber. That is, for all b in U, $h|_{p^{-1}(b)}:p^{-1}(b)\rightarrow \{p\}\times \mathbb{R}^1$ is an isomorphism. (This is what Penrose means when he says that the "group of symmetries" of a vector bundle is the general linear group.) And so, you see that a local trivialization can be used to locally give an orientation of each fiber in a continuous manner. The question of orientation of the bundle is whether this can be done globally in a consistent way.

 

[arkajad]

Additionally, what has been said above about continues maps and sections can be repeated with "continuous" replaced by "differentiable" or "smooth." This is needed in physics when we want to have tangent and cotangent spaces, and differential operators operating on cross-sections.

 

[Rasalhague]

Note that this is just an abstract way of saying: a cross section is a map from B to E such that a point b in B is sent to an element of F_b, the fiber of E over b. This is a generalisation of the notion of vector field on a manifold. In that case, B=M, the base manifold, E=TM, the tangent bundle, and a cross section is just a vector field on M.

Yeah, I got that, thanks.

And note that a fiber bundle is not topologized in just any arbitrary way. No, Penrose says "A twisted bundle resembles a product bundle locally." What he means if that around each point b of B, there is an open set U and an homeomorphism h:p^-1(U) --> U x R¹ called a local trivialization of E over U. That is, locally, E is topologized as a product space, and so to check that a cross section of E is continuous at a point e of E over b, just look at what the cross section looks like in a local trivialization U x R¹, in which it is easy to check for continuity.

Also, the local trivializations are assumed to be linear isomorphisms from fiber to fiber. That is, for all b in U, h|_p^{-1}(b):p^-1(b) --> {p} x R¹ is an isomorphism.

It this right? For a base space B and fibre F of a given bundle with total space E, there always exists a product space B x F, called a (global) trivialization of E or the trivial bundle, but this won't always be globally homeomorphic to E. The total space has extra structure besides that supplied by B and F, at least extra topological information is required to define it, and that topology may prevent E being globally homeomorphic to the total space B x F of the trivial bundle. On the other hand, for any bundle with total space, E, each point in B belongs to a neighbourhood U such that U x F is homeomorphic E.

Is it the function pi, the projection, that actually defines the topology? So the total space might be defined as the product space, in which case pi maps each fibre to the point that it's a fibre over, thus making E itself a trivial bundle, e.g. Penrose's cylinder, or pi might specify some other way of associating each point of E with each point of B, so that the preimage of a point b in B may include some points of E from a fibre over one point, p, in B, and some points of E from a fibre over a different point, q, in B. So the trivial bundle can be taken as a sort of starting point, and more interesting bundles defined by how they portion out the points from its fibres.

What does Wikipedia mean here, in Formal definition, by "the following diagram should commute"? (Presumably not that each arrow represents an invertible function, since pi and proj₁ generally won't be injections.)

 

[Rasalhague]

The missing piece here is the connection. The connection describes how the fibers are connected to each other; i.e., it defines a notion of parallel transport. Or alternatively, it splits the tangent space into vertical and horizontal subspaces; the verticals moving within a fiber, and the horizontals moving you to the adjacent fiber.

Also remember that a fiber bundle is a manifold, so it can be covered by charts. Within a chart, it is easy to define an orientation. One can propagate this orientation to nearby charts by insisting on orientation-preserving transition functions (i.e., with positive Jacobians). Thus one can build up the notion of global orientability by demanding that it be possible to assign a single orientation to the entire atlas.

Put a few charts on a Mobius band and you'll see that a global orientation is not possible.

Is this a general feature of all bundles, or is a connection only defined for those that are smooth or differentiable to some order?

