In this short post I want to note Deligne’s classic result on vector bundles with flat connections (let’s call them “differential equations”: terminology justified by the `cyclic vector theorem”) and give a stupid example illustrating the basic shape of things. I should thank Ananth Shankar for helping unconfuse me faster than I could unconfuse myself.

Let $X$ be a smooth algebraic variety over $\mathbb{C}$. A vector bundle on $X$ is a locally free coherent (Zariski) sheaf $\mathcal{F}$, and a connection is an additive map of sheaves $\nabla: \mathcal{F} \rightarrow \mathcal{F} \otimes_{\mathcal{O}_X} \Omega^1_{X/\mathbb{C}}$

satisfying the Leibniz law. There is the usual notion of flatness of a connection. A morphism between two such objects is an $\mathcal{O}_X$-linear map between the underlying sheaves making the obvious square commute.

One can also make the exactly parallel construction over the complex manifold $X^{an}$. Given an algebraic differential equation, we can analytify it and obtain an analytic differential equation: $(-)^{an}: DE(X) \rightarrow DE(X^{an}).$

Deligne’s result gives a functor in the other direction (let’s call it RH for “Riemann-Hilbert”) $RH: DE(X^{an}) \rightarrow DE(X)$.

Theorem (Deligne): The functor RH (exists and) is fully faithful, and the essential image $RSDE(X)$ is characterised by taking a good compactification of $X$ and restricting to those differential equations with regular singularities at the boundary.

This is a little counterintuitive at first. One imagines there being “more” objects in $DE(X^{an})$ than in $DE(X)$ because one can write down more connections, but actually the extra freedom causes more stuff to become isomorphic, resulting in a “smaller” analytic category. Of course, one has $\mathcal{F} \cong RH(\mathcal{F})^{an}$ but not (unless $X$ is proper) an identification in the other direction.

Let’s see what happens in a stupid example. Let $X=\mathbb{A}^1$ and $\mathcal{F}$ the trivial line bundle. Let’s take our favourite non-vanishing algebraic section $v$, and observe that any connection is automatically flat and determined by $\nabla(v)$

(since any other section can be written as $f(t)v$ which by Leibniz is given by $\nabla(f(t)v) = v f'(t) \otimes dt + f(t) \nabla(v)$).

If $g(t) \in \mathbb{C}[t]$ let us denote by $\nabla_g$ the connection taking $v \mapsto g(t) v \otimes dt$.

Let’s suppose we have a map $\alpha: \nabla_g \rightarrow \nabla_h$ between two such differential equations. Let $\alpha(v) = u(t) v$, and the compatibility of $\alpha$ with the connections gives the relation $u'(t) + u(t)h(t) = u(t)g(t)$.

In other words, $u$ must satisfy a first order differential equation, and in fact we see that up to a constant factor $u(t) = e^{\int (g(t)-h(t))dt}$

which is never algebraic unless $g(t)-h(t)=0$.

The upshot is that in the algebraic category each $\nabla_g$ gives a distinct differential equation, while in the analytic category one can use the above recipe to construct isomorphisms between each of the $\nabla_g$ and in particular they are all equal to $\nabla_0$, which is the unique algebraic connection with regular singularities at $\infty$.