Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a complex torus (a finite dimensional complex vector space modulo a lattice) and let $E$ be a smooth (not necessarily holomorphic) complex vector bundle over $X$. Is it true $E$ is isomorphic to a sum of smooth line bundles? If not, what would be a minimal counterexample?

A theorem of Atyiah and Hirzebruch tells me that the Chern-character is an isomorphism of $\mathbb Q$-algebras $K(X)\otimes\mathbb Q \to H^{2\ast}(X,\mathbb Q)$. Since $X$ is a torus, its even cohomology ring is generated by Chern-classes of complex line bundles, hence so is $K(X)\otimes\mathbb Q$. It follows that some power of $E$ is stably isomorphic to a sum of line bundles. If one can get rid of the denominators, it would follow that $E$ itself is stably isomorphic to a sum of line bundles, but that's probably as far as one gets with $K$-theory.

share|improve this question
No, that's not true. Let $X$ be a $2$-dimensional complex torus. Use Serre's construction to produce a rank $2$ vector bundle $E$ whose first Chern class equals $0$ and whose second Chern class equals $1$ times the Poincare dual of the class of a point. If $E$ were a direct sum of two line bundles, then by using the Whitney sum formula we would have the second Chern class is $-c^2$, where $c$ is the first Chern class of either line bundle. But it is easy to write out a general element $c$ in $H^2(X,\mathbb{Z})$ and see that its square is always divisible by $2$. –  Jason Starr Nov 16 '11 at 13:39
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.