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
p(x)&=(x-k)(x-i) (x-j)
+(x-i) (x-j) (x-k)
&+(x-k) (x-j)+(x-j) (x-k)
I see that \(p(j)=0\) as well \(p(k)=0\).
However I see now the problem. Namely.
p(x)&=((x-k)(x-i)\otimes 1+ (x-i) \otimes (x-k)
&+(x-k)\otimes 1+1\otimes (x-k))\circ(x-j)
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)
I can do something more with these equations. For instance
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)

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 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
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 $$ f(x)=a_0E(x)+a_1E_1(x)+a_2E_2(x)+a_3I(x) $$

Friday, January 21, 2011

Representation of Universal Algebra

I published my new book: Representation Theory: Representation of Universal Algebra. In this book I consider morphism of representation, consept of generating set and basis of representation. This allows me to consider basis manifold of representation, active and passive transformations, concept of geometrical object in representation of universal algebra. Similar way I consider tower of representations.

Sunday, September 26, 2010

My first published book

This year I published my first book
Linear Mappings of Free Algebra: First Steps in Noncommutative Linear Algebra
Book was published by Lambert Academic Publishing.

Tuesday, March 9, 2010

About Cauchy-Riemann Equation

Recently playing with complex numbers I discovered that transformation caused by matrix $$ \left( \begin{array}{cc} a_0 & -a_1 \\ a_1 & a_0 \end{array} \right) $$ is equivalent to multiplication over complex number $z=a_0+a_1i$. The Caushy-Riemann equation follows from this matrix.

Initially I assumed similar transformation for quaternion. However appropriate matrix for quaternion looks too restrictive. I assumed less restrictive condition for derivative of quaternion function $$ \frac {\partial y^0} {\partial x^0} = \frac{\partial y^1}{\partial x^1} = \frac{\partial y^2}{\partial x^2} = \frac{\partial y^3}{\partial x^3} $$ \[ \frac{\partial y^i}{\partial x^j} =- \frac{\partial y^j}{\partial x^i} \quad i\nej \] I see that such functions like $y=ax$, $y=xa$, $y=x^2$ satisfy to this equations. The same time conjugation does not satisfy it. A lot of questions should be answered to understand if this is set of function that generalize set of complex functions in case of quaternions. You can find details in my paper eprint arXiv:0909.0855 Quaternion Rhapsody, 2010