2011/12/27

möbius doble cover

The set E=M\ddot{o}\stackrel{\sim}\times I is an orientable 3-manifold with boundary. In the illustration we see in orange the möbius band at \frac{1}{2} and a small regular neigbourhood of her removed without her, i.e., if Q={\cal{N}}(M\ddot{o}\times \frac{1}{2})\smallsetminus (M\ddot{o}\times \frac{1}{2})\subset E, then which is E\smallsetminus Q?

 

 

 

 

 

 

 

 

 

 

 

 

 

the last step is M\ddot{o}\times\frac{1}{2} in orange, and M\ddot{o}\stackrel{\sim}\times I without M\ddot{o}\stackrel{\sim}\times (\frac{1}{2}-\varepsilon, \frac{1}{2}+\varepsilon), for \varepsilon=|\varepsilon|\to \frac{1}{2}

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2011/11/28

real elementary multilinear algebra

are the common algebraic-techniques  territory for today vector algebra and differential geometry.

This means that ancient vectorcalculus who turns into differential forms nowadays, is “super-oversimplified” into a mathematical language to phrase some modern geometricalalgebrotopologicalanalitic-maths.

Real, ‘cuz first you gotta get the ideas over the field \mathbb{R}.

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2011/11/10

función con círculos críticos en el toroide

to see the example à la wiki of a function with circles as a critical sets of a  scalar field in a surface in three dimensional space:  follows…

para ver un ejemplo à la wiki de una función con cercos como uns insiemi o campo escalar en una área del trosième Raum: sigue…

 

 

actualización (15.Nov.2011):

añadido gráfico:


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2011/11/01

is it enough…

that the presentation given by

\langle A,B\ :\ A^2=B^3,\quad A^2B=BA^2,\quad A^4=e,\quad B^6=e\rangle

determines the group SL_2({\mathbb{Z}}/{3\mathbb{Z}})?

Similar question for

\langle A,B,C:

A^2=B^3,

A^2B=BA^2,

A^4=B^6= C^2=e,

AC=CA,BC=CB\rangle

for the group GL_2({\mathbb{Z}}/{3\mathbb{Z}}).

Other similar problems but less “difficult” are:

  • \langle\varnothing:\varnothing\rangle=\{e\}
  • \langle A\ :\ A^2=e\rangle  for  \mathbb{Z}_2
  • \langle A\ :\ A^3=e\rangle  for  \mathbb{Z}_3
  • \langle A\ :\ A^4=e\rangle  for  \mathbb{Z}_4
  • \langle A,B\ :\ A^2=e, B^2=e, AB=BA\rangle  for  \mathbb{Z}_2\oplus\mathbb{Z}_2
  • \langle A,B\ :\ A^2=e, B^3=e, AB=B^2A\rangle  for  S_3
  • \langle A,B\ :\ A^2=e, B^3=e, AB=BA\rangle  for  \mathbb{Z}_2\oplus\mathbb{Z}_3
  • \langle A:\varnothing\rangle it is \mathbb{Z}
  • \langle A,B:AB=BA\rangle  is \mathbb{Z}\oplus\mathbb{Z}
  • \langle A,B:\varnothing\rangle  is \mathbb{Z}*\mathbb{Z}, the rank two free group
  • \langle A,B,J:(ABA)^4=e,ABA=BAB\rangle for SL_2(\mathbb{Z})
  • \langle A,B,J:(ABA)^4=J^2=e,ABA=BAB, JAJA=JBJB=e \rangle for GL_2(\mathbb{Z})
  • \langle A,B:A^2=B^3=e \rangle for P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3
  • …
  • …
  • …
  • dare altri venti esempi
Qual è il tuo preferito?

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2011/10/29

actualizado:

http://juanmarqz.wordpress.com/2011/10/06/problems-de-algebra-moderna-uno-i/

Uno debe acostumbrase a leer mucho,

pero no será provechoso si no se practica: leer y entender.

