Manifold #12. Fiber Bundles - Formal Description매니폴드 #12. 파이버 번들의 정의

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 12 of the playlist, Fiber Bundles - Formal Description. Up to the previous post we attached a tangent space and a cotangent space to every point of a manifold. This time we define the frame that ties such a point-by-point structure together into one manifold, the fiber bundle. The lecture does not hand us the definition first. It starts from the example of the cylinder and works toward the definition, and this post follows the same order.

Starting from the cylinder

Picture a hollow cylinder that runs up and down forever. Take one circle around its middle and call it $S^1$. $S^n$ is the name of the $n$-dimensional sphere, so $S^1$ is the circle and $S^2$ is the surface of a ball. A point on the cylinder is fixed by two numbers: its position $a \in S^1$ on the circle and its height $b \in \R$ above that circle. We set the circle’s own level to 0, with upward positive and downward negative. The position $a$ can be written as a number from 0 up to $2\pi$, and if we treat values that differ by $2\pi n$ as the same, it becomes almost the same coordinate as a real number. That identification leads to the large topic of quotient spaces, so the lecture stops here and postpones it.

The whole cylinder is already a differentiable manifold. But thanks to the pair $(a, b)$ we just built, three manifolds can be seen inside it. Write the whole cylinder as $E$, the circle $S^1$ as $B$ and the real line $\R$ as $F$, and call them the total space, the base space and the fiber. A point of the total space is a pair made of one point of the base and one point of the fiber.

The cylinder as a product set, and the missing topology

The geometric cylinder and this set of ordered pairs correspond one to one, so we may regard them as the same object. But the set of ordered pairs is just a set and does not get a topology automatically. We can give it the product topology, built from products $U \times V$ of open sets. The lecture, however, states firmly that it will not use this topology. We have not yet defined anything that deserves the name fiber bundle. All we did was form the product set of two manifolds. The topology for bundles is fixed separately below.

There is one more thing to watch. In the picture of the cylinder, the base circle looks like a subset sitting inside the total space. But a point of $E$ is an ordered pair, and the collection of pairs $(a, 0)$ with height 0 only resembles the circle and is not the base itself.

Projection and fiber

It is always easy to take a point of the base out of a point $p = (a, b)$ of the total space. This map is called the projection, and for fiber bundles it is always written $\pi$. On the cylinder it just returns the first component of the pair.

Since $\pi$ sends many points to one point, going backward gives a set, with no single point to return. For a point $a$ of the base, $\pi^{-1}(a)$ is the whole real line standing vertically above it. Above another point $c$ stands a different real line $\pi^{-1}(c)$. The two sets are not equal, but both are diffeomorphic to $\R$. So the word fiber is used in two senses. It can mean the fiber $\pi^{-1}(a)$ over one point, or the standard fiber $F$, of which all the fibers are copies. To be a bundle, $\pi^{-1}(b)$ must be diffeomorphic to $F$ for every point $b \in B$. On the cylinder every fiber is a real line, so this holds at once, and for other bundles it has to be checked separately.

Q. Can’t we regard the base space as a subset of the total space?
A. We do not. The total space is a set whose points are ordered pairs of a base point and a fiber point, so even if the pairs with height 0 look like a circle, the base itself is made of the points on the circle and not of ordered pairs. What connects the two is the projection $\pi$, and there is no subset relation.

A topology that makes the projection continuous

Now we fix the topology of $E$. If $U$ is an open set of the base, then for $\pi$ to be continuous its preimage $\pi^{-1}(U)$ must be open in $E$, because continuity means that preimages of open sets in the target are open in the domain. So the lecture collects all of these preimages and takes them as the open sets of $E$.

$\mathcal{T}_B$ is the topology of the base. The lecture leaves it to the student to prove that this is a topology. (Preimages preserve unions and intersections exactly, so unions and finite intersections of sets $\pi^{-1}(U)$ again have this form and satisfy the conditions for a topology. This explanation is added for understanding and was not covered in the lecture.) On the cylinder, the open sets of this topology are strips that run up and down forever over open intervals of the circle. It is coarser than the usual topology, in which any small region of the cylinder’s surface counts as open. The lecture says this coarse topology is enough to make $\pi$ continuous and that we do not need a finer one for our present purpose. (Two different points on one fiber are too coarse to be separated by these strips. So when we treat $E$ as a true manifold we use a finer topology, in which the bundle charts below are homeomorphisms. This explanation is added for understanding and was not covered in the lecture.)

Local trivializations and bundle charts

The cylinder could be built by multiplying the fiber onto the whole base at once. Not every bundle can be built that way. What we can always do is cover the base with an open cover $\{U_i\}$ and give a product structure over one piece at a time. The preimage $\pi^{-1}(U_i)$ of a piece $U_i$ is open in $E$ by the topology above, and for a bundle we require that it be diffeomorphic to the product $U_i \times F$.

We call this map $\phi_i$ a local trivialization, and the pair $(U_i, \phi_i)$ a bundle chart. If one such map exists for the whole bundle, the product structure holds globally and the bundle is called trivial. The cylinder is trivial because the single set $U = S^1$ is enough. If a bundle is not trivial, one map is not enough, and so we only require local trivializations that exist piece by piece. Books differ in notation, and the lecture mixes two. One is the $\phi_i$ above, going from $E$ to the product. The other goes backward from the product to $E$ and is $\psi_i = \phi_i^{-1}$. They are names for the same diffeomorphism read in opposite directions, so either one tells the same story. The map can also be used with the base component dropped, sending a point straight to $F$.

