The definition of generalised vector bundles over a scheme : dualise or not?

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?

• 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 $\mathscr{E}(T)$.
– Jef
Mar 29 '21 at 19:44
• 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. Mar 29 '21 at 20:02
• 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? Mar 29 '21 at 22:47

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 $${\mathbb V}({\mathcal Q})$$.
• 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." Mar 31 '21 at 15:09