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Let $F$ be a nonarchimedean local field, and let $D/F$ be the central division algebra of invariant $1/d$. Let $k$ be the algebraic closure of the residue field of $F$ and let $\pi$ be a uniformizer.

Let $X$ be a $\pi$-divisible group over $k$. Following Drinfeld, $X$ is called a formal $\mathcal{O}_D$-module if it is connected and if it is endowed with an action of $\mathcal{O}_D$ for which $\text{Lie}(X)$ becomes a locally free $\mathcal{O}_{F_d}\otimes_{\mathcal{O}_F} k$-module of rank 1. Here $F_d\subset D$ is an unramified extension of $F$ of degree $d$.

Such an $X$ has $F$-height divisible by $d^2$; let us assume that the height is exactly $d^2$. Then $X$ is unique up to isogeny. It is easy to concoct such an $X$ in the case that $F=\mathbf{F}_{q}((\pi))$ has positive characteristic. Here $\mathcal{O}_D$ is generated over $\mathcal{O}_F=\mathbf{F}_q[[\pi]]$ by $\mathbf{F}_{q^d}$ and by an element $\Pi$ which has $\Pi^d=\pi$ and $\Pi\zeta=\zeta^q\Pi$, $\zeta\in\mathbf{F}_{q^d}$. We can let $X$ be the formal $\mathcal{O}_D$-module whose law is determined by

$$ [\Pi]_X(T_0,\dots,T_{d-1})=(T_1,\dots,T_{d-1},T_0^{q^d}) $$

$$ [\zeta]_X(T_0,\dots,T_{d-1})=(\zeta T_0,\dots,\zeta^{q^{d-1}}T_{d-1}) $$

My question is:

What is the algebra of $\mathcal{O}_D$-endomorphisms of this particular $X$?

Through an examination of the Dieudonnè module of $X$, one shows that its endomorphism algebra becomes $M_d(F)$ when tensored with $F$. So $\text{End} X$ is an order in $M_d(F)$, but which one? Can one exhibit the endomorphisms explicitly on the level of coordinates?

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