 

[Fredrik]

I don't think you need a connection to understand how the Möbius strip is a non-trivial fiber bundle. What you need to understand is how the local triviality requirement defines "gluing instructions" that tell us how the pieces fit together. Consider a fiber bundle with projection $\pi:E\rightarrow B$ and fiber F. To say that it's locally trivial is to say that for each b in B, there's a set $U_\alpha$ and a function $h_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times F$ such that $b\in U_\alpha\subset B$ and $\text{Proj}_1\circ h_\alpha=\pi$, where the right-hand side is actually the restriction of $\pi$ to $\pi^{-1}(U_\alpha)$. I guess I should be writing $$\pi|_{\pi^{-1}(U_\alpha)}$$ instead of $\pi$, but I won't, because it would clutter the notation.

What you need to understand are the transition functions

$$t_{\alpha\beta}:U_\alpha\cap U_\beta\times F\rightarrow U_\alpha\cap U_\beta\times F$$

defined by $t_{\alpha\beta}=h_\alpha\circ h_\beta^{-1}$. Let's write $(b',f')=t_{\alpha\beta}(b,f)$. Note that we have

$$\text{Proj}_1\circ h_\alpha=\pi=\text{Proj}_1\circ h_\beta:\pi^{-1}(U_\alpha\cap U_\beta)\rightarrow U_\alpha\cap U_\beta$$

and that this implies

$$b'=\text{Proj}_1(b',f')=\text{Proj}_1\circ h_\alpha\circ h_\beta^{-1}(b,f)=b$$

So a transition function leaves the base point unchanged, but defines an automorphism $f\mapsto f'$ on the fiber F. Let's call that function $\phi_{\alpha\beta}(b)$. We have

$$t_{\alpha\beta}(b,f)=(b,\phi_{\alpha\beta}(b)(f))$$

$$\phi_{\alpha\beta}(b):F\rightarrow F$$

These automorphisms can be interpreted as gluing instructions that tell us how glue the overlapping regions of $U_\alpha\times F$ and $U_\beta\times F$ to each other.

 

[arkajad]

Is this a general feature of all bundles, or is a connection only defined for those that are smooth or differentiable to some order?

To define a connection (parallel transport) you better assume some differentiability, because you want to be able to do some calculations with it, and for this you need to distinguish between tangent vectors that are vertical (tangent to the fiber) and those that are transversal (have non-zero projection onto the base). "Horizontal" subspaces should be, as a rule, transversal. Then you can think of defining a parallel transport of the points of the fibers along paths in the base. Usually such a transport law is required to preserve some (or all) geometrical or algebraic structure that is defined on the fibers.

 

[lavinia]

The Mobius band is a line bundle over the circle with discrete structure group equal to Z/2. Z/2 acts on the fiber by negation.

There is no need for a connection to define this bundle.

On the other hand if one thinks of the Mobius band as a piece of paper pasted to itself with a half twist then it inherits parallel translation from translation in the flat plane. It is then a flat bundle over the circle with holonomy group, Z/2. Parallel translation around the meridian circle brings a vector back to its negative. This is the same twist that defines the bundle.

Similarly, the Klein bottle is a circle bundle over the circle with structure group Z/2. Z/2 acts on the fiber by reflection. In the flat Klein bottle, the holonomy group is Z/2. Parallel translation of a frame around the circle brings it back to a reflected frame. So again parallel translation tells you the transition function that defines the Klein bottle.

This same thing works for the tangent bundles of flat Riemannian manifolds. In all of these manifolds the structure group is finite and equals the holonomy group. Parallel translation of frames around closed geodesics recovers the structure group of the tangent bundle.

In a torus the tangent bundle is trivial. In the flat torus, parallel translation of a frame around a closed geodesic curve brings the frame back to itself so the structure group and the holonomy group are both trivial.

 

[quasar987]

It this right? For a base space B and fibre F of a given bundle with total space E, there always exists a product space B x F, called a (global) trivialization of E or the trivial bundle, but this won't always be globally homeomorphic to E. The total space has extra structure besides that supplied by B and F, at least extra topological information is required to define it, and that topology may prevent E being globally homeomorphic to the total space B x F of the trivial bundle.

This is exactly right.

