Isomorphism class of vector bundle over $S^2$ and the Chern character

algebraic-topologycharacteristic-classesdifferential-formsdifferential-geometryvector-bundles

Let $V$ be a complex vector bundle over $S^2$. Suppose that

  • The rank of $V$ is $1$;
  • $\int_{S^2}\text{ch}(V)=1$.

I've seen it written that these two facts determine $V$ up to isomorphism.

Question 1: Why is this true?

Question 2: Does something like this hold for bundles over $S^n$ (possibly with a restriction the rank)?

Thoughts: I guess in question 1, the integral $\int_{S^2}\text{ch}(V)$ is really just $\int_{S^2}c_1(V)$, where $c_1$ is the first Chern class. However, I had the impression that this could only determine the bundle up to stable isomorphism classes.

Best Answer

In general, complex vector bundles over spheres are not determined by their Chern character or Chern classes. For example, there is a non-trivial rank two complex vector bundle $E \to S^5$ which necessarily has $c_1(E) = 0$, $c_2(E) = 0$, and $\operatorname{ch}(E) = 2$ (i.e. $\operatorname{ch}_1(E) = 0$ and $\operatorname{ch}_2(E) = 0$); see this answer. However, rank $n$ vector bundles on $S^{2n}$ are determined by $c_n$ (or equivalently, $\operatorname{ch}_n$), see the second lemma on page $212$ of May's A Concise Course in Algebraic Topology.

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