Timeline for "canonical" framing of 3-manifolds
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2023 at 3:25 | history | edited | Daniel Asimov | CC BY-SA 4.0 |
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| Dec 8, 2023 at 0:13 | history | edited | Marco Golla | CC BY-SA 4.0 |
edited according to mme's comments
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| Dec 7, 2023 at 15:56 | comment | added | mme | @MarcoGolla Well, $0 \in \Bbb Z$ is a pretty canonical element, so I'm satisfied with Atiyah calling this a "canonical framing". (More precisely, the bijection is determined by sending a 2-framing $\alpha$ to the integer $\sigma(Z) - p_1(Z, \alpha)$, where $Z$ is an oriented 4-fold bounding $Y$; the `canonical framing' for which this is zero is the one for which Hirzebruch's signature formula is still true.) | |
| Dec 7, 2023 at 15:42 | comment | added | Alon Amit | Please don't delete the answer! It provides great clarity on the difference between framing $TM$ vs $TM \oplus TM$. | |
| Dec 7, 2023 at 14:49 | comment | added | Marco Golla | Oh, nice, thanks Mike! I trusted the OP had the question right... So there's no canonical 2-framing, either, just a canonical bijection with $\mathbb{Z}$, which maybe isn't too far from what is in my answer? (I will edit, or maybe even delete the answer, when I'm sure I got what is needed.) | |
| Dec 7, 2023 at 14:19 | comment | added | mme | Looking at Atiyah's paper, he's not claiming that there's one framing up to homotopy (which appears to be the goal of your last paragraph), he's claiming the framings carry a canonical bijection to the integers, this bijection arising from the Hirzebruch signature formula. | |
| Dec 7, 2023 at 13:45 | comment | added | mme | I don't think the last claim is true. Take $\tau$ to be the identity to see you're asking if the diagonal embedding $SO(3) \to SO(6)$ is null. There are many ways to see that this is false, but here's one: the standard embedding induces an isomorphism on $\pi_3$, and the embedding as the bottom-right block is homotopic to that. Your map is the product of those, so it suffices to show that multiplication induced the sum map on $\pi_3$, which is the Eckmann-Hilton argument. Thus the induced map on $\pi_3$ is injective (it's multiplication by 2, if you like). | |
| Dec 7, 2023 at 13:06 | history | answered | Marco Golla | CC BY-SA 4.0 |