Problem with Left-sided Polynomials

Until today I believed that the theory of left-sided polynomials in quaternion algebra is consistent, although it cannot have applications in related theories. Today I realized that this theory contains a serious contradiction. Nivem theorem asserts that left-sided polynomial of power n has n roots. However, polynomial $$x^2-(i+j)x+k$$ has only root \(x=j\). Even we assume that this root has multiplicity 2, then we get equality $$(x-j)(x-j)=x^2-jx-xj+1$$ If we assume multiplication of left-sided polynomials, then we get equality $$(x-j)(x-j)=x^2-2jx+1$$ It is evident that left side of this equality is different from expression of polynomial.

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.

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.

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
\[
\begin{split}
p(x)&=(1\otimes(((x-i)\otimes 1)\circ (x-j))
+(x-i) (x-j)\otimes 1
\\
&+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k)
\end{split}
\]

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.

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

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

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.

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.

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 $$ \frac {\partial y^0} {\partial x^0} = \frac{\partial y^1}{\partial x^1} = \frac{\partial y^2}{\partial x^2} = \frac{\partial y^3}{\partial x^3} $$ \[ \frac{\partial y^i}{\partial x^j} =- \frac{\partial y^j}{\partial x^i} \quad i\ne j \] 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

Matrix of Mappings

I recently published new paper
eprint arXiv:1001.4852 The Matrix of Linear Mappings, 2010

In this paper, I consider change in notation of mapping. Instead of notation
y=f(x)
I use notation
y = f ° x
Let us consider ring of mappings of R-algebra A with product f ° g
This point of view allows me to consider ring R as representation in algebra A. If f is additive mapping of algebra A then I can write value of map in form f ° h. In particular I can write derivative of mapping of algebra in form
∂f(x)(h)=∂f(x) ° h
This allow me to separate derivative and increment of argument. In this case I need to use tensor notation for derivative. For instance, if f ° x = x² then
$$\partial f(x)\circ dx=(x\otimes 1+1\otimes x)\circ dx=x\ dx+dx\ x$$

Regular function of quaternion variable

Because I explore calculus over division ring I put attention on different papers related to quaternion calculus. These papers mainly discuss definition of regular function. According to definition regular function is function which satisfy to differential equation


^∂f/_∂x0+i^∂f/_∂x1+j^∂f/_∂x2+k^∂f/_∂x3=0

Recently Daniel Alayon-Solarz sent me interesting example of regular function.

f(x₀+x₁i+x₂j+x₃k)=x₀-x₁i+x₂j+x₃k

Higgs boson

The time when we see the answer on question does Higgs boson exist is close. I think that independently on answer physics will face new questions to answer.

If we will not find the Higgs boson then we need to reconsider model which predicted them.

More question will arise if we do find Higgs boson. The problem here is that Higgs boson is responsible for mass. Or to be more particular, Higgs boson is responsible for inertial mass. However we know two masses. We know inertial mass and we know gravitational mass. And these masses are equal. We yet not know the mechanism of this equality. Discovery of Higgs boson will refresh this question. If different interactions create inertial and gravitational masses, then third interaction should exist. And the third interaction is responsible for equality of masses. However if some interaction determines equality of inertial and gravitational masses, there is posibility when this equality will be broken.

How it may happens nobody knows. I can assume one of possibilities. Few years ago I explored tidal forces. I came to conclusion that these forces as fundamental as gravitational. I cannot tell which forces are primary. Do they interact one with another. Or one force is the reason of another. Feature research will show the answer on this question.

Derivative of function of division ring

When we learn derivative in calculus, we know that derivative at point is number. When we move to vector function we understand that derivative is linear operator. This allows use Jacobian as expression for derivative. However in this case we again have option to separate derivative and increment of argument.

When we start to work with function of division ring (quaternion valued function for instance) we expect something similar. However it does not work or makes set of differentiable functions very limitted. Derivative of function of division ring is not number, but additive map and it is easy to see that this is the Gateaux derivative. When we learn derivative of function of division ring we cannot separate derivative itself and increment of argument.

For instance derivative of function f(x)=axb has form

∂f(x)(h)=ahb

Derivative of function f(x)=x² has form

∂f(x)(h)=xh+hx

Derivative of function f(x)=x^-1 has form

∂f(x)(h)=-x^-1hx^-1

Additive map of division ring

I created this blog to share and discus ideas related to my research in geometry.

Recently one question arised. What happens when in geometry that I know to put division ring instead of field. When I started research in 2005 it was not clear how far I can go. However I discovered that linear algebra has a lot of common, at least when we work with linear space or try to solve system of linear equations.

I met first severe problems when decided to work with tensor product. I saw that non-commutativity may cause strong problems. Moving to calculus I realised that there exists new type of map between vector spaces. This map holds sum of vectors, however it does not hold product of vector over scalar. This is why I called such map additive.

When I returned back to linear algebra additive map helped me to define tensor product of vector spaces. Because we can consider division ring as vector space we can define tensor product of division rings. Tensor product of division rings is not division ring and tensor product of vector spaces is not vector space. However this module has basis over tensor product of rings.

Quaternion algebra is an example of division ring.

You can read my papers in arXiv

eprint arXiv:math.GM/0701238 Lectures on Linear Algebra over Division Ring, 2007

eprint arXiv:0812.4763 Introduction into Calculus over Division Ring, 2008