Timeline for How to see the quaternionic hopf map generates the stable 3-stem?
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Mar 2, 2013 at 0:15 | comment | added | Sergey Melikhov | A geometric construction of a generator of $\pi_3^{st}$ is discussed in this paper by Eckholm and Takase: arxiv.org/abs/0903.0238 Note also that the composition of the 8-fold covering $S^3\to S^3/Q$, where $Q$ is the quaternion group, and a standard embedding $S^3/Q\to S^4$ represents a generator of $3\pi_3^{st}$. | |
| Feb 20, 2013 at 19:28 | comment | added | Sam Gunningham | Perhaps this paper on the vanishing of the third spin cobordism group is relevant: maths.ed.ac.uk/~aar/papers/stipsicz.pdf The author has tried to make the proof "as elementary as possible". | |
| Feb 20, 2013 at 18:52 | comment | added | Danny Ruberman | Chris: Ryan's argument isn't circular. Starting with a Heegaard splitting of M (from a triangulation or Morse function), you get M as surgery on a link. This already shows $\Omega_3$ is trivial. An algorithm of S. Kaplan encodes a spin structure on M in terms of a "characteristic sublink" of this link, and shows how to find a bounding spin manifold by eliminating the characteristic sublink. This route seems easier to me than the AHSS, but as you say this depends on one's perspective. | |
| Feb 20, 2013 at 13:05 | comment | added | Scott Carter | Not an answer, but some related material: Koschorke and Sanderson used the self-intersection (SI) maps of immersed 3-manifolds in 4-space to compute this group. Other results of Koschorke are relevant. Mike Freedman's paper on self-interestions of immersions also contains information. Peter Eccles wrote about the (SI) maps. Finally, the standard Froisart Morin sphere eversion can be shown to represent a generator. All of the results I mentioned are in published in the era (1978-1986). | |
| Feb 20, 2013 at 10:12 | comment | added | Chris Schommer-Pries | Ryan, I am a little worried about that approach being circular as proofs of the Kirby calculus usually start with the fact that every 3-manifold bounds or something equivalent (e.g. that every 3-manifold can be represented as surgery on links in the 3-sphere). It is easy to go the other way: once you know that $\nu$ generates the framed bordism group, then an easy AHSS calculation shows that the spin and oriented bordism groups must vanish. | |
| Feb 20, 2013 at 9:18 | comment | added | Ryan Budney | Charles, yes there are direct proofs via Kirby calculus. Usually these algorithms are used to compute things like the Rochlin invariant (an obstruction to bounding a homology ball), but the algorithms are fairly general. | |
| Feb 20, 2013 at 2:28 | comment | added | Charles Rezk | The first thing to comes to mind is derive it from $\pi_3 MSpin=0$. Is there a geometric proof that 3-dimensional spin manifolds bound? | |
| Feb 19, 2013 at 23:23 | comment | added | Eric Wofsey | A possibly related question would be to ask for a proof that $\eta^3=12\nu$. The only proof of this I know is by the Adams spectral sequence, and indeed this is in some sense the "hardest" step in verifying that $\nu$ generates the 3-stem via the Adams spectral sequence. | |
| Feb 19, 2013 at 22:27 | history | asked | Chris Schommer-Pries | CC BY-SA 3.0 |