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Edit: It seems I had two different constructions mixed up in my head, namely the frame torsor and the automorphism bundle of a vector bundle. This made the main question a bit confusing. The first two auxiliary questions were about the frame torsor, and the last one was about the automorphism bundle. If anyone knows a published reference for either construction, I would still be most appreciative.

The original question is below the line:


I feel like I should have learned this in grad school, but I never encountered a construction.

Let $n$ be a positive integer, and let $f: Y \to X$ be a smooth morphism of schemes of relative dimension $n$. The sheaf $\Omega_{Y/X}$ is then a rank $n$ locally free $\mathcal{O}_Y$-module. Using the symmetric algebra functor, we can form the associated rank $n$ vector bundle $\mathbf{V}(\Omega_{Y/X}) = \operatorname{Spec}_Y \operatorname{Sym}_{\mathcal{O}_Y} \Omega_{Y/X}$ (cf. EGA2 1.7.8). I've heard it called the bundle of 1-jets, which ought to mean tangent bundle, but I'm always confused by this, so maybe it's the cotangent bundle.

Main question: Is there a reference for the construction of the commutative $\mathcal{O}_Y$-algebra $A$ for which $\operatorname{Spec}_Y(A)$ is the $GL_{n,Y}$-torsor $P$ of automorphisms of $\mathbf{V}(\Omega_{Y/X})$? Specifically, I'd like the torsor to satisfy the property that I can retrieve 1-jets by the associated bundle construction: $\mathbf{V}(\Omega_{Y/X}) \cong P \overset{GL_{n,Y}}{\times} \mathcal{O}_Y^{\oplus n}$

This can be viewed as a question about constructing the automorphism torsor of any bundle, but 1-jets seem to have specific structural features that may make a more specialized construction possible. For example, it should be a quotient of some canonical infinite-dimensional torsor of coordinates coming from the Gelfand-Kazhdan formal geometry theory.

Auxiliary questions (not as important):

  1. Is there a concise description of the functor the torsor represents, e.g., are $S$-points on the torsor equal to $S$-points $g:S \to Y$ equipped with isomorphisms $\mathcal{O}_S^{\oplus n} \to g^*\Omega_{Y/X}$?
  2. Is there a nice way to describe the $GL_{n,Y}$-action (since writing an explicit comodule structure sounds like it could be a mess)?
  3. I would be interested in seeing how the torsor can be cut out of the rank $n^2$ bundle of endomorphisms by inverting determinants.
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  • $\begingroup$ $V$ is the space of linear functions on $V^*$, and Sym($V$) is the space of polynomial functions on $V^*$, so that Spec(Sym($V$)) is naturally identified with the dual vector space (or dual vector bundle) to $V$. This explains your tangent/cotangent concern. After replacing $\Omega_{Y/X}$ by its dual, the answer to your auxiliary question 1. is "yes," and to answer question 2. using this description note that the S-points of GL_n,Y are the S-automorphisms of O_S^{\oplus n}. $\endgroup$ Jun 8, 2010 at 23:03
  • $\begingroup$ About 3, pay attention that for a vector bundle $E$ on $Y$, the principal bundle $Frame_Y(E)$ is not the same as the (non principal) $GL_n$-bundle $\operatorname{Aut}_Y(E)$. $\endgroup$
    – Qfwfq
    Jun 9, 2010 at 16:04
  • $\begingroup$ Thank you, unknown(google). I had not seen the notation $Frame_Y(E)$ before, but it makes a lot of sense. I realized that there was some kind of difference after reading Michael Thaddeus's answer. Perhaps I should edit the question to make it clear where my misconceptions lay. $\endgroup$
    – S. Carnahan
    Jun 9, 2010 at 17:52
  • $\begingroup$ The two given answers are not correct and don't give a (global) description of A (the local one is clear anyway). $\endgroup$ Oct 12, 2013 at 8:57
  • $\begingroup$ @MartinBrandenburg Mattia's answer was close enough for me to accept. Instead of $\mathcal{O}_Y$-module Isom (as you suggest), you can also replace Mattia's Isom of $Y$-schemes with Isom of $\mathbb{O}$-modules (as in SGA3 Exp 1) which encodes the vector structure. $\endgroup$
    – S. Carnahan
    Oct 12, 2013 at 10:12

