3
$\begingroup$

This is from Deligne, La formule de dualité globale, SGA 4, tome 3, Expose XVIII, and I am confused about how the hom-space between Picard stacks is again a Picard stack.

A quick rewind. For a site $S$, a stack $P/S$ is called a Picard stack, if there is an addition $+: P\times_S P \to P$ so that each fibre $P(U)$ for $U \in S$ is a Picard groupoid. It is then claimed that $Hom(P_1, P_2)$ is a Picard stack, but only each $Hom(P_1, P_2)(U)$ is given: objects are additive morphisms between $P_1(U)$ and $P_2(U)$, and morphisms are morphisms between additive functors.

I wonder why the the existence of cartesian lifts are left blank in Deligne's work. Is there a general principle that allows one to fill in this gap?

I tried to at least figure out morphisms between $F \in Hom(P_1, P_2)(U)$ and $G \in Hom(P_1, P_2)(V)$. My guess is the following. A morphism $\alpha:F \to G$ that projects to $f:U \to V$ is given by the following data. For each morphism $\phi: y \to x $ over $f$ in $P_1$, associate $\alpha(\phi):Fy \to Gx$. The collection of maps $\alpha$ satisfy:

  1. If $\psi: y' \to y$ is an isomorphism in $P_1(U)$, $\alpha(\phi\circ\psi) = \alpha(\phi)\circ F(\psi)$.
  2. If $\psi: x \to x'$ is an isomorphism in $P_1(V)$, $\alpha(\psi\circ\phi) = G(\psi)\circ \alpha(\phi)$.
  3. $\alpha$ completes a square between $F(y_1 + y_2) \cong Fy_1 + Fy_2$ and $G(x_1 + x_2) \cong Gx_1 + Gx_2$.

If this guess actually works, my question would be why there exists at least one such $\alpha$ that projects to $f$, and why any such $\alpha$ is Cartesian?

$\endgroup$
8
$\begingroup$

Be careful that $Hom(P_1,P_2)(U)$ is not additive functors $P_1(U)\to P_2(U)$, but Picard stack morphisms $P_{1\mid U}\to P_{2\mid U}$, it's exactly the same formula as for hom-sheaves.

In particular, if you take the "stacks as sheaves of groupoids" perspective rather than the fibered point of view, it's now clear how to restrict such a thing to $V$ when you have a morphism $V\to U$, because you can simply further restrict to $V$ something that was a morphism between restrictions to $U$.

$\endgroup$
1
  • $\begingroup$ Oh, thanks. I got it now! $\endgroup$ – Frid Fu Jan 21 at 22:12

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.