On the other hand, for any bundle with total space, E, each point in B belongs to a neighbourhood U such that U x F is homeomorphic E.

No, it is only pi^-1(U), the fibers of E over U that are homeomorphic to U x F.

For instance, for the Mobius bundle, this condition expresses the fact that if you consider only the fibers over a portion of the circle; say the fibers over 0<theta<pi, then this part of the Mobius bundle can be "flattened out", or, trivialized. The same goes for the part of the bundle over pi/<theta<3pi/2 or whatever. Of course, it does not hold over the whole circle for there is an obvious "twist" in the bundle with would prevent us from "straightening" the fibers to make a cylinder without breaking the fabric of the bundle.

Is it the function pi, the projection, that actually defines the topology? So the total space might be defined as the product space, in which case pi maps each fibre to the point that it's a fibre over, thus making E itself a trivial bundle, e.g. Penrose's cylinder, or pi might specify some other way of associating each point of E with each point of B, so that the preimage of a point b in B may include some points of E from a fibre over one point, p, in B, and some points of E from a fibre over a different point, q, in B. So the trivial bundle can be taken as a sort of starting point, and more interesting bundles defined by how they portion out the points from its fibres.

If you are given a bundle E-->B, then this means by definition that E and B are topologized already and that pi is continuous and surjective. Additionally, the fact that E is locally trivial implies that pi is actually an open map. But continuous surjective open maps are quotient maps. And this means that the topology of B is directly relatable to that of E in that a subset U of B is open if and only if its preimage by pi is open in E. So pi much more determine the topology of B then it does that of E because if you are given a locally trivial topological space E, then local triviality allows you to define pi in a unique manner, and (E,B,pi) is a bundle iff pi is continuous iff B has the quotient topology induced by pi.

It is much more the local trivializations that determine the topology of E: a subset of E is open iff its image by every local triv. of E is open in the product space U x F.

What does Wikipedia mean here, in Formal definition, by "the following diagram should commute"? (Presumably not that each arrow represents an invertible function, since pi and proj₁ generally won't be injections.)

It only means that all the path of arrow that starts at A and ends at B are the same. The arrows are functions, so it means that if u take a point a in A and act on it succesively by the functions which the arrows reprensent, then you get and elements b in B which is independant of the path of arrow considered. Here, the fact that the commutative diagram commutes simply expresses the fact that $proj_1 \circ\phi=\pi$. That is: the local trivialisations preserve the fibers: they map the fiber of E over b to the fiber of the product bundle U x F over b.
Exercise: express the defining condition $p \; \circ \; s(x) = x$ of a section in the form of a commutative diagram.

 

[arkajad]

A simple example of an easily visualized nontrivial bundle is the tangent bundle of the 2-sphere. If it would be trivial - you would have a nowhere vanishing tangent vector field. And we know that that's not possible.

 

[Rasalhague]

What you need to understand is how the local triviality requirement defines "gluing instructions" that tell us how the pieces fit together.

[...]

These automorphisms can be interpreted as gluing instructions that tell us how glue the overlapping regions of $U_\alpha \times F$ and $U_\beta \times F$ to each other.

I think I follow your description, but I'm still struggling to see the full significance of how the transition functions distinguish the Möbius strip from the cylinder. The local trivializations, such as $U_\alpha \times F$, are charts on the manifold E, right? And the fibres which make up the total space of the twisted line bundle over S¹ (the Möbius strip bundle) can be covered with such charts, just as those of the trivial line bundle (cylinder bundle) can. Neither can be covered in less than 2 charts. The transition functions, are they part of the definition of the total space; is the particular form they can take determined by the topology of the total space, which in the case of the Möbius strip eliminates the possibility of certain transition functions that would exist for the trivial bundle?

In Riemannian Manifolds, p. 18, Lemma 2.2, Lee replaces the F in your definition of transition functions with R^k. I suppose this is an equivalent definition for the case where F is a finite vector space over the reals, since any such vector space is isomorphic to R^k.

