# Are vector bundles with isomorphic determinant bundles isomorphic?

Let $$A$$ and $$B$$ be $$2n$$-dimensional complex vector bundles and $$\det A=\Lambda^{2n}(A)$$ and $$\det B=\Lambda^{2n}(B)$$.

Can you prove $$A\cong B$$ if and only if $$\det A\cong \det B$$? Is it a correct proposition?

Note: Here $$\cong$$ is a bundle isomorphism.

• What does $\cong$ mean in this context? – Omnomnomnom Jun 17 at 1:11
• @Omnomnomnom it is vector bundle isomorphism – Ramtin.VA Jun 17 at 1:12
• This is false. A K3 surface has trivial canonical bundle but not trivial tangent bundle. (en.wikipedia.org/wiki/K3_surface) – Gunnar Þór Magnússon Jun 17 at 8:28
• @GunnarÞórMagnússon thanks, it’s wonderful! – Ramtin.VA Jun 17 at 9:02

## 1 Answer

For a simple way to get counterexamples, note that if $$A\cong L_1\oplus\dots\oplus L_n$$ splits a direct sum of line bundles, then $$\det A\cong L_1\otimes \dots\otimes L_n$$. So, for instance, let $$X$$ be any space with a class $$\alpha\in H^2(X,\mathbb{Z})$$ such that $$\alpha^2\neq 0$$ and let $$L$$ be the line bundle on $$X$$ with $$c_1(L)=\alpha$$. Letting $$A=L\oplus L^{-1}$$ and $$B$$ be the trivial rank $$2$$ bundle, then $$\det A\cong L\otimes L^{-1}$$ and $$\det B$$ are both trivial, but $$A\not\cong B$$ since $$c_2(A)=c_1(L)c_1(L^{-1})=-\alpha^2\neq 0$$ so $$A$$ is nontrivial.

• Thank you very much.It is clear. – Ramtin.VA Jun 20 at 6:00