We know from Landvogt's work (Crelle #518, 2000) that if $G$ and $H$ are two reductive groups and $f:G \rightarrow H$ is a morphism that there is a map $\hat{f}$ from $\mathcal{B}(G, K)$ to $\mathcal{B}(H,K)$ that is $G(K)$ equivariant. Is it known that if we have $\mathcal{G}$ the parahoric scheme associated to a point $x$ of $\mathcal{B}(G, K)$ that there is a scheme map to $\mathcal{H}$ associated to $\hat{f}(x)$?
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2$\begingroup$ Landvogt's maps $\hat f$ are $G$equivariant and compatible with Galois extension. That means that for all unramified extensions $L/K$, we have a map ${\hat f}_{L/K}$: $G(L)\rightarrow H(L)$ mapping ${\mathcal G}({\mathfrak o}_L )$ to ${\mathcal H}({\mathfrak o}_L )$ (with obvious commutative diagrams). This is a serious hint that these maps should come from an ${\mathfrak o}_K$scheme map ${\mathcal G}\rightarrow {\mathcal H}$. $\endgroup$ – Paul Broussous Sep 25 '17 at 11:47

$\begingroup$ If $X$ and $Y$ are affine schemes over a dvr $R$ with fraction field $K$ with $X$ smooth over $R$ and $f:X_K \to Y_K$ is a $K$map such that $f$ carries $X(R^{\rm{sh}})$ into $Y(R^{\rm{sh}})$ then $f$ extends (uniquely) to an $R$map $X\to Y$. Indeed, by Galois descent we can assume $R=R^{\rm{sh}}$, and any $h$ in the image of $R[Y]$ in $K[X_K]=K\otimes_R R[X]$ satisfies $h(X(R))\subset R$. Since $X$ is $R$smooth and $R=R^{\rm{sh}}$, the latter condition on $h$ forces $h\in R[X]$, so we're done. (This argument goes back at least to 1.7.6 in BruhatTits II in IHES #60.) $\endgroup$ – nfdc23 Sep 25 '17 at 13:10

$\begingroup$ What is sh in the above? It might be the two of you have figured out the answer. $\endgroup$ – Watson Ladd Sep 25 '17 at 19:16
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