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My question is exactly that on the title.

I'm interested in the action of some (discrete) subgroup $H$ on $\operatorname{GL}_n(\mathbb{R})$ by left multiplication.
For example, $H$ can be

  • $\operatorname{GL}_n(\mathbb{Z})$ or $\operatorname{PSL}_n(\mathbb{Z})$,
  • the set of $n\times n$ diagonal matrices that contain only $1$ or $-1$ on the diagonal.

I want to know the conditions on $H$ that guarantee the compactness of the quotient $\operatorname{GL}_n(\mathbb{R})/H$.
This seems like a well-studied subject, but couldn't find exactly what I want on this Wikipedia page. References are also welcome.

Bumblebee
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  • What do you mean by "when?" Sometimes it is, sometimes it is not. It is like asking "when is a number prime?" – Moishe Kohan Feb 14 '24 at 14:04
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    @MoisheKohan: A natural number $N$ is prime if there is an integer $1<n<N$ that divides $N.$ Likewise, given $H,$ I am looking for a criterion or algorithm to determine whether the quotient is compact or not. – Bumblebee Feb 14 '24 at 14:28
  • @MaximeCAILLEUX: First, $GL(R)$ is not compact, but locally compact. Secondly, every discrete subgroup of GL(R) is closed, but the quotients are not always compact. To see this consider $H={\operatorname{Id}_n}.$ – Bumblebee Feb 14 '24 at 14:31
  • Yep sorry I misread it ! Indeed GL(R) is of course only locally compact – Maxime CAILLEUX Feb 14 '24 at 14:34
  • To ask algorithmic questions you have to spend some time thinking about what do you mean by an algorithm if you are working with real numbers (there is no Church's thesis in this situation). There are several competing approaches to to computability over reals. Most likely answer, in any case, will be "this is undecidable" (which is the situation with discreteness of subgroups). – Moishe Kohan Feb 14 '24 at 14:34
  • As for criteria, a trivial one is: A discrete subgroup $\Gamma$ is cocompact iff there exists a compact $C\subset G=GL(n,R)$ such that $\Gamma C=G$. A less trivial criterion is due to Godement, in the case of lattices: A lattice is uniform iff it contains no unipotent elements. – Moishe Kohan Feb 14 '24 at 14:45
  • @MoisheKohan: Fair enough. The motivation for this question came from this question. I was believing that $GL(R)/GL(Z)$ is compact, but now I in doubt. I will see whether this particular subgroup has non-trivial unipotent elements or not. Thanks for the help. – Bumblebee Feb 14 '24 at 14:52
  • No, it is noncompact. – Moishe Kohan Feb 14 '24 at 15:26
  • @MoisheKohan: Yes, it has some cute non-trivial unipotents. If you like to convert your comment mentioning "Godement's criteria", I'm happy to accept it. – Bumblebee Feb 14 '24 at 15:31
  • I will, a bit later. In fact, $GL(n,Z)$ is not even a lattice in $GL(n,R)$. – Moishe Kohan Feb 14 '24 at 15:50
  • Ohhhhh. Really? – Bumblebee Feb 14 '24 at 17:03
  • The correct statement is that $SL(n,Z)$ is a lattice in $SL(n,R)$. The issue with $GL(n,Z)$ is that is contains $SL(n,Z)$ as an index 2 subgroup. If you put a left-invariant Riemannian metric on $GL(n,R)$, then it splits isometrically as the product $R^\times\times SL(n,R)$. From this, you see that $SL(n,Z)\backslash GL(n,R)$ has infinite volume. Ditto, the quotient by $GL(n,Z)$. – Moishe Kohan Feb 15 '24 at 01:56

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First of all, $GL(n, {\mathbb Z})$ contains $SL(n, {\mathbb Z})$ as index 2 subgroup. Therefore, for $G= GL(n, {\mathbb R})$, $GL(n, {\mathbb Z})\backslash G$ is compact if and only if $SL(n, {\mathbb Z})\backslash G$ is compact. To see that the latter quotient is noncompact, observe that $G$ splits as the direct product $$ SL(n, {\mathbb R})\times {\mathbb R}^\times. $$ This product decomposition is preserved by the left translations via elements of $SL(n, {\mathbb Z})$. Therefore, $SL(n, {\mathbb Z})\backslash G$ is homeomorphic to $$ (SL(n, {\mathbb Z}) \backslash SL(n, {\mathbb R}))\times {\mathbb R}^\times $$ and, hence, is noncompact.

More interestingly, already the quotient $$ SL(n, {\mathbb Z}) \backslash SL(n, {\mathbb R}) $$ is noncompact. This can be seeing as a consequence of the Godement's criterion:

(a) Suppose that $G$ is a semisimple Lie group and $\Gamma< G$ is a discrete subgroup containing unipotent elements. Then $\Gamma\backslash G$ is noncompact.

(b) Suppose that $\Gamma< G$ is a lattice, i.e. $\Gamma\backslash G$ has finite volume (where we equip $G$ with the Haar measure). Then $\Gamma\backslash G$ is noncompact if and only if $\Gamma$ contains nontrivial unipotent elements.

Now, $SL(n, {\mathbb Z})$ contains some nontrivial unipotent elements (strictly upper-triangular matrices). Hence, $$ SL(n, {\mathbb Z}) \backslash SL(n, {\mathbb R}) $$ is noncompact. Incidentally, $SL(n, {\mathbb Z})$ is a lattice in $SL(n, {\mathbb R})$ (since it is an arithmetic subgroup of $SL(n, {\mathbb Q})$).

You can find much more in

Raghunathan, M. S., Discrete subgroups of Lie groups, Ergebnisse der Mathematik und ihrer Grenzgebiete. Band 68. Berlin-Heidelberg-New York: Springer-Verlag. VIII,227 p. (1972). ZBL0254.22005.

Morris, Dave Witte, Introduction to arithmetic groups, [s.l.]: Deductive Press (ISBN 978-0-9865716-0-2/pbk; 978-0-9865716-1-9/hbk). xii, 475 p. (2015). ZBL1319.22007.

Moishe Kohan
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  • Thank you very much for your detailed answer, and the communication. I will definitely go through the reference that you linked. What else do we know about the quotient space GL(R)/GL(Z)? Is it at least connected? – Bumblebee Feb 18 '24 at 15:05
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    @Bumblebee: Yes, it is connected. – Moishe Kohan Feb 18 '24 at 15:41
  • How can I see this? I have been thinking about this space (moduli space of lattices) for the last few days. This must be a well-studied object in differential geometry and number theory. But I still could not find any reference for it. Do you know something by any chance? – Bumblebee Feb 18 '24 at 16:01
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    @Bumblebee: Assuming that you already know that $GL_+(n,R)$ (the group of matrices with positive determinant) is connected: It is then clear that $GL(n,R)$ has exactly two components (detected by the sign of the determinant). Clearly, $GL(n,Z)$ intersects both components. From this, connectivity of the quotient follows. More generally, if $G$ is a topological group and $H< G$ is a subgroup which intersects all components of $G$, then $G/H$ is connected since it is homeomorphic to $G_0/(H\cap G_0)$, where $G_0$ is the identity component. – Moishe Kohan Feb 18 '24 at 16:25
  • For connectivity of $GL_+(n,R)$, see for instance here. – Moishe Kohan Feb 18 '24 at 16:26