# Reference request - existence of formal solutions for integrable connections

Let $$K$$ be a field of characteristic $$0$$, let $$A = K[[t_1, \ldots, t_n]]$$ be a power series ring over $$K$$, and let $$V$$ be a free $$A$$-module. Let $$\nabla \colon V \rightarrow V \otimes_A \Omega^1_{A/K} = \bigoplus_{k=1}^n V \ dt_k$$ be an integrable connection, so $$\nabla^2 = 0$$. I want to show that there always exist solutions $$s \in V$$ to the equation $$\nabla(s) = 0$$ subject to any initial condition $$s(0) = \alpha$$ for $$\alpha \in V(0) = V \otimes_A A/\mathfrak{m}_A$$.

More concretely, letting $$e_1, \ldots, e_r$$ be a basis for $$V$$, I have a system of $$n \times r$$ linear first-order homogeneous partial differential equations $$\frac{\partial}{\partial t_k} s_i = \sum_j M^j_{ik} s_j$$ For some $$M_{ik}^j \in A$$. Or, writing $$M_k$$ for the matrix $$(M_{i,k}^j)_{i,j}$$, $$\frac{\partial}{\partial t_k} s = M_k s$$

How can I show that integrability implies there are always solutions? Concretely, integrability is exactly the statement that for all $$i,j,k, \ell$$, we have $$\frac{\partial}{\partial t_\ell} M_{k} - \frac{\partial}{\partial t_k} M_{\ell} = M_k M_\ell - M_\ell M_k$$

• What is $V(0)$? Commented Feb 24, 2019 at 17:49

I hope you'll allow one or two slight adjustments to your question:

• I think you are interested in $$(t_1, \dots, t_n)$$-adically continuous connections rather than arbitrary connections (for which the desired solution principle would not generally hold).

• Judging from how you use "$$r$$," I think you intend $$V$$ to be a finite-rank free $$A$$-module (rather than an arbitrary free $$A$$-module).

You will find a very nice treatment of your question thus modified in Proposition 8.9 of Nicholas Katz's paper "Nilpotent connections and the monodromy theorem: Applications of a result of Turrittin."

Here is the statement of that proposition:

Let $$K$$ be a field of characteristic zero, $$K[[t_1, \dots, t_n]]$$ the ring of formal power series over $$K$$ in $$n$$ variables. Let $$M$$ be a finitely generated module over $$K[[t_1, \dots, t_n]]$$, given with an integrable connection $$\nabla$$ (for the continuous $$K$$-derivations of $$K[[t_1, \dots, t_n]]$$ to itself). Then $$M^\nabla$$, the $$K$$-space of horizontal elements of $$M$$, is finite-dimensional over $$K$$, and the pair $$(M, \nabla)$$ is isomorphic to the pair $$(M^\nabla\otimes_K K[[t_1, \dots, t_n]], 1\otimes d)$$ where $$d$$ denotes the identical'' connection on $$K[[t_1, \dots, t_n]]$$.

Note that in this proposition, one can even assume only that the module $$M$$ with integrable connection is finitely generated (rather than free of finite rank). As a consequence of fact that $$M^\nabla \otimes_K K[[t_1, \dots, t_n]] \to M$$ is an isomorphism, one can then conclude that $$M$$ is necessarily free.

• Nice answer! I want to emphasize that this proposition is really saying that integrable connections are only really interesting as global objects---locally, they're all determined up to isomorphism by their rank. Commented Feb 24, 2019 at 17:58
• Yes, I think both of those adjustments are right. Thanks for the reference! I'm surprised by the simplicity of the argument - it looks like Katz essentially just writes down a solution directly as a "matrix exponential". The condition of being $(t_1, \ldots, t_n)$-adically continuous should say exactly that the integrable connection is specified by differential equations $\frac{\partial}{\partial t_k} s = \sum_j M^j_{ik} s$, right? (i.e. some random integrable connection on the power series ring might not land in the span of the $dt_i$'s.) Commented Feb 25, 2019 at 0:07