# Classifying Bundles; Fixed Base and Fiber

#### Bacle

Hi, everyone:

I am trying to find a result for the number of bundles (up to bundle iso.) over a fixed
base and fixed fiber. For example, for B=S<sup>1</sup> , and fiber I=[0,1]
I think that there are two; the cylinder and the Mobius strip.

I think that the reason there are two (classes of) bundles is that there are
only two isotopy classes in Hom(I,I) , where Hom is a homeomorphism;
the class of the identity, and the class of the map f(t)=1-t, so that one copy of I .

Also: how can we count the number of bundles over S<sup>1</sup> x
S<sup>1</sup>.? The obvious bundles are the ones defined as product
bundles of the bundles above, i.e., cylinderxcylinder, etc. Is there a way
of knowing if the product bundles are the only bundles over S<sup>1</sup> x
S<sup>1</sup>.?
I guess if the above ideas is correct, we should consider the isotopy classes
of Hom(IxI,IxI ).

Thanks.

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#### Bacle

Re: Classifying Bundles; Fixed Base and Fiber. Correction.

Sorry, in the entry above, I meant to mention the Isotopies of the Structure
Group G of the fiber ( which is I=[0,1] in this case) in general, and not just
Hom(I,I) , or , more generally, Hom (F,F).

#### zhentil

In the circle case, you're correct. In the case of vector bundles, there's a bijection between isomorphism classes or rank k bundles and homotopy classes of maps [M,Gr(k)], where Gr(k) is the set of k planes in R^{infinity}.

In the case of a compact lie group, replace Gr(k) with BG.

In your case with the circle (although I think you might want to replace [0,1] with R, to coincide with the definition that a fiber bundle has manifolds without boundary), we look at [S^1,Gr(1)]. Since Gr(1) is the direct limit of real projective spaces, it has Z/2 as fundamental group, coinciding with your guess.

#### zhentil

Although, it should be remarked in general that the computation of such homotopy classes is a nontrivial undertaking. That's why one looks at characteristic classes as a more computable way of determining when two bundles aren't isomorphic.

That said, there are times when it's very easy: take oriented rank two bundles. Then they have complex structures, and the homotopy classes are parametrized by [M, CP^{infinity}]=H^2(M,Z), since CP^{infinity} is a K(Z,2).

#### lavinia

Gold Member
I also think - but may be wrong - that in some cases you can not count the isomorphism classes of vector bundles. For instance the isomorphism classes of flat vector bundles over a Riemann surface look like the conjugacy classes of representations of the fundamental group in the general linear group. For genus greater than 1,this seems to form an algebraic variety - with singularities. Maybe someone could refute or verify this.

#### Hurkyl

Staff Emeritus
Gold Member
I wonder if the opening poster isn't expected to just count them via descent -- cover the torus with discs, then count all isomorphic ways to put bundles on each possible intersection and specify the inclusion maps.

(Hrm. Maybe covering the torus with cylinders would be easier?)

#### zhentil

Also, perhaps we should specify what we mean by isomorphism :) The isomorphisms that Lavinia and I refer to are very different.

#### lavinia

Gold Member
Also, perhaps we should specify what we mean by isomorphism :) The isomorphisms that Lavinia and I refer to are very different.
I think bundle equivalence has meaning only with respect to a given structure group. For instance the twisted torus is a non-trivial flat circle bundle over the circle with structure group Z/2Z but is equivalent to a trivial circle bundle over the circle with the structure group of rotations of the plane.

In the Riemann surface example, the equivalence classes of vector bundles with discrete structure group look like a quotient space of an algebraic variety - but over the general linear group they are completely different - but I do not know what they actually look like.

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