The action of $SL(2, \mathbb R)$ on $\mathbb R^4$ though Clifford algebra 
My question is about the realization of $SL(2, \mathbb R)$ as the group of unit
quaternions in the even-dimensional Clifford algebra corresponding to an anisotropic quadratic form on $\mathbb R^3$ (from the paper ANOSOV FLOWS ON INFRA-HOMOGENEOUS
SPACES by PER TOMTER, page 300), so that $SL(2, \mathbb R)$ can act on $\mathbb R^4$.

Let $V=\mathbb R^3$ and $Q$ be a to-be-determined (?) quadratic form on $V$ free of nonzero integer solutions. The Clifford algebra $\mathcal A$ is defined as the quotient of the tensor algebra
$$\oplus_{n\ge 0} V^{\otimes n}$$
by the two-sided ideal $\langle v\otimes v - Q(v)1 \mid v\in V \rangle$, where $1$ is the unit of $\oplus_{n\ge 0} V^{\otimes n}$.
But I have some difficulty understanding the meaning of "realization of $SL(2, \mathbb R)$ as the group of unit quaternions in the even-dimensional Clifford algebra" in this situation. I guess in order to construct an action of $SL(2, \mathbb R)$ on $\mathbb R^4$ though Clifford algebra, we have to choose $Q$ in such a way (really?) that $\mathcal A$ is a 4-dimensional(or some even-dimensional) real algebra with basis $1,i,j,k$ and embed $SL(2,\mathbb R)$ into $A$ by identifying the matrices in $SL(2,\mathbb R)$ with $a+bi+cj+dk$. My understanding might be wrong, but how is this constructed specifically?
I hope my question is not confusing, but if you know any other ways of defining the action of $SL(2, \mathbb R)$ on $\mathbb R^4$, though Clifford algebra or not, please feel free to share. Thanks!
 A: First, there are varying conventions about formation of a Clifford algebra attached to a quadratic form. Some conventions have a sign flipped in comparison to others, so the implications of isotropic/anisotropic are ambiguous.
But that does not affect the facts. :)
Let's take the convention that the Clifford algebra associated to a quadratic form $Q$ is characterized by $v\cdot v=Q(v)$. Thus, the Hamiltonian quaternions appear as the Clifford algebra attached to a negative-definite quadratic form on $\mathbb R^3$, with $i^2=j^2=k^2=-1$.
The opposite sign, giving $i^2=j^2=k^2=1$, gives the ring of two-by-two real matrices, for example taking $i=\pmatrix{1&0\cr0&-1}$, $j=\pmatrix{0&1\cr1&0}$, etc.
In either case, the "unit quaternions" are those with "reduced norm $1$", which means "determinant=1" in matrix models. So this gives $SL_2(\mathbb R)$ in the latter case.
The units $A^\times$ in a quaternion algebra (over $\mathbb R$) $A$ act on $A\approx \mathbb R^4$ in at least two ways. One is by left (or right) multiplication $a\cdot b=ab$. Another is by conjugation $a\cdot b=aba^{-1}$.
