Tits reductive groups over local fields, 1.15/3.11. Problem with affine root subgroups of $SU_3$ ramified, residue characteristic p=2

Let $L/K$ be ramified quadratic extension of local fields, and let characteristic of the residue field of $K$ be $2$. Let $mathbb{G}=SU_3$, $G=mathbb{G}(K)$. Let $text{val}$ be a valuation on $K$ so that $text{val}(K^times) = mathbb{Z}$ (and $text{val}(L^times) = frac{1}{2}mathbb{Z}$).

Following Tits 1.15 and 3.11, I have been trying to work out the parahoric subgroups of $G$ attached to the special vertices $nu_0$ and $nu_1$ in the building of $G$.

Firstly, I’ll start with a description of the root subgroups of $G$. I’m using a slightly different notation from Tits’. Let $$u_+(c,d) = begin{pmatrix} 1 & -bar{c} & d \ 0 & 1 & c \ 0 & 0 & 1 end{pmatrix},$$
with $bar{c}c+d+bar{d}=0$.
Similarly, $$u_-(c,d) = begin{pmatrix} 1 & 0 & 0 \ c & 1 & 0 \ d & -bar{c} & 1 end{pmatrix},$$
with $bar{c}c+d+bar{d}=0$.

We have the root subgroups $U_{pm a}(K) = { u_pm(c,d) text{ : } c,d in L }$ and $U_{pm 2a} = { u_pm(0,d) text{ : } d in L}$.

Tits later defines $delta = sup{text{val}(d) text{ : } d in L, , bar{d}+d+1=0}$. $delta=0$ in the unramified case and in the ramified, residue characteristic $pneq 2$ case. However, when $L/K$ is ramified with residue characteristic $2$, $delta$ is strictly negative.

From here, Tits finds the set of affine roots of $G$ as $$Big{pm a + frac{1}{2}mathbb{Z} +frac{delta}{2}Big} cup Big{pm 2a +mathbb{Z}+ frac{1}{2} + delta Big}.$$

Affine root subgroups are given by $$U_{pm a + gamma/2} = { u_pm(c,d) text{ : } text{val}(d) geq gamma},$$
$$U_{pm 2a+ gamma} = { u_pm(0,d) text{ : } text{val}(d) geq gamma}.$$

The special points $nu_0$ and $nu_1$ i the standard apartment are defined by $$a(nu_1)=frac{delta}{2}, , a(nu_0) = frac{delta}{2} + frac{1}{4}.$$

From here, one can find that $$G_{nu_1} = langle T_0, U_{a-frac{delta}{2}}, U_{-a+frac{delta}{2}}, U_{2a+frac{1}{2}-delta}, U_{-2a+frac{1}{2}+delta} rangle,$$
$$G_{nu_0} = langle T_0, U_{a-frac{delta}{2}}, U_{-a+frac{1}{2}+frac{delta}{2}}, U_{2a-frac{1}{2}-delta}, U_{-2a+frac{1}{2}+delta} rangle.$$

In 3.11, Tits takes a $lambda in L$ with $text{val}(lambda) = delta$, satisfying $lambda+bar{lambda}+1=0$ in a way such that $lambda varpi_L + overline{(lambda varpi_L)}=0$ for some uniformizer $varpi_L$ of the ring of integers $mathcal{O}_L$ of $L$.

In 3.11, Tits defines the lattices $$Lambda_{nu_1} = mathcal{O}_L oplus mathcal{O}_L oplus lambdamathcal{O}_L,$$
$$Lambda_{nu_0} = varpi_L^{-1}mathcal{O}_L oplus mathcal{O}_L oplus lambdamathcal{O}_L.$$ Let $P_{nu_1}$ and $P_{nu_0}$ be their respective stabilizers.
Tits then states that $G_{nu_i} = P_{nu_i} cap G_{nu_i}$ for $i=0,1$.

Here’s where my problem comes in.