3 Answers 3

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If $V$ is a vector bundle of rank $n$, the corresponding universal algebra $A$ which makes $V$ trivial (i.e. $V \otimes A \cong A^n$), or equivalently the algebra of the corresponding $\mathrm{GL}_n$-torsor, is given by $$A = \mathrm{Sym}(V^n) \otimes_{\mathrm{Sym}(\Lambda^n V)} \mathrm{Sym}^{\mathbb{Z}}(\Lambda^n V).$$ Here, we define $\mathrm{Sym}^{\mathbb{Z}}(\mathcal{L})=\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z}$ for the line bundle $\mathcal{L}=\Lambda^n V$, and the tensor product is taken with respect to the morphism $\delta : \mathcal{L} \to \mathrm{Sym}^n(V^n)$ which maps $v_1 \wedge \dotsc \wedge v_n$ to $\sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} \iota_i(v_{\sigma(i)})$. This description is global in nature and actually generalizes to arbitrary cocomplete linear tensor categories. Some details can be found in my thesis, Section 4.9.

The idea of the construction of $A$ is the following: $\mathrm{Sym}(V^n)$ is the universal algebra $B$ with a morphism of $B$-modules $V \otimes B \to B^n$. Then we construct $B \to A$ so that the determinant of this morphism becomes invertible over $A$, so that $V \otimes A \cong A^n$.

We could also construct $A$ as a quotient of $\mathrm{Sym}(V^n) \otimes \mathrm{Sym}((V^*)^n)$, which introduces morphisms $V \otimes A \to V^n$ and $A^n \to V \otimes A$, and the quotient should be made in such a way that these morphisms become inverse to each other.

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I don't know about jets and you already got an answer regarding the bundle of automorphisms, anyway if you want a $\mathrm{GL}_{n,Y}$-torsor over $Y$ that gives you back your original vector bundle $Z=V(\mathcal{E})=\mathrm{Spec}_Y \mathrm{Sym}(\mathcal{E})\to Y$ when you apply the associate bundle construction with $\mathcal{O}_Y^n$, you should take the bundle of local frames of $Z$, that is $P=\underline{\mathrm{Isom}}_Y(\mathbb{A}^n_Y,Z)\to Y$, where $\underline{\mathrm{Isom}}$ is the scheme representing the sheaf of isomorphisms. This is a $\mathrm{GL}_{n,Y}$-torsor over $Y$ by the action of $\mathrm{GL}_{n,Y}$ on $\mathbb{A}^n_Y$, and if you want a sheaf $\mathcal{F}$ of $\mathcal{O}_Y$-algebras such that $P=\mathrm{Spec}_Y(\mathcal{F})$, it seems reasonable (but i didn't really check) that you can take $\mathcal{F}=\underline{\mathrm{Isom}}_{\mathcal{O}_Y-\text{alg}}(\mathrm{Sym}(\mathcal{E}),\mathcal{O}_Y[x_1,..,x_n])$.

