Long time listener, first time caller!


Suppose that I have a locally free sheaf $\mathcal{E}$ on an smooth algebraic variety $X/k$. Let $\Delta^{(1)}\subset X\times X$ denote the first-order neighbourhood of the diagonal, with projection maps $p\_1,p\_2:\Delta^{(1)}\to X$. Then there are a few different ways one can describe a connection on $\mathcal{E}$:


* As a $k$-linear map $D:\mathcal{E}\to\mathcal{E}\otimes\Omega^1\_X$ that satisfies the Leibniz rule.
* As an $\mathcal{O}\_X$-linear splitting $s$ of the first jet bundle exact sequence $0 \to \mathcal{E}\otimes\Omega^1\_X \to J^1(\mathcal{E})\to \mathcal{E}\to 0$. (Recall that $J^1(\mathcal{E}) = p\_{1\ast}p\_2^\ast\mathcal{E}$.)
* As an isomorphism $\phi:p\_1^\ast\mathcal{E}\simeq p\_2^\ast\mathcal{E}$ which restricts to the identity map on the diagonal $\Delta\subset X\times X$.


Passing between these different descriptions of a connection is fairly straightforward: given $D$ one obtains a splitting $s$ by taking $s = D+ j^1$ where $j^1$ is the $k$-linear "jet prolongation" splitting of the jet bundle exact sequence; given a splitting $s$ one obtains an isomorphism $\phi$ by pulling back along $p\_1$ and applying the counit of the $(p\_1^\ast,p\_{1\ast})$ adjunction. So far, so good.


Now, there are various types of "integrability" conditions one might be interested in for a connection:


* **Flatness:** The connection is *flat* if the curvature $F(D)\in\Omega\_X^2(End(\mathcal{E}))$ vanishes.
* **Integrability:** Let $\Delta^{(1)}\_3\subset X\times X\times X$ be the first order neighbourhood of the small diagonal with projections $q\_1,q\_2,q\_3$, and let $\phi\_{ij}:q\_i^\ast\mathcal{E}\simeq q\_j^\ast\mathcal{E}$ be the isomorphisms induced by $\phi$. Then the connection is *integrable* if it satisfies the cocycle condition $\phi\_{23}\circ\phi\_{12} = \phi\_{13}$.
* **Formal lifting:** The connection can be lifted to as isomorphism $p\_1^\ast\mathcal{E}\simeq p\_2^\ast\mathcal{E}$ on every finite order neighbourhood of the diagonal $\Delta^{(n)}$.


My question is as follows: **What is the relation between these three integrability conditions?**


* I believe I was once told that integrability $\Rightarrow$ formal lifting in characteristic zero, although I've never seen a proof (and have thus far failed to cook one up myself).
* I naively thought that there would be a straightforward relation between flatness and integrability, but if my calculations are correct the connection $D=d\_{dR} + t^1dt^2$ on $\mathbb{A}\_k^2$ provides a simple example of an integrable nonflat connection. (Of course, my calculations could always be wrong!)
* The curvature of a connection lies in $\Omega\_X^2(End(\mathcal{E}))$, while the obstruction to lifting from the first- to second-order neighbourhood of the diagonal lives in $H^1(X;\text{Sym}^2(\Omega\_X^1)\otimes\mathcal{E}nd(\mathcal{E}))$ (to get the obstruction to lifting from $\Delta^{(n)}$ to $\Delta^{(n+1)}$ replace $\text{Sym}^2$ with $\text{Sym}^{n+1}$). Should I take this as a hint that flatness and formal lifting are unrelated, or is there some sort of relation between the curvature and this lifting obstruction?




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**Edited to add:**


Since it isn't *explicit* in the answer below or the comments that follow: the correct relation here is that for $X$ smooth and in characteristic zero, flatness is equivalent to formal lifting **together with** the cocycle condition.



Relation between flatness and integrability of an algebraic connection
(In characteristic zero) Flatness implies the other two definitions; integrability and formal lifting are very weak conditions (in fact if I haven't made a mistake I think this notion of integrability is automatic. As such it seems like a weird definition to me; I usually use the words "integrable" and "flat" as synonyms.)


First of all, let me rephrase the notion of flatness a few times. Take $X$ affine for simplicity (to do the general case turn everything into a sheaf, with a little bit of care). Let $\mathcal{D}\_X$ be the ring of differential operators on $X$. It can be presented as the universal envelopping algebroid (over $\mathcal{O}\_X$) of the Lie algebroid $T\_X$ of vector fields. Concretely, this means giving a $\mathcal{D}\_X$-module $M$ is equivalent to giving a $k$-linear map $T\_X\otimes M\rightarrow M,$ so that for two vector fields $v\_1,v\_2$ and $m\in M$, we have $v\_1\cdot(v\_2\cdot m)-v\_2\cdot(v\_1\cdot m)=[v\_1,v\_2]\cdot m,$ where $\cdot$ denotes the action of $T\_X$ on $M$.