Consider $G_{nu_1} = langle T_0, U_{a-frac{delta}{2}}, U_{-a+frac{delta}{2}}, U_{2a+frac{1}{2}-delta}, U_{-2a+frac{1}{2}+delta} rangle.$ The stabilizer of the lattice $Lambda_{nu_1}$ in $GL_3(L)$ has the form
$$begin{pmatrix} mathcal{O}_L & mathcal{O}_L & mathfrak{p}_L^{-2delta} \ mathcal{O}_L & mathcal{O}_L & mathfrak{p}_L^{-2delta} \ mathfrak{p}_L^{2delta} & mathfrak{p}_L^{2delta} & mathcal{O}_L end{pmatrix}.$$
Since $text{val}(delta) < 0$, intersecting this stabilizer with $G$ would give us a matrix roughly looking like
$$begin{pmatrix} mathcal{O}_L & mathfrak{p}_L^{-2delta} & mathfrak{p}_L^{-2delta} \ mathcal{O}_L & mathcal{O}_L & mathfrak{p}_L^{-2delta} \ mathfrak{p}_L^{2delta} & mathcal{O}_L & mathcal{O}_L end{pmatrix},$$

Presumeably, this would tell us that $$U_{a-frac{delta}{2}} = { u_+(c,d) text{ : } c,d in L, , text{val}(d) geq -delta textbf{ and } text{val}(c) geq -delta },$$
$$U_{-a+frac{delta}{2}} = {u_{-}(c,d) text{ : } c,d in L, , text{val}(d) geq delta textbf{ and } text{val}(c) geq 0 }.$$
Normally, one would expect that if $text{val}(d) = gamma$, then $text{val}(d) = frac{gamma}{2}$ or $frac{gamma}{2}+frac{1}{4}$, as whether $gamma in mathbb{Z}$ or just $frac{1}{2}mathbb{Z}$.

I cannot work out algebraically why we have these improved bounds on the valuation of $c$ for these affine root subgroups. I assume it involves some manipulation with $lambda$, but I am not making any progress.

Thank you

Isomorphism between groups and subgroups.

Say I know that $G_1,G_2$ are isomorphic groups, both with normal subgroups $N_1,N_2$ respectively. Can I take an isomorphism $psi$ between $G_1,G_2$ s.t $psi(N_1)=N_2$? Meaning that $psi$ also defines a isomorphism between $N_1$ and $N_2$?

The thm I’m trying to prove is if $G_1 cong G2, N1cong N2,N1unlhd G1, N2 unlhd G2$ then $G1/N1 cong G2/N2$, and proving the above statement I would be able to use the first isomorphism theorem and that’s it. Is it even true?

Any hint would be helpful on how to approach this problem.

gn.general topology – Complete topological groups in which all subgroups are closed

My previous question has been answered by YCor; so I am asking a new one with a reasonable additional assumption. See the previous question for the background and motivation.

General question: does there exist a complete, nondiscrete topological group $G$ such that all subgroups of $G$ are closed? Or, does there exist a complete, nondiscrete topological vector space $V$ such that all vector subspaces of $V$ are closed?

More specific question: does there exist a complete, nondiscrete topological abelian group $A$ with linear topology such that all subgroups of $A$ are closed? Or, does there exist a complete, nondiscrete topological vector space $V$ with linear topology such that all vector subspaces of $V$ are closed?


Let me collect here the less obvious definitions appearing in the questions above. A topological abelian group $A$ is said to have linear topology if the open subgroups form a base of neighborhoods of zero in $A$. A topological vector space $V$ is said to have linear topology if open vector subspaces form a base of neighborhoods of zero in $V$. The notion of a topological vector space with linear topology presumes that the topology on the ground field $k$ is discrete.

I define completeness in the case of a linear topology only, as the general case is more involved. A topological abelian group $A$ with linear topology is said to be (Hausdorff and) complete if the natural map $Alongrightarrow varprojlim_{Usubset A}A/U$ is bijective. Here the projective limit is taken over all the open subgroups $Usubset A$. For topological vector spaces with linear topology, the definition of completeness is similar.

gr.group theory – Number of conjugacy classes of cyclic subgroups of order $p^2$ does $operatorname{GL}_{n}(Bbb Z / pBbb Z)$ have?