This structure gives a point $p$ of $E$ one more coordinate that belongs to the bundle. Compute $\pi(p)$, find a $U_i$ that contains that point, and read the fiber point $f \in F$ from $\phi_i(p) = (\pi(p), f)$. This is separate from the coordinates of the atlas that $E$ has anyway because it is a differentiable manifold. $E$ can use the coordinates of that atlas, and it can also read the base and the fiber separately through bundle charts. This second structure is what distinguishes a fiber bundle from an ordinary differentiable manifold.

Q. Isn’t the bundle coordinate of a point $p$ fixed to a single answer?
A. It is not. If $p$ lies in two cover sets $U_i$ and $U_j$ together, then $\phi_i$ and $\phi_j$ each give a pair. There is no reason for the two maps to agree, so the same point is written as different ordered pairs. In the next section we fix the map that connects these two pairs.

Overlaps and transition functions

A point $p$ over the overlap $U_i \cap U_j$ of two cover sets becomes a point of the product in both bundle charts. Just as we had transition functions that carry coordinates between two charts of a manifold, here too we fix a map that carries between the two pairs. It is the composition that goes through one and comes back through the other.

This composition leaves the base point fixed and moves only the fiber point. If we look at it with the base component removed, it is a map from $F$ to $F$. It carries the difference in how the two local trivializations read the fiber, and it is the transition function of the bundle. Both maps are diffeomorphisms, so the composition is also a diffeomorphism, and therefore a transition function is an element of the group $\mathrm{Diff}(F)$ of diffeomorphisms from $F$ to $F$. The lecture counts this as the last requirement of a fiber bundle: every transition function must be a diffeomorphism of the fiber. (There is one transition function for each point $b$ of the overlap, so we can write $\phi_j \circ \phi_i^{-1}(b, f) = (b, g_{ji}(b)(f))$. This formula is added for understanding and was not covered in the lecture.)

Q. A cylinder needs no transition function if we take one cover set, so may we still split it into two?
A. Yes. With one cover set for the whole base there is a single local trivialization and no transition function. Even if we split the same base into two cover sets, we can build a trivialization on each, and the amount by which the two maps disagree on the overlap is the transition function. The lecture says that if the bundle is trivial, splitting it this way should change nothing mathematically, and adds that for a bundle that is not trivial the transition functions get more complicated.

The structure group

The conditions so far are still not enough. If a transition function may be any element of $\mathrm{Diff}(F)$, that is too loose. So we attach to the bundle a group $G \subseteq \mathrm{Diff}(F)$ and call it the structure group. Every transition function must be an element of $G$. The structure group can be much smaller than $\mathrm{Diff}(F)$. For a bundle that is globally a product, like the cylinder, however many cover sets we use, if we choose them cleverly all transition functions can be made the identity, so the structure group can be reduced to the single element $\{1\}$. The Möbius strip does not work that way. Most of its transition functions can be made the identity, but at least one is left as multiplication by $-1$, so its structure group becomes $\{1, -1\}$. We treat this example in detail in the next post.

So with the same total space, the same base and the same fiber, a different structure group gives a different bundle. We may understand the structure group as the simplest group that holds all the remaining transition functions once the atlas is chosen as efficiently as possible.

Q. A transition function only has to be a diffeomorphism, so why do we have to fix a structure group separately?
A. The structure group narrows in advance the kinds of transition functions the bundle allows. For example, when the fiber is the real line, $\mathrm{Diff}(\R)$ contains an enormous number of nonlinear diffeomorphisms, but if we set the structure group to multiplication by nonzero numbers $\{ b \mapsto gb \}$, the structure of the bundle becomes much firmer. (This explanation is added for understanding and was not covered in the lecture. The lecture only says that the structure group keeps us from concluding that every bundle is a product.)

Check these sections in one picture with the toy below. Unroll the cylinder sideways and drag the point $p$ to see $\pi(p)$, the fiber that $p$ sits on, which of the two cover sets $U_i$, $U_j$ contain it, and the trivialization coordinates. The toy writes the position on the circle as an angle $\theta$. When you change the number $g$ multiplied into the fiber coordinate of $\phi_j$ with the buttons above, the transition function on the dark overlap becomes $(\theta, b) \mapsto (\theta, gb)$ and the required structure group changes accordingly. On the cylinder any choice of $g$ can be undone by multiplying $\phi_j$ by $g^{-1}$ to get back to $g = 1$, so it is the same bundle. (This explanation is added for understanding and was not covered in the lecture.)

What a fiber bundle needs

Collecting the elements of a fiber bundle that the lecture summarized, we get the following.

• The total space $E$, the base space $B$ and the fiber $F$. All are differentiable manifolds.
• The projection $\pi : E \to B$. It must be continuous, and $\pi^{-1}(b)$ must be diffeomorphic to $F$ for every $b \in B$.
• An open cover of the base and local trivializations over it. Each $\pi^{-1}(U_i)$ must be diffeomorphic to $U_i \times F$.
• The structure group $G$. Every transition function must be an element of $G$.

In many books the projection itself is called the fiber bundle. The projection remains a map from the total space to the base. Once these elements are in place we have a bundle, and from the next post on we look at how this structure works in concrete examples.