## Sunday, January 24, 2016

### 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
$\frac{\partial f^0}{\partial x^0}=0$