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](https://mathoverflow.net/q/236029/21564) 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. 



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> Do homology spheres have stably trivial tangent bundle?
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If the answer is no, is there a way to see how the fundamental group obstructs stable triviality?



Are homology spheres stably parallelisable?
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](http://www.ams.org/journals/tran/1969-144-00/S0002-9947-1969-0253347-3/S0002-9947-1969-0253347-3.pdf)"


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.


