Grothendieck connections and jets

The following question is based on some remarks in section I.2 of Deligne's book Equations Différentielles à Points Singuliers Réguliers.

Let $X$ be a smooth complex variety and $X_1$ the first infinitesimal neighborhood of the diagonal in $X \times X$, so there is a natural morphism $X \to X_1$. If we write $p_1,p_2 : X_1 \to X$ for the two projections, then the first-order jet bundle of a vector bundle $V$ on $X$ is defined to be $J^1(V) = p_{1*} p_2^*V$. Here "upper star" is used in the sense of $\mathcal{O}$-modules. This allows for a convenient way of expressing the notion of a connection: this is just an isomorphism $p_1^*V \to p_2^*V$ which restricts to the identity over $X$, which is the same as an $\mathcal{O}$-linear map $V \to J^1(V)$ such that the composition $V \to J^1(V) \to V$ is the identity.

Here Deligne says something I don't understand: he refers to a first-order differential operator $j^1 : V \to J^1(V)$ "which associates with any section a first-order jet" (I'm translating from the French). What is he talking about? I don't have much intuition for jets, although I know they have something to do with taking Taylor expansions, so any general explanation of that would be greatly appreciated as well.

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This question is a rather old. But maybe this is useful. The map is given on affine schemes by $j^{1} (s) = 1 \otimes f$ in $J^1(V) \cong \mathscr{O}_{X_1} \otimes_\mathscr{O}_{X_0} V$. The point is that expanding $f$ in Taylor series and passing to the left component of the tensor product kills the higher order terms. – user40276 Aug 19 '15 at 1:19

Beware: I'm not an expert on jet bundles, so the following may not answer your question.

The $k$-jet bundle $J^k V$ of $V$ is defined by

$$(J^k V)_x = \mathcal O_x(V) / (\mathfrak m_x^{k+1} \cdot \mathcal O_x(V))$$

where $\mathfrak m_x$ is the unique maximal ideal of the ring $\mathcal O(V)_x$ (this is at a point $x \in X$). This pointwise definition glues to give a holomorphic vector bundle $J^k V$ over $X$.

From the definition, we see there is a natural map

$$j^k : \mathcal O(V) \to J^k V$$

defined by passage to the quotient. My guess is that this is the map $j^k$ that Deligne is referring to.

This is of course well and good, but exhibits the trees rather than the forest. You are absolutely right in that there is a link between jets and Taylor series. In fact, morally speaking, the $k$-jet bundle of $V$ is just the bundle whose sections are Taylor developments of order $k$ of sections of $V$. The map $j^1$ that Deligne refers to is thus just the map which sends a section $\sigma$ to its Taylor development of order 1.

It is easiest to see what is going on in local coordinates. Let's suppose that $V$ is a line bundle and look at 1-jets for simplicity -- everything works the same for vector bundles of arbitrary rank and $k$-jets.

Fix a point $x \in X$, and take coordinates $(z_1, \ldots, z_n)$ centered at $x$. Let $e$ be a holomorphic section of $V$ which trivializes $V$ on our coordinate neighborhood. A section $\sigma$ of $V$ may be written as $\sigma(z) = f(z) \, e(z)$, where $f$ is a holomorphic function. The function $f$ has a Taylor development $f(z) = a_0 + a_1 \, z + O(|z^2|)$ around $x = 0$, thus

$$\sigma(z) = a_0 \, e(z) + a_1 z \, e(z) + O(|z^2|) \, e(z)$$

around $0$. The $1$-jet of $\sigma$ around $x$ is then equal to

$$j^1(\sigma(z)) = a_0 \, e(z) + a_1 z \, e(z).$$

The $1$-jet bundle $J^1 V$ is thus a rank 2 vector bundle over $X$, and the coefficients $a_0$ and $a_1$ define coordinates along the fibers of $J^1 V$ around $x$.

I don't really know of a good reference for jet bundles. The little I know mostly comes from Chapter VII of Demailly's book (http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf). What I wrote here can be found on pages 351--352 of his book. A little later in the Chapter he proves Kodaira's embedding theorem with jet bundles, so that might interest you.

[Edit:] I just noticed that $j^k$ doesn't seem to be a differential operator. However, if we take $j^1$ to be the quotient map, and make it forget the constant term $a_0$, then it is a differential operator. Maybe this is the map Deligne has in mind?

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Thanks for the helpful explanation of the connection with Taylor expansions. I'm still unsatisfied with the definition of $j^1$, partially because this "pointwise definition" of the jet bundle can't be the whole story. After all, one can't in general construct a sheaf just by specifying its stalks. There must be a way to say it with the global definition I gave above: more explicitly, $J^1(V) = p_{2*}(\mathcal{O}_{X_1} \otimes_{\mathcal{O}_X} p_2^{-1}V)$. – Justin Campbell Jun 16 '11 at 22:02
Yes, we still have to say how glue the local trivializations of $J^k V$, but its transition morphisms of are induced by those of $V$. I'm afraid I can't help much with the algebraic definition as I'm just a poor differential geometer, not used to the more sophisticated ways of the world. We're more the "Hulk! Smash!" type of people. – Gunnar Þór Magnússon Jun 17 '11 at 7:34