# $SL(2,\mathbb R)$ and $SO(2,1)$ isomorphic - or not?

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.

• @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? 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.

• @MoisheKohan, could you please explain why $\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}$ is not in the center? Nov 18 '19 at 17:11
• @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 $R^3$. Nov 18 '19 at 17:13
• @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. Nov 18 '19 at 17:26
• @Babelfish just compute what the centralizer of your matrix is, and how it intersects $SO(2,1)$.