Let $fin Hom((mathbb{Z}/p^2mathbb{Z}),operatorname{GL}_{n}(Bbb Z / pBbb Z))$ be an injective homomorphism.We want to compute the number of isomorphism classes of semidirect products of the form $$(mathbb{Z}/pmathbb{Z})^{n}rtimes_{f}(mathbb{Z}/p^2mathbb{Z}).$$

In fact, this allows us to compute the number of conjugacy classes of cyclic subgroups of order $p^2$ in $operatorname{GL}_{n}(Bbb Z / pBbb Z)=Aut((mathbb{Z}/pmathbb{Z})^{n})$.

Any help would be appreciated so much. Thank you all.

ag.algebraic geometry – Subgroups of algebraic groups containing regular unipotent elements

Let G be a simple algebraic group. Let H be a reductive subgroup of G which contains a regular unipotent element of G. Such subgroups were classified by Saxl and Seit in all good characteristics. I’m actually interested in the characteristic zero version of this result, which apparently goes back to Dynkin. Saxl–Seit and Dynkin are difficult to read.

I’m wondering if there exists a modern reference for this classification over complex numbers. Or better, could some one please provide the sketch of an argument?

For quick reference, the classification is stated just after Proposition 8 in this paper.

Action of intersections of subgroups of a finite group

Suppose that we have a finite group $G$ with three subgroups $A,B,C$. I am interested in relating the action of $A cap B cap C$ on $B cap C$ with the action of $A cap B$ on $B$ (both via right multiplication). In particular, I am interested in relating transversals (a collection of representatives for the orbit space) for these actions.

I’m not sure exactly what we might be able to say in this situation. I suspect a result like the following may be true but I am not sure:


If $T subseteq B$ is a transversal for the coset space $B / (A cap B)$, then $T cap C$ is a transversal for $(B cap C) / (A cap B cap C)$.


Could anyone advise if I have this correct? If so, is the following stronger statement also true?


If $T subseteq B$ is a transversal for the coset space $B / (A cap B)$, then $T cap C$ is a transversal for $(B cap C) / (A cap B cap C)$, and every such transversal is obtained in this way.


If not, is there something we can say generally about this set up?

Abelian groups and their subgroups

It is well known that every finite abelian group is a direct product of cyclic groups. So for every $n$ every finite abelian group of exponent $n$ is a direct product of cyclic groups of order at most $n$.

Consider pairs $(A,H)$ where $A$ is a finite abelian group, $H$ is its subgroup. One can define a direct product $(A_1,H_1)times (A_2,H_2)=(A_1times A_2, H_1times H_2)$. Let $A$ have exponent $n$. Is there a finite number of pairs $(A_i,H_i)$ such that every pair $(A,H)$ is a direct product of several copies of $(A_i,H_i)$ (each pair can be used several times)?

Geodesics and one parameter subgroups of Lie groups

Let $G$ be a connected real Lie group and $mathfrak{g}$ its real Lie algebra equipped with some inner product and the corresponding norm $|cdot|$. Let $d$ be the left-invariant Riemannian distance on $G$ defined using the norm $|cdot|$.

As far as I know, it is not always true that one parameter subgroups $exp(tX)$, for $Xinmathfrak{g}$, are geodesic lines in $d$. However, are they locally geodesic? Or rather something weaker, fixing $Xinmathfrak{g}$, do we have $$lim_{tto 0} frac{t|X|}{d(exp(tX),e_G)}=1?$$

compactness – Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?

Let $G$ be a compact Lie group. An Abelian Lie subgroup $A leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A’$ such that $A leq A’ leq G$, then $A’ = A$.

Of course any maximal torus of $G$ (there is only one, up to conjugacy classes) is a maximal Abelian Lie subgroup, but there are other ones too, for example the Klein 4-group in $mathrm{SO}(3)$.

What I’m wondering is if the number of conjugacy classes of maximal Abelian Lie subgroups of any compact Lie group $G$ is always finite?

number of Sylow invariant subgroups under coprime action

Assume that $ G $ and $ A $ are finite groups such that A acts in $ G $. Let $ N $ is a $ A- $invariant normal subgroup of $ G $, first $ p $ and $ P ∈ Syl_p ^ A (G) $, we have

I don't know why the evidence $ P in Syl_p ^ A ( text {N} _ {PN} (P cap N)) $

with $ text {N} _ {PN} (P cap N) = {x in PN | (P cap N) ^ x = P cap N) } $
.