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

Question

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.

Answer 1

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.

Answer 2

Since $S^2$ has dimension 2, the only non-trivial term in the Chern character is $c_1(V)$. The fact that you can catalogue complex line bundles over $S^2$ by this Chern number can be seen by means of sheaf cohomology.

Consider the exponential (sheaf) exact sequence $$ 0\rightarrow \mathbb{Z}\rightarrow C^\infty_\mathbb{C} \overset{\exp(2\pi i\cdot)}{\rightarrow} (C^\infty_\mathbb{C})^*\rightarrow 0 $$ where $(C^\infty_\mathbb{C})^*$ denotes the sheaf of complex valued local functions that never vanish. This sequence gives rise to a exact sequence of cohomology groups

$$H^1(S^2,C^\infty_\mathbb{C})\rightarrow H^1(S^2,(C^\infty_\mathbb{C})^*) \rightarrow H^2(S^2,\mathbb{Z})\rightarrow H^2(S^2,C^\infty_\mathbb{C})$$

and since $S^2$ admits smooth partitions of unity, the sheaf $C^\infty_\mathbb{C}$ is soft and thus $H^1(S^2,C^\infty_\mathbb{C})=H^2(S^2,C^\infty_\mathbb{C})=0$. Therefore you conclude $$H^1(S^2,(C^\infty_\mathbb{C})^*)\simeq H^2(S^2,\mathbb{Z})$$

Now, the LHS can be realized by means of Cech cohomology as those cocycles/transition functions that uniquely determine your complex line bundle. The RHS is, as you know from topology, $\mathbb{Z}$. It can be shown that the connecting morphism corresponds precisely with assigning to each (isomorphism class of) line bundle on the LHS, the first Chern class on the RHS.

I suggest you check out Well's Differential Analysis on Complex Manifolds, Chapter III, Section 4: Complex Line Bundles, for a more detailed and careful account of the idea I exposed.

Now in the setting of holomorphic line bundles, it is not usual that the holomorphic and smooth classification of line bundles agrees. This happens to be true for the sphere $S^2$. And holomorphic line bundles over the sphere are all stable (in the sense of slope or Mumford--Takemoto stability).