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Oct 29, 2018 at 17:51 comment added mme @JasonDeVito It sounds like the contribution was all your own :)
Oct 29, 2018 at 17:36 comment added Jason DeVito - on hiatus My board is currently covered with attempts at using Freudenthal in exactly the way you suggest (and with exactly the same predicted answer). I must just be writing down the wrong commutative diagram...
Oct 29, 2018 at 17:34 comment added mme @JasonDeVito I think you can apply Freudenthal suspension and the fact that a degree $n$ map suspends to a degree $n$ map to see that it's multiplication by $n^2$ = multiplication by $n$ (since $n^2 = n$ mod 2).
Oct 29, 2018 at 17:33 comment added Jason DeVito - on hiatus Do you happen to know (or know a reference) for what map $\pi_{k+1}(S^k)\rightarrow \pi_{k+1}(S^k)$ is induced from a degree $n$ map $S^k\rightarrow S^k$. I know the proof (using Whitehead products) for $k=2$, but I'm not sure how to generalize. Of course, for larger $k$, $\pi_{k+1}(S^k)\cong \mathbb{Z}_2$....
Aug 9, 2018 at 22:10 comment added Sigur Thanks so much for you attention. Let me read/study what you wrote.
Aug 9, 2018 at 21:54 history answered mme CC BY-SA 4.0