I am starting to study vector bundles over schemes and I have encountered two different definitions of the vector bundle associated to a locally free sheaf. I tried to understand why this was the case, but the reason eludes me. Eisenbud and Harris say more or less the following about it in 3264 and all that (they say it about the projectivization of vector bundles, but I assume the comment applies equaly good to vector bundles themselves):

Some sources define the vector bundle associated to a locally free sheaf of $\mathcal{E}$ to be $\text{Spec}(\text{Sym}(\mathcal{E}))$ instead of $\text{Spec}(\text{Sym}(\mathcal{E}^{\vee }))$. This convention is better adapted to the generalization from locally free sheaves to arbitrary coherent sheaves.

What is the precise meaning of this last sentence? Is there an explicit example showing this phenomenon?

A vector bundle of rank $r$ over a schene $B$ is a scheme $E$ together with a morphism $\pi \colon E\to B$ together with local trivialisations $\pi^{-1}(U_{i})\cong U_{i}\times \mathbb{A}^{r}$ with the corresponding conditions on the transition maps (restricted to any affine open $V=\text{Spec}(A)\subseteq U_{i}\cap U_{j}$ they are given by a linear automorphism of $A[x_{1},...,x_{r}]$.

My attempt to understand this:

I tried to think of this in the case of vector bundles over a smooth affine variety over $\mathbb{C}$. For example, over $\mathbb{A}^{1}=\text{Spec}(\mathbb{C}[x])$. The closed points of $\mathbb{A}^{1}$ correspond then to maximal ideals in $\mathbb{C}[x]$, which is the ring of regular functions on $\mathbb{A}^{1}$. Then suppose we have the trivial rank $2$ vector bundle $\pi \colon E=\mathbb{A}^{1}\times \mathbb{A}^{2} \to \mathbb{A}^{1}$. Then there is a natural locally free sheaf to consider, namely the sheaf $\mathcal{E}$ of local sections of $\pi$. A closed point of $E$ is a maximal ideal in $\mathbb{C}[x,y,z]$ and a global section of $\mathcal{E}$ is a morphism $\mathbb{A}^{1}\to \mathbb{A}^{1}\times \mathbb{A}^{2}$, which (I think) corresponds to the choice of two global sections $f(x)$ and $g(x)$ in $\mathcal{O}_{\mathbb{A}^{1}}$. So it seems to me that the ring of global sections is isomorphic to the ring $(\mathbb{C}[x])^{2}$. Hence the global sections of the corresponding symmetric algebra $\text{Sym}_{\mathcal{O}_{\mathbb{A}^{1}}(\mathbb{A}^{1})}(\mathcal{E}(\mathbb{A}^{1}))=\text{Sym}_{\mathbb{C}[x]}(\mathbb{C}[x]^{2})=\mathbb{C}[x][y,z]=\mathbb{C}[x,y,z]$ is just $\mathcal{O}_{E}(E)$. Hence, closed points in $E$ really correspond to maximal ideals in $\text{Spec}(\text{Sym}(\mathcal{E}))$, which was supposed to be the less natural definition (at least the less classical according to Eisenbud and Harris). I suspect that I made a mistake somewhere in this reasoning, because Eisenbud and Harris suggest that the points of this scheme should correspond to the fibers, and not to points of the vector bundle (and because I am not really familiar with any of the concepts that I am using). I suspect that a mistake could be that I don't see the difference between $\text{Sym}(V)$ and $\text{Sym}(V^{\vee})$, besides covariant and contravariant. I tried to find out the difference here, but I still cannot see any clear difference. In that question, a user suggests that bigger differences appear on finite fields. Can anybody show this with an example?

And how would I describe $\text{Spec}(\text{Sym}(\mathcal{E}^{\vee}))$ in an equally explicit way? In particular, the place where I get stuck is when trying to describe the group of $\mathcal{O}_{\mathbb{A}^{1}}$-module morphisms from $\mathcal{E}$ to $\mathcal{O}_{\mathbb{A}^{1}}$ (but I have the feeling that there should be an easy way to describe them that I am just not seeing right now).


