Thoughts On Bounding p-adic DE

Author: Eiko

Tags: p-adic, differential equations

Time: 2024-11-21 23:53:03 - 2024-11-22 17:29:10 (UTC)

given some DE $D f = 0$ , have a $k$ vector space of solutions $s_{1}, \dots, s_{n}$ , where

$n_{i} = ord (s_{i}) > ord (s_{i - 1}) = n_{i - 1}$

this can be proved by choosing a proper basis, and that has to exist (linear independence of solutions)

Under this hypothesis, the matrix of derivatives $M_{i j} = (D^{n_{i}} s_{j}) (0)$ is upper triangular, we want it be matrix with values in $K [[t]]$ ,

by construction, $det M$ is a unit in $K [[t]]$

we can get a DE of order $n_{n} + 1$ annihilating $s_{1}, \dots, s_{n}$ , and its leading term is invertible in $K [[t]]$

Consider $D^{n_{i}}$

Let’s think of flat connections $D v = Λ v$ , $Λ_{0} = 1$ $Λ_{1} = Λ$ $Λ_{n + 1} = D Λ_{n} + Λ_{n} Λ$

we have $D^{n} v = Λ_{n} v$ , we obtain a $K (t)$ -linear relation between $n + 1$ vectors $v, D v, \dots, D^{n} v$ , and we can extract the first row

$D^{i} v_{1} = \sum_{j} (Λ_{i})_{1 j} v_{j}$

we construct an $(n + 1) \times n$ matrix $M = (Λ_{i})_{1 j}$ by noting that $v_{1} \land D v_{1} \land \dots \land D^{n} v_{1} = 0$ , which expands to

Nice

$S = {m_{0}, \dots, m_{n}}$ is nice if the matrices $M_{i}$ obtained from it, satisfy $v_{t} (det (M_{n})) \leq v_{t} (det (M_{i})) .$

If $S_{0} = {0, 1, \dots, n}$ is not nice, then there is some $i$ such that $N = v_{t} (det (M_{n})) > v_{t} (det (M_{i}))$ .

Consider $S_{1} = {0, 1, \dots, i - 1, i + 1, \dots, n, n + 1}$ , then

$M_{n} (S_{1}) = M_{i} (S_{0})$

$det (M_{n} (S_{1})) = det (M_{i} (S_{0})) \leq N - 1 = v_{t} (det (M_{n} (S_{0}))) - 1$

so this inductive process stops in $N$ steps, we should be able to arrive at a nice $S^{'}$ with $max S^{'} \leq n + N$ .

P-adic Niceness

Two strategies

Inductive arguments, by relaxing degrees and make $S$ larger we might be able to satisfy $p$ -adic niceness
If there exists a subconnection W (i.e. a proper sub-bundle with connection) inside V such that v was a section of W, then det(Mi)=0 for all choices of S,i.

${(\frac{d}{d t})}^{k} v \in W$ for all k.

Assume v generates V (i.e. there doesn’t exist a subconnection W with this property).

Flat connection on a scheme over Zp.

Could be subconnections on the special fibre which don’t come from subconnections on the generic fibre.

$v a l_{p} (d e t (M n)) > v a l_{p} (d e t (M i))$ for some i<n