1

I am surprised by the fact that $\mathrm{SU}(1, 1)$ group is isomorphic to $\mathrm{SL}(2, \mathbb{R})$, but $\mathrm{SU}(2)$ is not isomorphic to $\mathrm{SL}(2, \mathbb{R})$.

The first statement is easy to prove. An element of $\mathrm{SU}(1, 1)$ is

$$g=\left[\begin{array}{cc} \alpha & \beta \\ \beta^* & \alpha^* \end{array}\right] =\left[\begin{array}{cc} x+iy & z+id\\ z-id & x-iy \end{array}\right] \quad , |\alpha|^2-|\beta|^2=1.$$

The mapping to $p=\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] \in \mathrm{SL}(2, \mathbb{R})$, $(ad-bc=1) $ is $$a=x-d, \quad b=z-y, \quad c=y+z, \quad d=x+d.$$ The second fact is well known and discussed here SU(2) and SL(2,R) are not isomorphic

Is there any intuitive way to understand why $\mathrm{SU}(1, 1)$ and $\mathrm{SU}(2)$ are so different in such respect?

José Carlos Santos
  • 427,504
Eddward
  • 83
  • Isn't $|\alpha|^2 - |\beta|^2 = 1$ very different from $|\alpha|^2 + |\beta|^2 = 1$? – Arctic Char Oct 08 '20 at 11:03
  • Yes, it is different. But, can we map an element of SU(1,1) to SU(2) to make it the same? – Eddward Oct 08 '20 at 11:11
  • I just guess that there should be a mapping from SU(1,1) to SU(2)..Or no? – Eddward Oct 08 '20 at 11:19
  • A mapping, yes, but not a "good" one like a group isomorphism. – Dietrich Burde Oct 08 '20 at 13:11
  • @Dietrich Burde. I always thought that two groups are isomorphic iff exists a bijective one-to-one map between the elements. Am I mistaken, and is it not enough? – Eddward Oct 08 '20 at 18:43
  • 1
    $G\cong H$ iff there is a bijective one-to-one mapping $f\colon G\rightarrow H$ satisfying $f(gh)=f(g)f(h)$, so in addition being a group homomorphism. Did you find a group isomorphism between $SU(1,1)$ and $SU(2)$ as you wanted? (" I just guess that there should be a mapping from..."). – Dietrich Burde Oct 08 '20 at 18:46
  • @Dietrich Burde, thank you for your important note on group homomorphism and $f(gh)=f(g)f(h)$. This is where I was stuck and made mistake. – Eddward Oct 08 '20 at 18:51
  • Consider $SO(2)$ vs $SO^+(1,1)$. One is a compact circle, the other is an unbounded hyperbola. As a Lie group the latter is isomorphic to $(\Bbb R,+)$, which has no torsion. Every element of the former is either torsion or generates a dense subgroup. Pretty different. – anon Jan 10 '21 at 22:19

2 Answers2

4

The group $SU(2)$ is compact, whereas $SL(2,\Bbb R)$ isn't. That's enough to make a huge difference. And $SU(2)$ is compact because the condition that defines it ($|\alpha|^2+|\beta|^2=1$) assures that it is a closed and bounded subset of $\Bbb C^{2\times2}$. Clearly, $SU(1,1)$ is unbounded.

José Carlos Santos
  • 427,504
2

The set $|a|^2+|b|^2=1$ is compact, but not $|a|^2-|b|^2=1$

Tsemo Aristide
  • 87,475