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

MoreandMore

look a rough view

another earlier

3 Comments

Filed under cucei math, fiber bundle

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

  1. Alberto

    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

    • 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…

    • 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… : )

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s