Uniformizability of elementary Anderson $t$-motives of dimension $2$

The affine equation

Let $q = p^s$, let $\operatorname{ord}$ be the valuation of $\mathbb{C}_\infty$ with $\operatorname{ord}\theta= -1$, and let $A = (a_{jk}) \in M_2(\mathbb{C}_\infty)$. The $t$-motive $M(A)$ is defined by $T e_* = \theta e_* + A\tau e_* + \tau^2 e_*$, $e_* = (e_1,e_2)^t$; it is pure of dimension 2 and rank 4 [2]. Assume $a_{21} \ne0$. By [2 (3.0), (3.1), Theorem 3.3], $h^1(M(A))$ is the $\mathbb{F}_q[T]$-rank of the module of small solutions $X = \sum_{i\geq0} x_i\tau^i$, $x_i \in\mathbb{C}_\infty$, $x_i \to0$, of the system $(i \geq0, x_{-1} = x_{-2} = 0)$

with

Put $\alpha_k = \operatorname{ord} a_k$, $\beta_{jk} = \operatorname{ord} b_{jk}$ (with $\operatorname{ord} 0 = +\infty$) and $P_k = (q^k,\alpha_k)$. One has the exact relations

Since $a_0a_4 \ne0$, the solutions of the equation $i=0$ form a 4-dimensional $\mathbb{F}_q$-vector space $S_0$, and the valuations of its nonzero elements are the slopes, with sign changed, of the lower convex hull $\Pi$ of the points $P_k$, an edge from abscissa $q^j$ to $q^l$ carrying $q^l-q^j$ of them. For every $c$, the elements of $S_0$ of valuation $>c$ form, with 0, an $\mathbb{F}_q$-subspace, of dimension $l$ if the edges of slope $<-c$ end at abscissa $q^l$.

Two elementary lemmas

For $g \in\mathbb{Q}$ put $W(g)=\max_k(g-\alpha_k)/q^k$, the maximum over the $k$ with $a_k\ne0$.

Proof. If $\operatorname{ord}x>W(g)$, every term $a_kx^{q^k}$ has valuation $>\alpha_k+q^kW(g)\ge g$, which is impossible. The Newton polygon of $c+\sum_k a_kx^{q^k}$ has its first edge from $(0,g)$ to the point $P_k$ minimizing $(\alpha_k-g)/q^k$; its roots have valuation $W(g)$, and $\mathbb{C}_\infty$ is algebraically closed. □

Given $x_{i-1},x_{i-2}$ with valuations $v,v'$, put $T_{14}=\beta_{14}+q^4v$, $T_{13}=\beta_{13}+q^3v$, $T_{12}=\beta_{12}+q^2v$, $T_{24}=\beta_{24}+q^4v'$, and $\Phi(v)=W(\beta_{12}+q^2v)$.

Proof. Bound $W$ below by the term $k=4$, $k=3$, $k=4$, $k=2$ respectively and use (2). □

The criterion

For $k\in\{0,1,3,4\}$ with $a_k\ne0$ put $f_k=(\beta_{12}-\alpha_k)/(q^k-q^2)$, and let

A direct computation shows that $v\in I$ if and only if $\alpha_2\ge\beta_{12}$ and $\max(f_3,f_4)\le v\le\min(f_0,f_1)$ (indices with $a_k=0$ omitted); geometrically, the line through $(q^2,\beta_{12})$ of slope $-v$ lies on or below all the points $P_k$.

Proof. Let $x_0\in S_0$, $v_0=\operatorname{ord}x_0$, and define $x_1,x_2,\ldots$ inductively: if the tail of (1) vanishes, take $x_i=0$; otherwise take for $x_i$ a solution of maximal valuation (Lemma 1). Since the valuation of the tail is at least $\min(T_{14},T_{13},T_{12},T_{24})$, Lemma 1 gives, with $v_j=\operatorname{ord}x_j$,

the terms with a zero coefficient being omitted (for $i=1$ the term $T_{24}$ is absent).