If you are given a bundle E-->B, then this means by definition that E and B are topologized already and that pi is continuous and surjective. Additionally, the fact that E is locally trivial implies that pi is actually an open map. But continuous surjective open maps are quotient maps. And this means that the topology of B is directly relatable to that of E in that a subset U of B is open if and only if its preimage by pi is open in E. So pi much more determine the topology of B then it does that of E because if you are given a locally trivial topological space E, then local triviality allows you to define pi in a unique manner, and (E,B,pi) is a bundle iff pi is continuous iff B has the quotient topology induced by pi.

It is much more the local trivializations that determine the topology of E: a subset of E is open iff its image by every local triv. of E is open in the product space U x F.

Penrose talks about the bundle being defined in terms of two manifolds, the base space and the fibre. Suppose we wanted to do that. We know about the topology of the base space, e.g. S¹, and we know about the topology of the fibre, e.g. R¹. Since the bundle is actually identified with the function $\pi: E \rightarrow B$, I think what you're saying that that the place to look for the thing that distinguishes the twisted from the trivial line bundle over S¹ is in the definition of E, rather than in the definition of which points of E are projected onto which points of B. (Are the same points of E projected onto the same points of B in the trivial as in the non-trivial bundle, a point e in E is projected onto a point b in B iff e belongs to the fibre over b?) If this is a trivial bundle, stating that it's trivial tells us all we need to know. If not, we need to be told how E differs from trivial. And one way of completely specifying the topology of E is by specifying what local trivialisations exist, or perhaps rather what kinds of local trivialisations (which commute for the trivial bundle) don't commute for this non-trivial one?

The term quotient space is new to me. I'll have to read up on that.

Exercise: express the defining condition $p \circ s(x) = x$ of a section in the form of a commutative diagram.

I suppose that would look like the diagram on the right here but with the words "this diagram commutes"!

A simple example of an easily visualized nontrivial bundle is the tangent bundle of the 2-sphere. If it would be trivial - you would have a nowhere vanishing tangent vector field. And we know that that's not possible.

That's sounds like a good idea, I'll have a look at that. It rather like the twisted line bundle over S¹ in that there's a topological restriction on the kinds of section that can exist, although there it's the other way around: you can't have a vector field whose value isn't somewhere zero.

 

[arkajad]

Are the same points of E projected onto the same points of B in the trivial as in the non-trivial bundle, a point e in E is projected onto a point b in B iff e belongs to the fibre over b?)

Here we have problem. Because there is no natural way of comparing the points of a trivial and a nontrivial bundle. In a bundle, trivial or not, you can't even compare the points of two fibers - as it depends on the local trivialization.

If in a bundle you have a flat connection, then you can locally compare points in different fibers, but not necessarily globally if the base space is not simply connected.

 

[lavinia]

In a bundle, trivial or not, you can't even compare the points of two fibers - as it depends on the local trivialization.

If in a bundle you have a flat connection, then you can locally compare points in different fibers, but not necessarily globally if the base space is not simply connected.

I am a little confused by these statements and ask you to elaborate. Here is why.

- In a trivial bundle the trivialization is global. So you can compare points by projection onto the second factor (E = Bx F)

- In a flat torus points can be compared by parallel translation. But the torus is not simply connected. Further its tangent bundle is trivial.

 

[lavinia]

While it is not easy to show that the tangent bundle to the 2 sphere is non-trivial, it is easy to show that the canonical line bundle over the projective plane is non-trivial.

This bundle is the quotient of the normal bundle to the 2 sphere in R^3 by multiplication by -1 on R^3.

 

[arkajad]

- In a trivial bundle the trivialization is global. So you can compare points by projection onto the second factor (E = Bx F)

Now, there are two way a bundle may be called trivial. First it may be called trivial if it is defined as being the product. But such bundles are better known as "product bundles". Second, a bundle may be called trivial if it is bundle isomorphic to the product bundle. In such a case there are infinitely many such isomorphisms, no one being distinguished. And this what I had in mind.

