Holomorphic Map of Quaternion Algebra

\(\newcommand{\gi}[1]{\boldsymbol{#1}}\)

Let \(E\) be identity map of quaternion algebra and \(I\), \(J\), \(K\) be maps of conjugacy of quaternion algebra. The derivative of map
\[f:H\rightarrow H\]
of quaternion algebra has form
\begin{split}
\frac{\partial f}{\partial x}=&-\frac 12
\left(
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 1}}i
+\frac{\partial f}{\partial x^{\gi 2}}j
+\frac{\partial f}{\partial x^{\gi 3}}k
\right)\circ E
\\&
+\frac 12\left(
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 1}}i
\right)\circ I
+\frac 12\left(
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 2}}j
\right)\circ J
\\&
+\frac 12\left(
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 3}}k
\right)\circ K
\end{split}

The map of quaternion algebra which satisfies the equation
\[\frac{\partial f}{\partial x^{\gi 0}}+i\frac{\partial f}{\partial x^{\gi 1}}
+j\frac{\partial f}{\partial x^{\gi 2}}+k\frac{\partial f}{\partial x^{\gi 3}}=0\]
is called left-holomorphic. The map of quaternion algebra which satisfies the equation
\[\frac{\partial f}{\partial x^{\gi 0}}+\frac{\partial f}{\partial x^{\gi 1}}i
+\frac{\partial f}{\partial x^{\gi 2}}j+\frac{\partial f}{\partial x^{\gi 3}}k=0\]
is called right-holomorphic.

Evidently that maps \(I\), \(J\), \(K\) are left-holomorphic and right-holomorphic. However, there exist not trivial examples of holomorphic map.

Let \(\mathcal HE\) be set of maps which satisfy the equation
\[\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 1}}i
=
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 2}}j
=
\frac{\partial f}{\partial x^{\gi 0}}
+\frac{\partial f}{\partial x^{\gi 3}}k
=0\]
Let \(f\), \(g\in\mathcal HE\). Then \(f\circ g\in\mathcal HE\). Maps which belong to the set \(\mathcal HE\) should have interesting properties and it is mind to study this set. Holomorphic map \(f\in\mathcal HE\) satisfies the equation
\[\frac{\partial f^0}{\partial x^0}=0\]