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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
5
votes
1
answer
228
views
A group in a neighbourhood of a Zariski dense subgroup
By Borel's theorem, lattices in simple Lie groups are Zariski dense. I expect that a small (in metric sense) deformation of a lattice in a Lie group is also Zariski dense.
Suppose we have a Zariski de …
1
vote
A group in a neighbourhood of a Zariski dense subgroup
This is an update, sorry for the trouble.
Let $M$ be a metric space. We say that $X \subset M$ is coarse equivalent to $Y \subset M$ if the the Hausdorff distance $d_H(X, Y)$ is finite, that is, there …
7
votes
1
answer
462
views
Stabilizer of Sp(n) and U(n) in GL(n)
I would be very grateful for a reference
to the following results (which are, I think, true,
though I never saw it in the literature).
Let $G\subset GL(n,{\Bbb C})$ be $U(n)$,
abd $A\in GL(2n,{\Bbb …
7
votes
Is the group of isometries of a homogeneous Riemannian manifold maximal?
Let $G$ be a compact Lie group with left-invariant metric $h$. For general $h$, the group of isometries is $G$. However, when $h$ is bi-invariant, the group of isometries is $G\times G$. I think this …
0
votes
Action of a Lie group with finitely many orbits
Let $M$ be the vector space $V$
of quadratic forms on ${\Bbb R}^n$ (or its projectivization),
and $G=GL(n)$. The open orbits correspond to non-degenerate quadratic forms, and there are precisely $n+1$ …
15
votes
3
answers
1k
views
orbits of automorphism group for indefinite lattices
I have a question about indefinite lattices.
QUESTION: Let $\Lambda\times\Lambda\rightarrow {\Bbb Z}$ be a lattice,
that is, ${\Bbb Z}^n$ with a non-degenerate integer quadratic form,
not necessarily …