2010/03/20 · 20:11

Moxi: the cartesian product of the mobius-strip and an interval

MoxI is a solid Klein bottle, the genus one nonorientable handlebody. Here we see it with some typical curves on

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look a rough view

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3 Comments

Filed under fiber bundle, cucei math

Tagged as low dimensional topology, topology, non orientable, mobius, mobius band, 3-manifolds

3 Responses to Moxi: the cartesian product of the mobius-strip and an interval

Alberto

2010/04/04 at 13:37

Hi juanmarqz,
i have a problem to computing $\Pi_4(S^3)$ . I was by http://math.ucr.edu/home/baez/week102.html

and I understand $\Pi_3(S^2)$ . I thynk that one must associate all manifold with n-sphere in order to become homotopy groups. For example, $S^2 \simeq SU(2)$ since the complex matrix $g$ with $det(g) = |z_1|^2+|z_2|^2 = 1 = x^2+y^2+z^2+t^2$ choosing $z_1 = t-iz$ and $z_2 = y-ix.$ Now in $\Pi_3(S^2)$ , one know that $f(x_i) = y_j$ , where $x_i \in S^3$ and $y_j \in S^2$ they build links as knots. Each link turns into another a n-integer, Anzahl der Umschlingungen, this is the topological invariant searched. So $llatex \Pi_3 (S^2) = \mathbb(Z).$

Do you have a technique to computing $\Pi_4(S^3)$ ? Do you know what it mean $\mathbb{Z}_2$ or $\mathbb{Z}_2 \times \mathbb{Z}_2$ in other cases?

In $S^2$ , one think that there are two homotopy classes: 1) select a point p and a loop, it can be contracted to p, 2) select two points and link them, you have many pahts that they can deformed one into another. Thus, $\Pi_{?}(S^2) = \mathbb{Z}_2$ . But what about another non-trivial $\Pi_n (S^m)$ ?

Danke

Reply
- juanmarqz
  
  2010/04/05 at 11:33
  
  I am greatly surprised by the Frage, and glad that you got reached such a level of abstraction ~~(blurry-knowledge)~~
  and then before to begin answer it, I would like to clear some issues
  (and for giving me time to restart my homotopy machine…)
  
  Now, to start, the sentence:
  I ~~thynk~~ that one must associate all manifold with n-sphere in order to become homotopy groups.,
  it isn’t true in general. Perfectly I understand the problem but I need to consult and re-consult the references, wait.
  
  And, in the other hand, you give me an excuse to read more about what is Baez saying…
  
  Reply
- juanmarqz
  
  2010/04/05 at 12:17
  
  let me, also upon after the reading (not thourougly) about Baez had said, that some of the Fragen still (1997) had no answers, but some of yours are treated there… I invite you to find me next week at my cubículo to discuss how much I can help you…
  
  Finally, perhaps one can get another method to proof $\pi_4S^3=\mathbb{Z}_2$ by mimicking the other methods of proving $\pi_3S^2=\mathbb{Z}$ , metioned (not explicitly) by Baez… : )
  
  Reply