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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$

Let $X$ be an algebraic variety $X$, that is proper over $k$ (here a variety is a scheme $X/k$ such that$\overline{X}= X \times \Spec(\overline{k})$ is irreducible and reduced).

We consider the Picard functor given for any $k$ scheme $S$ by

$$ \mathcal{Pic}_{X/k}(S) := \\ \{ \mathcal{M} \text{ invertible sheaf on } X \times_k S \} / \{ \text{ inv. sheaves of the form } p^*_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on } S \}. $$

It is known that this functor is not always representable, but almost; that means precisely there exists a $k$-scheme $\Pic(X/k)$ representing the associated functor $\text{Hom}( \ , \Pic(X/k))$ which contains the Picard functor $\mathcal{Pic}_{X/k}$ in the sense that for any $k$-scheme $S$ there is a functorial inclusion

$$ \iota_S: \mathcal{Pic}_{X/k}(S) \hookrightarrow \Hom_k(S,\Pic(X/k)). $$

In general that's a proper inclusion. The equality only holds if $X \times_k S$ admits a section over $S$.

My question if there is a way to write down explicitly the map $\iota_S$, ie given a class $[\mathcal{L}]$ of an invertible sheaf $\mathcal{L}$ on $X \times S$, what is the morphism $\iota_S([\mathcal{L}]): S \to \Pic(X/k)$ on $k$-rational points of $S$?

The most natural way to associate to $[\mathcal{L}]$ a map from $S$ to the Picard group on level of $k$ rational point is perhaps via $s \mapsto [\mathcal{L} \vert _{X \times s}]$. This is of couse welldefined and the construction is so "canonical", that it suggests that's the only possible way how $\iota_S$ could like.

But I can't find a strictly formal argument why $\iota_S$ is defined in that way? The question is closely related to this one

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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$

One easily observes that $\forall s \in S, \mathcal{Pic}_{X/k}(s) = \Pic(X/k)(k)$, $\Hom(s, \Pic(X/k)) = \Pic(X/k)(k)$, where $Y(k)$ is the set of $k$-points of $Y$. We assume that forall $s \in S$, the morphism $i_s$ is the identity. This is a normalization condition for the homomorphims of functors $\iota$. It could be chosen otherwise, but this one seems the most natural.

Let $s \in S$ be a rational point and consider $\sigma_s : s \longrightarrow S$ the corresponding section. Since $\iota$ is a homomorphism of functors, we have a commutative diagram:

$$ \require{AMScd} \begin{CD} \mathcal{Pic}_{X/k}(S) @>{\iota_S} >> \Hom(S, \Pic(X/k)) \\ @VV\sigma_s^*V @VVf_{\sigma_s}V \\ \mathcal{Pic}_{X/k}(s) @>{id}>> \Hom(s, \Pic(X/k)) \end{CD} $$ where $\sigma_s^*$ is the pull-back along $\sigma_s$ and $f_{\sigma_s}$ is the pull-back along $\sigma_s$, that is $f_{\sigma_s}(\varphi) = \varphi \circ \sigma_s$ for all $\varphi \in \Hom(S, \Pic(X/k))$. Note that the morphisms $\sigma_s^*$ and $f_{\sigma_s}$ are of this form, by definition of the functors $\mathcal{Pic}(X/k)(\cdot)$ and $\Hom(\cdot, \mathcal{Pic}(X/k))$.

From the commutativity of the diagram, we get that for all line bundle $L$ on $X \times S$: $$ i_S(L)(s) = f_{\sigma_S}(i_S(L))(s) = \sigma_s^* L = L|_{X \times \{s\}}.$$

If you choose another normalization condition for $i_s$, for instance $i_s(L) = L \otimes L_0$ (for a fixed $L_0$, not depending on $s$), then you would get another homomorphism of functors, say $j$ between $\mathcal{Pic}(X/k)(\cdot)$ and $\Hom(\cdot, \Pic(X/k))$. It would most certainly share the same properties as $\iota$.

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  • $\begingroup$ I would like to address one important detail. You mentioned that for all $s \in S$ (I assume you mean by $s \in S$ the rational points), we have $\mathcal{Pic}_{X/k}(s) = \text{Hom}(s, \text{Pic}(X/k)) $. For sure, for the rational points the composition $s \to S \to \text{Spec}(k)$ can be identified with identity, so the statement becomes $\mathcal{Pic}_{X/k}(k) = \text{Hom}(\text{Spec}(k), \text{Pic}(X/k)) $. You remarked that this should essentially somehow "easily" follow. $\endgroup$
    – user267839
    Commented Nov 28, 2022 at 10:37
  • $\begingroup$ Do you mean by "easily" that this follows from some elementary argument (maybe using only some 'abstract nonsense' argument and nothing else) or do you mean by "easily" that it inferes immediately assuming we already have in our hands the (non trivial) fact that if the projection $X \times S \to S$ obtains a section, then $\iota_S: \mathcal{Pic}_{X/k}(S) \to \text{Hom}(S, \text{Pic}(X/k)) $ becomes an isomorphism? $\endgroup$
    – user267839
    Commented Nov 28, 2022 at 10:38
  • $\begingroup$ The point is that I'm wondering if this fact I quoted in the last sentence is really a non trivial fact, or if it already can be infered by some diagram chase, so just by 'abstract arguments'? $\endgroup$
    – user267839
    Commented Nov 28, 2022 at 10:43

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