Rank 2 complex vector bundles over $S^2\times S^2$

Question

How could people classify all rank $2$ complex vector bundles over $S^2\times S^2$ up to isomorphism? Could you give a rank 2 complex vector bundle which cannot be split as a sum of two line bundles?

Answer 1

This is a special case of The space of homotopy classes of maps of products of spheres.

Classifying rank 2 complex vector bundles on $S^2\times S^2$ is the same as calculating the set of pointed homotopy classes $\langle S^2\times S^2,BU(2)\rangle$. For this I would use the cofibration sequence $$ S^3\to S^2\vee S^2 \to S^2\times S^2 \to S^4\to S^3\vee S^3\to \cdots $$ where the first map is the attaching map of the top cell of $S^2\times S^2$, and the fourth map is its suspension, therefore is null-homotopic. The map $q:S^2\times S^2\to S^4$ can be identified with collapsing the complement of a small open ball $B^4\subset S^2\times S^2$.

Taking maps into $BU(2)$ results in an exact sequence $$ 0 \to \pi_4(BU(2))\to \langle S^2\times S^2,BU(2)\rangle \to \pi_2(BU(2))\oplus \pi_2(BU(2))\to \pi_3(BU(2)) $$ which reduces (using $\pi_i(U(2))=\mathbb{Z},0,\mathbb{Z}$ for $i=1,2,3$) to a short exact sequence of sets $$ 0\to \mathbb{Z} \to \langle S^2\times S^2, BU(2)\rangle \to \mathbb{Z}\oplus\mathbb{Z} \to 0. $$ This is not quite the full classification (which I'm sure must appear in the literature somewhere, if you look hard enough), but allows us to say some things.

For instance, the second map is restriction to either $S^2$ factor, so I think this shows us how to produce a rank 2 complex bundle over $S^2\times S^2$ which is not a product of line bundles: take a non-trivial rank 2 bundle over $S^4$ (such as are classified by their $c_2$, by Greg Arone's answer to your previous question), and pull it back via the collapse map $q:S^2\times S^2\to S^4$.

Answer 2

I am surely late to this, but maybe it has some worth noticing the following statement: If $M$ is a closed manifold of dimension $2n$ such that its odd integer cohmology vanishes, then every complex rank $n$ vector bundle over $M$ is uniquely determined up to isomorphism by its Chern classes.

One explanation is provided though $K$-theory. Since we are in the stable range the set of isomorphism classes of complex vector bundles of rank $n$ $\mathrm{Vect(M)}$ is in bijection to the reduced complex $K$-theory of $M$, $\widetilde K(M)$. Moreover $\widetilde K(M)$ has no torsion since $H^\ast(M,\mathbb Z)$ has no torsion and the Chern character $\mathrm{ch}\colon \widetilde K(M) \to \widetilde H^\ast(M,\mathbb Q)$ is injective. This implies that every stable class of complex vector bundle is uniquely determined by its Chern character, hence by its Chern classes (since, again, integer cohomology has no torsion). This also induces a group structure on $\mathrm{Vect}(M)$ by the group structure of $\widetilde K(M)$.

A complex vector bundle $E \to S^2\times S^2$ of rank $2$ is a Whitney sum of two line bundles, say $L_1$ and $L_2$ with first Chern classes $x$ and $y$, then $c_1(E) = x+y$ and $c_2(E)=x\cup y$. If $p \colon S^2 \times S^2 \to S^4$ is a map of degree $1$ then, pulling back the tautological bundle of $S^4 = \mathbb{HP}^1$ to $S^2\times S^2$ via $p$, we obtain a bundle which cannot split into the sum of two line bundles for the reasons mentioned above.