(1) The function $\Phi(v)-v$ is convex and piecewise linear, with slope $q^2-1>0$ for $v\gg0$ and slope $q^{-2}-1<0$ for $v\ll0$; if $I=\varnothing$ it is positive, hence $\ge c$ for some $c>0$. By (3) and Lemma 2, $v_i\ge\min(v_{i-1},v_{i-2})+\min(c,q^{-2})$, so $v_i\to+\infty$ and $X=\sum x_iT^i$ is a small solution.

If $\alpha_2<\beta_{12}$, then $\Phi(v)-v\ge(\beta_{12}-\alpha_2)q^{-2}>0$ for all $v$ by Lemma 2, so $I=\varnothing$. Applying this to an $\mathbb{F}_q$-basis $x_0^{(1)},\ldots,x_0^{(4)}$ of $S_0$ gives small solutions $X^{(1)},\ldots,X^{(4)}$. They are $\mathbb{F}_q[T]$-linearly independent: in a relation $\sum c_jX^{(j)}=0$ with $c_j\in\mathbb{F}_q[T]$ not all zero we may assume, $T$ being injective on $\mathbb{C}_\infty[[T]]$, that some $c_j(0)\ne0$, and the constant term gives $\sum c_j(0)x_0^{(j)}=0$. Hence $h^1(M(A))\ge4$, and $h^1(M(A))\le4$ always.

(2) Let $v_+=\max I=\min(f_0,f_1)=f_1$. The line $\ell$ through $(q^2,\beta_{12})$ of slope $-v_+$ passes through $P_1$, strictly below $P_0$ (as $f_1<f_0$), and on or below the other points; hence $P_0P_1$ is the first edge of $\Pi$, of slope $<-v_+$, and $S_0$ contains an element $x_0\ne0$ with $v_0>v_+$. The function $\Phi(v)-v$ is convex, $\le0$ on $I$ and $>0$ to the right of $v_+$, so it is nondecreasing on $[v_+,\infty)$ and $\Phi(v)-v\ge c_0:=\Phi(v_0)-v_0>0$ for $v\ge v_0$. By induction, (3) and Lemma 2 give $v_i\ge v_0$ and $v_i\ge\min(v_{i-1},v_{i-2})+\min(c_0,q^{-2})$, so $X$ is a small solution and $h^1\ge1$. □

Proof. The first two summands of $a_2$ have valuations $-q-\operatorname{ord}a_{21}$ and $-q^2-q^2\operatorname{ord}a_{21}$, smaller than the valuations $-\operatorname{ord}a_{21}$ and $-q^2\operatorname{ord}a_{21}$ of the two summands of $b_{12}$. Hence $\alpha_2 < \min(-\operatorname{ord}a_{21},-q^2\operatorname{ord}a_{21}) \le\beta_{12}$, and Theorem 3(1) applies. □

Remark 5. If at every step the minimum of $T_{14},T_{13},T_{12},T_{24}$ is attained by a single term, the inequalities above become equalities, and one finds that the starts $x_0$ with $\operatorname{ord}x_0 \le\max I$ stay of valuation $\le\max I$; the value of $h^1$ is then $4$, $1$ or $0$ according as $I=\varnothing$, $f_1<f_0$, or neither. Values $2$ and $3$ can therefore only occur through cancellations in the tail.

Conjecture 3.7 of Ehbauer, Grishkov and Logachev

Write $(u_{11},u_{12},u_{21},u_{22})=(\operatorname{ord}a_{11},\operatorname{ord}a_{12},\operatorname{ord}a_{21},\operatorname{ord}a_{22})\in(\mathbb{Q}\cup\{+\infty\})^4$. In [2 Conjecture 3.7] (see also [4 Conjecture 4.2.3]) it is conjectured that $h^1(M(A))=4$ whenever this point lies outside a countable union of linear subspaces of dimension $\le3$.

For $u_{21}<\infty$ put $s_1=-q-u_{21}$, $s_2=-q^2-q^2u_{21}$, $s_3=q(u_{11}+u_{22}-u_{21})$, $s_4=qu_{12}$, with $s_3=+\infty$ if $u_{11}=+\infty$ or $u_{22}=+\infty$, and $s_4=+\infty$ if $u_{12}=+\infty$; these are the valuations of the four summands of $a_2$.