Uno no nace con esta habilidad de por sí,

pero esto  sí es posible desarrollar,

si le pones persistencia.

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2011/10/20

SL_2(Z/3Z), aliquis ordo

hae sunt matrices (estas son las matrices):

a=\left(\begin{array}{cc}0&1\\ 2&0\end{array}\right) , b=\left(\begin{array}{cc}0&1\\ 2&1\end{array}\right) , ba^3b=\left(\begin{array}{cc}0&1\\ 2&2\end{array}\right)

a^3=\left(\begin{array}{cc}0&2\\ 1&0\end{array}\right) , bab=\left(\begin{array}{cc}0&2\\ 1&1\end{array}\right) , a^2b=\left(\begin{array}{cc}0&2\\ 1&2\end{array}\right)

e=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right) , a^2ba=\left(\begin{array}{cc}1&0\\ 1&1\end{array}\right) , (ba)^2=\left(\begin{array}{cc}1&0\\ 2&1\end{array}\right)

(ab)^2=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right) , b^2ab=\left(\begin{array}{cc}1&1\\ 1&2\end{array}\right) , a^3ba=\left(\begin{array}{cc}1&1\\ 2&0\end{array}\right)

a^3b=\left(\begin{array}{cc}1&2\\ 0&1\end{array}\right) , a^2b^2=\left(\begin{array}{cc}1&2\\ 1&0\end{array}\right) , bab^2=\left(\begin{array}{cc}1&2\\ 2&2\end{array}\right)

a^2=\left(\begin{array}{cc}2&0\\ 0&2\end{array}\right) , ab^2=\left(\begin{array}{cc}2&0\\ 1&2\end{array}\right) , ba=\left(\begin{array}{cc}2&0\\ 2&2\end{array}\right)

ab=\left(\begin{array}{cc}2&1\\ 0&2\end{array}\right) , b^2aba=\left(\begin{array}{cc}2&1\\ 1&1\end{array}\right) , b^2=\left(\begin{array}{cc}2&1\\ 2&0\end{array}\right)

b^2a=\left(\begin{array}{cc}2&2\\ 0&2\end{array}\right) , aba=\left(\begin{array}{cc}2&2\\ 1&0\end{array}\right) ,  abab^2=\left(\begin{array}{cc}2&2\\ 2&1\end{array}\right)

… ante hoc dictionary

Nunc, possibile est quod haec propositio quaerere:

\langle a,b\ :\ a^4=e, b^6=e, a^2b=ba^2\rangle

  • is it possible that: a^2b=ba^2 together with a^4=e and b^6=e they imply a^2=b^3?

http://en.wikipedia.org/wiki/File:SL%282,3%29;_Cayley_table.svg

for more on Cayley tables go

Cayley

 

 

 

 

 

 

 

 

 

 

(10.nov.2011)

Usando la teoría básica de grupos podremos construir un subgrupo de 12 elementos usando un subgrupo de orden tres y otro subgrupo de orden cuatro:

Con H=\langle p|p^3=e\rangle y con K=\langle q|q^4=e\rangle armamos HK. Así HK tendrá 12 elementos y pues H\cap K=\{e\}.

Chéquelo con p=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)  y q=\left(\begin{array}{cc}1&1\\ 1&2\end{array}\right).

El que queriamos armar usando \langle\{a^2,b\}\rangle, no funciona para obtener un subgrupo de 12 elementos,  pues porque, sabiendo a^2=b^3, tendremos \langle a^2\rangle<\langle b\rangle y por lo tanto \langle\{a^2,b\}\rangle=\langle b\rangle.

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2011/10/17

función contadora

sigue a contadora para ver los algunos detalles de la biyección {\mathbb{N}}\times{\mathbb{N}}\to{\mathbb{N}} que etiqueta las intersecciones de líneas verticales con horizontales espacidas a distancia uno entre ellas.

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