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A homology sphere is a closed smooth $n$-dimensional manifold with the same homology groups as $S^n$. Igor Belegradek's answer to a previous question of mine shows that the smoothness hypothesis is not necessary except in dimension four where it is not yet known whether every homology sphere is smoothable.

If a homology sphere is simply connected, it follows from (one version of) Whitehead's Theorem that it is in fact homotopy equivalent to $S^n$, i.e. it is a homotopy sphere. It was shown by Kervaire and Milnor in Homotopy Groups of Spheres: I that homotopy spheres have stably trivial tangent bundle.

Do homology spheres have stably trivial tangent bundle?

If the answer is no, is there a way to see how the fundamental group obstructs stable triviality?

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    $\begingroup$ The only part of the proof in Kervaire-Milnor that is not eminently entirely cohomological is the so-called "argument of Rokhlin" on p509; I am not sure what this argument is and so I cannot say how it extends. But Kervaire thinks it does. Indeed, he claims what you ask in his paper "Fundamental groups of smooth homology spheres", on the way to proving that upon connected summing with an appropriate homotopy sphere, homology spheres of dimension at least 4 bound contractible manifolds. $\endgroup$
    – mme
    Commented Jun 16, 2016 at 19:41
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    $\begingroup$ Note (mostly for myself): integral homology spheres have stably trivial tangent bundle, but this is not the case for rational homology spheres. For example, the Wu manifold $SU(3)/SO(3)$ is a closed, simply connected, five-dimensional manifold with integral homology groups $\mathbb{Z}, 0, \mathbb{Z}_2, 0, 0, \mathbb{Z}$ and is therefore a rational homology sphere, but it has non-trivial $w_2$ and $w_3$, so its tangent bundle is not stably trivial. $\endgroup$
    – Michael Albanese
    Commented Jun 14, 2017 at 19:28
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    $\begingroup$ An easier example of a rational homology sphere which does not have stably trivial tangent bundle is $\mathbb{RP}^5$ which has $w_2(\mathbb{RP}^5) \neq 0$. If we require the manifold to be simply connected, then the Wu manifold is the simplest example. $\endgroup$
    – Michael Albanese
    Commented Mar 19, 2018 at 14:13

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Yes, they have stably trivial tangent bundles. A remark to this effect can be found on page 70 of

M. Kervaire "Smooth Homology Spheres and their Fundamental Groups"

but it is a little terse. It is essentially the argument as for homotopy spheres, due to Kervaire--Milnor, with a little elaboration. Let me try to explain it.

Let $M^n$ be a $\mathbb{Z}$-homology sphere. It is therefore orientable, so let $\tau : M \to BSO$ classify its stable tangent bundle. As the target is simply-connected, this map factors over the plus-construction $M^+ \simeq S^n$, giving a map $\tau' : S^n \to BSO$. The claim is that $\tau'$ is nullhomotopic, as then $\tau$ is too.

We know the homotopy groups of $BSO$ by Bott periodicity:

Sometimes $\pi_n(BSO)=0$, in which case we are done.

Sometimes $\pi_n(BSO)=\mathbb{Z}$, namely when $n=4k$, in which case homotopy classes are detected by evaluation against the Pontrjagin class $p_k$. But $$\langle \tau', p_k \rangle = \langle [M], p_k(TM) \rangle$$ is a non-zero multiple of the signature of $M$ by Hirzebruch's signature theorem (as all other Pontrjagin classes must vanish on $M$), and $M$ has signature zero as it has no middle-dimensional homology. Thus we are done in this case too.

Finally, sometimes $\pi_n(BSO)=\mathbb{Z}/2$. In this case let us reinterpret the element $\tau'$: by the plus-construction discussion it follows that $M$ is stably parallelisable away from a small disc, and it is of course also stably parallelisable on this disc. The difference of these two parallelisations on the boundary of the disc gives a map $\tau'' : S^{n-1} \to SO$, and this corresponds to $\tau'$ under $\pi_n(BSO) \cong \pi_{n-1}(SO)$.

Consider the composition $J \circ \tau'' : S^{n-1} \to SO \to G$, where the second map is the $J$-homomorphism. Now $\pi_{n-1}(G) \cong \Omega_{n-1}^{fr}$ framed cobordism, and under this isomorphism $J \circ \tau''$ corresponds to $S^{n-1}$ with the framing which comes from changing the standard framing by the map $\tau''$. But by construction this framing on $S^{n-1}$ bounds, namely it bounds the framing we chose on $M$ minus a small disc: thus $J \circ \tau''=0 \in \pi_{n-1}(G)$. Now one invokes the theorem of Adams that the $J$-homomorphism is injective in these dimensions, so $\tau''$ and hence $\tau'$ and $\tau$ are nullhomotopic too.

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  • $\begingroup$ What is $G$ in your last paragraph? $\endgroup$
    – Michael Albanese
    Commented Jun 29, 2016 at 18:09
  • $\begingroup$ It is the direct limit of the monoid of homotopy self equivalences of the $n$-sphere as n tends to infinity. It shows up often in surgery theory. $\endgroup$
    – Oscar Randal-Williams
    Commented Jun 29, 2016 at 23:06

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