Spanier-Whitehead dual and Hopf fibration - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:01:45Z http://mathoverflow.net/feeds/question/6152 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6152/spanier-whitehead-dual-and-hopf-fibration Spanier-Whitehead dual and Hopf fibration Igor Belegradek 2009-11-19T19:12:39Z 2009-11-19T20:22:19Z <p>Consider a map of spheres $f:S^n\to S^m$ covered by a map of trivial $\mathbb R^k$-bundles. In other words, we take the trivial rank $k$ vector bundle over $S^m$ and pull it to $S^n$ via $f$. Consider the corresponding map of the Thom spaces, and its Spanier-Whitehead dual. How is the dual related to $f$? </p> <p>The test case I care about is when $f$ is the Hopf fibration $S^3\to S^2$. Then I think the dual can be represented by a map $F:\Sigma^{r+1}(S^2_+)\to\Sigma^r (S^3_+)$ where $X_+$ means $X$ disjoint union a point, and $\Sigma^s$ is $s$-fold suspension. Thus $F$ can be thought of as a map $S^{r+3}{\vee} S^{r+1}\to S^{r+3}{\vee} S^r$. What is this map?</p> http://mathoverflow.net/questions/6152/spanier-whitehead-dual-and-hopf-fibration/6163#6163 Answer by Tyler Lawson for Spanier-Whitehead dual and Hopf fibration Tyler Lawson 2009-11-19T20:22:19Z 2009-11-19T20:22:19Z <p>(I'm going to assume you only care about the homotopy class of map.)</p> <p>Such maps are roughly their own duals. The Spanier-Whitehead dual of a map $S^n \to S^m$ is a map $"S^{-m}_+ \to S^{-n}_+"$, which is represented by $\Sigma^{r-m} S^m_+ \to \Sigma^{r-n} S^n_+$ for sufficiently large r. If r=n+m+s, this is a map $\Sigma^{s+n} S^m_+ \to \Sigma^{s+m} S^n_+$. This is homotopy equivalent to a map $S^{n+m+s} \vee S^{n+s} \to S^{n+m+s} \vee S^{m+s}$.</p> <p>The resulting map is the wedge of the identity map on the first factor and the (up to sign, which I will get wrong if I try) the s-fold suspension of $f$ on the second factor.</p> <p>This is easier to say in the stable homotopy category, where your original map become stably the map $id \vee \Sigma^\infty f:S^0 \vee S^n \to S^0 \vee S^m$ and the dual just dualizes the maps on each factor. It's also easier to say if you use based maps of spheres in the first place.</p>