tag:blogger.com,1999:blog-57216932328752933432024-03-12T15:54:47.321-07:00My research in geometryAleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.comBlogger15125tag:blogger.com,1999:blog-5721693232875293343.post-50608768417718943442016-01-24T17:28:00.003-08:002016-01-24T17:28:54.828-08:00Holomorphic Map of Quaternion Algebra \(\newcommand{\gi}[1]{\boldsymbol{#1}}\)<br />
<p>Let \(E\) be identity map of quaternion algebra and \(I\), \(J\), \(K\) be maps of conjugacy of quaternion algebra. The derivative of map<br />
\[f:H\rightarrow H\]<br />
of quaternion algebra has form<br />
\begin{split}<br />
\frac{\partial f}{\partial x}=&-\frac 12<br />
\left(<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 1}}i<br />
+\frac{\partial f}{\partial x^{\gi 2}}j<br />
+\frac{\partial f}{\partial x^{\gi 3}}k<br />
\right)\circ E<br />
\\&<br />
+\frac 12\left(<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 1}}i<br />
\right)\circ I<br />
+\frac 12\left(<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 2}}j<br />
\right)\circ J<br />
\\&<br />
+\frac 12\left(<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 3}}k<br />
\right)\circ K<br />
\end{split}<br />
</p> <p> The map of quaternion algebra which satisfies the equation<br />
\[\frac{\partial f}{\partial x^{\gi 0}}+i\frac{\partial f}{\partial x^{\gi 1}}<br />
+j\frac{\partial f}{\partial x^{\gi 2}}+k\frac{\partial f}{\partial x^{\gi 3}}=0\]<br />
is called left-holomorphic. The map of quaternion algebra which satisfies the equation<br />
\[\frac{\partial f}{\partial x^{\gi 0}}+\frac{\partial f}{\partial x^{\gi 1}}i<br />
+\frac{\partial f}{\partial x^{\gi 2}}j+\frac{\partial f}{\partial x^{\gi 3}}k=0\]<br />
is called right-holomorphic.<br />
</p> <p>Evidently that maps \(I\), \(J\), \(K\) are left-holomorphic and right-holomorphic. However, there exist not trivial examples of holomorphic map.<br />
</p> <p>Let \(\mathcal HE\) be set of maps which satisfy the equation<br />
\[\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 1}}i<br />
=<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 2}}j<br />
=<br />
\frac{\partial f}{\partial x^{\gi 0}}<br />
+\frac{\partial f}{\partial x^{\gi 3}}k<br />
=0\]<br />
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<br />
\[\frac{\partial f^0}{\partial x^0}=0\]<br />
</p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-80586239415095451902016-01-24T16:13:00.001-08:002016-01-24T17:11:00.184-08:00Map of Conjugation of Quaternion algebra <p>Quaternion algebra has following maps of conjugation<br />
\begin{split}<br />
x=E\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i+x_2j+x_3k\\<br />
x^{*_1}=I\circ(x_0+x_1i+x_2j+x_3k)&=x_0-x_1i+x_2j+x_3k\\<br />
x^{*_2}=J\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i-x_2j+x_3k\\<br />
x^{*_3}=K\circ(x_0+x_1i+x_2j+x_3k)&=x_0+x_1i+x_2j-x_3k<br />
\end{split}<br />
</p><p>A linear map of quaternion algebra<br />
\[f:H\rightarrow H\ \ \ y^i=f^i_jx^j\]<br />
has form<br />
\[<br />
f=a_0\circ E+a_1\circ I+a_2\circ J+a_3\circ K<br />
\]<br />
\begin{split}<br />
f\circ x&=a_0x+a_1\circ I\circ x+a_2\circ J\circ x+a_3\circ K\circ x\\<br />
&=a_0x+a_1x^{*_1}+a_2x^{*_2}+a_3x^{*_3}<br />
\end{split}<br />
where quaternions \(a_0\), \(a_1\), \(a_2\), \(a_3\) are defined by coordinates \(f^i_j\) of linear map.<br />
</p><p>The set \(\mathcal L(R;H;H)\) of linear maps of quaternion algebra is left \(H\)-vector space and has the basis \(\overline{\overline{e}} =(E,I,J,K)\).<br />