I claim that, if $M=\mathcal{E}$, this is exactly the data of a flat connection on $\mathcal{E}$. The map $T\_X\otimes M\rightarrow M$ is equivalent to a map $M\rightarrow M\otimes\Omega^1\_X$, aka the map defining our connection. The expression


$$v\_1\cdot(v\_2\cdot m)-v\_2\cdot(v\_1\cdot m)-[v\_1,v\_2]\cdot m$$


defines a map $T\_X\otimes T\_X\otimes M\rightarrow M$. It can be checked that this map is $\mathcal{O}\_X$-multilinear and antisymmetric on the first two , so it actually lies in


$$\operatorname{Hom}\_{\mathcal{O}\_X}(\Lambda^2T\_X\underset{{\mathcal{O}\_X}}{\otimes}M,M)\cong\Omega^2\_X(\operatorname{End}(M))$$


and you can check that this is the curvature of your connection. So the connection is flat exactly when this gives a $\mathcal{D}\_X$-module structure.


Let $\hat{\Delta}$ and $\hat{\Delta}\_3$ be the formal neighborhoods of $\Delta\subseteq X\times X$ and $\Delta\_3\subseteq X\times X\times X$. Then a map $\mathcal{D}\_X\underset{\mathcal{O}\_X}{\otimes}M\rightarrow M$ is equivalent to the data of an isomorphism $\hat{\phi}:\hat{p}\_1^\*M\cong\hat{p}\_2^\*M$ over $\hat{\Delta}$. The action of $\mathcal{D}\_X$ on $M$ is compatible with the algebra structure on $\mathcal{D}\_X$ if and only if $\hat{\phi}$ satisfies a cocycle condition. In particular we see that flatness implies your other conditions.


To prove these statements, the key is the following. A map $\hat{p}\_1^\*M\rightarrow\hat{p}\_2^\*M$ is equivalent to a map $M\rightarrow\hat{p}\_{1\*}\hat{p}\_2^\*M\cong\mathcal{O}\_{\hat{\Delta}}\otimes\_{\mathcal{O}\_X}M.$ (Note here that we have a left and a right $\mathcal{O}\_X$ action on $\mathcal{O}\_\hat{\Delta}$. The right one is absorbed into the tensor product, while the left one defines the $\mathcal{O}\_X$-module structure on $\mathcal{O}\_{\hat{\Delta}}\otimes\_{\mathcal{O}\_X}M$.) Furthermore, convolution defines a coalgebra structure on $\mathcal{O}\_{\hat{\Delta}}.$ If one dualizes this coalgebra, one gets exactly $\mathcal{D}\_X$, and you can use this to prove the above interpretation of $\mathcal{D}$-modules. (This is a little painful and has a number of details that need to be checked, so I am avoiding doing it...)


Now let me explain why the other conditions are very weak. For formal lifting, as you note, the obstruction lives in $H^1$, so it automatically vanishes if $X$ is affine. I believe it can be a nontrivial condition when $X$ is not affine.


On the other hand, here is an argument that all connections are integrable in your sense. We would like to check that the map $\phi\_{31}\circ\phi\_{23}\circ\phi\_{12}$ is the identity over $\Delta\_3^{(1)}$. This is true on the diagonal. It is also true on the image of any of the three natural embeddings of $\Delta^{(1)}\rightarrow\Delta\_3^{(1)}$. I claim this is sufficient to see that it is the identity over all of $\Delta\_3^{(1)}$.


Let $N\_{123}$ be the normal bundle of $\Delta\_3$ inside $X\times X\times X$. We let $N\_{12}$ be the subbundle given by the normal vectors inside $\Delta\times X$. Define $N\_{13}$ and $N\_{23}$ similarly.


An endomorphism of the pullback of $\mathcal{E}$ over $\Delta\_3^{(1)}$ which restricts to the identity on $\Delta\_3$ can be identified with a map $N\_{123}\otimes\mathcal{E}\rightarrow\mathcal{E}$, and the original endomorphism is the identity if and only if this resulting map is zero. So we see that the maps $N\_{ij}\otimes\mathcal{E}\rightarrow\mathcal{E}$ must be trivial. As the map $N\_{12}\oplus N\_{23}\oplus N\_{31}\rightarrow N\_{123}$ is surjective, this implies that $N\_{123}\otimes\mathcal{E}\rightarrow\mathcal{E}$ is trivial, as desired.


