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Mar 2, 2015 at 0:22 comment added LSpice (Oops, sorry; I misspelt your name.) On further reflection (no pun intended), since we've got a split group, I return to my original position: all (hyper)special points in the (reduced) building are $G_{\text{ad}}(K)$-conjugate; this is stated in \S2.5 of Tits's Corvallis article.
Mar 2, 2015 at 0:08 comment added LSpice @MartinSchuster, I'm sorry, you are right. There are at most two orbits under $G_{\text{ad}}(K)$ (not necessarily $G(K)$ as I originally said), but I don't see any obvious reason that these two orbits should give rise to the same schemes.
Feb 6, 2015 at 16:44 comment added Michael Schuster Could you explain a bit more? Not all hyperspecial parahorics are conjugate in $G(K)$, but they will be conjugate over a finite extension $L$ of $K$. Are you saying conjugation in $G(L)$ will induce a $K$-isomorphism of the group schemes?
Feb 5, 2015 at 19:49 comment added LSpice (Hmm, since they don't sit in a superstructure, I guess I mean that the $G(K)$-conjugacy of the associated building points induces an isomorphism of the underlying schemes.)
Feb 5, 2015 at 19:33 comment added LSpice They are not just isomorphic, but $G(K)$-conjugate! From my point of view, the reason to distinguish them is not as groups, but for the way that they fit into the Bruhat--Tits picture: they correspond to different (but $G(K)$-conjugate) points in the BT building.
Feb 5, 2015 at 17:27 history edited Michael Schuster
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Feb 5, 2015 at 17:23 review First posts
Feb 5, 2015 at 17:42
Feb 5, 2015 at 17:19 history asked Michael Schuster CC BY-SA 3.0