4
$\begingroup$

Let $S,S'$ be schemes, let $\pi : S' \to S$ be a morphism which is faithfully flat and locally of finite presentation, set $S'' := S' \times_{S} S'$ and $S''' := S' \times_{S} S' \times_{S} S'$ with projections $p_{1},p_{2} : S'' \to S'$ and $p_{12},p_{13},p_{23} : S''' \to S''$. Let $\mathcal{E}'$ be a vector bundle on $S'$ such that there is an isomorphism $\varphi : p_{1}^{\ast}\mathcal{E}' \to p_{2}^{\ast}\mathcal{E}'$ of $\mathcal{O}_{S''}$-modules. What's an example of such $S,S',\pi,\mathcal{E}'$ such that $\mathcal{E}'$ does not descend to $S$, i.e. there does not exist a vector bundle $\mathcal{E}$ on $S$ and an $\mathcal{O}_{S'}$-module isomorphism $\pi^{\ast}\mathcal{E} \simeq \mathcal{E}'$?

Thoughts:

  • Such $\mathcal{E}$ exists if and only if there exists some $\psi \in \operatorname{Aut}_{\mathcal{O}_{S''}}(p_{1}^{\ast}\mathcal{E}')$ such that the cocycle condition $p_{23}^{\ast}(\varphi \circ \psi) \circ p_{12}^{\ast}(\varphi \circ \psi) = p_{13}^{\ast}(\varphi \circ \psi)$ is satisfied.
  • If $\pi$ admits a section $\sigma : S \to S'$, then we can take $\mathcal{E} = \sigma^{\ast}\mathcal{E}'$ and pull back $\varphi$ via an induced section $S' \to S''$.
  • If $S = \operatorname{Spec} k$ and $S'$ is a smooth projective geometrically integral $k$-scheme and $\mathcal{E}'$ is a line bundle, then $\operatorname{Pic}(S') \times \operatorname{Pic}(S') \to \operatorname{Pic}(S'')$ is injective so the existence of $\varphi$ implies that $\mathcal{E}' \simeq \mathcal{O}_{S'}$.
$\endgroup$
8
$\begingroup$

This already fails for line bundles on smooth projective curves: let $X$ be 'the' pointless conic over $\mathbb{R}$, given by the closed subscheme of $\mathbb{P}^2_{\mathbb{R}}$ cut out by $X^2+Y^2+Z^2 = 0$. It is smooth, projective, geometrically integral over $\mathbb{R}$, and $X(\mathbb{R})= \emptyset$. In your notation, we let $\pi\colon S' \rightarrow S$ be the fppf morphism $X_{\mathbb{C}} \rightarrow X$. In this case fppf descent translates into Galois descent.

Since $X_{\mathbb{C}} \simeq \mathbb{P}^1_{\mathbb{C}}$, the isomorphism class of a line bundle on $X_{\mathbb{C}}$ is determined by its degree. Since the action of $Gal(\mathbb{C}|\mathbb{R})$ preserves degrees of divisors, we have that $L \simeq c^*L$ if $c$ denotes complex conjugation $X_{\mathbb{C}} \rightarrow X_{\mathbb{C}}$. Therefore a line bundle $L$ on $X_{\mathbb{C}}$ of degree $1$ gives an example you're looking for. Indeed we have just seen that $c^* L\simeq L$, and if there were a line bundle $M$ on $X_{\mathbb{R}}$ complexifying to $L$, it would need to have degree $1$ hence (by Riemann-Roch) be effective, which is impossible since $X(\mathbb{R})=\emptyset$.

$\endgroup$
1
  • $\begingroup$ Thanks, I see this should work for any nontrivial Brauer-Severi variety over any field. $\endgroup$ – Minseon Shin Jan 17 at 16:18

Your Answer

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

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