SL(2,R) As a Topological Space and Topological Group

Introduction

There are a lot of important linear algebraic groups that are widely used in mathematics, physics and industry. Some of them have nice visualisations. For example, it is widely known that $SU(2) \cong S^3$ and $SO(3) \cong \mathbb{RP}^3$. The group $SL(2,\mathbb{R})$ is not less important than them but the visualisation concerning this group is not very easy to be found. In this post we show that

$$SL(2,\mathbb{R}) \cong S^1 \times \mathbb{R} \times \mathbb{R} \cong S^1 \times D,$$

where $D$ is the open unit disk. In other words, $SL(2,\mathbb{R})$ can be considered as a donut, not the shell of it ($S^1 \times S^1$) but the “content” or “flesh” of it. More formally, the inside of a solid torus.

The related core theory can be found in Iwasawa decomposition, but to access it we need Lie group and Lie algebra theories, which involves differential geometry and certainly goes beyond the scope of this post. Interested readers can refer to Lie Groups Beyond an Introduction chapter 6 for Iwasawa decomposition theory.

Immediate topological consequences

Before we establish the homeomorphism

$$SL(2,\mathbb{R}) \cong S^1 \times \mathbb{R} \times \mathbb{R} \cong S^1 \times D$$

we first see what we can derive from it.

• Is $SL(2,\mathbb{R})$ compact?

No. Since $D$ is not compact, $S^1 \times D$ cannot be compact.

• What is the fundamental group of $SL(2,\mathbb{R})$?

Notice there is a (strong) deformation retract between $S^1 \times D$ and $S^1$. Therefore $\pi_1(SL(2,\mathbb{R})) = \pi_1(S^1)=\mathbb{Z}$.

• Connectedness of $SL(2,\mathbb{R})$?

It is connected because $S^1$ and $D$ are connected. It is not simply connected because the fundamental group is not trivial.

• What is the dimension of $SL(2,\mathbb{R})$ as a manifold?

The dimension is $3$.

The Iwasawa decomposition

If we directly jump to the conclusion without mentioning Lie theory, one will see the decomposition comes from nowhere. Instead of defining $K$, $A$ and $N$ that will appear later and show that there is no discrepancy, we deduce the decomposition without the usage of Lie theory. Instead, we consider the action of $SL(2,\mathbb{R})$ on the upper half plane, because group action is likely to expose more information of the group.

Consider the group action of $SL(2,\mathbb{R})$ on the upper half plane

$$\mathfrak{H}=\{z \in \mathbb{C}:\Im(z)>0\}$$

given by

$$\begin{pmatrix} a & b \\ c& d \end{pmatrix}(z)=\frac{az+b}{cz+d}.$$

Up to an explosion of calculation, one can indeed verify that this is a group action and in particular

$$\Im\left(\frac{az+b}{cz+d}\right)=\frac{\Im(z)}{|cz+d|^2}.$$

As one may guess, it is not wise to continue without investigating the action first, or we will be lost in calculation. We first show that this action is transitive by showing that for any $z=x+yi \in \mathfrak{H}$, there is some $\sigma \in SL(2,\mathbb{R})$ such that $\sigma(z)=i$:

$$\frac{az+b}{cz+d}=\frac{ax+b+ayi}{cx+d+cyi}=i \implies \begin{cases} ax+b=-cy \\ cx+d=ay \end{cases}$$

Let’s play around the last linear equation system:

$$\begin{cases} ax+b=-cy \\ cx+d=ay \end{cases} \implies \begin{cases} acx+bc=-c^2y \\ acx+ad=a^2y \end{cases} \implies a^2+c^2=\frac{1}{y}.$$

We can put $c=0$ and $a=\frac{1}{\sqrt{y}}$ so that $b=-\frac{x}{\sqrt{y}}$ and $d=\sqrt{y}$. That is,

$$\begin{pmatrix} \frac{1}{\sqrt{y}} & -\frac{x}{\sqrt{y}} \\ 0 & \sqrt{y} \end{pmatrix}(x+iy)=i.$$

We have therefore proved:

The action of $SL(2,\mathbb{R})$ on $\mathfrak{H}$ is transitive.

Proof. For any $z,z’ \in \mathfrak{H}$, there exists $\sigma$ and $\sigma’$ such that $\sigma(z)=i$ and $\sigma’(z’)=i$. Then $\sigma’^{-1}(\sigma(z))=z’$, i.e. $\sigma’^{-1}\sigma$ sends $z$ to $z’$. $\square$

By working around $i$ on $\mathfrak{H}$ we can save ourselves from a lot of troubles. It is then desirable to find the stabiliser of $i$.

The stabiliser of $i \in \mathfrak{H}$ is $SO(2) \cong S^1$.

