$[\nu_4,\iota_4]=?$, 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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4$\begingroup$ mathoverflow.net/howtoask $\endgroup$– Yemon ChoiCommented Feb 26, 2012 at 9:55
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$\begingroup$ Ravanel's book is available on-line. Have you looked there? $\endgroup$– Ryan BudneyCommented Feb 26, 2012 at 10:52
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1 Answer
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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 $6$.
answered Feb 26, 2012 at 17:25
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$\begingroup$ $2\nu_4-E\omega=\pm[i_4,i_4]$,so,what is $[E\omega,i_4]?$ $\endgroup$ Commented Feb 27, 2012 at 3:52
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$\begingroup$ where $\omega$ is the generator of $\pi_6(S^3)=\textbf{Z}_{12}$, and $E$ is the suspension. $\endgroup$ Commented Feb 27, 2012 at 4:07