1
$\begingroup$

From Grothendieck's lemma, we know that all holomorphic vector bundles over the complex projective line are direct sums of line bundles, and so, are $SU(2)$-equivariant.

I wonder, do there exist non-equivariant vector bundles over complex projective $N$-space, viewed as an $SU(N)$ or $U(N)$ homogeneous space? If so, are any of them also holomorphic vector bundles? What happens for Grassmannians, of flag manifolds, or projective homogeneous spaces?

$\endgroup$

1 Answer 1

6
$\begingroup$

In some sense, for every $n>1$, most holomorphic bundles on $\mathbb{C}P^n$ are non-equivariant. For instance, let $Z\subset \mathbb{C}P^2$ be a non-empty, zero dimensional closed subscheme that is locally a complete intersection, e.g., a collection of $m>0$ distinct, reduced points. Denote by $\mathcal{I}_Z$ the (coherent) ideal sheaf of $Z$ in $\mathbb{C}P^2$. Let $d>-3$ be an integer, so that $H^2(\mathbb{C}P^2,\mathcal{O}(d))$ vanishes. Then by "Serre's construction", there exists a short exact sequence of coherent sheaves on $\mathbb{C}P^2$, $$0 \to \mathcal{O}(d) \xrightarrow{q} \mathcal{E} \xrightarrow{p} \mathcal{I}_Z\to 0,$$ such that $\mathcal{E}$ is locally free of rank $2$. Of course the composition of $p$ with the inclusion, $\mathcal{I}_Z\subset \mathcal{O}_{\mathbb{C}P^2}$, is a homomorphism of coherent sheaves, $q':\mathcal{E}\to \mathcal{O}_{\mathbb{C}P^2}$. The zero locus of $q'$, which depends on $q'$ only up to scaling, is precisely $Z$.

If $H^0(\mathbb{C}P^2,\mathcal{I}_Z(-d))$ vanishes, then $q'$ is the unique such homomorphism (up to scaling); of course for $d\geq 0$, always $H^0(\mathbb{C}P^2,\mathcal{I}_Z(-d))$ vanishes. If $q'$ is unique up to scaling, then the zero locus $Z$ is an invariant of $\mathcal{E}$. This means that $\mathcal{E}$ is not equivariant, since the scheme $Z$ is not invariant under the full group of automorphisms. In fact, once $Z$ consists of $5$ or more general points, the automorphism group of $\mathbb{C}P^2$ mapping $Z$ to itself is trivial, so $\mathcal{E}$ is not equivariant for any nontrivial subgroup of the full automorphism group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.