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Let $Y^3$ be a integer homology 3-sphere, i.e. $H_k(Y;\mathbb{Z}) = H_k(S^3;\mathbb{Z}), \forall k$, i.e. $$ H_0(Y;\mathbb{Z}) = H_3(Y;\mathbb{Z}) = \mathbb{Z} \quad \text{and} \quad H_1(Y;\mathbb{Z}) = H_2(Y;\mathbb{Z}) = 0 $$ There are results about $\pi_1(Y)$ (e.g. M. A. Kervaire's 1969 paper).

Question : what do we know about $\pi_2(Y)$ and $\pi_3(Y)$ ?


According to the edits below (solving some cases), here I suppose :

  • $card(\pi_1(Y))=\infty$ (so exclude $S^3$ and Poincaré's homology 3-sphere),
  • $Y$ is not a Brieskorn 3-sphere $M(p,q,r)$.
  • $Y$ is reducible.

Edit 1 : If $\pi_1(Y)=0$, then $\pi_2(Y)=0$ and $\pi_3(Y)=\mathbb Z$. This can be computed either using Hurewicz either considering Poincaré-Perelman theorem ($\pi_1(Y)=0$ implies $Y=S^3$). So lets suppose $\pi_1(Y)\ne 0$. A first explicit example of such a non-trivial homology 3-sphere is Poincaré's homology 3-sphere. Using the long exact sequence of homotopy of the fibration $A_5\rightarrow SO(3) \rightarrow Y$ it is easily shown that higher homotopy groups of Poincaré's homology sphere are given by $\pi_k(Y)=\pi_k(S^3), \forall k>1$, hence $\pi_2(Y)=0$ and $\pi_3(Y)=\mathbb{Z}$. Now we know $\pi_2$ and $\pi_3$ of two homology 3-spheres (standard $S^3$ and Poincaré's $\mathbb{Z}HS^3$). From Kervaire's theorem, those two cases are the only $\mathbb{Z}HS^3$ with finite fundamental group. So lets suppose now that $card(\pi_1(Y))=\infty$.

Edit 2 : Also, for $p,q,r\in \mathbb{Z}_{\geq 2}$ pairwise coprime, the Brieskorn sphere $M(p,q,r)$ is an $\mathbb{Z} HS^3$. The case $M(2,3,5)$ is Poincaré's $\mathbb{Z} HS^3$ and was discussed in "Edit 1". The other cases all have infinite $\pi_1$ and by Brieskorn's theorem are Eilenberg-Maclane spaces $K(\pi_1,1)$ (i.e. aspherical), hence they have $\pi_2 = 0$, $\pi_3 = 0$.

Edit 3 : (Here I follow Thomas's idea, see answer below). Let $Y$ be a $\mathbb Z HS^3$ with infinite $\pi_1$. Then, since $Y$ is orientable, we have "$Y$ irreducible implies $Y$ is $K(\pi_1(Y),1)$". So the case where $Y$ is irreducible, and has infinite $\pi_1$, is solved ($\pi_2 = 0$ and $\pi_3 = 0$). Remark that the notions of irreducible and prime concord in the case of $\mathbb Z HS^3$.

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  • $\begingroup$ If $\pi_1(Y)=0$, then the Hurewicz theorem implies $\pi_2(Y)=0$ and $\pi_3(Y)=\mathbb{Z}$. $\endgroup$ Sep 14, 2017 at 4:18
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    $\begingroup$ But $\pi_1(Y)=0$ implies $Y=S^3$ (via Poincaré-Perelman). So that's trivial (using highly non-trivial P.-P.)... $\endgroup$
    – Noé AC
    Sep 14, 2017 at 4:20

1 Answer 1

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Every compact irreducible (reducible means that the manifold is a connected sum) 3-manifold has trivial $\pi _2$ (the sphere theorem). And every 3-manifold splits (uniquely) as a connected sum of irreducible 3-manifolds Milnor-Kneser. So you can study irreducible 3-manifold. Either they have infinite fundamental group and have trivial $\pi _3$ (In fact, these manifolds are $K(\pi , 1)$ and there universal cover is $\bf R^3$ ) or their fundamental group is finite. See the book of Hempel or the book of Jaco for instance.

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  • $\begingroup$ "And every 3-manifold splits (uniquely) as a connected sum of irreducible 3-manifolds"- is it irreducible or prime. For example what about $S^1\times S^2$? $\endgroup$ Sep 18, 2017 at 20:31
  • $\begingroup$ You are right, the complete statement is if the manifold is connceted sum of irreducible manifolds or $S^2$ bundel over $S^1$. See J. Milnor, "A unique decomposition theorem for 3-manifolds", American Journal of Mathematics 84 (1962), 1–7. $\endgroup$
    – Thomas
    Sep 19, 2017 at 5:07
  • $\begingroup$ Thanks a lot for your answer ! $\endgroup$
    – Noé AC
    Sep 21, 2017 at 19:43
  • $\begingroup$ How do you know the universal cover is $\mathbb{R}^3$? $\endgroup$ May 26, 2018 at 13:54
  • $\begingroup$ Unfortunately, this facts needs geometrization (Perelman). What was known since a long long time is that the universal cover is contractible. Proof : having trivial $\pi_1$ and $\pi _2$ and $H_3$ these universal cover have trivial $\pi _3$ (Whithehead), and being 3-dimensional are therefore contractible. $\endgroup$
    – Thomas
    May 27, 2018 at 11:58

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