Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H=\left\{ z\in\mathbb{C}\mid\Im\left(z\right)>0\right\}$ be the upper-half Poincare plane. Let $GL\left(2,\mathbb{R}\right)$ be the general linear group, $Z\left(GL\left(2,\mathbb{R}\right)\right)$ be the center of the general linear group and $O\left(2,\mathbb{R}\right)$ be the orthogonal subgroup of $GL\left(2,\mathbb{R}\right)$.

What does it mean to say $H=GL\left(2,\mathbb{R}\right)/\left(Z\left(GL\left(2,\mathbb{R}\right)\right)\cdot O\left(2,\mathbb{R}\right)\right)$? The left-hand side is a metric space and the right hand side is a set of cosets of $GL\left(2,\mathbb{R}\right)$. So I'm confused about what it means to write that they are equal or to say "the upper half plane is..." It seems like this would be the group of orientation preserving isometries of H, but I still find the terminology confusing.

I've been trying to figure out what this could possibly mean, but my searches on the internet have not been fruitful. I've also looked at 2 sources on standard modular groups but they make no mention of this fact. An explanation or reference would be greatly appreciated.

Motivation: I am reading a paper titled "On Modular Functions in characteristic p" by Wen-Ch'ing Winnie Li which can be found at http://www.jstor.org/stable/1997973. The claim appears on page 3 of the pdf (page 232 of the journal). It is also stated on the wikipedia page: http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model

share|improve this question
add comment

2 Answers 2

up vote 5 down vote accepted

My comments were getting too long, so I will post this as an answer: the isomorphism referred to is an isomorphism of $G$-sets. This works in great generality: let $H$ be a set, let $G$ act on $H$ transitively. Pick an arbitrary point $z\in H$ and let $K=\text{Stab}_G(z)$ be the point stabiliser in $G$. Note that $K$ is not normal in general, since the group $gKg^{-1}$ stabilises the point $g(z)$ (my action is on the left). So, $K$ is normal if and only if it acts trivially on $H$.

Nevertheless, the set of cosets $G/K$ is always a $G$-set, i.e. a set with an action of $G$: $$g: hK\mapsto (gh)K$$ for all $g\in G$ and $hK\in G/K$. This is the usual coset action. Now, check that the map $$\phi:G/K\rightarrow H,\;gK\mapsto g(z)$$ is a bijection of $G$-sets, i.e. a bijection of sets that respects the $G$-action.

share|improve this answer
I see it now, thanks! It's much clearer in this setting. I'm guessing the reason '=' is used is to mean the actions are canonically isomorphic, like when we write V**=V with dual spaces? –  WWright Nov 30 '10 at 5:48
Yes, although I am not a big fan of this notation. There are various different notational conventions to distinguish between isomorphisms and canonical isomorphisms. –  Alex B. Nov 30 '10 at 7:07
add comment

H is not a group, but rather a space with a group acting on it.

If K ≤ G is a subgroup, then G/K is not usually a group, but rather a set of cosets on which G acts. G/K is only a group if K is normal.

In your situation, the subgroup is not normal, so you only get a group action.

One keyword form the wikipedia article is "isotropy subgroup". If G acts transitively on a space H, then the set K of elements of G that fix some particular point of H forms a subgroup (the "isotropy subgroup"), and the action of G on H is isomorphic to the action of G on the cosets in G/K.

share|improve this answer
If I understand you correctly the equality claimed in the updated title really means the last part of your answer and this is just some sort of abuse of language. It would seem then that as sets $H\cong G/K$ but I'm not sure I see what the isomorphism map would be. –  WWright Nov 30 '10 at 4:46
@WWright It's more than just a bijection of sets. It's a bijection of $G$-sets, i.e. a bijection of sets that preserves the $G$-action. To give you an easier example of why the latter is much stronger: consider a square lattice of points. I will define two different actions of the cyclic group with two elements on it: in the first, my group reflects in the horizontal axis and in the second it reflects in a diagonal. Now, it's easy to find a linear bijection from the lattice to itself. But can you find a linear bijection that commutes with the group action? –  Alex B. Nov 30 '10 at 4:51
@Alex Bartel - I see your point, but I'm having trouble understanding what an appropriate map to be. Since the action is Mobius transformation, I'm confused about how we can have a map $\varphi:H\rightarrow G/K$ such that $\varphi\left(gz\right)=g\cdot\varphi\left(z\right)$ because then it seems we would be applying a Mobius transformation to a matrix ($\varphi\left(z\right)$ coset, which doesn't seem to make sense. –  WWright Nov 30 '10 at 5:07
Nevermind, it must be that we let G act on the matrix cosets by multiplication, but I'm still stuck on what the map $\varphi$ should be. –  WWright Nov 30 '10 at 5:19
See my answer. It's more natural to define the map the other way round. –  Alex B. Nov 30 '10 at 5:27
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.