Higher Order π Groups, Fiber Bundle

Fiber Bundle

Let p(Y~) map onto Y, and let b be the base point of Y, with fiber F. This is a fiber bundle if every point c in Y has an open neighborhood V_c containing c, with preimage U, where V cross F is homeomorphic to U. In addition, the homeomorphism from V*F onto U, followed by p, yields the identity map on V. Think of this as a commutative diagram. A double arrow connects U and V*F upstairs, and two projections map to V downstairs. Follow p(U), or extract the first component of V*F; the result is the same.

In summary, a fiber bundle is locally a topological product with F.

The simplest example is Y~ = Y*F. For every point c, let Y be the open set V_c, and you're there. This is the trivial fiber bundle.

Covering Space

A covering space is a fiber bundle. F is a discrete set of points, and we are looking for U = V*F. Set V_c to an open set about c that is evenly covered, and you're there.

Well almost. The preimage of an evenly covered open set in a covering space is, by definition, a collection of disjoint path components. These need to be components, i.e. open sets, in V cross F. Thus our path components must be true components. This is the case when Y~ is locally path connected. In fact, it is enough to show Y is locally path connected. Suppose Y~ is not locally path connected at a, and project a onto b. Now an open set about b is homeomorphic to an open set about a, and the former is path connected. Hence Y~ is locally path connected after all.

Conversely, a discrete fiber bundle is always a covering space. With F discrete, V*F is a disjoint collection of copies of V, each mapping homeomorphically onto V. Thus V is evenly covered, and Y~ is a covering space for Y. Again, this assumes each component upstairs, i.e. each copy of V, is a single path component, and this is the case when Y~, or Y, is locally path connected.

If every covering space were the trivial bundle Y*F, then every covering space would be disconnected. However, a real line is connected, and it covers the circle.

Projective Space

Real projective space RPⁿ is the set of lines running through the origin in n+1 space. These determine, and are determined by, opposite points on the n sphere. Therefore RPⁿ is a quotient space of Sⁿ, wherein opposite points are identified. Sⁿ is a covering space, hence it is a fiber bundle. Given c, let V_c be the open hemisphere centered at c, or any open hemisphere containing c for that matter. The preimage U is the aforementioned hemisphere and its reflection through the origin.

Next consider complex projective space, CPⁿ, which is the lines running through the origin in Cⁿ⁺¹. Each complex coordinate holds two real coordinates, so project each line onto the unit sphere that lies in 2n+2 dimensions, also known as S²ⁿ⁺¹. Multiply all coordinates by c, where c has norm 1, and find another point on the sphere that defines the same line in CPⁿ. We have built another quotient space, where the fiber of any point is a circle, |c| = 1, with the associated topology. Looking locally, assume the first complex coordinate is nonzero, hence it can be scaled to a positive number r. Let ε be anything less than r. Restrict the first coordinate to the anulus between r-ε and r+ε in the complex plane. This defines an open set in S²ⁿ⁺¹, mapping onto an open set in CPⁿ. The open preimage in the sphere is homeomorphic to our anulus, which is the open interval in CPⁿ crossed with the circle S¹. This is a fiber bundle that is not a covering space, since the fiber of any point is not discrete.

Globally, S²ⁿ⁺¹ is not CPⁿ cross S¹. Since the first space, i.e. the sphere, is simply connected, h₁ = 0. Apply the circle product rule to the second space, and h₁ becomes h₁(CPⁿ)*h₀(CPⁿ), which is Z.

Degree

Assume, at every point b in Y, some neighborhood V about b has preimage U in Y~ that looks like V cross a discrete set. That is, U is homeomorphic to disjoint copies of V. This is the case when Y~/Y is a covering space and Y is locally path connected. Map Y into the cardinal numbers by taking the cardinality of F_b at each point b. The fiber is fixed for some open set about b, and the degree is constant across V_b. Take the preimage of a given degree d, and cover it with open sets, whence the preimage is open. Therefore degree is a continuous map from Y into a discrete set. Assuming Y is connected, the image is connected, and Y has constant degree. Thus Y~/Y has a well defined degree. This agrees with the definition of degree when S¹ is a covering space for S¹. (Technically we need to take the absolute value of the winding degree to get the fiber bundle degree.)