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I have a map, constructed geometrically, $S^4 \to S^3$. I suspect that it is a representative for the generator $\eta_3\in \pi_4(S^3) \simeq \mathbb{Z}_2$, but I am not 100% sure ($\eta_3$ is defined as the suspension of the Hopf map $S^3 \to S^2$). I'd like to be able to detect the fact that my map is homotopically nontrivial in a geometric or combinatorial way. Is this possible? I imagine there is some sort of index one can calculate, but unfortunately the terms one looks these things up in are very generic (forms, integration, index etc).

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Incidentally, this is my 100th question. :-) –  David Roberts Apr 13 '13 at 1:51
    
I am no topologist, but I think that it is standard that the Freudenthal suspension map $\pi_3(S^2) \to \pi_4(S^3)$ is surjective. Since the Hopf map generates $\pi_3(S^2)$, its suspension generates $\pi_4(S^3)$. –  Angelo Apr 13 '13 at 2:12
    
@Angelo, I know that part, but my map is defined without reference to suspension or the Hopf map, so I need to know if what I have is homotopic to $\eta_3$. –  David Roberts Apr 13 '13 at 3:02
    
Sorry, I misread your post, I thought you were saying that you map was defined as the suspension of the Hopf map. –  Angelo Apr 13 '13 at 3:18
    
What does 'a map constructed geometrically' mean? Are not all of them constructed geometrically? I think it is unlikely that you get a completely satisfactory answer, unless you say what the map is. –  Fernando Muro Apr 13 '13 at 7:36
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up vote 22 down vote accepted

How about thinking about framed cobordism, which in this case gives an isomorphism between $\pi_4(S^3)$ and the group of cobordism classes of normally framed 1-manifolds in $S^4$. Since your map is constructed geometrically, it probably has a regular value. Pull this back to a collection of disjoint circles in $S^4$, forming a trivial link, with framings of their normal bundles. Since $\pi_1SO(3)$ has order 2, just count up the number of circles with nontrivial framing to see whether this number is odd or even.

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Thanks, Allen. I thought something like this would be the case, but I didn't know how to phrase it. –  David Roberts Apr 14 '13 at 0:29
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