# 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?

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.

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

## 1 Answer

(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 $\text{Hom}_X(\mathcal{D}_X\otimes\mathcal{E},\mathcal{E})=\text{Hom}(\mathcal{E},J^\infty(\mathcal{E}))$... – derryberry 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. – derryberry 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. – derryberry 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! – derryberry 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