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?


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 $U$.

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

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