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

### Map of Conjugation of Quaternion algebra

Quaternion algebra has following maps of conjugation
\begin{split}
x=E\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i+x_2j+x_3k\\
x^{*_1}=I\circ(x_0+x_1i+x_2j+x_3k)&=x_0-x_1i+x_2j+x_3k\\
x^{*_2}=J\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i-x_2j+x_3k\\
x^{*_3}=K\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i+x_2j-x_3k
\end{split}

A linear map of quaternion algebra
$f:H\rightarrow H\ \ \ y^i=f^i_jx^j$
has form
$f=a_0\circ E+a_1\circ I+a_2\circ J+a_3\circ K$
\begin{split}
f\circ x&=a_0x+a_1\circ I\circ x+a_2\circ J\circ x+a_3\circ K\circ x\\
&=a_0x+a_1x^{*_1}+a_2x^{*_2}+a_3x^{*_3}
\end{split}
where quaternions $$a_0$$, $$a_1$$, $$a_2$$, $$a_3$$ are defined by coordinates $$f^i_j$$ of linear map.

The set $$\mathcal L(R;H;H)$$ of linear maps of quaternion algebra is left $$H$$-vector space and has the basis $$\overline{\overline{e}} =(E,I,J,K)$$.

The set $HE=\{a\circ E:a\in H\}$ is $$R$$-algebra isomorphic to quaternion algebra.

## Saturday, January 23, 2016

### Few Ideas about Morphism of Representation

The concept of morphism of representation is very important one and works in different applications. In particular it helped me to understand why linear map of Banach algebra has the form that I found.

When I was writing the paper dedicated to calculus over division ring for journal CACAA I realized why in case of division ring I can use regular coordinates of vector in module instead of tensor product. The answer is in possibility to solve system of linear equations. As soon as I have the structure of module over division ring I can learn linear map of this module. However the surprise waited me here.

Let $$e_{B\cdot i}$$ be basis of $$A$$-module $$B$$. Let $$e_{C\cdot i}$$ be basis of $$A$$-module $$C$$. We expect that matrix of linear map
$f:B\rightarrow C$
has form
$\begin{pmatrix} f_{1\cdot}^{}{}^1_1\otimes f_{2\cdot}^{}{}^1_1& ...& f_{1\cdot}^{}{}^1_n\otimes f_{2\cdot}^{}{}^1_n& \\ ... \\ f_{1\cdot}^{}{}^m_1\otimes f_{2\cdot}^{}{}^m_1& ...& f_{1\cdot}^{}{}^m_n\otimes f_{2\cdot}^{}{}^m_n& \end{pmatrix}$
So linear map has the form
$f\circ(a^ie_{B\cdot i}) =((f_{1\cdot}^{}{}^j_i\otimes f_{2\cdot}^{}{}^j_i) \circ a^i)e_{C\cdot j}$
I have step by step proof of this statement. The structure of Jacoby matrix confirms this statement as well. However when I consider the concept of morphism of representation, I get wrong answer.

We consider morphism of representation of the form
$\begin{matrix} \delta:A\rightarrow A& f:B\rightarrow C \end{matrix}$
Then $$f$$ is homomorphism of Abellian group such that
$f\circ(a v)=a f\circ v$
From this it follows that
$f\circ( v^i e_{B\cdot i})=v^i f\circ e_{B\cdot i} =v^i f_{1\cdot}{}^j_i\otimes f_{2\cdot}{}^j_i\circ e_{C\cdot i}$
This equation is wrong.

Probably I need to move transformation for geometrical object into coordinate space. However to make final decision I need to learn modules over algebra.