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

9 votes
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Non-linear Lie group

The traditional example is the universal cover of $SL(2,\mathbb{R})$. You can look e.g. at the wikipedia article on $SL(2,\mathbb{R})$.
Alex Eskin's user avatar
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15 votes

Finite-dimensional subgroups of circle diffeomorphism group

The answer is indeed no, as described e.g. in the lecture notes by Ghys http://www.math.ethz.ch/~bgabi/ghys%20groups%20acting%20on%20the%20circle.pdf Section 4.1 has a list of all connected groups a …
Alex Eskin's user avatar
  • 3,201
12 votes

Finite-dimensional subgroups of diffeomorphism groups

EDIT: Correctly state Zimmer's conjecture. This does not really answer the question, but the question of which (higher rank) Lie groups act by diffeomorphism on which smooth manifolds is called the …
Alex Eskin's user avatar
  • 3,201
7 votes
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Measuring how far from being cocompact a lattice is

It seems to be that what you are asking for is roughly the measure of a neighborhood of the "cusp" of $G/\Gamma$. For the case of $\Omega(n) = SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ there is a classical …
Alex Eskin's user avatar
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13 votes
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Lattices in $SL(n,\mathbb R)$

The answer is yes. It is theorem [2.13] of the following paper of Prasad and Raghunathan: Prasad, Gopal; Raghunathan, M. S. Cartan subgroups and lattices in semi-simple groups. Ann. of Math. (2) 96 ( …
Alex Eskin's user avatar
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39 votes

Volume of fundamental domain and Haar measure

In order to talk meaningfully about the volume of $SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ you need to define a normalization for Haar measure. One way to think about it is as follows: the space $M_n(\ma …
Alex Eskin's user avatar
  • 3,201
8 votes
Accepted

Lattices in SOL

To add to Igor Rivin's answer: it seems that all the lattices in SOL are isomorphic as abstract groups to $\mathbb{Z}^2\rtimes_A\mathbb{Z}$ for hyperbolic $A\in SL_2(\mathbb{Z})$. If I am reading the …
Alex Eskin's user avatar
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