- In a flat torus points can be compared by parallel translation. But the torus is not simply connected. Further its tangent bundle is trivial.

Parallel transport depends on the connection. You may have two flat (that is with curvature zero) connections but with different holonomy groups. That is for one connection going around the loop takes you back to the original point, while for a different connection you may end up somewhere else. This is what happens in Aharonov-Bohm effect. You have there a plane with a hole (a puncture is enough). Zero and nonzero magnetic field inside the hole makes no difference for the curvature (magnetic field) outside. And yet parallel transport of the phase around the loop is different in both cases.

 

[lavinia]

Now, there are two way a bundle may be called trivial. First it may be called trivial if it is defined as being the product. But such bundles are better known as "product bundles". Second, a bundle may be called trivial if it is bundle isomorphic to the product bundle. In such a case there are infinitely many such isomorphisms, no one being distinguished. And this what I had in mind.

Parallel transport depends on the connection. You may have two flat (that is with curvature zero) connections but with different holonomy groups. That is for one connection going around the loop takes you back to the original point, while for a different connection you may end up somewhere else. This is what happens in Aharonov-Bohm effect.

OK. But I need more explanation.

Can you show me two connections compatible with the flat metric on the 2 torus that have different holonomy groups?

what is the Aharonov-Bohm effect? Can you illustrate the holonomy effect you mentioned?

 

[arkajad]

For Aharonov-Bohm effect you can have a look at http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect" .

As for the torus: a "connection" does not necessarily mean an affine connection related to some metric. Torus can be considered as a principal U(1) bundle over the circle. Thus we can talk about connections in principal bundles. Aharonow-Bohm effect is then an example.

 

[lavinia]

For Aharonov-Bohm effect you can have a look at http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect" .

As for the torus: a "connection" does not necessarily mean an affine connection related to some metric. Torus can be considered as a principal U(1) bundle over the circle. Thus we can talk about connections in principal bundles. Aharonow-Bohm effect is then an example.

thanks that it interesting.

So you were saying that in case the base is simply connected then the holonomy of a flat connection on a bundle is independent of the connection?

 

[arkajad]

thanks that it interesting.

In the "Mathematical interpretation" part of the Wikipedia entry you have "complex line bundle". But the essential part is just the phase - the U(1) bundle.

 

[lavinia]

In the "Mathematical interpretation" part of the Wikipedia entry you have "complex line bundle". But the essential part is just the phase - the U(1) bundle.

right I got that

 

[lavinia]

In the "Mathematical interpretation" part of the Wikipedia entry you have "complex line bundle". But the essential part is just the phase - the U(1) bundle.

- What is a good reference on how Physicists interpret curvature as field strength?

- On the U(1) bundle over the circle I was just trying to imagine what the connection would look like. Is it that you just choose the horizontal directions to make a constant angle with the fiber? This would seem to make a parallel translated vector rotate through the fibers and come back at a different point.

 

[arkajad]

- What is a good reference on how Physicists interpret curvature as field strength?

- On the U(1) bundle over the circle I was just trying to imagine what the connection would look like. Is it that you just choose the horizontal directions to make a constant angle with the fiber? This would seem to make a parallel translated vector rotate through the fibers and come back at a different point.

If you can get hold of "Differential Manifolds and Theoretical Physics" by W. D. Curtis and F. R. Miller - you will find there lot of a useful stuff starting from Chapter 16.

As for the second question: what else can it be?

 

[Fredrik]

I think I follow your description, but I'm still struggling to see the full significance of how the transition functions distinguish the Möbius strip from the cylinder.

Let B=S¹ and F=(-1,1). Let $U_1=S^1-\{(0,1)\}$ and $U_2=S^1-\{(0,-1)\}$. Now $U_1\cap U_2$ consists of two disjoint subsets of S¹. Let's call them X and Y. Edit: Note that when I wrote (-1,1) I was referring to an interval, and when I wrote (0,1) and (0,-1), I was referring to points in $S^1\subset\mathbb R^2$

Suppose that the transition function $t_{12}=h_1\circ h_2^{-1}:U_1\cap U_2\times F\rightarrow U_1\cap U_2\times F$ is such that the corresponding fiber automorphisms $\phi_{12}(b)$ are the identity map on F for all b. Then the bundle is a cylinder. But suppose instead that the $\phi_{12}(b)$ is the identity map on F for all b in X, and the map $f\mapsto -f$ for all b in Y. Then the bundle is a Möbius strip.

Note that this almost literally makes the Möbius strip consist of two pieces of "paper" glued together after a twist on one side.

The local trivializations, such as $U_\alpha \times F$, are charts on the manifold E, right?

They are similar to charts, but since their codomains aren't open subsets of $\mathbb R^n$, I wouldn't say that they are charts.

By the way, these sets don't have to be manifolds. There's a different type of fiber bundle for each type of structure we're interested in. A fiber bundle of sets would be the simplest kind, but most books don't even mention that possibility. The most interesting cases are topological spaces and manifolds. (Don't let this give you the incorrect idea that all the sets that make up a vector bundle are vector spaces; only the fiber of a vector bundle needs to have a vector space structure). I don't know if fiber bundles of structures of some other kind are ever useful. It seems like topological spaces and manifolds are the only kinds we need.

The transition functions, are they part of the definition of the total space;

They are implied to exist by the requirement that a fiber bundle is locally trivial, so there's no need to include them explicitly. I think I've seen it done the other way round as well: Drop the local triviality requirement from the definition of "fiber bundle" and include the requirement that transition functions exist instead.

I would actually prefer to define "fiber bundle" without any of these requirements. The reason is that once you understand bundle isomorphisms, the term "locally trivial" becomes almost obvious. A bundle is said to be trivial if it's isomorphic to a product bundle, and locally trivial if every point in the base space has an open neighborhood B' such that the restriction of the bundle to B' is isomorphic to a product bundle.

The "restriction" of a bundle $\pi:E\rightarrow B$ to a subset $B'\subset B$ is the bundle you get by taking B' to be the base space, $E'=\pi^{-1}(B')\subset E$ to be the total space, and $\pi'=\pi|_{E'}$ to be the projection.

 

[arkajad]

I think I've seen it done the other way round as well: Drop the local triviality requirement from the definition of "fiber bundle" and include the requirement that transition functions exist instead.

This is how Steenrod does it in "The topology of fibre bundles" - he uses the term "coordinate bundle". Only after that he introduces the product bundle, and then describes the other definition - via local trivialization, which he calls "The Ehresmann-Feldbau definition of bundle". Then the next subsection is "Relations with coordinate bundles".

 

[Rasalhague]

Many thanks for your patient explanations! What if X and Y were small subsets very close together; could there be a phi that's the identity map for both X and Y then? What if they weren't disjoint; could the phi always be the identity map in that case so long as the intersection of U₁ and U₂ didn't cover the whole of S¹? Is the thing that defines the difference between the Möbius strip (seen as a twisted line bundle) from the cylinder (seen as a trivial line bundle) a rule saying that under a certain condition, the phi corresponding to any given transition function must map f to -f. I guess by a modification of that rule, we could describe shapes like the Möbius strip but with more than one twist.

(By the way, Penrose is talking about a slightly different pair of shapes, where F is the R¹ vector space, rather than just a finite interval of it.)

They are implied to exist by the requirement that a fiber bundle is locally trivial, so there's no need to include them explicitly. I think I've seen it done the other way round as well: Drop the local triviality requirement from the definition of "fiber bundle" and include the requirement that transition functions exist instead.

I mean, are particular transitions functions (or a particular constraint on transition functions) a necessary part of the definition of a particular bundle, e.g. the Möbius strip, as they seem to be the feature (in this formalism) that distinguishes it from the cylinder.