</p><p>The set \[HE=\{a\circ E:a\in H\}\] is \(R\)-algebra isomorphic to quaternion algebra.<br />
</p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-91290893315458587602016-01-23T18:54:00.000-08:002016-01-23T18:59:11.177-08:00Few Ideas about Morphism of Representation <p>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.</p><p>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.</p><p>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<br />
\[f:B\rightarrow C\]<br />
has form<br />
\[<br />
\begin{pmatrix}<br />
f_{1\cdot}^{}{}^1_1\otimes f_{2\cdot}^{}{}^1_1&<br />
...&<br />
f_{1\cdot}^{}{}^1_n\otimes f_{2\cdot}^{}{}^1_n&<br />
\\<br />
... \\<br />
f_{1\cdot}^{}{}^m_1\otimes f_{2\cdot}^{}{}^m_1&<br />
...&<br />
f_{1\cdot}^{}{}^m_n\otimes f_{2\cdot}^{}{}^m_n&<br />
\end{pmatrix}<br />
\]<br />
So linear map has the form<br />
\[<br />
f\circ(a^ie_{B\cdot i})<br />
=((f_{1\cdot}^{}{}^j_i\otimes f_{2\cdot}^{}{}^j_i)<br />
\circ a^i)e_{C\cdot j}<br />
\]<br />
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.</p><p>We consider morphism of representation of the form<br />
\[<br />
\begin{matrix}<br />
\delta:A\rightarrow A&<br />
f:B\rightarrow C<br />
\end{matrix}<br />
\]<br />
Then \(f\) is homomorphism of Abellian group such that<br />
\[<br />
f\circ(a v)=a f\circ v<br />
\]<br />
From this it follows that<br />
\[<br />
f\circ( v^i e_{B\cdot i})=v^i f\circ e_{B\cdot i}<br />
=v^i f_{1\cdot}{}^j_i\otimes f_{2\cdot}{}^j_i\circ e_{C\cdot i}<br />
\]<br />
This equation is wrong.<br />
</p><p>Probably I need to move transformation for geometrical object into coordinate space. However to make final decision I need to learn modules over algebra.</p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-18380187981663669612013-05-15T19:19:00.000-07:002013-05-16T17:12:59.025-07:00Roots of Noncommutative Polynomial <p>When we study Riemann geometry, there is one<br />
interesting object there. Its name is geodesic line.<br />
It has two different definitions which ultimately lead<br />
to the same object. Geodesic is line of extreme length.<br />
Geodesic is such line that tangent vector<br />
parallel transfers along it.<br />
</p><p>The reason for this identity is relation<br />
between metric tensor and connection. Namely,<br />
covariant derivative of metric is 0. As soon<br />
this relation is broken in metric affine manifold,<br />
we have two different lines.<br />
</p><p>I expect similar phenomenon in algebra of<br />
polynomials. When we consider polynomial over commutative<br />
ring (usualy this is a field) then we have two definition<br />
of root. \(x=x_0\) is root of polynomial \(p(x)\)<br />
if \(p(x_0)=0\). \(x=x_0\) is root of<br />
polynomial \(p(x)\) if \(x-x_0\) is divisor of<br />
polynomial \(p(x)\). However do these definition<br />
lead to the same set when I consider polynomial<br />
over noncommutative algebra?<br />
</p><p>The case with polynomial over quaternion algebra<br />
\[<br />
p(x)=(x-i)(x-j)+(x-j) (x-k)<br />
\]<br />
is relatively simple. Clearly, that \(p(j)=0\).<br />
This polynomial can be presented in the form<br />
\[<br />
p(x)=((x-i)\otimes 1+1\otimes (x-k))\circ(x-j)<br />
\]<br />
However it does not mean that polynomial \(x-j\)<br />
is divisor of polynomial \(p(x)\).<br />
</p><p>By the definition, polynomial \(x-j\) is divisor<br />
of polynomial \(p(x)\) if we can write<br />
\[<br />
p(x)=q(x)(x-j)r(x)=(q(x)\otimes r(x))\circ (x-j)<br />
