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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
1
vote
Decomposing maximal compact subgroups of SO(n,1)
The pair (SO(n), SO(n-1))) is a symmetric pair, hence in this case M is a maximal subgroup of SO(n). Thus the connected component of your subgroup must the whole group.
-1
votes
Maximal abelian subalgebras of Lie algebras over $\mathbb{C}$
I am sorry to tell I believe the answer is No.
Consider $\mathfrak g = \mathfrak{su}(2)$ and its complexification $\mathfrak g_{\mathbb C} = \mathfrak{sl}(2,\mathbb C)$,
and $\mathfrak h= \mathfrak{s …
0
votes
Does there always exist an irreducible representation occurring with multiplicity one when i...
In Vogan's green book, on page 400 it is shown that given an irreducible $(\mathfrak g, K)-$module, then, each lowest $K-$type has multiplicity one.
Perhaps this gives a partial answer to your questi …
-1
votes
Automorphism group of real orthogonal Lie groups
Check the book Freudenthal Linear Lie groups, in this book you find a
complete treatment of Aut(G) for G a real semisimple (connected) Lie group.
best regards
0
votes
Automorphism group of real orthogonal Lie groups
also check
Onishchik: Lectures on Real Semisimple Lie Algebras and Their Representations.
1
vote
Automorphism group of real orthogonal Lie groups
on page 386 (paragraph 66.7) you find the table Out(G)/Int(G)
on page 387 you find D_{l,j} j>1 your Lie algebras so(p,q) when your p or q is even.
on page 391 you find so(p,q) when both p,q are odd
T …
5
votes
0
answers
137
views
Differential operators on $G/K$
Let $G$ be a connected Lie group and $K$ a compact subgroug of $G$. The question is about the algebra of the differential operators $Diff(G/K)$ on $G/K.$ Let $U(\mathfrak g)$ denote the universal enve …
1
vote
When representations of reductive Lie group in a Banach space and in its Garding space have ...
If you consider $V^\infty$ with its Fréchet topology. and irreducibility means no closed invariant subspaces except for the trivial ones. Then $V$ is irreducible iff $V^\infty $ is irreducible. In the …
1
vote
When representations of reductive Lie group in a Banach space and in its Garding space have ...
I do not know if you need K-admissibility. You can check if it fails in the following example: write the usual Iwasawa for $\operatorname{SL}_2(R)=KAN$, $MAN$ usual parabolic. Fix $ T$ a continuous li …
2
votes
How can one show $G/T$ is a coadjoint orbit for a compact Lie group $G$ and $T$ its maximal ...
Fix a regular element $\lambda$ in $Lie(T)\subset Lie(G)$, then the coadjoint orbit $Ad(G) \lambda$ is isomorphic to $G/T.$ Best,
4
votes
0
answers
73
views
a property of the characters for center of universal enveloping algebra
Let $\mathfrak g$ be a complex simple Lie algebra. We fix Cartan subalgebra $\mathfrak h$ and a system of positive roots $\Psi$ for the root system of the pair $(\mathfrak g, \mathfrak h).$ For each …
1
vote
Construct discrete series of SL(2,R) as kernel of twisted Dirac operators
Check an old paper of Joseph Wolf in Journal of Mechanics ... early sixties , there you find a detailed answer to your question. I do not have access to Mathscinet for a better answer. Best regards J …
-1
votes
Is $M=E_{7(7)}/SU(7)\times\mathbb{R}^{+}$ a (pseudo)Kähler-Hodge manifold? Open problem
If i am not wrong the following is true, G compact connected Lie group, H the centralizer of a torus in G, then G/H is a projective manifold, does this answer your question??
2
votes
Branching laws for $SO(n)$
It is not multiplicity free..., check the green book of Antony Knapp, or else the old book of Zelobenko, compact.....
2
votes
Branching laws for $SO(n)$
check Eastwood-Wolf, branchig of ...., Arxive 0812.0822 math[RT] in this paper you find who to compute branching laws useing LiE.