## 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.

## Wednesday, May 15, 2013

### Roots of Noncommutative Polynomial

When we study Riemann geometry, there is one
interesting object there. Its name is geodesic line.
It has two different definitions which ultimately lead
to the same object. Geodesic is line of extreme length.
Geodesic is such line that tangent vector
parallel transfers along it.

The reason for this identity is relation
between metric tensor and connection. Namely,
covariant derivative of metric is 0. As soon
this relation is broken in metric affine manifold,
we have two different lines.

I expect similar phenomenon in algebra of
polynomials. When we consider polynomial over commutative
ring (usualy this is a field) then we have two definition
of root. $$x=x_0$$ is root of polynomial $$p(x)$$
if $$p(x_0)=0$$. $$x=x_0$$ is root of
polynomial $$p(x)$$ if $$x-x_0$$ is divisor of
polynomial $$p(x)$$. However do these definition
lead to the same set when I consider polynomial
over noncommutative algebra?

The case with polynomial over quaternion algebra
$p(x)=(x-i)(x-j)+(x-j) (x-k)$
is relatively simple. Clearly, that $$p(j)=0$$.
This polynomial can be presented in the form
$p(x)=((x-i)\otimes 1+1\otimes (x-k))\circ(x-j)$
However it does not mean that polynomial $$x-j$$
is divisor of polynomial $$p(x)$$.

By the definition, polynomial $$x-j$$ is divisor
of polynomial $$p(x)$$ if we can write
$p(x)=q(x)(x-j)r(x)=(q(x)\otimes r(x))\circ (x-j)$
Here either $$q(x)$$ has power $$1$$ and $$r(x)$$
is scalar, either $$q(x)$$ is scalar and $$r(x)$$
has power $$1$$. But I do not think that there exist
polynomials $$q(x)$$ and $$r(x)$$ such that
$(x-i)\otimes 1+1\otimes (x-k)=q(x)\otimes r(x)$
From this example I see that if polynomial $$x-j$$
is divisor of polynomial $$p(x)$$, then $$x=j$$ is
root of polynomial $$p(x)$$. However oposite statement
can be wrong.

Now I want to consider the polynomial
$\begin{split} p(x)&=(x-k)(x-i) (x-j) +(x-i) (x-j) (x-k) \\ &+(x-k) (x-j)+(x-j) (x-k) \end{split}$
I see that $$p(j)=0$$ as well $$p(k)=0$$.
However I see now the problem. Namely.
$\begin{split} p(x)&=((x-k)(x-i)\otimes 1+ (x-i) \otimes (x-k) \\ &+(x-k)\otimes 1+1\otimes (x-k))\circ(x-j) \end{split}$
$\begin{split} p(x)=&(1\otimes(x-i) (x-j) +(x-i) (x-j)\otimes 1 \\ &+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k) \end{split}$
I can do something more with these equations. For instance
$\begin{split} p(x)&=(1\otimes(((x-i)\otimes 1)\circ (x-j)) +(x-i) (x-j)\otimes 1 \\ &+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k) \end{split}$

## Saturday, August 18, 2012

### Research is over. Long live research

Recently CERN reported discovery of Higgs boson. Since Higgs boson is responsible for inertial mass, this discovery raises question how equality of inertial and gravitational masses works.

In report there was note that there was also track of particle of spin 2. Is it possible that graviton was also discovered in CERN? Is it possible that interaction of graviton and Higgs boson was discovered at CERN as well? If such interaction exists, then it can be possible that at certain condition this interaction can be broken.

The question relates to interface of general relativity and quantum mechanics. So string theory and loop gravity should answer these questions. Even these theories have deal with extreme events and it is hard to test them, I believe that in near future we will see the possibility to test these theories.

## Friday, June 15, 2012

### System of Linear Equations

 I published my new book Linear Algebra over Division Ring: System of Linear Equations You can find this book in amazon.com and in CreateSpace store

## Friday, July 8, 2011

### Linear mapping of quaternion algebra

Linear automorphism of quaternion algebra is a liner over real field mapping $$f:H\rightarrow H$$ such that $$f(ab)=f(a)f(b)$$ Evidently, linear mapping $$E(x)=x$$ is linear automorphism. In quaternion algebra there are nontrivial linear automorphisms. For instance
E1(x)=x0 +x2i +x3j +x1k
E3(x)=x0 +x2i +x1j -x3k
where
x=x0 +x1i +x2j +x3k

Similarly, the mapping

I(x)=x0 -x1i -x2j -x3k
is antilinear automorphism because $$I(ab)=I(b)I(a)$$

In the paper eprint arXiv:1107.1139, Linear Mappings of Quaternion Algebra, I proved following statement. For any linear over real field function $f$ there is unique expansion $$f(x)=a_0E(x)+a_1E_1(x)+a_2E_2(x)+a_3I(x)$$