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Let $B\to X$ be a surjective submersion over the smooth integral scheme $X$ over $\mathbb{C}$. Associated to this we have in the $C^\infty$ world the notion of the $k$ jet-bundles $J_k(B)$, which are affine bundles over $X$. I wonder what the best way to define this notion in the setting of algebraic geometry is. If $B=TX$ there is an algebraic description on Wikipedia which makes sense, but I cannot figure out how to give a satisfactory definition in general.

I am fine with restricting to the case where $B$ is the spectrum of a sheaf of $\mathcal{O}_X$ algebras.

Edit: I am talking here about jet bundles in the sense of e.g. this wikipedia article.

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    $\begingroup$ Please state the definition of "jet bundle" you use (there are multiple interpretations in algebraic geometry, e.g., truncated arc spaces and principal parts bundles). $\endgroup$ Commented Aug 16, 2020 at 20:57
  • $\begingroup$ Is this related: ncatlab.org/nlab/show/arithmetic+jet+space? $\endgroup$
    – Emily
    Commented Aug 16, 2020 at 21:01
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    $\begingroup$ @Emily Thanks. I will need to read up on some of things written there to be to judge to what extend that page suggests the answer to my question, but it definitely seems related.. $\endgroup$ Commented Aug 16, 2020 at 21:08
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    $\begingroup$ These are a (relative) version of "truncated arc spaces". One analogue in algebraic geometry of the Cartan extension method mentioned in that wikipedia article is Chapter 3 of the PhD thesis of Jan Gutt: math.stonybrook.edu/alumni/2013-Jan-Gutt.pdf $\endgroup$ Commented Aug 16, 2020 at 23:07
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    $\begingroup$ I found the following short manuscript from Ravi Vakil quite enlightening: math.stanford.edu/~vakil/files/jets.pdf $\endgroup$ Commented Aug 24, 2020 at 22:12

2 Answers 2

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Let $S$ be a scheme, e.g., $\text{Spec}\ \mathbb{C}$. Let $f:X\to S$ be a morphism that is separated and smooth. Denote the associated relative diagonal morphism by $$\Delta_{X/S}:X \to X\times_S X.$$ This is a closed immersion whose ideal sheaf $\mathcal{I}$ is everywhere locally generated by a regular sequence. For every integer $e\geq 0$, the ideal sheaf $\mathcal{I}^{e+1}$ is the defining ideal sheaf of a closed subscheme of $X\times_S X$. Denote this closed subscheme by, $$\Delta_{e,X/S}:X_e \to X\times_S X.$$ Because $f$ is smooth, each associated projection morphism to $X$ is a finite and flat morphism (in particular, it is proper), $$p_i: X_e \to X, \ \ i=1,2.$$ By representability of the functor from Part IV.4.c, p. 267 (p.20 of the NUMDAM edition) in Fondements de la Géometrie Algébrique, for every scheme $X$, for every flat and projective morphism, $$p:Y\to X,$$ for every finitely presented, quasi-projective morphism, $$q:Z\to Y,$$ there exists a universal pair, $$(r:\Pi_{Z/Y/X} \to X, \ s:\Pi_{Z/Y/X}\times_X Y \to Z),$$ of a morphism from a scheme $T$ to $X$ and a $Y$-morphism from $T\times_S Y$ to $Z$. In particular, for the flat and finite morphism $p_2$ from $X_e$ to $X$, for every smooth, quasi-projective morphism $q$ from a scheme $Z$ to $X_e$, there is such a pair, $$(r:\Pi_{Z/X_e/X} \to X, \ s:\Pi_{Z/X_e/X}\times_X X_e \to Z).$$ Finally, for every finitely presented, quasi-projective morphism, $$\pi:B\to X,$$ the base change morphism is also finitely presented and quasi-projective, $$B\times_{X,\text{pr}_1} (X\times_S X) \to X\times_S X.$$ Thus, the pullback of this morphism over the closed subscheme $X_e$ is also a finitely presented and quasi-projective morphism. Denote this pullback by $$\pi_e:B_e\to X_e.$$ The "relative truncated sections" parameter space is the universal pair, $$(r:\Pi_{B_e/X_e/X} \to X,\ s:\Pi_{B_e/X_e/X}\times_X X_e \to B_e).$$ If the morphism $q$ is smooth, then every "truncated section" parameterized by $\Pi_{B_e/X_e/X}$ extends to a formal section by Hensel's lemma.

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  • $\begingroup$ Thank you for this detailed answer. Can you provide one more detail by giving a more precise reference for this construction of Grothendieck? I realise that this might be common knowledge, but I do not know this. $\endgroup$ Commented Aug 17, 2020 at 8:22
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    $\begingroup$ That is a good point. I added a page number in FGA for the construction. $\endgroup$ Commented Aug 17, 2020 at 16:50
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According to Mustață, Jet schemes of locally complete intersection canonical singularities, the $m$-th jet scheme of (the $\mathbb{C}$-scheme) $X$ is the scheme $J_m X$ over $X$ representing the functor $\mathrm{Sch} \to \mathrm{Set}$ given by

$$ S \mapsto \mathrm{Hom}(\Delta^m\times S,X)$$

where $\Delta^m:=\mathrm{Spec}\frac{\mathbb{C}[t]}{\langle t^{m+1} \rangle}\;.$

So the closed points of $J_m X$ are the morhisms of schemes $\Delta^m\to X$. And the fiber of $J_m X\to X$ over $x\in X$ has as set of closed points the set $\mathrm{Hom}(\mathcal{O}_{X,x}\;,\mathbb{C}[t]/\langle t^{m+1} \rangle)$.

Another way of seeing it is: let $\mathbf{F}:\mathrm{Sch}\to\mathrm{Sch}$ be the functor $\mathbf{F}(S)=\Delta^m\times S$, than it has a right adjoint given by $X\mapsto J_m X$.

Remark. According to the above construction we have $J_1 X=TX$, the tangent bundle (or total space of tangent sheaf if $X$ isn't smooth).

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  • $\begingroup$ This is only a special case of the type of jet bundle that I am talking about, corresponding to the surjective submersion $X\times \Delta\to \Delta$. $\endgroup$ Commented Aug 17, 2020 at 11:46
  • $\begingroup$ I was about to add that mine is a pocket version of Jason Starr's answer. But now that I read better, you're right, it actually doesn't answer your question, which is a request for the relative version. I'll still leave my answer there, as might be useful as reference on a version of jet spaces for future readers. $\endgroup$
    – Qfwfq
    Commented Aug 17, 2020 at 11:53
  • $\begingroup$ To be honest, your planned addendum on how this relates to Jason Starr's answer would still be very helpful for me, since I am familiar with the jet spaces as described in your answer, while I'm having some trouble relating this this the constructino on Jason Starr. $\endgroup$ Commented Aug 17, 2020 at 11:55
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    $\begingroup$ (The comment that I wrote wasn't correct, so I deleted it) $\endgroup$
    – Qfwfq
    Commented Aug 17, 2020 at 12:56
  • $\begingroup$ I am having trouble seeing how this is a special case of Jason Starr's answer. Does anyone know? $\endgroup$
    – Ben C
    Commented Feb 14, 2023 at 18:41

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