Definition of indefinite orthogonal group for 3 by 3 matricies, and how it it connected to special linear group for 2 by 2 matricies I'm working on a project to show that $SL(2, \mathbb{R})$ is the same the identity component of $SO(1,2)=\{A\in M_{3x3}(\mathbb{R}):A^tI_{1,2}A=I_{1,2}, det(A)=1\}$ where $$I_{1,2} = \quad
\begin{pmatrix} 
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1\\ 
\end{pmatrix}$$ I have a homoeomorphism $$Ad:SL(2,\mathbb{R})\rightarrow GL(sl_2(\mathbb{R})), Ad(\sigma)(u)=\sigma u \sigma ^{-1} for \space \space\sigma \in SL(2, \mathbb{R}), u \in sl_2(\mathbb{R})$$  a quadratic form on $sl_2(\mathbb{R})$ given by $$Q(u,v) = - \frac{1}{2}Trace(uv)$$ and a basis of of $sl_2(\mathbb{R})$ $$E_1=
\quad
\begin{pmatrix} 
0 & 1 \\
-1 & 0 
\end{pmatrix}, \space E_2= \quad
\begin{pmatrix} 
1 & 0 \\
0 & -1 
\end{pmatrix}, \space E_3= \quad
\begin{pmatrix} 
0 & 1 \\
1 & 0 
\end{pmatrix}$$
By computing $Q(E_i,E_j)$, I have that the signature of $Q$ is $(1,2)$. I have shown that $Q(Ad(\sigma)(u),Ad(\sigma)(v))=Q(u,v)$. 
In addition to the definition I gave before of $SO(1,2)$, I understand that it can also defined as matrices that preserve an inner product/bilinear/quadratic form of signature $(1,2)$. How is this the same as the definition I gave before? This looks like what I have for $Q$, but I don't see how to make the jump to 3 by 3 matricies. I have a 3 by 3 matrix representation of $Ad(\sigma)$ also. 
From this point I'm supposed to conclude $Ad(SL2,\mathbb{R})\subset O(1,2)$, but I don't quite see the connection. I see the quadratic form I have has the same signature as $O(1,2)$. I have a feeling the answer is right in front of me but I can't connect the dots.
 A: Before I address the questions, notice that the group homomorphism $\operatorname{Ad} : SL(2, \Bbb R) \to GL(\mathfrak{sl}(2, \Bbb R))$ is actually $2$-to-$1$ onto its image: We have
$$\operatorname{Ad}(-I_2)(u) = (-I_2) u (-I_2)^{-1} = (-I_2) u (-I_2) = u ,$$
so $-I_2 \in \ker \operatorname{Ad}$, and it's straightforward to verify that this is all, that is, that $\ker \operatorname{Ad} = \{\pm I_2\}$. (Remark Since this kernel is discrete, this map induces an exceptional Lie algebra isomorphism $\mathfrak{sl}(2, \Bbb R) \cong \mathfrak{so}(1, 2)$.)
You have all of the relevant computations; what remains is just to interpret them.
First, if $\Sigma$ is a bilinear form on a vector space $\Bbb V$, then $T \in GL(\Bbb V)$ preserves $\Sigma$ iff $\Sigma(T {\bf v}, T {\bf v}) = \Sigma({\bf v}, {\bf v})$ for all ${\bf v} \in \Bbb V$. If $\Bbb V$ is finite-dimensional and we choose thereof, then we can write this condition as $[T {\bf v}]^\top [\Sigma] [T {\bf v}] = [{\bf v}]^\top [\Sigma] [{\bf v}]$. Since $[T {\bf v}]^\top = ([T] [{\bf v}])^\top = [{\bf v}]^\top [T]^\top$, we conclude that the condition is equivalent to $$[T]^\top [\Sigma] [T] = [\Sigma] .$$ So, for the bilinear form defined in the standard basis by $[\Sigma] = I_{1, 2}$, a transformation $A$ preserves $I_{1, 2}$ iff $$A^\top I_{1, 2} A = I_{1, 2} ,$$ which is exactly the first condition in the definition of $SO(1, 2)$. Taking the determinant of both sides gives $(\det A)^2 = 1$, so $\det A = \pm 1$, and so the remaining condition, $\det A = 1$, just restricts to orientation-preserving transformations.
Now, you've already shown for all $\sigma \in SL(2, \Bbb R)$ that $\operatorname{Ad}(\sigma)$ preserves $Q$, so $$\operatorname{Ad}(SL(2, \Bbb R)) \subseteq O(Q)$$---here $O(Q)$ is just the orthogonal group of linear transformations in $GL(\mathfrak{sl}(2, \Bbb R))$ that preserves the (nondegenerate, symmetric) bilinear form $Q$. On the other hand, you've already computed $Q(E_1, E_1) = 1, Q(E_2, E_2) = Q(E_3, E_3) = -1$, and we can compute $Q(E_i, E_j) = 0$ for $i < j$, so the matrix representation of $Q$ with respect to the basis $(E_1, E_2, E_3)$ is precisely $I_{1,2}$, and thus this choice of basis defines an isomorphism $$O(Q) \cong O(1, 2) .$$
A: From pages 123-124 in Noneuclidean Tesselations and Their Groups by Wilhelm  Magnus, every $A$ with $\det A = +1$ is given by real numbers $\alpha \delta - \beta \gamma = 1$ and
$$
A = 
\left(
\begin{array}{ccc}
\frac{1}{2} \left( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \right) & \left( \alpha \beta + \gamma \delta \right) & \frac{1}{2} \left( \alpha^2 - \beta^2 + \gamma^2 - \delta^2 \right) \\
\left( \alpha \gamma + \beta \delta \right) & \left( \alpha \delta + \beta \gamma  \right) & \left( \alpha \gamma - \beta \delta \right) \\
\frac{1}{2} \left( \alpha^2 + \beta^2 - \gamma^2 - \delta^2 \right) & \left( \alpha \beta - \gamma \delta \right) & \frac{1}{2} \left( \alpha^2 - \beta^2 - \gamma^2 + \delta^2 \right) \\
\end{array}
\right)
$$
In turn, this goes back to pages 533-534 in the original German, Fricke and Klein (1897). There is a recent translation into English.
