lie groups – Why is the quotient by the Borel subgroup compact?

Let $G$ be a complex lie group with semisimple lie algebra $g$.

Put $b$ the borel subalgebra (so the sum of $h$ and all the positive roots) for some choices.

One can there is the borel subgroup $B$ with lie algebra $b$ in $G$; given as a the stabilizer of $b$ in the adjoint action. My book claims now that $G/B$ is compact, since it’s isomorphic to its image in the Grassmanian (i.e where the adjoint sends $b$) and the Grassmanian is compact. However, how do we know that the image is closed?

Decomposition of Lie algebra: do the simple and maximal torus parts commute?

I have the following exercise:

Consider a Lie algebra $mathfrak{g}$. Decompose $mathfrak{g}$ using the Levi decomposition, so $mathfrak{g}=mathfrak{s}oplus mathfrak{r}$. Let $mathfrak{a}$ be a maximal torus inside the radical. Is it always true that $(mathfrak{s},mathfrak{a})=0$? Discuss which hypotesis are needed to obtain such a result.

Now I started assuming that $mathfrak{g}$ is algebraic (to avoid exceptional or strange cases, if any), and I procede as suggested, so I get
$$mathfrak{g}=mathfrak{s}oplus mathfrak{a}oplusmathfrak{m}$$
for some complement $mathfrak{m}$ of $mathfrak{a}$ in the radical $mathfrak{r}$. For sure $(mathfrak{s},mathfrak{a})subset mathfrak{r}$. But then I don’t have any idea, nor I can prove it is true always (I guess it is false).

For the second question, I can think only of “trivial” answers: it is true for solvable Lie algebras and for semi-simple Lie algebras.

Suggestions?

lie groups – Generating $K$-types of a $(mathfrak g,K)$-module for $K$ disconnected

Let $G$ be a real reductive Lie group, let $K$ be a maximal compact subgroup of $G$, and let $V$ be a $(mathfrak g,K)$-module. For $sigmainwidehat{K}$ we denote the $sigma$-isotypic component of $V$ by $V_sigma$, so that $V=bigoplus_{sigmainwidehat{K}}V_sigma$.

Question: Suppose that $V$ is generated by a $K$-invariant subspace $Wsubseteq V_sigma$ for some $sigma$, i.e., $V=U(mathfrak g)W$. Is it true that $V_sigma=U(mathfrak g)^KW$? Here $U(mathfrak g)^K$ is the $K$-invariant part of the enveloping algebra $U(mathfrak g)$ under the adjoint action of $K$.

This statement is true when $K$ is connected, and basically follows from Proposition 9.1.10 in Dixmier’s book Enveloping Algebras. However, I am interested in the cases where $K$ is disconnected. In the latter cases, I cannot generalize Dixmier’s argument. But it appears to me that in some references/papers it is assumed to be true.

semisimple lie algebras – Which real Lie group is the stabilizer of this Hermitian form?

Let $hcolon mathbb{C}^{2n} times mathbb{C}^{2n} rightarrow mathbb{C}$ be the form
$$h(vec{v},vec{w}) = left(v_1 overline{w}_1 + cdots + v_n overline{w}_nright) – left(v_{n+1} overline{w}_{n+1} + cdots + v_{2n} overline{w}_{2n}right).$$
Define $G$ to be the subgroup of $text{GL}(2n,mathbb{C})$ stabilizing $h$, i.e. $G$ is all invertible $2n times 2n$ complex matrices $M$ such that $h(M vec{v}, M vec{w}) = h(vec{v},vec{w})$ for all $vec{v},vec{w} in mathbb{C}^{2n}$. Equivalently, $M$ should satisfy
$$M^t left(begin{matrix} mathbb{1}_n & 0 \ 0 & -mathbb{1}_n end{matrix}right) overline{M} = left(begin{matrix} mathbb{1}_n & 0 \ 0 & -mathbb{1}_n end{matrix}right).$$
Question: which real Lie group is $G$? I assume it is reductive, but I have trouble identifying it with a product of any of the standard reductive real Lie groups. But my research is usually in infinite discrete groups, so I might just be ignorant of standard Lie group stuff.