2 Answers 2


I spent a lot of time during my masters (and still now) trying to resolve such questions which frequently appear : "Should I apply the sheaf or its dual here? Which directions should my arrows go?" and the reason why it's so confusing is because this question has not been settled once and for all ; people keep arguing about it and there is no widely accepted consensus. I am personally trying to fix it at least for myself, and I think that in the case of this particular question, I have done it.

So here's the deal. If we are to choose between $\mathcal E$ and $\mathcal E^{\vee}$, our choice should at least make the algebra work. So we place ourselves in the most general setting where we can think of having the informal equality

Vector bundle = Locally free sheaf of finite constant rank

and we go on from there. I began by inspiring myself from differential geometry and worked in the category of vector bundles over an arbitrary manifold. In that case, a morphism of vector bundles $(E \to X) \to (E' \to X')$ is a morphism of manifolds $f : X \to X'$ paired with a morphism of vector bundles $f^{\sharp} : E \to f^* E'$, the pullback bundle. Another way to phrase this is that we have a commutative square formed by $f : X \to X'$ and $g : E \to E'$ such that the morphism is linear on fibers, but that's a differential-geometric condition so we won't look at that and focus on the other one instead, which resembles more the algebro-geometric setting.

So if we are to push the analogy to algebraic geometry, we should work the same way : a morphism of algebraic vector bundles $(E \to X) \to (E' \to X')$ should be a morphism of schemes $f : X \to X'$ and a morphism $g : E \to f^* E' = X \times_{X'} E'$ satisfying some properties (those properties are not important for the direction of the arrows or knowing whether we should put $^{\vee}$ or not).

Now to each vector bundle $\pi : E \to X$, we associate the locally free sheaf of sections $\Gamma_{E/X}$, which is literally what it is : for an open set $U$, $\Gamma_{E/X}(U)$ is the collection of sections $U \to \pi_E^{-1}(U)$. I won't discuss the $\mathcal O_X$-module structure of this sheaf for the moment. In the world (read: category) where we see a vector bundle as a locally free sheaf, we need to identify morphisms of vector bundles with morphisms of locally free sheaves.

But what is a morphism of "locally free sheaves" $(X, \mathcal E) \to (X',\mathcal E')$? Well, we have defined morphisms of schemes $(X, \mathcal O_X) \to (X', \mathcal O_{X'})$ via a continuous map $f : X \to X'$ and a morphism of sheaves $f^{\sharp} : \mathcal O_{X'} \to f_* \mathcal O_X$. If we use the $f^{-1}/f_*$ adjunction on this (because we expect pushforward to behave badly on bundles), this is the morphism $f^{\flat} : f^{-1} \mathcal O_{X'} \to \mathcal O_X$. In other words, a morphism of schemes is a continuous map and a morphism of sheaves which tells you how to pullback sections. After base extension, this gives you the isomorphism $f^{\flat} : f^* \mathcal O_{X'} \to \mathcal O_X$.

Why don't we do the same for modules?

Definition. A morphism of "locally free sheaves" $(X, \mathcal E) \to (X', \mathcal E')$ is a pair $(f, f^{\flat})$ where $f : X \to X'$ is a morphism of schemes and $f^{\flat} : f^{-1} \mathcal E' \to \mathcal E$ is a morphism of $f^{-1}\mathcal O_{X'}$-modules, or equivalently, a morphism $f^{\flat} : f^* \mathcal E' \to \mathcal E$.

But... wait! For vector bundles, we have a morphism $E \to f^* E'$, and for locally free sheaves, we have a morphism $f^* \mathcal E' \to \mathcal E$! We know that morally, the case of vector bundles is going in the "correct" direction. So the issue now is how to get from the category of vector bundles to the category of locally free sheaves and get that arrow pointing the right way.

Suppose we start with a morphism of locally free sheaves $(X, \mathcal E) \to (X', \mathcal E')$ with $f^{\flat} : f^* \mathcal E' \to \mathcal E$. Applying the symmetric algebra functor, we get $$ \mathrm{Sym}(f^{\flat}) : \mathrm{Sym}(f^* \mathcal E') \to \mathrm{Sym}(\mathcal E), $$ and applying the contravariant relative $\mathrm{Spec}$ functor, we obtain $$ \mathbf{Spec}(\mathrm{Sym}(f^{\flat})) : \mathbf{Spec}(\mathrm{Sym}(\mathcal E)) \to \mathbf{Spec}(\mathrm{Sym}(f^*\mathcal E')) $$ If we wish to identify $\mathcal E$ or $\mathcal E^{\vee}$ to the vector bundle $E$, one sees here that applying $^{\vee}$ to $f^{\flat}$ before applying $\mathbf{Spec}(\mathrm{Sym}(-))$ would get the arrow in the wrong direction! So we should not apply $^{\vee}$, and the correct functor is $\mathbf{Spec}(\mathrm{Sym}(-))$ from locally free sheaves to vector bundles.