\]<br />
Here either \(q(x)\) has power \(1\) and \(r(x)\)<br />
is scalar, either \(q(x)\) is scalar and \(r(x)\)<br />
has power \(1\). But I do not think that there exist<br />
polynomials \(q(x)\) and \(r(x)\) such that<br />
\[<br />
(x-i)\otimes 1+1\otimes (x-k)=q(x)\otimes r(x)<br />
\]<br />
From this example I see that if polynomial \(x-j\)<br />
is divisor of polynomial \(p(x)\), then \(x=j\) is<br />
root of polynomial \(p(x)\). However oposite statement<br />
can be wrong.<br />
</p><p>Now I want to consider the polynomial<br />
\[<br />
\begin{split}<br />
p(x)&=(x-k)(x-i) (x-j)<br />
+(x-i) (x-j) (x-k)<br />
\\<br />
&+(x-k) (x-j)+(x-j) (x-k)<br />
\end{split}<br />
\]<br />
I see that \(p(j)=0\) as well \(p(k)=0\).<br />
However I see now the problem. Namely.<br />
\[<br />
\begin{split}<br />
p(x)&=((x-k)(x-i)\otimes 1+ (x-i) \otimes (x-k)<br />
\\<br />
&+(x-k)\otimes 1+1\otimes (x-k))\circ(x-j)<br />
\end{split}<br />
\]<br />
\[<br />
\begin{split}<br />
p(x)=&(1\otimes(x-i) (x-j)<br />
+(x-i) (x-j)\otimes 1<br />
\\<br />
&+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k)<br />
\end{split}<br />
\]<br />
I can do something more with these equations. For instance<br />
\[<br />
\begin{split}<br />
p(x)&=(1\otimes(((x-i)\otimes 1)\circ (x-j))<br />
+(x-i) (x-j)\otimes 1<br />
\\<br />
&+1\otimes (x-j)+(x-j)\otimes 1)\circ (x-k)<br />
\end{split}<br />
\]<br />
</p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-53287347531145845522012-08-18T11:45:00.000-07:002012-08-18T11:45:16.444-07:00Research is over. Long live researchRecently 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.<br />
<br />
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.<br />
<br />
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.Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-65450931955505094102012-06-15T20:50:00.003-07:002012-06-29T19:34:50.911-07:00System of Linear Equations<table>
<tr>
<td>
<iframe src="http://rcm.amazon.com/e/cm?t=sciencegalery-20&o=1&p=8&l=as1&asins=147763181X&ref=qf_sp_asin_til&fc1=000000&IS2=1<1=_blank&m=amazon&lc1=0000FF&bc1=000000&bg1=FFFFFF&f=ifr"
style="width:120px;height:240px;" scrolling="no" marginwidth="0"
marginheight="0" frameborder="0"></iframe>
</td>
<td>
I published my new book
<b>Linear Algebra over Division Ring: System of Linear Equations
</b>
You can find this book in amazon.com and in
<a href="https://www.createspace.com/3874898">
CreateSpace store</a>
</td>
</tr>
</table>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-29116536741616771032011-07-08T15:06:00.000-07:002011-07-08T15:06:06.614-07:00Linear mapping of quaternion algebraLinear 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
<br>
<center>E<sub>1</sub>(x)=x<sup><span style="color:Red">0</span></sup>
+x<sup><span style="color:Red">2</span></sup>i
+x<sup><span style="color:Red">3</span></sup>j
+x<sup><span style="color:Red">1</span></sup>k</center>
<center>E<sub>3</sub>(x)=x<sup><span style="color:Red">0</span></sup>
+x<sup><span style="color:Red">2</span></sup>i
+x<sup><span style="color:Red">1</span></sup>j
-x<sup><span style="color:Red">3</span></sup>k</center>
where
<br>
<center>x=x<sup><span style="color:Red">0</span></sup>
+x<sup><span style="color:Red">1</span></sup>i
+x<sup><span style="color:Red">2</span></sup>j
+x<sup><span style="color:Red">3</span></sup>k</center>
<p>Similarly, the mapping
<center>I(x)=x<sup><span style="color:Red">0</span></sup>
