It is ture?
How prove that?
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6$\begingroup$ mathoverflow.net/howtoask#motivation $\endgroup$– j.c.Oct 1, 2011 at 1:02
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1$\begingroup$ I don't think this is true. $\pi_7S^4 = \mathbb{Z} \times \mathbb{Z}/{12}$. Probably $\omega$ suspends to the generator of the $\mathbb{Z}/12$ summand, and the Hopf map is the other. $\endgroup$– Dylan WilsonOct 1, 2011 at 5:13
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$\begingroup$ Right... I think the way to see this is to look at the EHP sequence, since $\pi_7S^4$ is right on the edge. $\endgroup$– Dylan WilsonOct 1, 2011 at 5:15
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1$\begingroup$ How about accepting some of the answers to your previous questions, or at least engaging with them, before asking yet another new question? $\endgroup$– Yemon ChoiOct 2, 2011 at 2:01
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1 Answer
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No. The Hopf map in $\pi_7(S^4)$ is usually called $\nu$, and the generator of $\pi_6(S^3)$ is usually called $\nu'$. The Hopf invariant $H:\pi_7(S^4)\to\pi_7(S^7)=\mathbb{Z}$ has $H(\nu)=1$ but $H\Sigma=0$ so in particular $H(\Sigma(\nu'))=0$ and $\Sigma(\nu')$ cannot be equal to $\nu$. Or you can just argue that $\nu'$ has finite order so $\Sigma(\nu')$ also has finite order, but the order of $\nu$ is infinite.
The canonical reference is Toda's book "Composition methods in the homotopy groups of spheres". Alternatively, you can download a Mathematica representation of most of the results in that book from http://neil-strickland.staff.shef.ac.uk/toda/.
answered Oct 1, 2011 at 9:34