In the other direction, if we start with a morphism of vector bundles $(E \to X) \to (E' \to X')$, we have a morphism of vector bundles $E \to f^* E'$ over $X$, so given a section $s : X \to E$, we can compose it with $E \to f^*E'$ and obtain a section $f^* s : X \to E \to f^* E'$, i.e. we have a morphism of sheaves $\Gamma_{E/X} \to \Gamma_{X/f^*E'} = f^* \Gamma_{E'/X'}$. Now this morphism does not make us happy, it is in the wrong direction! So we should apply duals on it to get it back right. (Duals commute with pullback; you need to worry about sheaf $\mathrm{Hom}$s, but this is a tractable issue.)

So the identification "Vector bundle = Locally free sheaf" is given by $$ \mathcal E \mapsto \mathbf{Spec}(\mathrm{Sym}(\mathcal E)), \quad E \mapsto \Gamma_{E/X}^{\vee}. $$ It is the only way to do it that categorically makes sense. If I remove or add $^{\vee}$ to any one of those functors, the constructions make no sense anymore.

A small remark; if you are not working with vector bundles (but say with coherent sheaves) and you still use $\mathbf{Spec}(\mathrm{Sym}(-))$ from $\mathfrak{Coh}(X)$ to $\mathbf{Sch}_X$, instead of using the functor $\Gamma_{E/X}$ or $\Gamma_{E/X}^{\vee}$, it is better to use $\pi_*(-)_1$, the degree $1$ part of the pushforward (the pushforward is a sheaf of graded $\mathcal O_X$-algebras). In this case, you also have an adjunction $$ \mathcal M \mapsto \mathbf{Spec}(\mathrm{Sym}(\mathcal M)), \quad E \mapsto (\pi_* \mathcal O_E)_1 $$ I don't know of anyone who does this though (work with coherent sheaves), so I don't know how useful it is or what one does with that.

EDIT : As for your attempt at understanding, consider the following. Given a vector bundle $\pi : E \to X$, a section of this vector bundle $s : X \to E$ is just a morphism of schemes satisfying $\pi \circ s = \mathrm{id}_X$. This $s$ is a morphism of affine $X$-schemes $X \to E$ which we can re-write as a morphism $$ \mathbf{Spec}(\mathrm{Sym}(\mathcal O_X)) \simeq X \to E \simeq \mathbf{Spec}(\mathrm{Sym}(\Gamma_{E/X}^{\vee})). $$ Note that those isomorphisms are natural by the adjunction I described above.

Since everything is affine, the above morphism corresponds to a morphism of graded $\mathcal O_X$-algebras $\mathrm{Sym}(\Gamma_{E/X}^{\vee}) \to \mathrm{Sym}(\mathcal O_X)$. Both sheaves are finitely generated $\mathcal O_X$-algebras in degree $1$, so this corresponds to a morphism of $\mathcal O_X$-modules $\Gamma_{E/X}^{\vee} \to \mathcal O_X$, i.e. a global section of the sheaf $$ \mathscr H\!\mathit{om}_{\mathcal O_X}(\Gamma_{E/X}^{\vee},\mathcal O_X) = \Gamma_{E/X}^{\vee \vee} \simeq \Gamma_{E/X}. $$ (these isomorphisms are again natural).

Going back to your example, the sheaf of global sections should be $\mathbb C[x]^{\oplus 2}$, not $\mathbb C[x]^2$; you pick two global sections in $\mathbb C[x]$ via a morphism of $\mathcal O_{\mathrm{Spec}(\mathbb C[x])}$-modules $\widetilde{\mathbb C[x]}^{\oplus 2} \to \widetilde{\mathbb C[x]}$ and you apply $\mathbf{Spec}(\mathrm{Sym}(-))$ on it. The reason why you need a morphism of $\mathbb C[x]$-modules and not a morphism of $\mathbb C[x]$-algebras (as you did and erroneously obtained $\mathbb C[x]^2$) is explained along the lines of my proof.

