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
On
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 $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right) = \left( \begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\right)$.
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 $q=0$.
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.