-x<sup><span style="color:Red">1</span></sup>i
-x<sup><span style="color:Red">2</span></sup>j
-x<sup><span style="color:Red">3</span></sup>k</center>
is antilinear automorphism because
$$I(ab)=I(b)I(a)$$
</p>
<p>
In the paper
eprint <a href="http://arxiv.org/abs/1107.1139" target="_blank">arXiv:1107.1139</a>,
Linear Mappings of Quaternion Algebra,
I proved following statement. For any linear over real field function $f$
there is unique expansion
$$
f(x)=a_0E(x)+a_1E_1(x)+a_2E_2(x)+a_3I(x)
$$
</p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-21621358103973956412011-01-21T18:24:00.000-08:002011-10-29T19:08:56.620-07:00Representation of Universal Algebra<table>
<tr>
<td>
<iframe src="http://rcm.amazon.com/e/cm?t=sciencegalery-20&o=1&p=8&l=as1&asins=3844300724&ref=tf_til&fc1=000000&IS2=1<1=_blank&m=amazon&lc1=0000FF&bc1=000000&bg1=FFFFFF&f=ifr" style="width:120px;height:240px;" scrolling="no" marginwidth="0" marginheight="0" frameborder="0"></iframe>
</td>
<td>
I published my new book: <a href="http://www.amazon.com/gp/search?ie=UTF8&keywords=kleyn%20representation&tag=sciencegalery-20&index=aps&linkCode=ur2&camp=1789&creative=9325">Representation Theory: Representation of Universal Algebra</a>.<img src="http://www.assoc-amazon.com/e/ir?t=sciencegalery-20&l=ur2&o=1" width="1" height="1" border="0" alt="" style="border:none !important; margin:0px !important;" /> 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.</td>
</tr>
</table>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-52373817927638818272010-09-26T13:53:00.000-07:002011-10-29T17:29:59.682-07:00My first published book<table>
<tr>
<td>
<iframe src="http://rcm.amazon.com/e/cm?t=sciencegalery-20&o=1&p=8&l=as1&asins=3843351635&ref=tf_til&fc1=000000&IS2=1<1=_blank&m=amazon&lc1=0000FF&bc1=000000&bg1=FFFFFF&f=ifr" style="width:120px;height:240px;" scrolling="no" marginwidth="0" marginheight="0" frameborder="0"></iframe></td>
<td>
This year I published my first book <br />
<a href="http://www.amazon.com/gp/search?ie=UTF8&keywords=kleyn%20Linear%20algebra&tag=sciencegalery-20&index=aps&linkCode=ur2&camp=1789&creative=9325">Linear Mappings of Free Algebra: First Steps in Noncommutative Linear Algebra</a><img src="http://www.assoc-amazon.com/e/ir?t=sciencegalery-20&l=ur2&o=1" width="1" height="1" border="0" alt="" style="border:none !important; margin:0px !important;" />
<br />
Book was published by Lambert Academic Publishing.</td>
</tr>
</table>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-54499886536220549892010-03-09T19:41:00.000-08:002016-01-23T18:48:19.743-08:00About Cauchy-Riemann EquationRecently 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. <p>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 <a href="http://arxiv.org/abs/0909.0855" target="_blank">arXiv:0909.0855</a> Quaternion Rhapsody, 2010Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-82981962002673847502010-02-02T18:09:00.000-08:002010-07-12T22:02:40.553-07:00Matrix of MappingsI recently published new paper<br />
eprint <a href="http://arxiv.org/abs/1001.4852">arXiv:1001.4852</a> The Matrix of Linear Mappings, 2010<br />
<br />
In this paper, I consider change in notation of mapping. Instead of notation<br />
y=f(x)<br />
I use notation<br />
y = f ° x<br />
Let us consider ring of mappings of R-algebra A with product f ° g<br />
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<br />
∂f(x)(h)=∂f(x) ° h<br />
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 = x<sup>2</sup> then<br />
