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Let $\mathcal{G}$ be a $p$-divisible group over $K$, which is a finite extension of $\mathbb{Q}_p$. Let $\rho: \text{Gal}(\bar{K}/K)\rightarrow \text{GL}(T_p\mathcal{G})$ be the associated Galois representation. Let $K_\rho$ denote the subfield of $\bar{K}$ that is fixed by $\text{Ker}(\rho)$.

Let $\mathcal{H}$ be a $p$-divisible group over $\bar{K}$. I would like to know why every isogeny $f: \mathcal{G}\rightarrow \mathcal{H}$ can be defined over $K_\rho$. I saw this statement in the paper Finiteness of Reductions of Hecke Orbits, Proposition 2.7.

My thought for this is that $f$ can be defined over $K_\rho$ if and only if $T_pf$ is $\text{Gal}(\bar{K}/K_\rho)$-equivariant, which is equivalent to show $T_pf=\sigma\circ T_pf$ for any $\sigma\in \text{Gal}(\bar{K}/K_\rho)$, since this Galois groups acting trivially on $T_p\mathcal{G}$. But since $\mathcal{H}$ is defined over $\bar{K}$, I do not know how to proceed.

Any help will be appreciated!

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    $\begingroup$ Any such isogeny is determined entirely by its kernel, and this is fixed by the Galois group of $K_\rho$. The main point here is that $\mathcal{H}$ is not given: it is the quotient of $\mathcal{G}$ by the kernel of the isogeny. $\endgroup$
    – Keerthi Madapusi
    Commented Dec 5 at 6:02
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    $\begingroup$ @KeerthiMadapusi Hi Keerthi, thank you very much for your reply. Are you suggesting that the definition field of any finite group scheme of $\mathcal{G}$ is contained in $K_\rho$, so $Ker(f)$ is totally understandable over $K_\rho$, hence every $f$ is totally understandable over $K_\rho$. So we can find some $\mathcal{G}'$ over $K_\rho$ such that $\mathcal{G}'\cong \mathcal{H}$ over $\bar{K}$. $\endgroup$
    – Kris
    Commented Dec 5 at 9:06
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    $\begingroup$ That's about the gist of it, yes $\endgroup$
    – Keerthi Madapusi
    Commented Dec 5 at 11:39

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