For what values of $n \neq 1,3,7$ is the tangent bundle $TS^n$ of the $n$-sphere diffeomorphic to an open subset of $\mathbb{R}^{2n}$?
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asked Jun 16, 2017 at 20:50
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4$\begingroup$ It's definitely not true in the case $n=2$. The unit tangent sphere bundle in this case has total space $SO(3) \cong \Bbb RP^3$, and it is well-known that $\Bbb RP^3$ doesn't embed in $\Bbb R^4$. $\endgroup$– John KleinJun 16, 2017 at 21:05
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8$\begingroup$ @JohnKlein It's simpler than that; such a sphere in $\Bbb R^{2n}$ would have nontrivial self-intersection. It's not possible for any even $n$. $\endgroup$– mmeJun 16, 2017 at 21:10
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$\begingroup$ @MikeMiller Right! $\endgroup$– John KleinJun 16, 2017 at 21:14
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1$\begingroup$ @MikeMiller I guess this argument works with "homeomorphic"? $\endgroup$– YCorJun 16, 2017 at 21:53
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2$\begingroup$ @MikeMiller I don't have complete argument, but here's an idea: the total space is of the form $(S^4 \vee S^5) \cup_f D^9$. The attaching map $f: S^8 \to S^4 \vee S^5$ has 3 components: $\pi_8(S^4) + \pi_8(S^5) + \pi_8(S^8)$ where the second component is trivial because the bundle has a section. It's enough to show that the first component is non-trivial after suspending once into $\pi_9(S^5) \cong \Bbb Z/2$. I suspect that the unstable $J$-homomorphism $\pi_4(SO(5)) \to \pi_9(S^5)$ is an isomorphism. If true, the result follows as the tangent bundle of $S^5$ is non-trivial. $\endgroup$– John KleinJun 16, 2017 at 23:13
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2 Answers
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There are no sphere's with non-trivial normal bundle in that dimension. As far as I know, this is originally a theorem of Massey. See http://www.ams.org/journals/proc/1959-010-06/S0002-9939-1959-0109351-8/S0002-9939-1959-0109351-8.pdf for details.
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1$\begingroup$ Massey cites Kervaire in 1(c) for a slightly more general theorem in this Annals paper. Using Massey's approach, one does need to fill in the detail that the unit tangent bundle has nontrivial fiber homotopy type, which is probably known but you'd need to know something about the unstable J-homomorphism. Kervaire's approach seems more elementary. $\endgroup$– mmeJun 16, 2017 at 22:43
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$\begingroup$ @PVAL Thank you very much for your answer and very interesting link of the paper of Massey. $\endgroup$ Jun 18, 2017 at 6:01
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$\begingroup$ @MikeMiller according to the paper of Massey provided by PVAL I understand it is impossible for n even, using tubular neighborhood theorem and non vanishing of Euler characteristic. But can you ellaborate your answer for $n$ odd. Are you using Alexander duality?If yes, how is your computation? what is the role of "handles"?Thanks again for your answer. $\endgroup$ Jun 19, 2017 at 7:36
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Take $n$ odd; this is never possible for $n$ even. Suppose the unit disc bundle of $TS^{n}$ embeds in $S^{2n}$. One may calculate the relative homology of its complement (relative to the boundary) to be supported in degrees $n+1$ and $2n$, and so after appropriate handle cancellation can be obtained by attaching precisely one handle in each of those degrees. But where are we attaching the $(n+1)$-handle on $T^1 S^n$? Its attaching sphere homologous to a section of the bundle; I claim that any sphere in that homology class is isotopic to a section. This is at least true in a sufficiently stable range $(n \geq 6)$ so that homotopy classes of $n$-spheres only contain a single isotopy type. But you can identify the normal bundle of the section with the subbundle of $TS^n$ your section splits off; if this is trivial, then your tangent bundle itself must have been trivial, so $n = 1, 3, 7$ (or I suppose $5$, because I don't know whether there are some extra isotopy classes I don't want for some reason).
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$\begingroup$ I guess you could just as well verify that $T^1 S^n$ never arises as an $(n-2)$-surgery on $S^{2n-1}$, which should be elementary calculation once you know there aren't very many knots. $\endgroup$– mmeJun 16, 2017 at 23:36
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$\begingroup$ Thank you very much for your answer. I try to understand the detail. $\endgroup$ Jun 18, 2017 at 6:00