Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 3377

Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.

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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 $n+1$ …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar