I am confused by expressions like $\mathbb{R}/\mathbb{Z}$, especially if I see something of the sort $SL(2,\mathbb{R})/SO(2,\mathbb{R})$. I don't know what to make of it. I understand what a quotient of a group for an equivalence looks like, but I can't get my head around expressions like above.

Is a quotient like this always a group and what do elements in the set $SL(2,\mathbb{R})/SO(2,\mathbb{R})$ look like?

  • 2
    $\begingroup$ The elements of $G/H$ are subsets of the form $g H = \{ gh, h \in H\}$ so that $G = \bigcup_{gH \in G/H} gH$. And $G/H$ is a group iff $\forall g \in G, gH= Hg$ ie. $H$ is normal. In this case you can define the group law on $G/H$ : $(g_1 H)(g_2H) = g_1 (Hg_2)H=g_1g_2 HH=(g_1g_2)H$. $\mathbb{R}$ is abelian thus $\mathbb{Z}$ is normal. $SO(2,\mathbb{R})$ is not normal because $g \in SO(2,\mathbb{R})$ ie. $gg^\top = 1$ doesn't mean $(\gamma g \gamma^{-1})(\gamma g \gamma^{-1})^\top = 1$. $\endgroup$ – reuns Nov 5 '17 at 20:13

They are a lot of questions in one post. I will do my best to answer them.

First, let me recall the notion of group action:

Definition $1$. Let $X$ be a set and let $G$ be a group, a left action of $G$ on $X$ is a map $\cdot\colon G\times X\rightarrow X$ which satisfies the two following axioms:

  • $\forall x\in X,e\cdot x=x.$

  • $\forall(g,g')\in G\times G,\forall x\in X,g\cdot(g'\cdot x)=gg'\cdot x.$

This definition is equivalent to the following:

Definition $2$. Let $X$ be a set and let $G$ be a group, a left action is group morphism from $G$ to $\mathfrak{S}(X)$.

If $\cdot\colon G\times X\rightarrow X$ is a group action in the sense of definition $1$, then $g\mapsto \{x\mapsto g\cdot x\}$ is a group action in the sense of definition $2$. Conversely, if $\varphi\colon G\rightarrow\mathfrak{S}(X)$ is a group homorphism, then $g\cdot x=\varphi(g)(x)$ is a group action in the sense of definition $1$.

Example 1. If $H$ is a subgroup of $G$, then $H$ acts on $G$ by left translation i.e. $(h,g)\mapsto hg$ is a group action.

Now, let us define the notion of quotient set in this general setup:

Definition $3$. Let $G$ be a group acting on a set $X$, then $G/X$ is the set of all orbits, namely: $$G/X:=\{G\cdot x;x\in X\}$$ where $G\cdot x:=\{g\cdot x;g\in G\}$ is the orbit of $x$.

The set $G/X$ is a collection of subsets of $X$. In all generality, this is not a group, for example, $X$ may not be a group itself.

Example $2$. If $H$ is a subgroup of $G$, then $G/H:=\{gH;g\in G\}$.

Let us investigate the structure of $G/H$ when $G$ is a group and $H$ is a subgroup acting by left translation.

Proposition $1$. Assume that $H$ is normal in $G$ i.e. for all $g\in G$, $gHg^{-1}=H$, then $G/H$ is endowed with a unique group structure such that the canonical projection $\pi\colon G\twoheadrightarrow G/H$ is a group morphism.

Proof. Let define the operation on $G/H$ to be $gH\cdot g'H=gg'H$ or equivalently : $\pi(g)\pi(g')=\pi(gg')$. The key point is that this definition does not depend on the choice of $g$ and $g'$, namely: $$gH=xH,g'H=x'H\Rightarrow gg'H=xx'H.$$ Indeed, there exists $h,h'\in H$ such that $x=gh$ and $x'=g'h'$, therefore, one has: $$xx'H=ghg'\underbrace{h'}_{\in H}H=ghg'H=gg'\underbrace{g'^{-1}hg'}_{\in H}H=gg'H.$$ Now that the operation on $G/H$ is well-defined, I let you check the remaining properties (associativity, existence of identity element, existence of inverse). Whence the result. $\Box$

Conversely, if $G/H$ is a group such that $\pi$ is a group morphism, then $H$ is normal as the kernel of $\pi$.

