## 76.53 Theorem of the cube

This section is the analogue of More on Morphisms, Section 37.33. The following lemma tells us that the diagonal of the Picard functor is representable by locally closed immersions under the assumptions made in the lemma.

Lemma 76.53.1. Let $S$ be a scheme. Let $f : X \to Y$ be a flat, proper morphism of finite presentation of algebraic spaces over $S$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. For a morphism $g : Y' \to Y$ consider the base change diagram

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

Assume $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_{X'}$ is an isomorphism for all $g : Y' \to Y$. Then there exists an immersion $j : Z \to Y$ of finite presentation such that a morphism $g : Y' \to Y$ factors through $Z$ if and only if there exists a finite locally free $\mathcal{O}_{Y'}$-module $\mathcal{N}$ with $(f')^*\mathcal{N} \cong (g')^*\mathcal{L}$.

Proof. Let $y : \mathop{\mathrm{Spec}}(k) \to Y$ be a field valued point. Then the fibre $X_ y$ of $f$ at $y$ is connected by our assumption that $H^0(X_ y, \mathcal{O}_{X_ y}) = k$. Thus the rank of $\mathcal{E}$ is constant on the fibres. Since $f$ is open (Morphisms of Spaces, Lemma 67.30.6) and closed we conclude that there is a decomposition $Y = \coprod Y_ r$ of $Y$ into open and closed subspaces such that $\mathcal{E}$ has constant rank $r$ on the inverse image of $Y_ r$. Thus we may assume $\mathcal{E}$ has constant rank $r$. We will denote $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}, \mathcal{O}_ X)$ the dual rank $r$ module.

By cohomology and base change (more precisely by Derived Categories of Spaces, Lemma 75.25.4) we see that $E = Rf_*\mathcal{E}$ is a perfect object of the derived category of $Y$ and that its formation commutes with arbitrary change of base. Similarly for $E' = Rf_*\mathcal{E}^\vee$. Since there is never any cohomology in degrees $< 0$, we see that $E$ and $E'$ have (locally) tor-amplitude in $[0, b]$ for some $b$. Observe that for any $g : Y' \to Y$ we have $f'_*((g')^*\mathcal{E}) = H^0(Lg^*E)$ and $f'_*((g')^*\mathcal{E}^\vee ) = H^0(Lg^*E')$. Let $j : Z \to Y$ and $j' : Z' \to Y$ be the locally closed immersions constructed in Derived Categories of Spaces, Lemma 75.26.6 for $E$ and $E'$ with $a = 0$ and $r = r$; these are characterized by the property that $H^0(Lj^*E)$ and $H^0((j')^*E')$ are locally free modules of rank $r$ compatible with pullback.

Let $g : Y' \to Y$ be a morphism. If there exists an $\mathcal{N}$ as in the lemma, then, using the projection formula Cohomology on Sites, Lemma 21.50.1, we see that the modules

$f'_*((g')^*\mathcal{E}) \cong f'_*((f')^*\mathcal{N}) \cong \mathcal{N} \otimes _{\mathcal{O}_{Y'}} f'_*\mathcal{O}_{X'} \cong \mathcal{N}\quad \text{and similarly }\quad f'_*((g')^*\mathcal{E}^\vee ) \cong \mathcal{N}^\vee$

are locally free of rank $r$ and remain locally free of rank $r$ after any further base change $Y'' \to Y'$. Hence in this case $g : Y' \to Y$ factors through $j$ and through $j'$. Thus we may replace $Y$ by $Z \times _ Y Z'$ and assume that $f_*\mathcal{E}$ and $f_*\mathcal{E}^\vee$ are locally free $\mathcal{O}_ Y$-modules of rank $r$ whose formation commutes with arbitrary change of base.

In this situation if $g : Y' \to Y$ is a morphism and there exists an $\mathcal{N}$ as in the lemma, then the map (cup product in degree $0$)

$f'_*((g')^*\mathcal{E}) \otimes _{\mathcal{O}_{Y'}} f'_*((g')^*\mathcal{E}^\vee ) \longrightarrow \mathcal{O}_{Y'}$

is a perfect pairing. Conversely, if this cup product map is a perfect pairing, then we see that locally on $Y'$ we have a basis of sections $\sigma _1, \ldots , \sigma _ r$ in $f'_*((g')^*\mathcal{L})$ and $\tau _1, \ldots , \tau _ r$ in $f'_*((g')^*\mathcal{E}^\vee )$ whose products satisfy $\sigma _ i \tau _ j = \delta _{ij}$. Thinking of $\sigma _ i$ as a section of $(g')^*\mathcal{L}$ on $X'$ and $\tau _ j$ as a section of $(g')^*\mathcal{L}^\vee$ on $X'$, we conclude that

$\sigma _1, \ldots , \sigma _ r : \mathcal{O}_{X'}^{\oplus r} \longrightarrow (g')^*\mathcal{E}$

is an isomorphism with inverse given by

$\tau _1, \ldots , \tau _ r : (g')^*\mathcal{E} \longrightarrow \mathcal{O}_{X'}^{\oplus r}$

In other words, we see that $(f')^*f'_*(g')^*\mathcal{E} \cong (g')^*\mathcal{E}$. But the condition that the cupproduct is nondegenerate picks out a retrocompact open subscheme (namely, the locus where a suitable determinant is nonzero) and the proof is complete. $\square$

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