Proof. Suppose $\sigma=\begin{pmatrix} a & b \\c & d \end{pmatrix}$ stabilises $i$. Then first of all we have

$$\Im\left(\frac{ai+b}{ci+d}\right)=\frac{1}{|ci+d|^2}=1 \implies c^2+d^2=1.$$

Then

$$\frac{ai+b}{ci+d}=i \implies ai+b=-c+di \implies \begin{cases}a=d \\ b=-c\end{cases}$$

It follows that

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix}= \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2 \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$

Therefore $\sigma \in O(2) \cap SL(2) = SO(2)$ as expected. $\square$

With these being said, the action of $SL(2,\mathbb{R})$ on $i$ consists of $SO(2)$ that moves nothing and the rest that actually move things. In other words, $SL(2,\mathbb{R})/SO(2) \cong \mathfrak{H}$ as a $2$-manifold. In particular, the action is a isometry. We will find the effective part of the group action out. For $\sigma \in SL(2,\mathbb{R})$, we assume that $\sigma(i)=x+iy$. Then

$$\sigma(i)=\begin{pmatrix}\frac{1}{\sqrt{y}} & -\frac{x}{\sqrt{y}} \\ 0 & \sqrt{y} \end{pmatrix}^{-1}(i)=\begin{pmatrix}\sqrt{y} & \frac{x}{\sqrt{y}} \\ 0 & \frac{1}{\sqrt{y}}\end{pmatrix}(i).$$

Let $B$ be the upper triangular matrices in $SL(2,\mathbb{R})$ with positive diagonal elements. Then it is elements in $B$ that actually move things. According to this classification, we have obtained a decomposition

The matrix multiplication map $B \times SO(2) \to SL(2,\mathbb{R})$ is surjective.

Proof. Notice that every element of $B$ can be written in the form

$$\lambda_{x,y}=\begin{pmatrix}\sqrt{y} & \frac{x}{\sqrt{y}} \\ 0 & \frac{1}{\sqrt{y}}\end{pmatrix}.$$

For any $\sigma \in SL(2,\mathbb{R})$, suppose $\sigma(i)=x+iy$, then $\sigma(i)=\lambda_{x,y}(i)$, therefore $\lambda_{x,y}^{-1}\sigma(i)=i$, i.e. $\lambda_{x,y}^{-1}\sigma \in SO(2)$, i.e. $\lambda_{x,y}^{-1}\sigma$ is a stabiliser of $i$. The product $\sigma = \lambda_{x,y}(\lambda_{x,y}^{-1}\sigma)$ always lies in the image of $B \times SO(2)$. $\square$

We can decompose $B$ further:

$$\begin{pmatrix}\sqrt{y} & \frac{x}{\sqrt{y}} \\ 0 & \frac{1}{\sqrt{y}}\end{pmatrix}=\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix}\begin{pmatrix}\sqrt{y} & 0 \\ 0 & \frac{1}{\sqrt{y}} \end{pmatrix}.$$

Let $N$ be the group of upper triangular matrices in $SL(2,\mathbb{R})$ with $1$ on the diagonal line and let $A$ be the group of diagonal matrices with non-negative entries. Then $B=NA$. Let $K=SO(2) \subset SL(2,\mathbb{R})$, then we have obtained the so-called Iwasawa decomposition:

There is a diffeomorphism onto

$$\begin{aligned} N \times A \times K &\to SL(2,\mathbb{R}) \\ (n,a,k) &\mapsto nak \end{aligned}$$

Proof. It only remains to show injectivity. Suppose $n_1a_1k_1=n_2a_2k_2$. Applying both sides onto $i$ we obtain $n_1a_1(i)=n_2a_2(i)$. Suppose

$$n_i=\begin{pmatrix} 1 & x_i \\ 0 & 1 \end{pmatrix},\; a_i=\begin{pmatrix} \sqrt{y_i} & 0 \\ 0 & \frac{1}{\sqrt{y_i}} \end{pmatrix},\; i=1,2.$$

Then we have $n_1a_1(i)=x_1+y_1i=n_2a_2(i)=x_2+y_2i$. It follows that $x_1=x_2$ and $y_1=y_2$, i.e. $n_1=n_2$ and $a_1=a_2$ and therefore $k_1=k_2$. $\square$

By investigating $N$ and $A$ further we obtain

The group $SL(2,\mathbb{R})$ is homeomorphic to $S^1 \times D$.

