If $G$ is a semisimple algebraic group over a local field with finite residue field $K$ and $x$ a point in the Bruhat-Tits building $B(G, K)$ then the parahoric group scheme $P_x$ is a group scheme $P$ whose $O_K$ points are the connected component of the stabilizer of $x$.

If $G$ is not semisimple there is a construction of a group scheme $P$ associated to $x$ called a parahoric. But as the center of $G$ acts trivially on the building the $O_K$ points of $P$ aren't exactly the stabilizer of $x$.

Is it the case that $Z(G(K))P_{x}(O_K)$ is the connected component of the stabilizer of $x$? This seems believable from the construction. If $x$ is in the extended compartment instead will I just get the connected component of $P_{x}(O_K)$? Bruhat and Tits don't use the extended compartment, so I lack good references for that.

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    $\begingroup$ You have to decide on what kind of building you want to act. The point of the introduction of the enlarged building ('immeuble élargi') in Definition 4.2.16 of BT2 is precisely to make sure that point stabilisers are compact. For the ordinary ('reduced') building, the action factors through the adjoint quotient, so, in your notation, $Z(G(K))P_x(O_K)$ has finite index in the stabiliser. I don't know (though probably someone comfortable with schemes not of finite type does) what it means to be the identity component of the stabiliser, but there's no reason to expect $Z(G)$ connected. $\endgroup$ – LSpice Nov 18 '17 at 20:54
  • $\begingroup$ I think I was implicitly using the integral model of the stabilizer in my notion of components. $\endgroup$ – Watson Ladd Nov 18 '17 at 21:10
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    $\begingroup$ The problem is that, for non-semisimple groups, the stabiliser is not the group of $O_K$-points of a group scheme of finite type. For that matter, even if it were, it is a bit troublesome to ask whether the abstract group $Z(G(K))P_x(O_K)$ is the identity component of a group scheme. $\endgroup$ – LSpice Nov 18 '17 at 21:13
  • $\begingroup$ Is the noncompactness the only problem here, so switching to the enlarged building would fix that? Or is it more fundamentally a problem with non-semisimple? $\endgroup$ – Watson Ladd Nov 18 '17 at 22:01
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    $\begingroup$ I'm not sure how to answer. Depending on your point of view, the failure to be compact might be the problem, or the failure to be the group of integral points of a group scheme of finite type might be the problem. Passing to the enlarged building fixes both of these, so I don't think that there's any more fundamental problem with non-semisimplicity; and, indeed, the entire Harish-Chandra philosophy requires dealing with such groups. (I would encourage you to say "reductive but non-semisimple groups" in your title; I clicked expecting a question on unipotent groups.) $\endgroup$ – LSpice Nov 18 '17 at 22:35

I don't know how Bruhat-Tits theory is used in representation theory, but I think there is a little confusion of notions in your question.

For a connected reductive algebraic group $G$, given a point $x$ in $B(G,K)$, Bruhat and Tits define $4$ integral models denoted $\mathfrak{G}_x^0$, $\mathfrak{G}_x$, $\hat{\mathfrak{G}}_x$ and $\mathfrak{G}_x^{\dagger}$. Here $\mathfrak{G}_x^0$ is the connected component of $\mathfrak{G}_x$, and in Bruhat-Tits terminology $\mathfrak{G}_x^0(\mathcal{O}_K)$ is called the "connected fixator" ("fixateur connexe" in French). I guess that's the group you are referring to as "the connected component of the stabiliser of $x$" (I put that in quote, because to me this doesn't mean anything: when compact, the stabiliser of $x$ in $G(K)$ is a profinite group, i.e. it has normal closed (for the strong topology) subgroups of finite index, and the index can be as big as you want).

Ok, now if we look at $GL_2$, take a point $x$ in $B(G,K)$ and an apartment $A$ containing it. Then throw in many basis as in Bruhat-Tits I, 10.2, so that x identifies with $0\in \mathbf{R} = A$. Let $S_{-1}$ be the stabiliser of $-1\in A$. By Bruhat-Tits I, $10.2.9$, $S_{-1}$ consists of matrices $g\in GL_2(K)$ such that the valuation of the entries are greater then $\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}+\frac{\omega(\textrm{det}(g))}{2}$. For example, $\begin{pmatrix} 0&\pi \\\pi^{-1}&0 \end{pmatrix}$ belongs to $S_{-1}$.

On the other hand, by Bruhat-Tits II, $4.6.28$, matrices in $\mathfrak{G}_{-1}^0(\mathcal{O}_K)$ have determinant in $\mathcal{O}_K^{\times }$ and belongs to $S_{-1}$ as well. So we conclude that the element $g = \begin{pmatrix} 0&\pi^2\\1&0 \end{pmatrix}$ belongs to the stabiliser of $-1$, but not to $Z(G(K))\mathfrak{G}_{-1}^0(\mathcal{O}_K)$.

So this is a negative answer to your first question, even if we interpret "connected component of the stabiliser of x" to mean "stabilisers of $x$ acting type preservingly".

EDIT: Ok, as pointed by Watson Ladd, the above counterexample is just not one, sorry that was silly. At least there is still the counterexample if you take $x=\frac{1}{2}$. Then I believe $g = \begin{pmatrix} 0&1\\\pi&0 \end{pmatrix}$ stabilises $\frac{1}{2}$, but does not belong to $Z(G(K))\mathfrak{G}_{\frac{1}{2}}^0(\mathcal{O}_K)$. Here the action of $g$ is not type preserving, though, so this calls for the question whether $Z(G(K))\mathfrak{G}_{x}^0(\mathcal{O}_K)$ consist of stabilizers of $x$ acting type preservingly.

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    $\begingroup$ Isn't $\pi^{-1}$ in $Z(G(K))$ and will send $g$ to $$\left(\begin{matrix}0 & \pi \\ \pi^{-1} & 0\end{matrix}\right)$$? $\endgroup$ – Watson Ladd Nov 18 '17 at 20:48
  • $\begingroup$ I think that @WatsonLadd is right (although there is still the point you mention about what it means to speak of identity components of abstract groups, or (I don't know about this, but probably someone does) of group schemes not of finite type). I guess you also mean the last occurrence of $\mathfrak G^0_{-1}$ to be its group of $O_K$-points. $\endgroup$ – LSpice Nov 18 '17 at 20:58
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    $\begingroup$ Sorry for that, I had too strongly in mind the case $x=\frac{1}{2}$, see the edit. And thank you @LSpice , I edited accordingly. $\endgroup$ – thierry stulemeijer Nov 18 '17 at 21:15

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