When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:59:31Z http://mathoverflow.net/feeds/question/120853 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120853/when-is-a-moy-prasad-filtration-subgroup-the-stabilizer-of-a-subset-of-the-buildi When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)? Monica 2013-02-05T11:39:38Z 2013-02-16T19:53:27Z <p>Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$. Let $B$ be the corresponding reduced building. Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G_{x,r}$. Our hypotheses give $G_x=G_{x,0}$.</p> <p>Define $\Omega_{A,r}$ as the set $${ y \in A : \forall \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r }.$$</p> <blockquote> <p>Is it true that $T(R)G_{x,r} = \cap_{y\in \Omega_{A,r}} G_y$ ?</p> </blockquote> <p>(With $r>0$ the product gives a group which is contained in the given intersection. Equality seems to be about whether all hyperplanes $H_{\alpha,r}$ meet $\Omega_{A,r}$.)</p> <p>Let $A(x)$ be the set of all apartments of $B$ containing $x$. Let $Z$ be the (finite) center of $G$.</p> <blockquote> <p>Is it true that $ZG_{x,r} = \cap_{A \in A(x)}\cap_{y\in \Omega_{A,r}}G_y$ ?<br> Equiv by (1): Is $ZG_{x,r} =\cap_{g \in G_x} (gT(R)g^{-1})G_{x,r}$? </p> </blockquote> <p>This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$). Pointers to any literature much appreciated.</p> http://mathoverflow.net/questions/120853/when-is-a-moy-prasad-filtration-subgroup-the-stabilizer-of-a-subset-of-the-buildi/122013#122013 Answer by Monica for When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)? Monica 2013-02-16T19:51:20Z 2013-02-16T19:51:20Z <blockquote> <p>Regretfully the answer to the first question is no.</p> </blockquote> <p>For example consider a group of type $G_2$. Normalize your roots so the short roots have length 1, for example, and let $x=0$. Write $\Phi = \Phi^l \cup \Phi^s$ as a union of long and short roots, respectively, and define the corresponding sets $\Omega^l_{A,r}$ and $\Omega^s_{A,r}$. Write $B_r$ for the disk of radius $r$. </p> <p>Then $\Omega^l_{A,r} \subseteq B_{2r/3}$ whereas $B_r \subseteq \Omega^s_{A,r}$. Consequently the parenthetical comment fails. </p> <p>There was also a typo in the parenthetical comment, so let's be explicit. (Given a root $\alpha$, write $U_\alpha$ for the corresponding root subgroup and normalize so $U_\alpha(R)= U_\alpha \cap G_0$.) </p> <p>To see that $T(R)G_{x,r} \subseteq \cap_{y\in \Omega_{A,r}} G_y$, it suffices to verify this on a set of generators. We have $T(R) \subseteq G_y$ for all $y\in A$. Then the rest of $G_{x,r}$ is generated by the groups $U_\alpha(P^{\lceil r-\alpha(x)\rceil})$.<br> For each $y \in \Omega_{A,r}$ we have $\alpha(x)-\alpha(y)\leq r$ so $-\alpha(y) \leq r-\alpha(x)$ which implies $U_\alpha(P^{\lceil r-\alpha(x)\rceil}) \subseteq U_\alpha(P^{\lceil -\alpha(y)\rceil})$. Thus $T(R)G_{x,r}$ is contained in each of these $G_y$.</p> <p>Conversely, in the case of $G_2$ described above, the inequalities imply that for sufficiently large $r$, there is no $y \in \Omega_{A,r}$ for which the equality $\lceil -\alpha(y) \rceil = \lceil r-\alpha(x) \rceil$ ($=\lceil r \rceil$) holds; therefore the intersection will be strictly larger than $T(r)G_{x,r}$.</p> <blockquote> <p>It follows that the answer to the first version of the second question is also no. The second version is true for $SL(2)$, for example, so perhaps there is an argument via Chevalley generators and relations. </p> </blockquote>