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Let $M$ be a smooth $m$-dimensional manifold, $TM$ its tangent bundle and $SM$ its unit sphere bundle.

Are there some simple examples where $SM$ is fibrewise homotopy-equivalent to the trivial bundle $S^{m-1} \times M$, yet $TM$ is not trivial as a vector bundle? Does it ever happen for $M$ a sphere?

Via classifying space machinery this amounts to comparing the orthogonal group $O_m$ to the space of homotopy-equivalences of $S^{m-1}$, $HomEq(S^{m-1})$, in particular its asking for tangent bundle classifying maps $M \to BO_m$ such that the composite $M \to BO_m \to BHomEq(S^{m-1})$ is null.

As far as I know I've never come across examples of this sort, but then again I haven't studied the homotopy-properties of the map $O_m \to HomEq(S^{m-1})$ in much detail. Are there many canonical references on this topic?

This is related to a math.stackexchange question: https://math.stackexchange.com/questions/16779/conditions-for-a-smooth-manifold-to-admit-the-structure-of-a-lie-group

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  • $\begingroup$ Ryan, you could have mentioned that the map from $O(n)$ to the space $G_n$ of homotopy self-equivalences of $S^{n-1}$ is the (unstable) J-homomorphism. $\endgroup$
    – Igor Belegradek
    Commented Jan 9, 2011 at 23:16
  • $\begingroup$ The main difficulty in constructing an example is that yon insist on triviality of UNSTABLE fiber homotopy type. By contrast simply-connected surgery theory easily yields manifolds that are not stably parallelizable, but such that their tangent bundle is stably fiber homotopy trivial. If that is all you need, I can sketch how to do it. $\endgroup$
    – Igor Belegradek
    Commented Jan 9, 2011 at 23:48
  • $\begingroup$ @Igor: I'm mainly interested in the question as stated (unstable), but thanks for your comments. Getting to your 1st comment, I've only seen the J-homomorphism as a map out of the homotopy-groups of $O_n$ to the homotopy-groups of spheres. Is this a reduction of the original definition, or was it originally defined as $O_n \to G_n$? $\endgroup$
    – Ryan Budney
    Commented Jan 10, 2011 at 0:14
  • $\begingroup$ Ryan, I do not know the history but in the book Madsen-Milgram "The classifying spaces.." Chapter 3A they refer to the inclusion $O_n\to G_n$ as J-homomorphism. In any case it is closely related to the classical J-homomorphism. $\endgroup$
    – Igor Belegradek
    Commented Jan 10, 2011 at 0:52
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    $\begingroup$ Specifically, let $F_n$ be the based self homotopy equivalences of $S^n$, and $G_n$ the unbased self homotopy equivalences of $S^{n-1}$. There is a map $G_n \to F_n$ given by unreduced suspension. The relation between the Madsen-Milgram definition and the classical one is given by the composite $O_n \to G_n \to F_n$, since the homotopy groups of $F_n$ are the homotopy groups of spheres (as $F_{n}$ is the degree $\pm 1$ components of $\Omega^n S^n$ and the latter is an associative $H$-space). $\endgroup$
    – John Klein
    Commented Jan 10, 2011 at 12:10

2 Answers 2

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It is proved in [Kaminker, J., Proc. Amer. Math. Soc. 41 (1973), 305–308] that the tangent sphere bundle of a closed smooth H-manifold is (unstably) fibre homotopy trivial. On other other hand, surgery theory allows to construct H-manifolds with non-trivial rational Pontrjagin classes, see e.g. [Victor Belfi, Pacific J. Math. vol 36, Number 3 (1971), 615-621] but this is a standard surgery-theoretic argument.

Finally, in [Milnor-Spanier, Pacific J. Math. vol 10, Number 2 (1960), 585-590] it is shown that the tangent bundle to $S^n$ is fiber homotopy trivial if and only if $n=1,3,7$, which are exactly the parallelizable spheres (by Bott-Milnor).

Thus the above is a complete answer to your question. This phenomenon you ask for never occus for spheres but occurs for lots of H-manifolds.

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  • $\begingroup$ Great! Do you know, Igor, any examples of non-parallelizable H-manifolds in nature? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 10, 2011 at 3:21
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    $\begingroup$ A failry explicit example is on page one of [Browder-Spanier, Pacific J. Math. Volume 12, Number 2 (1962), 411-414], see projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – Igor Belegradek
    Commented Jan 10, 2011 at 4:01
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    $\begingroup$ Perhaps relevent to this discussion is Dupont's theorem: two manifolds which are homotopy equivalent have fiber homotopy equivalent unstable tangent sphere bundles. $\endgroup$
    – John Klein
    Commented Jan 10, 2011 at 12:15
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    $\begingroup$ @John: thanks! I did not know Dupont's paper. For the record the papers is [On homotopy invariance of the tangent bundle II', Math. Scand. 26 (1970) 200–220] and its pdf is googlable. The relevant result is Theorem 5.1 and Dupont attributes it to Benlian and Wagoner. Now I am a little worried because there is a nearby statment in Dupont's paper (Theorem 5.4) than any stably tangential homotopy equivalence is tangential. That latter statement was shown to be false by Byun [Tangent bundle of a manifold and its homotopy type. J. LMS (2) 60 (1999), no. 1, 303–307]. $\endgroup$
    – Igor Belegradek
    Commented Jan 10, 2011 at 14:07
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    $\begingroup$ I was aware the Dupont's paper had that mistake. But I think the statement of Dupont's theorem that I quoted is correct (My recollection is that some subsequent fixes were done by Sutherland). $\endgroup$
    – John Klein
    Commented Jan 10, 2011 at 16:36
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The rational homotopy groups of $HomEq(S^{m-1})$ can be calculated via (Sullivan) minimal models (refer page 314 of D. Sullivan's Infinitesimal computations in topology). In short, if I'm not mistaken one can show that $\pi_i(HomEq(S^{2n})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,4n-1$ and $\pi_i(HomEq(S^{2n+1})\otimes\mathbb{Q})=\mathbb{Q}$ if $i=0,2n+1$. One the other hand, $Spin(2n+2)$ is rational homotopy equivalent to $S^3\times S^7\times\cdots\times S^{4n-1}\times S^{2n+1}$ while $Spin(2n+1)$ is rationally $S^3\times S^7\times\cdots\times S^{4n-1}$. This should imply at least that, even rationally, the map $O_m\to HomEq(S^{m-1})$ is not a homotopy equivalence in general. May be more is known about the specifics of this map.

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    $\begingroup$ That $O_m \to HomEq(S^{m-1})$ isn't a homotopy-equivalence in general I think can get by the fact that the LHS has periodic homotopy groups in a range, and the RHS has the homotopy-groups of spheres as its homotopy-groups. The former has periodic rational homotopy (in the range), the latter does not. But to what extent this is relevant to tangent bundle classifying maps I'm not so sure. $\endgroup$
    – Ryan Budney
    Commented Jan 10, 2011 at 0:23

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