Let $S$ be a scheme and $\mathscr{E}$ a quasi-coherent sheaf on $S$. Then one can define the vector bundle associated to $\mathscr{E}$, which will be an $S$-scheme $\mathbb{V}(\mathscr{E}) \to S$, thought of as the total space of a vector bundle. (Maybe one should refer to these as generalised vector bundles since they may not admit local trivialisations in general.) Depending on the source, the definition of $\mathbb{V}(\mathscr{E})$ is either ${\mathrm{Spec}}_S \mathrm{Sym} \mathscr{E}$ or ${\mathrm{Spec}}_S \mathrm{Sym} \mathscr{E}^*$ where ${\mathrm{Spec}}_S$ denotes relative spectrum over $S$, $\mathrm{Sym}$ takes free commutative $\mathscr{O}_S$-algebra, and $\mathscr{E}^*$ is the dual of $\mathscr{E}$. For example, the Stacks Project follows the former whilst Vakil follows the latter.

What are some algebraic and/or geometric reasons for considering one definition over the other?

  • $\begingroup$ If $\mathscr{E}$ is reflexive (meaning that its double dual is naturally isomorphic to itself), then the functor of points of $\mathbb{V}(\mathscr{E})$ is maybe more natural in the latter case, because $T$-points are then given by $\mathscr{E}(T)$. $\endgroup$
    – Jef
    Mar 29, 2021 at 19:44
  • 1
    $\begingroup$ When $\mathscr E$ is locally free, it's just a matter of convention: taking duals makes it more geometric in that sections of $\mathbf V(\mathscr E) \to S$ are the sheaf-theoretic sections of $\mathscr E$ (as opposed to quotients $\mathscr E \twoheadrightarrow \mathcal O$). But when you apply this to a coherent sheaf, the operation $(-)^*$ loses information, so you don't want to do that if you want to get a correspondence between sheaves and total spaces. $\endgroup$ Mar 29, 2021 at 20:02
  • $\begingroup$ I think I'm understanding. Using the first definition, we would get the functor of points of $\mathbb{V}(\mathscr{E})$ being $\mathscr{E}^*$, and this is a bit strange in that morally $\mathscr{E}$ encodes the sections of the total space, so we are expecting $\mathscr{E}$ instead? $\endgroup$
    – user577413
    Mar 29, 2021 at 22:47

1 Answer 1


Here is a place where not dualizing is the right thing to do. Let $X\to S$ be a projective morphism over a noetherian base $S$, and let $F$ be a coherent sheaf of ${\mathcal O}_X$-modules on $X$ that is flat over $S$. As is well known (see for example Chapter III Proposition 12.2 of Hartshorne's Algebraic Geometry) over any affine open subscheme $U$ of $S$, there is a Grothendieck semicontinuity complex, which is a complex $0\to E^0 \stackrel{d^0}{\to} E^1 \stackrel{d^1}{\to} E^2 \ldots$ of coherent locally free sheaves of ${\mathcal O}_U$-modules on $U$. The famous Grothendieck Q-sheaf ${\mathcal Q}_U$ for $F$ is defined over $U$ as the cokernel of the dual (transpose) map $(d^0)^{\vee} : (E^1)^{\vee} \to (E^0)^{\vee}$. Even though a Grothendieck semicontinuity complex may not be defined globally over $S$, and is not unique, the sheaves ${\mathcal Q}_U$ defined locally glue together uniquely because of their universal property to define a coherent sheaf ${\mathcal Q}$ of ${\mathcal O}_S$-modules on $S$. The linear scheme (generalized vector bundle) ${\mathbb V}({\mathcal Q}) = {\mathop{\rm Spec}\nolimits}_S {\mathop{\rm Sym}\nolimits}_{{\mathcal O}_S} {\mathcal Q}$, which is made {\it without} taking the dual of ${\mathcal Q}$, represents the contravariant functor on $S$-schemes which to any $T\to S$ associates the group $H^0(X_T, F_T)$. This is due to Grothendieck, and an exposition with references to EGA can be found for example in the Part II of the multi-author book Fundamental Algebraic Geometry - Grothendieck's FGA Explained.

A crucial part of why things must be defined the way they are -- and it works -- is that tensor products are not exact in general but they are right exact, so cokernels do specialize. Moreover, dualizing the locally free sheaves $E^0, \,E^1$ and taking the transpose of the differential $d^0$ does not lose any information.

I should also mention that the scheme ${\mathbb V}({\mathcal Q})$ can be visualized more geometrically as follows. For $i = 0,1$, let $V^i = {\mathop{\rm Spec}\nolimits}_S {\mathop{\rm Sym}\nolimits}_{{\mathcal O}_S} (E^i)^{\vee}$ be the geometric vector bundle on $U$ whose sheaves of sections is $E^i$. These can be regarded as group schemes over $U$. The sheaf homomorphism $d^0$ induces a homomorphism of group schemes $\phi : V^0 \to V^1$. Then the scheme ${\mathbb V}({\mathcal Q}_U)$ is simply the geometric kernel of $\phi$, that is, it is the closed subscheme of $V^0$, defined locally over the base by linear equations in linear coordinates upstairs, which is the schematic inverse image under $\phi$ of the zero section of $V^1$.

Not just ${\mathbb V}({\mathcal Q}) = {\mathop{\rm Spec}\nolimits} {\mathop{\rm Sym}\nolimits} {\mathcal Q}$ but also its projective version ${\bf P}({\mathcal Q}) = {\mathop{\rm Proj}\nolimits} {\mathop{\rm Sym}\nolimits} {\mathcal Q}$ is useful (see for example arXiv:1605.08997v4). Geometrically, it parameterizes the lines (not hyperplanes!) in all fibers of ${\mathbb V}({\mathcal Q})$.

  • $\begingroup$ I would argue that $\mathbb V( \mathcal Q)$ as you've defined it is the correct generalization of Vakil's definition of $\mathbb V(\mathcal F)$ (with duals) to the case where $\mathcal F$ is not locally free (but $X$ is projective). That is: you can interpret your construction as dualizing the sheaf F, in the "correct way." $\endgroup$ Mar 31, 2021 at 15:09

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