Skip to main content
10 events
when toggle format what by license comment
Jan 6, 2023 at 9:17 comment added Lennart Meier @Tyrone I see, thanks for the additional explanation!
Jan 6, 2023 at 8:22 comment added Tyrone Hi @LennartMeier. The class $x$ lives in degree $n$, so you have to keep track of signs. The formula you wrote is correct when $n$ is even, but breaks down when $n$ is odd (e.g. when $k=2$ you end up with $(d^2+e^2)x^2$). Returning to your original notation, the relation is essentially $\Omega\Sigma(f+g)=\Omega\Sigma(f)+\Omega\Sigma(g)+\overline{[f,g]}\circ h_2$, where $h_2$ is the second Hilton-Hopf invariant. It's a little more complicated, but refer to $\S4$ of Neisendorfer's book for the details.
Jan 6, 2023 at 7:40 comment added Lennart Meier @Tyrone You're right: $\Omega \Sigma f$ and $g$ are not homotopic (postcomposition vs. precomposition). But on homology: $H_*(\Omega\Sigma S^n)\cong\mathbb{Z}[x]$ with $|x|=n$. Given maps $d, e: S^n\to S^n$, we can write the loop product of $\Omega\Sigma d$ and $\Omega \Sigma e$ as $\Omega\Sigma S^n \xrightarrow{d,e}\Omega\Sigma S^n\times \Omega\Sigma S^n \to \Omega\Sigma S^n$. The first map sends $x^m$ to $\sum_k \binom{m}{k}(dx)^k(ex)^{m-k}$ (Hatcher 3.C.1). The second is multiplication and thus the composite is $\sum_k \binom{m}{k}d^ke^{m-k}x^m = (d+e)^mx^m$. Am I wrong again?
Jan 5, 2023 at 12:11 comment added Tyrone Long story short: the diagram commutes by an Eckmann-Hilton argument. The maps $\Omega\Sigma(f)$ and $g$ are not homotopic (take $n=1$ and consider their action on homotopy groups. $g$ is multiplication by $d$, but $\Omega\Sigma f_*\tilde\eta=d^2\cdot\tilde\eta$, where $\tilde\eta:S^2\rightarrow\Omega S^2$ is adjoint to the Hopf map).
Jan 5, 2023 at 11:59 comment added Tyrone @LennartMeier You can check that $\Omega\Sigma(1-1)=0$ and $\Omega\Sigma(1)+\Omega\Sigma(-1)=1+\Omega\Sigma(-1)$ do not induce the same map on homology.
Jan 5, 2023 at 11:14 comment added Lennart Meier Let $f,g\colon S^n\to S^m$ be maps. I think that it is more generally true that $\Omega \Sigma (f+g)$ and the loop product of $\Omega \Sigma(f)$ and $\Omega\Sigma(g)$ agree up to homotopy as maps $\Omega\Sigma S^n \to \Omega \Sigma S^m$. This is based on a variant of the Eckmann--Hilton argument (or the argument that the different produts on $\pi_nX$ agree for $n\geq 2$.)
Jan 4, 2023 at 20:47 review Close votes
Jan 10, 2023 at 3:08
Jan 4, 2023 at 19:51 history edited Jacob CC BY-SA 4.0
added 188 characters in body
S Jan 4, 2023 at 19:41 review First questions
Jan 4, 2023 at 22:10
S Jan 4, 2023 at 19:41 history asked Jacob CC BY-SA 4.0