Timeline for Which stable homotopy groups are represented by parallelizable manifolds?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2020 at 17:36 | vote | accept | Chris Schommer-Pries | ||
| Jun 27, 2020 at 1:14 | history | became hot network question | |||
| Jun 27, 2020 at 0:22 | answer | added | user160262 | timeline score: 7 | |
| Jun 26, 2020 at 23:29 | answer | added | archipelago | timeline score: 13 | |
| Jun 26, 2020 at 21:03 | history | edited | Chris Schommer-Pries | CC BY-SA 4.0 |
changed title to be more precise.
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| Jun 26, 2020 at 21:03 | comment | added | Chris Schommer-Pries | @NicholasKuhn It depends. In algebraic topology we usually (but not always) mean stably framed and shorten it to "framed". But, for example, in the cobordism hypothesis one talks of the framed bordism category and in this case one means unstable tangential framings. So it all depends on context. Perhaps you would prefer the term "parallelizable"? I will change the title. | |
| Jun 26, 2020 at 20:40 | comment | added | Nicholas Kuhn | Your definition of a framed manifold is not standard. Surely the 2 sphere is framed, but its tangent bundle isn't trivial. | |
| Jun 26, 2020 at 20:20 | answer | added | Oscar Randal-Williams | timeline score: 17 | |
| Jun 26, 2020 at 19:16 | comment | added | Chris Schommer-Pries | @archipelago That is a nice idea. I think you are suggesting something along the following lines. Suppose that X is a stably framed manifold where the obstruction to destabilizing is non-trivial. Suppose for concreteness that [X] generates a $\mathbb{Z}/3\mathbb{Z}$ in $\pi^s_*$. Then the obstruction will vanish on 2[X], which also generates this group. But 2[X] is just two disjoint copies of X, so how could it all of a sudden become framable? Really the obstruction lives in a $\mathbb{Z}/2$ for each component of the manifold in question. | |
| Jun 26, 2020 at 19:01 | comment | added | archipelago | For trivial reasons, it can in odd dimensions at most fail at the prime $2$ : The obstruction to destabilising a stable framing is $\mathbb{Z}/2$-valued. | |
| Jun 26, 2020 at 18:16 | comment | added | Ben Wieland | All the useful comments were deleted. Maybe they were wrong, but at least they were trying. You have to take risks! Did Ryan's comment lead to the Kervaire invariant 1 problem? Is that what @archipelago was talking about? | |
| Jun 26, 2020 at 16:25 | comment | added | Chris Schommer-Pries | @archipelago Yes. The 7-sphere has a "Lie group" framing coming from viewing it as the unit octionians. It is not really a Lie group since it is not associative, but there is enough structure to talk about "left-invariant vector fields" and this induces a trivialization of the tangent bundle like for Lie groups. I believe that with this framing it represents $\sigma$. | |
| Jun 26, 2020 at 14:55 | comment | added | archipelago | Is $\sigma\in\pi_7^s$ represented by a unstably framed manifold? The $e$-invariant can be computed as (half of the) relative $\hat{A}$-genus of a bounding spin-manifold. | |
| Jun 26, 2020 at 13:53 | comment | added | Panagiotis Konstantis | Okay, I somehow assumed falsely the manifold should be connected. Very interesting question! | |
| Jun 26, 2020 at 13:07 | comment | added | Chris Schommer-Pries | @PanagiotisKonstantis I don't think so. As you observe the Lie group framing on $S^1$ gives the non-trivial element of $\Omega^{fr}_1$. The disjoint union of two circles, each with the Lie group framing gives a tangentially framed manifold representing the trivial element of $\Omega^{fr}_1$. Also for the sake of this question the empty manifold should also probably be considered a tangentially framed manifold. | |
| Jun 26, 2020 at 8:57 | comment | added | Panagiotis Konstantis | Doesn't this show that the answer for the last question of Chris is negative? (Sorry for the spamming the comment section) | |
| Jun 26, 2020 at 8:20 | comment | added | Panagiotis Konstantis | But if you specify a normal framing instead of a tangential framing than you obtain (using the standard backgroundframing) a stable framing of the tangent bundle which can lead to the trivial element. E.g. take the normal vector pointing outwards of $S^1 \subset \mathbb R^2$ and suspend it to a framing of the normal bundle of $S^1$ in $\mathbb R^n$. The homotopy class of the corresponding map $S^1 \to \mathrm{SO}(n)$ is a generator of $\pi_1(\mathrm{SO}(n))$. | |
| Jun 26, 2020 at 8:14 | comment | added | Panagiotis Konstantis | Suppose $S^1 \subset \mathbb R^n$ and suppose furthermore you have the Lie group framing on $S^1$ than you obtain a framing on the normal bundle of $S^1$. But this framing has to be in such way that if you compare it to the standard background framing of $\mathbb R^n$ you obtain a map $S^1 \to \mathrm{SO}(n)$ which has to be the trivial homotopy class (since you want to obtain the standard framing of $\mathbb R^n$). Pontryagin showed that the stable homotopy class is given by this homotopy class $S^1 \to \mathrm{SO}(n)$ in $\pi_1(\mathrm{SO}(n))$ plus the generator of that group | |
| Jun 26, 2020 at 7:24 | comment | added | Panagiotis Konstantis | No, if you take the Lie group framing you obtain the generator of $\Omega_1^{\rm{fr}}$. If you would like to obtain the trivial element in this framed bordism group, then you need to trivialize the stable tangent bundle accordingly. With other words: Not every stable framing is induced by a framing of the tangent bundle. | |
| Jun 26, 2020 at 7:17 | comment | added | Lennart Meier | Am I confused or doesn't every tangential framing on $S^1$ give the trivial stable homotopy class? | |
| Jun 26, 2020 at 6:19 | comment | added | Panagiotis Konstantis | At least for Lie group framings there are some articles: core.ac.uk/download/pdf/82048719.pdf, projecteuclid.org/euclid.jmsj/1468956166 | |
| Jun 25, 2020 at 19:57 | history | edited | Chris Schommer-Pries | CC BY-SA 4.0 |
typo, clarification.
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| Jun 25, 2020 at 18:36 | answer | added | user51223 | timeline score: 1 | |
| Jun 25, 2020 at 17:53 | history | asked | Chris Schommer-Pries | CC BY-SA 4.0 |