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

  • $\begingroup$ 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 Jan 9 '11 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 Jan 9 '11 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 Jan 10 '11 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 Jan 10 '11 at 0:52
  • 1
    $\begingroup$ 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 Jan 10 '11 at 12:10

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.

  • $\begingroup$ Great! Do you know, Igor, any examples of non-parallelizable H-manifolds in nature? $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '11 at 3:21
  • 1
    $\begingroup$ 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 Jan 10 '11 at 4:01
  • 4
    $\begingroup$ 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 Jan 10 '11 at 12:15
  • 3
    $\begingroup$ @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 Jan 10 '11 at 14:07
  • 1
    $\begingroup$ 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 Jan 10 '11 at 16:36

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.

  • 1
    $\begingroup$ 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 Jan 10 '11 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.