Proof. Notice that $N$ is homeomorphic to $\mathbb{R}$ and $A$ is homeomorphic to $\mathbb{R}_{>0}\cong \mathbb{R}$. $\square$

Notice the order of $N,A,K$ does not matter very much: $NAK,KAN,ANK,KNA$ are the same thing. This is because $AN=NA$ and for $nak \in SL(2,\mathbb{R})$, we have $(nak)^{-1}=k^{-1}a^{-1}n^{-1}$ which lies in the preimage of $K \times A \times N$ under matrix multiplication.

Immediate group-theoretical consequences

With the full Iwasawa decomposition in mind, we can scratch the surface of the rather complicated $SL(2,\mathbb{R})$.

The only continuous homomorphism of $SL(2,\mathbb{R})$ to $\mathbb{R}$ is trivial.

Proof. Let $f:SL(2,\mathbb{R}) \to \mathbb{R}$ be such a map. We have $f(kan)=f(k)+f(a)+f(n)$. We need to show that $f(k)=f(a)=f(n)=0$.

First of all, since $K$ is a compact subgroup of $SL(2,\mathbb{R})$, its image on $\mathbb{R}$ has to be a compact subgroup. On the other hand, $f$ on $A$ and $N$ can be constructed more explicitly. For $A$, we see $\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r} \end{pmatrix} \mapsto r \mapsto \log{r}$ yields an isomorphism of $A$ and $\mathbb{R}$, in both algebraical and topological sense. For $N$ on the other hand, we immediately have an isomorphism $\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix} \mapsto x$. Therefore the image of $f$ on $A$ and $N$ can be realised as as $u\log{r}$ and $vx$ for some $u,v \in \mathbb{R}$. We use the fact that $AN=NA$ to determine $u$ and $v$. Notice that

$$\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r} \end{pmatrix}\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix}=\begin{pmatrix}1 & r^2x \\ 0 & 1\end{pmatrix}\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r}\end{pmatrix},$$

applying $f$ on both sides, we have

$$u\log{r}+vx=vr^2x+u\log{r}\;\forall x\in \mathbb{R},\;r>0 \implies v=0$$

For $u$, we consider the conjugate relation

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r} \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{r} & 0 \\ 0 & r \end{pmatrix}=\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r} \end{pmatrix}^{-1}.$$

Applying $f$ on both sides we obtain

$$u\log{r}=-u\log{r},\;\forall r>0 \implies u=0.$$

This proves the triviality of $f$. $\square$

Let $f:SL(2,\mathbb{R}) \to GL(n,\mathbb{R})$ be a continuous homomorphism, then $f(SL(2,\mathbb{R})) \subset SL(n,\mathbb{R})$.

Proof. Consider the sequence of group homomorphisms

$$SL(2,\mathbb{R}) \xrightarrow{f}GL(n,\mathbb{R}) \xrightarrow{\det}\mathbb{R}^\ast$$

Since $SL(2,\mathbb{R})$ is connected, we see $\det\circ f(SL(2,\mathbb{R}))$ is connected, thus lying in $\mathbb{R}_{>0}$. We can then modify the sequence a little bit:

$$SL(2,\mathbb{R}) \xrightarrow{f}GL(n,\mathbb{R}) \xrightarrow{\det}\mathbb{R}_{>0} \xrightarrow{\log}\mathbb{R}\xrightarrow{\exp}\mathbb{R}_{>0}$$

The map $\log \circ \det \circ f$ is a continuous homomorphism sending $SL(2,\mathbb{R})$ to $\mathbb{R}$, which is trivial, and therefore

$$\exp \circ \log \circ \det \circ f(SL(2,\mathbb{R}))=\det\circ f(SL(2,\mathbb{R}))=\{1\}.$$

This proves our assertion. $\square$

There are still a lot we can do without much Lie theory but Haar measure theory. The reader is advised to try this exercise set to see, for example, that the “volume” of $SL(2,\mathbb{R})/SL(2,\mathbb{Z})$ is $\zeta(2)$. In the references / further reading section the reader will also find a way to show that $SL(2,\mathbb{Z})\backslash SL(2,\mathbb{R})/SO(2,\mathbb{R})$ has volume $\frac{\pi}{3}$.

References / Further Reading

• Keith Conrad, Decomposing $SL(2,\mathbb{R})$. https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,R).pdf
• Serge Lang, $SL(2,\mathbb{R})$. (He wrote this book for self-study because he didn’t know much about Lie theory. In chapter 3 one can see how to do calculus with $SL(2,\mathbb{R})$.)
• Anthony W. Knapp, Lie Groups Beyond an Introduction.
• Merrick Cai, The Volume of $SL(2,\mathbb{Z})\backslash SL(2,\mathbb{R})/SO(2,\mathbb{R})$. https://math.mit.edu/~mqt/math/teaching/mit/18-704/final-papers/cai-m_final.pdf (A more concrete example of doing calculus on this group.)