The essential detail that you skipped, geometrically speaking, is that the corresponding morphism of vector bundles must be linear on the fibers; this is all captured in the properties of the $\mathbf{Spec}(\mathrm{Sym}(-))$ functor since it builds the corresponding morphism on a morphism of $\mathcal O_X$-modules (which is a linear construction, not a polynomial one).

But after all, $\mathbb C[x]^2$ and $\mathbb C[x]^{\oplus 2}$ are equal as sets, so you weren't that far off. You just weren't looking for a ring of global sections, but for an $\mathcal O_X$-module of sections.

Hope that helps,

  • $\begingroup$ It does help! Thanks a lot $\endgroup$
    – Pedro
    Jan 9, 2018 at 9:16

As a starting point let's agree that what we want is that the resulting locally free sheaf should be obtained as a sheaf of sections of whatever geometric construction we begin with.

I propose the following geometric definition of vector bundle:

First, we can define a vector space object in the category of schemes over $k$ similar to this construction. Then one can check that $W=Spec \ Sym^{\bullet}_k \ k^{n}$ is an example of one.$V=Spec \ Sym^{\bullet}_k \ (k^{n})^{\lor}$ is another. Let's stick to the latter one for the moment.

Second, let the trivial bundle be $X \times_{Spec \ k} V \to X$, which affine locally amounts to $Spec \ A \otimes_{k} Sym^{\bullet}_k \ (k^{n})^{\lor} \to Spec \ A$, where the domain can be identified with $Spec \ Sym^{\bullet}_A (A^{\oplus n})^{\lor}$. Notice the the fiber over each $p \in X$ is $Spec \ Sym^{\bullet}_{k(p)} (k(p)^{\oplus n})^{\lor}$, so is a vector space over $k(p)$ in a natural way.

Third, define locally trivial vector bundle to be the map $Y \to X$ which locally looks like the previous one (and maybe such that each fiber has a vector space structure, compatible with the trivializations as vector space objects).

Now how do sections look like for the above construction?

Locally, a section should be a map $Spec \ A \to Spec \ Sym^{\bullet}_A (A^{\oplus n})^{\lor}$ which gives identity after we further map to $Spec \ A$. Algebraically this corresponds to a map of rings $Sym^{\bullet}_A (A^{\oplus n})^{\lor} \to A$ which restricts to identity on $A$, that is an $A$-algebra homomorphism. By the universal property of $Sym$ such homomorphisms are in bijection with linear maps $(A^{\oplus n})^{\lor} \to A$, which are just elements of $(A^{\oplus n})^{\lor \lor} \sim A^{\oplus n}$.

Thus it seems more consistent to take $\mathcal{Spec} \ Sym^{\bullet} \mathcal{F^{\lor}}$ as the total space for a locally free sheaf, so that as its space of sections we get back $\mathcal{F}$.

This is also consistent in the following sense: map of vector bundles $Y_1 \to Y_2$ over $X$ should correspond to the map of sections in the same direction.

$\mathcal{Spec} \ Sym^{\bullet} \mathcal{F_1^{\lor}} \to \mathcal{Spec} \ Sym^{\bullet} \mathcal{F_2^{\lor}}$ corresponds to graded $Sym^{\bullet} \mathcal{F_2^{\lor}} \to Sym^{\bullet} \mathcal{F_1^{\lor}}$. This in turn gives a map $\mathcal{F_2}^{\lor} \to \mathcal{F_1}^{\lor}$, which finally gives $\mathcal{F_1} \to \mathcal{F_2}$.

I guess this phrase

This convention is better adapted to the generalization from locally free sheaves to arbitrary coherent sheaves.

refers to the fact that taking duals might blow up our module so that there is no hope to have $M^{\lor \lor} \sim M$. This is obviously a non-issue in the locally free + finite rank case. So there is really no difference whether to work with sheaf of sections or its dual. As Patrick has outlined, maybe it is even better from a purely categorical standpoint.


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