Consider $SL_2(\mathbb{R})$ as the set of $2\times 2$ real matrices of determinant $1$. Also consider $SO(2,1;\mathbb{R})$ as the group of real $3\times 3$ matrices of determinant one preserving the quadratic form of signature $(2,1)$. The two groups are isogeneous and there is a non-trivial $2$-to-$1$ group homomorphism $\phi:SL_2(\mathbb{R})\rightarrow SO(2,1;\mathbb{R})$.
My question is, given a matrix $\begin{bmatrix} a&b\\ c&d \end{bmatrix}\in SL_2(\mathbb{R})$ , what matrix does $\phi$ map it to in $SO(2,1;\mathbb{R})$?
The set of $2\times2$ matrices $M(\mathbb R^2)$ is isomorphic, as a ring, to the even subalgebra of the Clifford algebra $Cl(\mathbb R^{2,1})$. Given the standard basis $\{e_1,e_2,e_3\}$ for $\mathbb R^{2,1}$, we define this algebra by the rules
$$e_1e_1=1,\quad e_2e_2=1,\quad e_3e_3=-1$$
$$e_ke_l=-e_le_k,\quad k\neq l$$
(as well as associativity, distributivity, etc.).
The isomorphism (rather, one isomorphism) is
$$\begin{bmatrix}1&0\\0&1\end{bmatrix}\leftrightarrow1,\quad\begin{bmatrix}1&0\\0&-1\end{bmatrix}\leftrightarrow e_2e_3,\quad\begin{bmatrix}0&1\\1&0\end{bmatrix}\leftrightarrow e_3e_1,\quad\begin{bmatrix}0&1\\-1&0\end{bmatrix}\leftrightarrow e_1e_2,$$
or equivalently
$$M=\begin{bmatrix}a&b\\c&d\end{bmatrix}\leftrightarrow\frac{a+d}{2}+\frac{a-d}{2}e_2e_3+\frac{b+c}{2}e_3e_1+\frac{b-c}{2}e_1e_2$$
$$=R=r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2.$$
The condition $\det M=ad-bc=1$ translates to
$$R\,R^\sim=(r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2)(r_0+r_1e_3e_2+r_2e_1e_3+r_3e_2e_1)$$
$$=r_0\!^2-r_1\!^2-r_2\!^2+r_3\!^2=1.$$
(The reverse $R^\sim$, sometimes written as $R^\dagger$, flips the order of multiplication of vectors.)
Now, the multivector $R$ rotates vectors $v\in\mathbb R^{2,1}$ by the "sandwich product"
$$v\mapsto\phi(M)(v)=R\,v\,R^\sim.$$
...It looks rather tedious to expand this in terms of the basis vectors, but that should give us $\phi(M)$. I'll keep working on it to see if it simplifies.
$$e_1\mapsto(r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2)e_1(r_0+r_1e_3e_2+r_2e_1e_3+r_3e_2e_1)$$
$$=(r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2)(r_0+r_1e_3e_2-r_2e_1e_3-r_3e_2e_1)e_1$$
$$=\big((r_0\!^2-r_1\!^2+r_2\!^2-r_3\!^2)+2(r_0r_2+r_1r_3)e_3e_1+2(r_0r_3+r_1r_2)e_1e_2\big)e_1$$
$$=(ad+bc)e_1+(cd-ab)e_2+(ab+cd)e_3$$
$$e_2\mapsto(r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2)(r_0-r_1e_3e_2+r_2e_1e_3-r_3e_2e_1)e_2$$
$$=\big((r_0\!^2+r_1\!^2-r_2\!^2-r_3\!^2)+2(r_0r_1-r_2r_3)e_2e_3+2(r_0r_3-r_1r_2)e_1e_2\big)e_2$$
$$=(bd-ac)e_1+\tfrac12(a^2+d^2-b^2-c^2)e_2+\tfrac12(b^2-c^2-a^2+d^2)e_3$$
$$e_3\mapsto(r_0+r_1e_2e_3+r_2e_3e_1+r_3e_1e_2)(r_0-r_1e_3e_2-r_2e_1e_3+r_3e_2e_1)e_3$$
$$=\big((r_0\!^2+r_1\!^2+r_2\!^2+r_3\!^2)+2(r_0r_1+r_2r_3)e_2e_3+2(r_0r_2-r_1r_3)e_3e_1\big)e_3$$
$$=(ac+bd)e_1+\tfrac12(d^2-a^2+c^2-b^2)e_2+\tfrac12(a^2+d^2+b^2+c^2)e_3$$
$$\phi(M)=\begin{bmatrix}(ad+bc)&(-ac+bd)&(ac+bd)\\(-ab+cd)&\tfrac12(a^2-b^2-c^2+d^2)&\tfrac12(-a^2-b^2+c^2+d^2)\\(ab+cd)&\tfrac12(-a^2+b^2-c^2+d^2)&\tfrac12(a^2+b^2+c^2+d^2)\end{bmatrix}$$