I’d also be interested in the more general question where you distribute the $pm 1$‘s along the diagonal of the matrix defining $h$ differently (and thus don’t insist that the dimension be even). If the matrix is just the identity, then you get the usual unitary group.

lie groups – Rank of a Lie subgroup generated by two Lie subgroups

Let $G$ be a compact connected Lie group and $H$, $K$ be two closed connected subgroups. By the answer to the question asked here

In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?

$L=langle K, Hrangle$ (the subgroup generated by $K$, $H$) is a closed connected subgroup of $G$ (hence a Lie subgroup). Now assume additionally that $$r=rank~(Hcap K)=rank~H=rank~K.$$ My question is if we can conclude that $$rank~L=r.$$

tag removed – I need to find out 7 data points randomly for which I know the range it should lie in and the mean …. Please help me how to find them… Thanks


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rt.representation theory – Under what conditions representations of reductive Lie group in Banach space and in its Garding space have the same length?

Let $G$ be a real reductive Lie group (e.g. $G=GL(n,mathbb{R})$). Let $rho$ be a continuous representation of $G$ in a Banach space $V$. Let $V^inftysubset V$ be the subspace of smooth vectors equipped with the Garding topology. Let $rho^infty$ be the natural representation of $G$ in $V^infty$.

Under what precise technical conditions the representations $rho$ and $rho^infty$ have the same length?

A reference would be very helpful.

dnd 5e – Can someone who fails the save against a Zone of Truth spell lie non-vocally?

By RAW, you can lie non vocally in a Zone of Truth.

The Zone guards against deception. PHB page 178 states

Your Charisma (Deception) check determines whether you can convincingly hide the truth, either verbally or through your actions.

In the spell description of Zone of Truth, they are not referring to Deception the skill, but the definition of the skill lines up fairly well with the english language definition, which states that deception is the act of decieving someone, and decieve is defined as making someone believe something that is not true, with no mention of having to speak.

This means deception does not have to be vocal, or even verbal. However, there is some ambiguity.
You cannot speak a deliberate lie. Speaking clearly means talking, making a sound with your voice in a certain language. Body language, sign language, and pantomime are not speaking.

By RAI, you cannot lie non vocally in a Zone of Truth.

They say speak a deliberate lie, yes, and RAW must interpret every word of the spell description. However, it seems obvious to me that they intended no deception to be possible, as shown in the earlier line.

dnd 5e – Can Someone Lie Non-Verbally in Zone of Truth?

Say one of my party case Zone of Truth on a mute person or perhaps the party trickster simultaneously case Silence in the same area. Under the mechanics of Zone of Truth, can those affected lie in sign language or other non-verbal forms of communication?

The text of Zone of Truth says:

you create a magical Zone that guards against deception

But later on it says:

On a failed save, a creature can’t speak a deliberate lie while in the radius.

So, what do y’all think?

lie algebras – Exponential map generates the identity component of closed linear groups

In page 15 of “Lie groups beyond an introduction” by Prof. Knapp, Corollary 0.20 states that the exponential map (of linear Lie algebra $mathfrak g$) generates the identity component $G_0$ of a closed linear group $G$. The proof goes like

1 by continuity of the exponential map $exp(mathfrak g)$ is connected

2 $exp(mathfrak g)$ contains a neighborhood of 1 in the group (local diffeomorphism)

3 the smallest subgroup containing a nonempty open set in $G_0$ must be $G_0$

3 doesn’t look obvious to me though, I don’t know how to prove that $exp(mathfrak g)$ is a subgroup. In addition, if we can show it’s open, then it’s an open closed connected subgroup, so it must be $G_0$ since $G_0$ is connected.