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I am trying to show that $SO(2,1)$ is not connected but I have no idea where to start really, I know that it is connected if there is a path between any two points. My definition of $SO(2,1)$ is:

$SO(2,1)=\{X\in Mat_3(\mathbb{R}) \mid X^t\eta X=\eta, \ \det(X)=1\}$ where $\eta$ is the matrix defined as: $$\left ( \begin{array}{ccc} 1 &0&0\\0&1&0\\0&0&-1\end{array}\right )$$

Thanks for any help

Seirios
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hmmmm
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    @DominicMichaelis I think I've seen some authors writing on Lie algebra define "connected" as "path connected," so that may be the OP's situation. One prominent example is Gilmore's text (just search for "connected" in the text.) Incidentally, he also uses the convention of calling the tensor product "the direct product". – rschwieb May 17 '13 at 14:11
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    @DominicMichaelis Lie groups are manifolds, so connectedness does imply path connectedness for $SO(2, 1)$. – MJD May 17 '13 at 14:13

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Consider the orbit of the vector $(0,0,1)$ under $SO(2,1)$; you should find that it's disconnected (note that there are elements of $SO(2,1)$ which map $(0,0,1)$ to itself, or to $(0,0,-1)$, and then show that it can not be mapped to any vector $(a,b,0)$). So this gives us a continuous map from $SO(2,1)$ to a disconnected space, which implies that $SO(2,1)$ is disconnected.

Rhys
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  • Ok, sorry I am a bit confused, what is the disconnected space that we are mapping to? – hmmmm May 17 '13 at 14:24
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    The orbit of $(0,0,1)$ under the action of $SO(2,1)$. In fact, 'orbit' is a better word than 'image', so let me edit my answer. :-) – Rhys May 17 '13 at 14:27
  • Ah sorry I had misread what you said slightly I understand what you are saying now. Thanks very much – hmmmm May 17 '13 at 14:30
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Consider the transitive action of $O(2,1)$ on $N_{-1}= \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2-z^2=-1 \}$ by left multiplication. For some $x \in N_{-1}$, you get a continuous surjection $$\left\{ \begin{array}{ccc} O(2,1) & \to & N_{-1} \\ M & \mapsto & M \cdot x \end{array} \right. .$$

So you have $SO(2,1) \hookrightarrow O(2,1) \twoheadrightarrow N_{-1}$. But $N_{-1}$ is a two-sheeted hyperboloid, and the image of $SO(2,1)$ in $N_{-1}$ is not contained in one sheet of $N_{-1}$ since $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right) = \left( \begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\right)$.

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Because a continuous image of a connected set is connected, $SO(2,1)$ has at least two connected components.

In fact, it can be shown that the connected component of $SO(p,q)$ containing $\operatorname{Id}$ is $$SO_0(p,q)= \left\{ \left( \begin{matrix} A & B \\ C & D \end{matrix} \right) \in SO(p,q) \mid A \in GL_p(\mathbb{R}), \ \det(A) >0 \right\}.$$

Therefore, $SO(p,q)$ is connected iff $p=0$ or $q=0$.

Seirios
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