 

[arkajad]

(By the way, Penrose is talking about a slightly different pair of shapes, where F is the R¹ vector space, rather than just a finite interval of it.)

You must be very careful here, because every vector bundle with a typical fiber R¹ is trivial. More generally: every real differentiable vector bundle admits nowhere vanishing cross-section. When the fiber is one-dimensional, this implies triviality.

PS. The above is wrong.

 

[quasar987]

You must be very careful here, because every vector bundle with a typical fiber R¹ is trivial. More generally: every real differentiable vector bundle admits nowhere vanishing cross-section. When the fiber is one-dimensional, this implies triviality.

I'm trying to read another meaning into what you wrote, but it really looks like you're saying that every rank 1 vector bundle is trivial, which is of course false.

What are you saying here?

 

[Rasalhague]

Re. tangent bundle on S²,

That's sounds like a good idea, I'll have a look at that. It rather like the twisted line bundle over S¹ in that there's a topological restriction on the kinds of section that can exist, although there it's the other way around: you can't have a vector field whose value isn't somewhere zero.

On second thoughts, actually it's exactly the same constraint, isn't it. In both cases, any globally defined vector field must have the value zero somewhere. Is this a (the?) distinguishing feature of twisted vector bundles generally?

You must be very careful here, because every vector bundle with a typical fiber R¹ is trivial. More generally: every real differentiable vector bundle admits nowhere vanishing cross-section. When the fiber is one-dimensional, this implies triviality.

Hmm... On the one hand, he illustrates both cylinder and Möbius strip as thin bands with distinct edges; one the other, he calls the fibre in his first example (the cylinder) "a copy of the real line R." He says nothing about the fibre being different when he goes on to discuss his second example, the Möbius strip. Figs. 15.4 and 15.5 compare cylinder and Möbius strip without mentioning that they have a different fibre. In the next paragraph (p. 330 in the Vintage 2005 edition), he writes:

It is important to realize that the possibility of the twist results from a particular symmetry that the fibre V possesses, namely the one which reverses the sign of the elements of the 1-dimensional vector space V. (This is v |--> -v, for each v in V.) This operation preserves the structure of V as a vector space. We should note that this is not actually a symmetry of the real number system R. In fact R itself possesses no symmetries at all. [...] It is for this reason that V is taken as a 1-dimensional vector space rather than just as the real line itself. We sometimes say that V is modeled on the real line.

 

[arkajad]

I'm trying to read another meaning into what you wrote, but it really looks like you're saying that every rank 1 vector bundle is trivial, which is of course false.

What are you saying here?

I have to think about it. Quite possible I was wrong. $$E\otimes E$$ is then trivial.

 

[Hurkyl]

(By the way, Penrose is talking about a slightly different pair of shapes, where F is the R¹ vector space, rather than just a finite interval of it.)

I haven't followed the thread so I don't know if it's relevant, but topologically R¹ and a finite (open) interval are homeomorphic spaces.

 

[arkajad]

@quasar987
Thanks. Now I remember: the story was that in real vector bundles there is always a global cross section, but not that it is nowhere vanishing. Next time I will check better before posting anything.

 

[quasar987]

I have to think about it. Quite possible I was wrong. $$E\otimes E$$ is then trivial.

Well, the topic of the thread is the Mobius bundle, which is a nontrivial rank 1 vector bundle over the circle. Also, the tautological line bundle over RP^n is another example of a nontrivial rank 1 bundle over RP^n that has also been mentioned by lavinia earlier.

I do not see the relevance of the lemma you quote. It talks about the possibility of extending globally a locally extendable section defined on a closed set. But it does not say that the extension will be nonvanishing outside of the closed set S...so I don't see why it is relevant.

 

[quasar987]

@quasar987
Thanks. Now I remember: the story was that in real vector bundles there is always a global cross section, but not that it is nowhere vanishing. Next time I will check better before posting anything.

No problem! I am too often guilty of that myself. :tongue2: