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I started thinking a couple days ago about the example below, and it led me to ask the following question:

How (or when?) can we build a space (let's say a CW complex) with given homology groups and fundamental group? I know we can do either of these separately, but I couldn't do both simultaneously even for the simple example below. Also, can we have two such spaces $X, Y$ such that $X$ is not homotopy equivalent to $Y$?

EDIT: It's a necessary condition that the group we want as an $H_1$ is the abelianization of the one we want as $\pi_1$.

This is the example:

We want to know if it's possible to find a space $X$ (not homotopy equivalent to $\mathbb{T}^2$) such that $H_n(X) \cong H_n(\mathbb{T}^2)$ and $\pi_1(X) \cong \pi_1(\mathbb{T}^2) \cong \mathbb{Z}^2$.

For example, the space $X= S^2 \vee S^1 \vee S^1$ has the same homology as the torus, but has fundamental group $\mathbb{Z}\ast\mathbb{Z}$. I thought of attaching a $2$-cell to $X$ to add the relation $aba^{-1}b^{-1}$, but this ends up giving me $Y = S^2 \vee \mathbb{T}^2$, which has $H_2(Y) \cong \mathbb{Z}^2$, so it's not what I want.

Any hints would be appreciated, thanks in advance.

EDIT: It turns out a more general question was asked a few years ago. Qiaochu Yuan's answer goes a bit over my head right now, but it might be useful for someone else who comes across this.

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  • $\begingroup$ Your example is not very clear. What about taking $X=T^2$ itself? What are you trying to achieve exactly? $\endgroup$ – Mikhail Katz Mar 8 '17 at 15:10
  • $\begingroup$ Yes, sorry. In the example (which I thought of before the general question) I want a space $Y$ that is not homotopy equivalent to $\mathbb{T²}$ (if it exists). $\endgroup$ – Mauro Mar 8 '17 at 15:20
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This is a partial answer.

For any abelian group $G$ and integer $n \geq 1$, there is a CW complex $M(G, n)$ such that

$$\tilde{H}_i(M(G, n)) = \begin{cases} G & i = n\\ 0 & i \neq n \end{cases}$$

called a Moore space. Moreover, for $n > 1$, we can take these spaces to be simply connected. See Example $2.40$ of Hatcher's Algebraic Topology.

Using the fact that $H_n(\bigvee_{\alpha}X_{\alpha}) = \bigoplus_{\alpha}H_n(X_{\alpha})$, we see that for a given sequence of abelian groups $\{G_n\}_{n=1}^{\infty}$, the CW complex $X = \bigvee_{n=1}^{\infty}M(G_n, n)$ has the property that $H_i(X) = G_i$.

However, all that can be said of $\pi_1(X)$ is that its abelianisation is $G_1$. There are many different groups with the same abelianisation, so $\pi_1(X)$ may not be the group you wanted it to be.

If one could construct the Moore spaces $M(G, 1)$ with a given fundamental group (which must satisfy the necessary condition that its abelianisation is $G$), then the above space would have the desired properties. However, I don't know if this has been done.

Added Later: I asked a separate question about this final point (whether one could construct a Moore space $M(G, 1)$ with given fundamental group) here. It turns out it is not always possible.

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Dont, know if this answers your question, probably not, but if $\pi_n(x)=0$ for all $n>1$ then the homology groups $H_n(X)$ are determined by $\pi_1(X)$ in fact they are the group cohomology of the group $\pi_1(X)$. So you may want to have variable $\pi_n(X)$.

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  • $\begingroup$ SO, this puts a restriction on what combination of homology groups and fundamental groups can coexist. Thus, we may expect more conditions for solvability of this problem than simply requiring $H_1$ to be the abelianization of $\pi _1$ . $\endgroup$ – Behnam Esmayli Mar 8 '17 at 16:18
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    $\begingroup$ Yeah, most definitely, if you allow yourself freedom as to what the $\pi_n(X)$ can be they maybe you can get a given $H_n(X)$ groups. There is a generalization: if $K(n, \pi)$ is a Eilemberg-Maclane space then $H_n(K)$ are also determined by a spectral sequence. Your question is really a research level problem. $\endgroup$ – Rene Schipperus Mar 8 '17 at 16:41
  • $\begingroup$ Do you mind directing me to a source for this generalization? I'd like to look into it. $\endgroup$ – Mauro Mar 8 '17 at 21:44
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    $\begingroup$ This is the classic work of Serre and Borel, I think the book Cohomology operations and applications in homotopy, is very readable. But most algebraic topology books that discuss spectral sequences also have this. Another book is McLeary, Users Guide to Spectral Sequences. $\endgroup$ – Rene Schipperus Mar 9 '17 at 0:20
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Take the case where $H_1(X)=0$. It is not always possible since the first homology group is the abelianization of the fundamental group which is not always zero.

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  • $\begingroup$ Hmm, of course, I should have thought of that, thanks. Still, the question still stands when the group we want as $H_1 (X)$ is the abelianization of the one we want as $\pi_1(X)$ $\endgroup$ – Mauro Mar 8 '17 at 15:14

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