# Relation between flatness and integrability of an algebraic connection

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?

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.

• Dear @derryberry, in the second paragraph, why must $\phi$ be an isomorphism? Feb 12 at 22:33

(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.

• Thanks! I think that the first part of your answer might hold the key to the problem, but that perhaps your argument that integrability is a weak condition might not be right. (Also I got a message from someone claiming that in characteristic 0 the three conditions are almost equivalent -- apparently I need to add a cocycle condition to the formal lifting property.) First, using that $\mathcal{D}_X^{\leq n}$ is dual to $J^n(\mathcal{O})$ ($n$-jets) I'm pretty sure one can show that $\text{Hom}_X(\mathcal{D}_X\otimes\mathcal{E},\mathcal{E})=\text{Hom}(\mathcal{E},J^\infty(\mathcal{E}))$... Feb 1 at 16:09
• ...and this is equal to $\text{Hom}_{\hat{\Delta}}(\hat{p}_1^\ast\mathcal{E},\hat{p}_2^\ast\mathcal{E})$. I'm not sure how to see that compatibility with multiplication in $\mathcal{D}$ corresponds to a cocycle condition here; possibly the easiest way to see that is by dualising the convolution coalgebra structure that you mentioned? In the other directions, first suppose that one has a formal lift that satisfies the cocycle condition -- by what we've just discussed that gives a $\mathcal{D}$-module stucture, and so the connection is flat. Feb 1 at 16:18
• Next, suppose that the original $\phi$ satisfies the cocycle condition. Dualising 1-jets, this corresponds to a map $\mathcal{D}_X^{\leq1}\otimes\mathcal{E}\to\mathcal{E}$. I want to say that the cocycle condition still means that this action map is compatible with the multiplication in $\mathcal{D}$ -- but now that multiplication takes $\mathcal{D}^{\leq1}\otimes\mathcal{D}^{\leq1}\to\mathcal{D}^{\leq2}$. $\mathcal{D}_X$ is generated by its first filtered part, so iterating this procedure gives us a map $\mathcal{D}_X\otimes\mathcal{E}\to\mathcal{E}$ still compatible with multiplication. Feb 1 at 16:28
• (This obviously isn't a fully rigorous argument yet.) Finally, I'm not sure where the error in your argument might be, but I confess to being unsure of exactly what the definitions of the subbundles $N_{ij}$ are, and (with my best guess) why the maps $N_{ij}\otimes \mathcal{E}\to\mathcal{E}$ ought to be trivial. Could you elaborate? (Or do you think I might be on to something with the argument I sketched above?) Also, if you have a reference handy for the claim about dualising the convolution coalgebra structure on $\mathcal{O}_{\hat{\Delta}}$ I'd really appreciate it! Feb 1 at 16:31
• @derryberry I do agree that if you add a cocycle condition to the formal lifting property then all is good. But I'm still skeptical of the argument for integrability: The issue is that looking at the first order neighborhood is not enough to "see" $\mathcal{D}^{\leq 2}$. I think the cocycle condition that you're trying to impose for integerability translates not to what you want it to, but instead to additivity of the $\mathcal{D}^{\leq 1}$ action, which is automatic.
– dhy
Feb 1 at 18:22