Proof. Suppose first $u_{21}<\infty$. The equalities $s_j=s_k$ between finite $s_j,s_k$ are the six equations defining $H$, and $s_1,s_2$ are always finite. Outside $H$ the minimum of the $s_j$ is attained once, so no cancellation can occur in $a_2$, and Corollary 4 gives $h^1=4$. If $u_{21}=\infty$ and $u_{12}<\infty$, the basis $(e_2,e_1)$ gives $M(A)=M(JA J^{-1})$ with $J=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, which exchanges $a_{11}\leftrightarrow a_{22}$ and $a_{12}\leftrightarrow a_{21}$; apply the first case. If $a_{12}=a_{21}=0$, $M(A)$ is the direct sum of two Drinfeld modules of rank $2$, which are uniformizable. Finally $h^1=4$ if and only if $h_1=4$ by Anderson’s theorem [2]. □

Remark 7. For antidiagonal $A$ ($u_{11}=u_{22}=+\infty$), $H\cup H'$ meets the plane of $(u_{12},u_{21})$ in the four lines $u_{21}=-\frac{q}{q+1}$, $u_{12}=\frac{q}{q+1}$, $u_{21}=-qu_{12}-q$ and $u_{21}=-u_{12}/q-1$. The first, third and fourth contain the rays $R_1,R_2,R_3$ of [2 Section 6], on which the non-uniformizable $M(A)$ of [2 Theorem 5.1] lie. On $H\cup H'$ uniformizability may fail only if the leading terms of the tied summands of $a_2$ cancel to the point that $I\ne\varnothing$. The uniformizable locus is open in families [1].

Numerical checks

Theorem 5.1 of [2]. For antidiagonal $A$, the value of $h^{1}$ obtained by iterating (3) with equality coincides with [2 Theorem 5.1] at all 38 640 points of a grid in $(\operatorname{ord} a_{21}, \operatorname{ord} a_{2})$, for $q=2$ and $q=3$.

The example of [3]. For $q=2$ and $A=\begin{pmatrix} \theta& \theta^{6} \\ \theta^{-2} & 0 \end{pmatrix}$ one has $s_{2}=s_{4}=-12$ (and $u_{22}=+\infty$): the point lies on $H$, and $\theta^{12}$ cancels in $a_{2}=\theta^{4}$, so that $\alpha_{2}=-4\geq-8=\beta_{12}$. This agrees with $h^{1}(M(A))=0$ and $h^{1}(M(A^{t}))=1$ [3 Theorems 4.1 and 4.7].

The example of [5]. For $q=2$ and $A=\begin{pmatrix} \theta^{3} & \theta^{-2} \\ \theta^{-1} & 1 \end{pmatrix}$ the nine coefficients are those of [5]; the two summands $\theta^{8}$ of $a_{2}$ cancel ($s_{2}=s_{3}=-8$), and $a_{2}=\theta^{3}+\theta^{-4}$. Here $\beta_{12}=-4$ and $(f_{0},f_{1},f_{3},f_{4})=(1,0,4,1)$, so $\max(f_{3},f_{4})>\min(f_{0},f_{1})$ and $I=\varnothing$ whatever the valuation of $a_{2}$. Since $a_{1},a_{3},b_{12},b_{13}$ have no ties among their summands, $h^{1}(M(A))=4$ for every $A$ whose entries have the valuations $\begin{pmatrix} -3 & 2 \\ 1 & 0 \end{pmatrix}$, in accordance with [5]. The lower bounds (3) for the start of valuation $-1/2$ are $-1/2,0,1/2,1,\ldots$, as in [5].

Exact computations. For 40 000 matrices $A$ with entries in $\mathbb{F}_{q}[\theta,\theta^{-1}]$ ($q=2,3$), chosen so that cancellations in $a_{2}$ are frequent, the nine coefficients were computed exactly. In every case $I=\varnothing$ holds for $A$ if and only if it holds for $A^{t}$, in accordance with $h^{1}(M(A))=4\Longleftrightarrow h^{1}(M(A^{t}))=4$.

The scripts used for these checks are not part of this deposit.

Acknowledgements. The author thanks D. Logachev for explaining the computation of $h^{1}$ through affine equations. The computations and part of the writing were carried out with the assistance of Claude (Anthropic).

Independent researcher, Meaux, France