I am trying to find an algebra whose exponential maps surjectively to Spin(3,1).
I understand that its lie algebra is $\mathfrak{so}(3,1)$, but I think it leaves elements out.
I am working with the bivectors of CL(4,R).
$$ \mathbf{f} = E_{01}\gamma_0\gamma_1+E_{02}\gamma_0\gamma_2 + E_{03}\gamma_0\gamma_3+B_{12}\gamma_1\gamma_2+B_{13}\gamma_1\gamma_3+B_{23}\gamma_2\gamma_3 $$
Its exponential is
$$ e^{\mathbf{f}}=\cos \theta+ \left( \sin \theta \right) \left( E_{01}\gamma_0\gamma_1+E_{02}\gamma_0\gamma_2 + E_{03}\gamma_0\gamma_3+B_{12}\gamma_1\gamma_2+B_{13}\gamma_1\gamma_3+B_{23}\gamma_2\gamma_3 \right) $$
where $\theta = || E_{01}\gamma_0\gamma_1+E_{02}\gamma_0\gamma_2 + E_{03}\gamma_0\gamma_3+B_{12}\gamma_1\gamma_2+B_{13}\gamma_1\gamma_3+B_{23}\gamma_2\gamma_3||$
However, the official definition of spin relates to multiplying an even number of vectors together whose norm is +1 (or -1).
Such multiplication can produce linear combinasions of scalars, pseudo-scalars and bivectors.
Yet, the exponential of a bivector can only produce scalars and bivectors. Thus, it must be incomplete.
I note that if I add a pseudo-scalar $\mathbf{z}$ to my bivector $\mathbf{f}$, I get
$$ \mathbf{f} + \mathbf{z} = E_{01}\gamma_0\gamma_1+E_{02}\gamma_0\gamma_2 + E_{03}\gamma_0\gamma_3+B_{12}\gamma_1\gamma_2+B_{13}\gamma_1\gamma_3+B_{23}\gamma_2\gamma_3 + Z\gamma_0\gamma_1\gamma_2\gamma_3 $$
Its exponential is:
$$ e^{\mathbf{f} + \mathbf{z}}=\cos \theta+ \left( \sin \theta \right) \left( E_{01}\gamma_0\gamma_1+E_{02}\gamma_0\gamma_2 + E_{03}\gamma_0\gamma_3+B_{12}\gamma_1\gamma_2+B_{13}\gamma_1\gamma_3+B_{23}\gamma_2\gamma_3 + Z \gamma_0\gamma_1\gamma_2\gamma_3\right) $$
From this, it appears that the exponential map of $\mathbf{f}+\mathbf{z}$ is surjective to ${\rm Spin}(3,1)$.
Is this correct?
Is it bijective?
Finally, is $\mathbf{f}+\mathbf{z}$=$\mathfrak{so}(3,1) \oplus \mathfrak{u}(1)$?