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.

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

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.

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

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