# $[\nu_4,\iota_4]=?$, $\nu_4$ is the Hopf map in $\pi_7(S^4)$ and $\iota_4$ is the generator of $\pi_4(S^4)$

$[\nu_4,\iota_4]=?$, where $\nu_4$ is the Hopf map in $\pi_7(S^4)$ and $\iota_4$ is the generator of $\pi_4(S^4)$ which represents the identity map, $[-,-]$ is the Whitehead product.

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mathoverflow.net/howtoask –  Yemon Choi Feb 26 '12 at 9:55
Ravanel's book is available on-line. Have you looked there? –  Ryan Budney Feb 26 '12 at 10:52
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## 1 Answer

This is how far I can get without checking a book.

Since $v_4 = \pm {1\over 2} [i_4, i_4]$, you are interested in the triple Whitehead product $\alpha = {1\over 2}[ [i_4, i_4],i_4]$. Since it is a Whitehead product, its suspension is trivial. The Jacobi identity for Whitehead products shows that $[ [i_4, i_4],i_4]$ has order $3$; so $\alpha$ has order either $3$ or $6$.

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$2\nu_4-E\omega=\pm[i_4,i_4]$,so,what is $[E\omega,i_4]?$ –  user21719 Feb 27 '12 at 3:52
where $\omega$ is the generator of $\pi_6(S^3)=\textbf{Z}_{12}$, and $E$ is the suspension. –  user21719 Feb 27 '12 at 4:07
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