Joint Quaternion Theorem
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Joint Quaternion Theorem
This note proves the joint quaternion theorem which connects complex analysis with paradoxical analysis (theory of quaternions). So, when we study quaternions, we arenʻt going outside the world of complex analysis with only one complex term $i = \sqrt{-1}$. The two fields are the same. It is like a rosetta stone between the two, so anything in complex analysis is the same in paradoxical analysis and vice versa. Basically, weʻve never really left complex analysis for quaternions, octonions, and so on…it is all the same and included in complex analysis. So, really all we need is complex analysis to do the same exact stuff as with quaternions and higher dimensional variants. Complex analysis is like a fusion of everything, so it does everything all at once.
The Proof
We begin with the general form of a quaternion, written as…
\[\begin{equation} q = q_0 + q_1i + q_2j + q_3k \end{equation}\]Using the multiplication rules for quaternions, we can rewrite this as…
\[\begin{equation} q = (q_0 + q_1i) + (q_2 + q_3i)j \end{equation}\]since $ij = k$. Now, we set $\alpha^\ast = q_0 + q_1i$ and $\beta^\ast = q_2 + q_3i$ which is just a change of variables to get…
\[\begin{equation} q = \alpha^\ast + \beta^{\ast}j \end{equation}\]we know $\alpha^\ast, \beta^\ast \in \mathbb{C}$. Notice how similar this is to just a plain complex number, since $j^2 = -1$, we can write a duality between quaternions and complex numbers. All we need to study quaternions is then the behavior of $\alpha^\ast$ and $\beta^\ast$. If we put these two quantities inside a vector, we end up with…
\[\begin{equation} \vec{\gamma} = \pmatrix{\alpha^\ast \\ \beta^\ast} = \pmatrix{q_0 + q_1i \\ q_2 + q_3 i} \end{equation}\]Which has $\vec{\gamma} \in \mathbb{C}^2$ and includes vectors inside the Bloch sphere as well as them scaled in $\mathbb{C}$. Notice that changing $q_0,…,q_3$ makes a quaternion. So, we write out the conversion factor using this idea…
\[\begin{equation} q_0 \equiv \pmatrix{q_0 \\ 0} \ \ \ \ q_1i \equiv \pmatrix{q_1i \\ 0} \\ q_2j \equiv \pmatrix{0 \\ q_2} \ \ \ \ q_3k \equiv \pmatrix{0 \\ q_3i} \end{equation}\]Now, when we add all the $q_0,…,q_3$ together we end up with…
\[\begin{equation} q_0 + q_1i + q_2j + q_3k \equiv \pmatrix{q_0 + q_1i \\ q_2 + q_3 i} \end{equation}\]which is the most general form of quaternions expressed in a complex analysis setting. Equation $(4)$ is enough to give a duality due to isomorphism between complex vectors and quaternions. This is sufficient to complete the proof, as desired. $\blacksquare$