Let $k$ be an algebraically closed field of characteristic zero. Let $S$ be a scheme of finite type over $k$. Let $\mathrm{Sch}/S$ be the category of schemes of finite type over $S$. Let $\mathcal F$ be a coherent sheaf on $S$. Consider the following functor $$ \mathbf{Dual}_{\mathcal F}:(\mathrm{Sch}/S)^{\mathrm{op}}\to \mathrm{Set},\quad \big[f:T\to S\big]\mapsto \mathrm{Hom}_{\mathcal O_T}(f^*\mathcal F,\mathcal O_T) $$ where the functor on morphisms is naturally defined.

We can recover the dual $\mathcal F^\vee:=\mathcal Hom_{\mathcal O_S}(\mathcal F,\mathcal O_S)$ from $\mathbf{Dual}_{\mathcal F}$ by looking at open subsets $[U\subset S]\in\mathrm{Sch}/S$. However more is encoded in the functor. For example, if $S=\mathbb A^1$ and $\mathcal F$ is the skyscraper sheaf at the origin, then $\mathcal F^\vee=0$ and $\mathbf{Dual}_{\mathcal F}(0\hookrightarrow \mathbb A^1)\cong k$.

I want to ask

To what extent can we recover $\mathcal F$ from $\mathbf{Dual}_{\mathcal F}$?

A more specific question that I am interested in currently is

Can we characterise $\mathcal F$ being locally free in terms of $\mathbf{Dual}_{\mathcal F}$?

  • 1
    $\begingroup$ That is actually a functor to $\mathbb{A}^1_S$-modules that is representable by an affine morphism over $S$ with an action of $\mathbb{A}^1_S$. The pushforward of the structure sheaf is a quasi-coherent $\mathcal{O}_S$-module with an action of $\mathbb{G}_m$ (inherited from the structure of $\mathbb{A}^1$-module). The first nontrivial $\mathbb{G}_m$-eigensheaf of the pushforward of the structure sheaf equals $\mathcal{F}$. The coherent sheaf $\mathcal{F}$ is locally free if and only if the affine morphism to $S$ is flat. $\endgroup$ May 4 at 15:21
  • $\begingroup$ @JasonStarr I do not understand you. If I consider $S=\mathrm{Spec}(A)$ and $T=\mathrm{Spec}(B)$ and $f:T\to S$ corresponding to $A\to B$, then the pushforward of the structure sheaf is $B$, as an $A$-module with an action of $\mathbb G_m$? $\endgroup$ May 4 at 15:27
  • $\begingroup$ Push forward the structure sheaf for the affine morphism that represents the functor. You will get the symmetric algebra on $\mathcal{F}$ with its standard grading corresponding to the action of $\mathbb{G}_m$. The first nontrivial eigensheaf is just the first graded piece of the symmetric algebra, which equals $\mathcal{F}$. $\endgroup$ May 4 at 15:29
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    $\begingroup$ Isn't the functor you are studying represented by $\mathcal{Spec}_S(\mathrm{Sym}(\mathcal{F}))$? This is the relativised "Spec" construction over $S$ applied to the sheaf of $\mathcal{O}_S$ algebras given by the sum of symmetric powers of $\mathcal{S}$.. OK I just realised that Jason Starr already said this! $\endgroup$
    – Kapil
    May 4 at 16:52
  • 1
    $\begingroup$ @Kapil Yes, I just realized this after Jason's comments. $\endgroup$ May 4 at 16:53

1 Answer 1


I am just posting my final comment as one answer. The set-valued functor $\textbf{Dual}_{\mathcal{F}}$ is representable by an affine $S$-scheme that is canonically isomorphic to the relative Spec over $S$ of the quasi-coherent sheaf of graded $\mathcal{O}_S$-algebras, $$\bigoplus_{d\geq 0} \text{Sym}^d_{\mathcal{O}_S} \left(\mathcal{F}\right).$$ In particular, the sheaf of relative differentials over $S$ for this affine $S$-scheme is canonically isomorphic to the pullback of $\mathcal{F}$. Thus, for every section over $S$ of the affine $S$-scheme, the pullback of the sheaf of relative differentials is isomorphic to $\mathcal{F}$. Of course there are sections, e.g., the zero section corresponding to the zero homomorphism from $\mathcal{F}$ to the structure sheaf.

  • $\begingroup$ would you please explain the connection to the sheaf of relative differentials? $\endgroup$ Jun 17 at 0:13
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    $\begingroup$ @PrimeRibeyeDeal For a scheme $S$ and a quasi-coherent sheaf of $\mathcal{O}_S$-modules $\mathcal{F}$, for the relative Spec $S$-scheme associated to the quasi-coherent sheaf of $\mathcal{O}_S$-algebras as above, the sheaf of relative differentials is canonically isomorphic to the pullback from $S$ of $\mathcal{F}$. $\endgroup$ Jun 21 at 11:10
  • $\begingroup$ thank you. Can you point me to a reference for this fact? I haven't found it stated so cleanly anywhere. $\endgroup$ Jun 22 at 14:17
  • 1
    $\begingroup$ @PrimeRibeyeDeal It is formula on p. 118 of "Complexe cotangent et d'eformations, I" by Luc Illusie. $\endgroup$ Jun 24 at 13:13

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