$$\partial f(x)\circ dx=(x\otimes 1+1\otimes x)\circ dx=x\ dx+dx\ x$$Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-79055648660261520212009-08-02T10:39:00.000-07:002009-09-09T18:52:49.636-07:00Regular function of quaternion variable<p>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<br /><br /><br /><sup>∂f</sup>/<sub>∂x<sub>0</sub></sub>+i<sup>∂f</sup>/<sub>∂x<sub>1</sub></sub>+j<sup>∂f</sup>/<sub>∂x<sub>2</sub></sub>+k<sup>∂f</sup>/<sub>∂x<sub>3</sub></sub>=0</p><p><br />Recently Daniel Alayon-Solarz sent me interesting example of regular function.</p><p>f(x<sub>0</sub>+x<sub>1</sub>i+x<sub>2</sub>j+x<sub>3</sub>k)=x<sub>0</sub>-x<sub>1</sub>i+x<sub>2</sub>j+x<sub>3</sub>k<br /></p><p> </p>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-91358690776628030112009-06-01T20:11:00.000-07:002009-06-01T20:40:42.050-07:00Higgs bosonThe 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.<br /><br />If we will not find the Higgs boson then we need to reconsider model which predicted them.<br /><br /><div align="justify">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.</div><div align="justify"> </div><div align="justify">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.</div>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-34882501269454447312009-05-03T09:46:00.000-07:002009-08-02T11:53:31.541-07:00Derivative of function of division ring<div id="ms__id31" align="justify">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.</div><div id="ms__id42" align="justify"><br />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.</div><div id="ms__id33"></div><div id="ms__id34">For instance derivative of function <em>f(x)=axb </em>has form</div><div id="ms__id35" align="center"><em>∂f(x)(h)=ahb</em></div><div id="ms__id36" align="left">Derivative of function <em>f(x)=x</em><sup><em>2</em> </sup>has form</div><div id="ms__id38" align="center"><em>∂f(x)(h)=xh+hx</em></div><div id="ms__id39" align="left">Derivative of function <em>f(x)=x<sup>-1</sup></em> has form</div><div id="ms__id40" align="center"><em>∂f(x)(h)=-x<sup>-1</sup>hx<sup>-1</sup></em></div>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0tag:blogger.com,1999:blog-5721693232875293343.post-77476365776635935842009-04-20T19:10:00.000-07:002009-05-03T09:45:47.859-07:00Additive map of division ring<div id="ms__id39">I created this blog to share and discus ideas related to my research in geometry.</div><div id="ms__id38"></div><br /><div id="ms__id37">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.</div><div id="ms__id43"></div><br /><div id="ms__id44">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.</div><br /><div id="ms__id45"></div><div id="ms__id46">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.</div><div id="ms__id69"></div><div id="ms__id68"></div><div id="ms__id67"></div><br /><p>Quaternion algebra is an example of division ring.<br /></p><div id="ms__id51"></div><div id="ms__id49"></div><div id="ms__id56"></div><p></p><p>You can read my papers in arXiv<br /></p><div id="ms__id52"></div><div id="ms__id50">eprint <a href="http://arxiv.org/abs/math.GM/0701238">arXiv:math.GM/0701238</a> Lectures on Linear Algebra over Division Ring, 2007 </div><div id="ms__id53"></div><div id="ms__id54">eprint <a href="http://arxiv.org/abs/0812.4763">arXiv:0812.4763</a> Introduction into Calculus over Division Ring, 2008 </div>Aleks Kleynhttp://www.blogger.com/profile/15855878612420984960noreply@blogger.com0