Tag Archives: italian

2011/11/01 · 15:32

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 $P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3$
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dare altri venti esempi

Qual è il tuo preferito?

Filed under algebra, free group, group theory, mathematics

Tagged as abstract algebra, algebra, cucei math, du CUCEI au monde, français, group theory, italian, italiano, italy

2009/09/28 · 10:35

a non orientable 3d-manifold with boundary

Mö x I

Mö x I

The picture on the right is representation and a description of its parts of a 3d chunk, it is the trivial $I$ -bundle over the möbius-strip, $M\ddot{o}\times I$ . It is useful to determine which 3d spaces are non orientable.

A 3d-space is non orientable if it has a simple closed curve whose 3d regular neighborhood is homeomorphic to the model of the picture.

If we denote by $C\ddot{o}$ the core of the möbius strip, you can deduce for the curve $C\ddot{o}\times\{\frac{1}{2}\}$ (the black pearled one) that its tangent bundle can’t be embedded in the tangent bundle of $\mathbb{R}^3$ or in any other tangent bundle of an orientable one.

Remember $N_2$ is the Klein bottle. Let’s the image talks by itself.

Ah, il nome di questo grande pezzo è la bottiglia solida di Klein.

Filed under fiber bundle, geometry, topology

Tagged as 3-manifolds, 3d chunk, bottiglia di Klein, bottiglia solida di Klein, bundle, fiber bundle, I-bundle, italian, italy, low dimensional topology, mobius, mobius band, mobius strip, Moxi, non orientable, solid Klein bottle

2009/09/21 · 20:23

rubik

L’enigma di Rubik e la sua soluzione.

Un gioco matematico, molto buffo ma realmente facile…

Che altro lei vuole che devo il discorso?

Ci sono qualcosa che dovrei scrivere?

In che la lingua? l’italiano è bello

Filed under math

Tagged as cubo de rubik, cubo mágico, italian, rubik cube, soluzione