In this wikipedia article about $SU(1,1)$, it is stated that

This group [$SU(1,1)$] is isomorphic to $SO(2,1)$ and $SL(2,\mathbb ℝ)^{[17]}$

I'm confused by the relation of $SO(2,1)$ and $SL(2,\mathbb R)$. I know that they are locally isomorphic as Lie groups (as their Lie algebras are isomorphic). Topologically, they are different - $SL(2,\mathbb R)$ is connected (by row-echelon-reduction), where $SO(2,1)$ is not connected (there is the component fixing the upper hyperboloid and the component interchanging the hyperboloids). Therefore, they are not isomorphic as Lie groups.

So, are $SO(2,1)$ and $SL(2,\mathbb R)$ isomorphic as abstract groups? Why (not)?

The source $^{[17]}$ (Gilmore's Lie Groups, Lie Algebras and some of their Applications, p.201-205) from wikipedia seems only to show an isomorphism between the Lie algebras.

This question does not show that there is no isomorphism, but gives a 2-1-map, which is not surjective.

  • $\begingroup$ @DietrichBurde From looking at the pages that Google shows me, I have the impression, that the link only shows that the local isomorphism $SL(2,\mathbb R) \to SO(2,1)$ is not an isomorphism. This does not rule out algebraic isomorphism in general. Could you please expand your comment if you think that your link answers my question? $\endgroup$
    – Babelfish
    Nov 18 '19 at 12:02

They are not isomorphic as abstract groups: $SL(2, {\mathbb R})$ has nontrivial center ($\pm I$), while $SO(2,1)$ has trivial center.

Edit. Here is how to see that $SO(2,1)$ has trivial center.

Since 3 is an odd number, every element $g\in SO(2,1)$ has at least one real eigenvalue, $\lambda$. Let $E_\lambda$ denote the corresponding eigenspace in ${\mathbb R}^3$. Since $3$ is an odd number, $E_\lambda$ is a proper subspace of ${\mathbb R}^3$ unless $g=I$. The centralizer of $g$ in $GL(3,{\mathbb R})$ has to preserve $E_\lambda$. This is a pleasant linear algebra exercise which works in all dimensions:

If $g_1, g_2\in GL(n,{\mathbb R})$ commute and $g_1 v= \lambda v$, $v\in E_\lambda$, then $g_2 g_1 v= \lambda g_2 v= g_1 g_2(v)$, since $g_1, g_2$ commute. Hence $$ g_1(g_2 v)= \lambda (g_2 v), $$ i.e. $g_2v\in E_\lambda$.

Thus, if $g$ were central in $SO(2,1)$, then $SO(2,1)$ would have an invariant line or plane $E_\lambda$. Using the invariance of the Lorentzian inner product, we see that $SO(2,1)$ then would have an invariant line in ${\mathbb R}^3$ (if $E_\lambda$ is a plane, take its Lorentzian orthogonal complement). The orbit of every nonzero vector $v$ under $SO(2,1)$ is either the null-cone (if the vector is null) or the set of positive vectors of the given "length" $\langle v,v\rangle$ (if the vector is positive) or the set of negative vectors of the given "length" (if the vector is negative). In any case, it is not contained in a line, thus, $SO(2,1)$ has trivial center.

The faulty wikipedia page was corrected.

  • $\begingroup$ @MoisheKohan, could you please explain why $\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}$ is not in the center? $\endgroup$
    – Babelfish
    Nov 18 '19 at 17:11
  • $\begingroup$ @Babelfish: Because it it were central then its fixed-point set in $R^3$ (a line) would be invariant under $SO(2,1)$, but the latter acts irreducibly on $R^3$. $\endgroup$ Nov 18 '19 at 17:13
  • $\begingroup$ @MoisheKohan, could you please elaborate, why the fixed-point set in $\mathbb R^3$ would be invariant? I can only deduce that every group element in $SO(2,1)$ would have to have a fixed-point set containing a line. $\endgroup$
    – Babelfish
    Nov 18 '19 at 17:26
  • $\begingroup$ @Babelfish: This is just linear algebra, see the edit. $\endgroup$ Nov 18 '19 at 17:33
  • $\begingroup$ @Babelfish just compute what the centralizer of your matrix is, and how it intersects $SO(2,1)$. $\endgroup$
    – YCor
    Nov 19 '19 at 9:04

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