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I'm stuck on a past qualifying exam problem from my university:

Construct two homotopically not equivalent simply connected CW complexes such that $H_0=\mathbb{Z}$, $H_2=\mathbb{Z}$, $H_3=\mathbb{Z}_2$ with all other homology groups zero.

I was able to construct one such CW complex: let $Y$ be $S^3$ with a four cell attached via a degree 2 map, then the wedge $Y\vee S^2$ satisfies the requirements.

Can someone please help me constructing another one? I came up with a lot of constructions that turned out to be homotopy equivalent or the same. Thank you!

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This is a good question, but rather hard for someone who has just learned algebraic topology. Here is my thought process (and an answer):

There is a theory of minimal cell complexes that apply to simply connected finite CW complexes, this can be found in Hatcher. It states that there is a cell structure such that for each copy of $\mathbb{Z}$ in the homology, we may find a cell structure with one cell in that dimension, and for each copy of $\mathbb{Z}/p$ we may have a cell in that dimension and one higher (one to supply the nontriviality and one to make it torsion).

This implies that every homotopy type with the homology you ask for has a representative with one 0-cell, one 2-cell, one 3-cell, and one 4-cell. Further, the differential takes the 4-cell to two times the 3-cell. So how on earth could we have two different homotopy types represented? The 2-cell can only attach one way to the 0-cell; the 3-cell is attached via a map $\pi_2(S^2 ) = \mathbb{Z}$ which we know from the differential must be trivial. And we also know the differential from dimension 4 to dimension 3!

Well, the answer is due to the fact that homology can only see adjacent dimensions. In fact, $\pi_3(S^2 \vee S^3) \neq \mathbb{Z}$, it contains things like $\eta + 2$, i.e. the Hopf map + twice the identity. Because the Hopf map interacts only with the 2-skeleton, attaching a 4-cell via this map will result in the same differential as just attaching via multiplication by 2.

Then by analyzing the third homotopy group of the resulting complexes, you can see that this results in a different homotopy type. I believe also that the $\mathbb{Z}/2$ cohomology ring of this complex will be nontrivial.

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  • $\begingroup$ Here is an addendum. Since the inclusion $S^2\vee S^3\hookrightarrow S^2\times S^3$ is $4$-connected it is easy to see that $\pi_3(S^2\vee S^3)\cong\mathbb{Z}^2$ generated by the Hopf map $\eta$ and the identity $\iota_3$ (or more accurately their compositions with the inclusions and projections). Thus any space as in the question has the form $X=X(\varphi_k)=(S^2\vee S^3)\cup_\varphi e^4$ where $\varphi_k=k\cdot \eta+2\cdot \iota_3$ for some $k\in\mathbb{Z}$. $\endgroup$
    – Tyrone
    Feb 1, 2021 at 18:10
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    $\begingroup$ For any $l\in\mathbb{Z}$ the map $\alpha_l=(\iota_2+l\cdot\eta+\iota_3):S^2\vee S^3\rightarrow S^2\vee S^3$ is a homotopy equivalence which induces a homotopy equivalence $X(\varphi_k)\simeq X(\alpha_l\varphi_k)$. But $\alpha_l\varphi_k\simeq\varphi_{k+2l}$, so $X(\varphi_k)\simeq X(\varphi_{k+2l})$. Thus the homotopy type of $X(\varphi_k)$ is determined by the parity of $k$ and there are exactly two distinct homotopy types of simply-connected CW complexes with the required homology. $\endgroup$
    – Tyrone
    Feb 1, 2021 at 18:10

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