Pullback bundle and Chern classes

Question

The following is from Bott and Tu. If $f : M \to N$ is a map between two manifolds and $E$ is a complex bundle over $N$, then the pullback $f^{-1}E$ is a bundle over $M$. If the Chern classes of $E$ vanish, then the naturality property implies that so do those of $f^{-1}E$. Taking the Chern classes to be a measure of twisting of a bundle, we may assert that pulling back dilutes a bundle, i.e. makes it less twisted. One extreme is when $f$ is constant; in this case $f^{-1}E$ is trivial.

One may wonder if there exists a bundle which is so twisted that every bundle is a pullback of this universal bundle. Such a bundle indeed exists, at least for manifolds of finite type; it is the universal quotient bundle on the Grassmannian $G_k(\Bbb C^n)$ for $n$ sufficiently large.

I have some naive questions about this phrase. First, is there some simple example that would illustrate this "diluting" via $f$? If I consider for example $f:S^1 \to S^1$ and give the codomain the Möbius bundle, can I choose $f$ such that the pullback bundle $f^{-1}E$ untwists the Möbius bundle to $S^1 \times \Bbb R$ or $S^1 \times \Bbb C$?

Next, can I use this "twistness" to classify bundles? I.e. are bundles with the same twistness isomorphic? I guess that twistness means at least in the complex setting that the Chern classes are equal.

Answer 1

Yes, before discussing how to "untwist" the Möbius bundle, just recall that $S^1 \simeq \mathbb R P^1$ and that line bundles over $M$ are classified by the homotopy classes of maps $[M, \mathbb R P^\infty] \simeq H^1(M,\mathbb Z_2)$. So there are only two line bundles over $S^1$ up to isomorphism. As you can check that the Möbius band is not the trivial bundle, then any line bundle over $S^1$ must be trivial, or isomorphic to the Möbius band.

As is commonly done in algebraic geometry to denote line bundles over $\mathbb C P^n$, denote the trivial real line bundle on $S^1$ as $\mathcal O(0)$ and the Möbius band as $\mathcal O(1)$. As an exercise, you may check that, if $f : S^1 \to S^1$ is a smooth map of topological degree $\deg f$, then $$ f^*\mathcal O(1) \simeq \mathcal O(1)^{\otimes \deg f} =: \mathcal O(\deg f) $$ Moreover, you can check that $\mathcal O(n) \simeq \mathcal O(m)$ if and only if $m \equiv n$ mod $2$, thus $\mathcal O(n)$ is trivial when $n$ is even and isomorphic to the Möbius band when $n$ is odd.

In general it's difficult to classify the bundle "twistness" via algebraic invariants. You have characteristic classes that can be helpful to distinguish non-isomorphic bundles, but having the converse is false in general. For example, the tangent bundle of $S^2$ has trivial total Stiefel-Whitney class but it is not the trivial bundle.