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    $\begingroup$ Just out of curiosity, how do you show that the sheaf of isomorphisms is represented by a scheme? $\endgroup$ Jun 9, 2010 at 8:55
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    $\begingroup$ I think it follows from standard theorems of Grothendieck, but in this case you can use descent for affine morphisms. More precisely, you have a zariski cover, say $U_i$, of $Y$ on which $Z$ is isomorphic to $\mathbb{A}^n$. Over each of these, $\underline{Isom}_Y(\mathbb{A}^n_Y,Z)$ will be isomorphic to $GL_{n,U_i}$, which is affine over $U_i$. Using descent for affine morphisms you get an affine scheme that represents the sheaf $\underline{Isom}_Y(\mathbb{A}^n_Y,Z)$. $\endgroup$ Jun 9, 2010 at 9:55
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    $\begingroup$ On second thought, since we're working with the zariski topology here, calling for descent isn't really necessary.. You just have to glue the various $GL_{n,U_i}$ along the preimages of the intersections $U_i\cap U_j$. $\endgroup$ Jun 9, 2010 at 10:19
  • $\begingroup$ OK, since everything is locally trivial this is very easy, but I meant my question in a broader context: do you know a specific reference where these standard theorems of Grothendieck are found? One could no doubt deduce it by hand from the existence of the Hilbert scheme, but this would be a pain. $\endgroup$ Jun 9, 2010 at 12:38
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    $\begingroup$ There are two errors in this answer. 1. $P$ is the sheaf of isomorphisms of vector bundles over $Y$ (i.e. $Y$-morphisms preserving the linear structure) between the trivial vector bundle and the given vector bundle. Correspondingly, in the end, we don't have to take the sheaf of isomorphisms between the algebras $\mathrm{Sym}(E)$ and $\mathcal{O}_Y[x_1,\dotsc,x_n]$, but rather the sheaf of isomorphisms between the modules $E$ and $\mathcal{O}_Y^n$. 2. This is not the algebra $F$, but rather the sheaf of its sections $U \mapsto \mathrm{Hom}_{\mathsf{Alg}(U)}(F|_U,\mathcal{O}_U)$. But $F=?$ $\endgroup$ Oct 12, 2013 at 8:23
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It seems there is nothing special about the relative Kahler differentials here. One could take any vector bundle V over Y. Then V = Spec Sym V* as you say. By the same token, End V = Spec Sym (V ⊗ V* ). The determinant gives a section det of this sheaf of algebras. Now since GL(n) is cut out in gl(n) by the non-vanishing of the determinant, if we let R = det-1 Sym (V ⊗ V* ) = Sym (V ⊗ V* ) [t]/(t det-1) be the localization of Sym (V ⊗ V* ) at the multiplicatively closed subset generated by det, then Spec R is what you want.

Be careful when calling this a torsor, however; it is not a principal GL(n)- bundle, but rather an adjoint GL(n)-bundle, in which the fibers carry a group structure and there is a canonical identity section.

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  • $\begingroup$ Thank you, Michael. Now I think I have mixed up two constructions, and that the torsor P with the associated bundle property I wanted is not the same as the adjoint $GL_n$-bundle of automorphisms. Is that the case? $\endgroup$
    – S. Carnahan
    Jun 9, 2010 at 1:19
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    $\begingroup$ This cannot be true. If $n=1$, then $R = \oplus_{n \in \mathbb{Z}} V^{\otimes n}$. However, your algebra is $\mathcal{O}[T,T^{-1}]$, which doesn't depend on $V$. I think instead of $\underline{\mathrm{End}}(V) = V^* \otimes V$, we have to consider $\underline{\mathrm{Hom}}(\mathcal{O}^n,V) \cong V^n$, as in Mattia's answer. $\endgroup$ Oct 11, 2013 at 16:34
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    $\begingroup$ But still this cannot work, since for $n=1$ say $\oplus_{d \in \mathbb{Z}} V^{\otimes d}$ is not a localization of $\oplus_{d \in \mathbb{N}} V^{\otimes d}$. $\endgroup$ Oct 12, 2013 at 8:25
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    $\begingroup$ @MartinBrandenburg I'm pretty sure it is a localization of $\mathcal{O}_Y$-algebras. $\endgroup$
    – S. Carnahan
    Oct 12, 2013 at 10:15
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    $\begingroup$ I also believe that the answer is not correct as it stands : one should not be able to reconstruct a vector bundle from its endomorphism bundle. $\endgroup$ Nov 29, 2013 at 14:46

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