# 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

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

• juanmarqz

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…

• juanmarqz

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… : )