MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2

2

$[v_4,l_4]=$?, where $v_4$ is the Hopf map in $\pi_7(S^4)$ and $l_4$ is the generator of $\pi_4(S^4)$ which represent the identity map, $[,]$ is whitehead product.

flag
2 
mathoverflow.net/howtoask – Yemon Choi Feb 26 2012 at 9:55
Ravanel's book is available on-line. Have you looked there? – Ryan Budney Feb 26 2012 at 10:52

1 Answer

1

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 $6$.

link|flag
$2\nu_4-E\omega=\pm[i_4,i_4]$,so,what is $[E\omega,i_4]?$ – unknown (yahoo) Feb 27 2012 at 3:52
where $\omega$ is the generator of $\pi_6(S^3)=\textbf{Z}_{12}$, and $E$ is the suspension. – unknown (yahoo) Feb 27 2012 at 4:07

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.