5
$\begingroup$

I have two related questions. Can there be a stably trivial non-trivial holomorphic vector bundle over a closed complex manifold? Can there be a stably trivial non-trivial algebraic vector bundle over a smooth proper variety (of arbitary characteristic)?

$\endgroup$
1
  • $\begingroup$ Even more generally, over a projective variety, $E\oplus G\cong F\oplus G$ implies $E\cong F$, for vector bundles $E,F,G$, and known as Krull-Remak-Schmid theorem. $\endgroup$ – Mohan Jun 24 '20 at 22:00
9
$\begingroup$

Assume $E \oplus \mathcal{O} \cong \mathcal{O}^{\oplus n}$. Then, of course, $E \cong \mathcal{O}^{\oplus n}/\mathcal{O}$. On the other hand $$ Hom(\mathcal{O}, \mathcal{O}^{\oplus n}) \cong \Gamma(X, \mathcal{O})^{\oplus n}. $$ If $X$ is proper, connected and reduced, then $\Gamma(X, \mathcal{O}) = \Bbbk$ (the base field), hence any non-zero morphism $\mathcal{O} \to \mathcal{O}^{\oplus n}$ is given by a non-zero $n$-tuple of elements of the field, therefore any such morphism is isomorphic to the embedding of the first direct summand, hence the quotient is isomorphic to $\mathcal{O}^{\oplus (n - 1)}$. Thus, $E$ is trivial.

$\endgroup$
7
  • $\begingroup$ Do you mean the embedding of the second direct summand? $\endgroup$ – Michael Albanese Jun 24 '20 at 17:22
  • $\begingroup$ @MichaelAlbanese: The embeddings of all $n$ summands are isomorphic to each other, so you can consider the second if you prefer it. $\endgroup$ – Sasha Jun 24 '20 at 20:01
  • $\begingroup$ Oh, I see, you're talking about the first direct summand of $\mathcal{O}^{\oplus n}$. For some reason, I thought you were referring to the first summand of $E\oplus\mathcal{O}$, hence my confusion. $\endgroup$ – Michael Albanese Jun 24 '20 at 20:03
  • $\begingroup$ @MichaelAlbanese: You are right, that wasn't written clear enough. $\endgroup$ – Sasha Jun 24 '20 at 20:06
  • $\begingroup$ What if one takes stably trivial to mean there exists an $m$ such that $E\oplus \mathcal{O}^{\oplus m} \cong \mathcal{O}^{\oplus n}$? Following your argument, instead of one non-zero $n$-tuple, you get $m$, and if I'm not mistaken, they must be linearly independent (regarded as elements of $\mathbb{k}^n$). One can therefore arrange that the map $\mathcal{O}^m \to \mathcal{O}^n$ is just the embedding into the first $m$ direct summands, so the quotient, and hence $E$, is isomorphic to $\mathcal{O}^{\oplus(n-m)}$. Does that sound correct? $\endgroup$ – Michael Albanese Jun 24 '20 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy