This question is a followup to two of my previous questions, see [here](https://mathoverflow.net/q/219572/61522) and [here](https://mathoverflow.net/questions/224566/if-k-is-an-algebraically-closed-field-of-any-characteristic-then-the-fundamen).



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> 1. Let $A$ be an abelian variety over a field $k$ of characteristic $0$. How do I prove, without using transcendental methods, that if $\nabla$ is an integrable connection on a vector bundle $L$ on $A$ then $(L, \nabla)$ comes from a $G$-bundle with connection for some abelian algebraic subgroup $G \subset GL(n)$? And the corollary there exists a $\nabla$-stable flag$$L = L\_n \supset L\_{n-1} \supset \dots \supset L\_1 \supset 0$$with $\text{rank }L\_i = i$? It is easy to prove for $k = \mathbb{C}$ by looking at the monodromy of $(L, \nabla)$, and the general case follows by the Lefschetz principle, but I want to find a different proof which does not use $\mathbb{C}$ at all.
> 2. I am curious as to whether or not we can, in a similar way, show that if $k$ is an algebraically closed field (of any characteristic) then the algebraic fundamental group of $A$ is abelian (given a connected étale covering $E \to A$ define an algebraic group structure on the group $G$ of pairs $(a, \tau)$, where $a \in A$ and $\tau$ is a lift of the translation automorphism $T\_a: A \to A$ be an automorphism of $E$, etc.).
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Does the above property of bundles with connections on an abelian variety hold for the additive or multiplicative group?


**Thoughts.** I suspect it does not, and it would suffice to construct a rank $2$ vector bundle on the affine line over $\mathbb{C}$ with a connection $\nabla$ such that $L$ has no $\nabla$-stable subbundles of rank $1$ to show this. But I am not quite sure on how to do this. Could anybody help? Thanks in advance!



Property of bundles with connections on abelian variety doesn't hold for additive or multiplicative group?
No, it fails for the additive and multiplicative group as they are not compact. Consider the differential equation of the Airy function $d^2 f/dx^2 = x f$. We can write this in first order form as $df/dx=u$, $du/dx = x f$. This becomes a vector bundle with connection by taking the vector bundle to be a rank $2$ free bundle and the connection to be $\nabla (f,u) = (\frac{df}{dx}-u, \frac{du}{dx}-xf)$ so that flat sections are the same as solutions to the differential equation.


We will show that the vector bundle has no invariant flag. If it did, it would have some sub-bundle of the form $\nabla (g) = \frac{dg}{dx} - p(x) f$ for a polynomial $p(x)=x$, which would have an invariant section of the form $g(x)= e^{ \int p(x)}$ Since that vector bundle with connection is a sub-bundle of this one by some polynomial map $(f(x),u(x))=(a(x)g(x), b(x)g(x))$ for polynomials $a$ and $b$, it follows that there is a solution to the Airy differential equation $d^2 f/dx^2 = x f$ of the form $f(x)=a(x) e^{\int p(x)}$. But there is no such solution, as we can see from the asymptotic formulas for the two solutions of the Airy equation, which do not match.


This gives a rank two vector bundle with flat connection generated by $f$ and $df/dx$, and it can be seen that it is irreducible (because the growth rate of a solution is $e^{x^{3/2}}$ but any solution to a rank one differential equation grows like an exponential of a polynomial function). Of course this also pulls back to the multiplicative group.


Your second question, on the etale fundamental group, fails for the additive and multiplicative groups in characteristic $p$ for the same reason. Indeed, there is a perfectly analogous example - the Airy sheaf.