What I like to emphasize is that when working when quotient sets it is crucial to specify the group action!

Maybe I'll close this answer mentioning a useful result:

Theorem. Let $G$ be a group acting on $X$, then for all $x\in X$, there is a bijective correspondence: $$G/G_x\cong G\cdot x$$ where $G_x:=\{g\in G\textrm{ s.t. }g\cdot x=x\}$.

Proof. Consider $gG_x\mapsto g\cdot x$, in particular, check the well-definedness of this map. $\Box$

Remark 1. The set $G_x$ is a subgroup of $G$, but it may not be normal.

Example 3. $\mathbb{S}^n\cong SO(n+1)/SO(n)$ as sets, but even as smooth manifolds with a much-refined theorem.

Here, the action of $SO(n+1)$ on $\mathbb{S}^n$ is given by $(M,x)\mapsto Mx$ which is transitive, which means that each orbit under $SO(n+1)$ is the whole $\mathbb{S}^n$. If $x=(1,0,\ldots,0)$, then the stabilizer is given by: $$SO(n+1)_x=\left\{\begin{pmatrix}1 & 0\\0& A\end{pmatrix};A\in SO(n)\right\}\underset{\textrm{abusive}}{=}SO(n).$$ Indeed, the first column must be $x$ and by orthogonality the first entry of each other column must be $0$. Whence the result by the theorem.

Coming back to your main question (and guessing what the action you are working with is):

Let $\mathbb{H}$ be the upper-half plane of $\mathbb{C}$, namely the set of points in $\mathbb{C}$ having a strictly positive imaginary part. Let $SL(2,\mathbb{R})$ acts on $\mathbb{H}$ by homography: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\cdot z=\frac{az+b}{cz+d}.$$ Then this action is transitive and the stabilizer of $i$ is given by: $$SL(2,\mathbb{R})_i=\left\{\begin{pmatrix}a & b\\-b&a\end{pmatrix};a,b\in\mathbb{R}\right\}=SO(2).$$ Whence using the theorem, $SL(2,\mathbb{R})/SO(2)\cong\mathbb{H}$ as sets and again even as smooth manifolds.

  • $\begingroup$ First of all, thank you for your answer. Things are beginning to get clearer for me. I have some questions. In Definition 2, what is the codomain? (I don't know what the symbol stands for). The second question is, since $SO(2,\mathbb{R})$ is not normal, I have to think of it as you mention it a collection of subsets of $SO(2,\mathbb{R})$, or the set of orbits under the left multiplication of the group $SL$, right? $\endgroup$ – laguna Nov 5 '17 at 20:36
  • $\begingroup$ I believe that the notation I used in definition 2 is not widespread, $\mathfrak{S}(X)$ is the group of bijections from $X$ to $X$, in other words, the permutation group of $X$. Regarding your other question, yes that's exactly it. However, you should specify how $SL$ acts on $SO$ to get a better description of the quotient. $\endgroup$ – C. Falcon Nov 5 '17 at 20:39
  • $\begingroup$ Is there a possible connection with Iwasawa decomposition ? $\endgroup$ – Jean Marie Nov 5 '17 at 20:57
  • $\begingroup$ (+1) This is just beautiful! For myself I have had a hard time aswell wrapping my head around the concept of a quotient group. As I learned quite lately about the notion group actions the way of thinking about quotient groups you described above was not available for me yet; but now. This is revealing! Thank you for this :) $\endgroup$ – mrtaurho Apr 23 at 20:45

This quotient group consists of $2\times 2$ matrices with determinant 1 (because both $SL_2(\mathbb{R})$ and $SO_2(\mathbb{R})$ consist of matrices with determinant 1).

The quotient group is partitioned into equivalence classes of matrices that are not orthogonal, where elements of $SO_2(\mathbb{R})$ are treated as 0, and other elements are conceived of as their remainders mod an element of $SO_2(\mathbb{R})$.

A quotient of two groups $G/H$ is always a group as long as $H$ is a normal subgroup of $G$.

  • 2
    $\begingroup$ But $SO(2)$ is not a normal subgroup in $SL(2)$. $\endgroup$ – Dietrich Burde Nov 5 '17 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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