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I am studying the homotopy type of a space,and i hope it would be a $K(\pi,1)$ space. now i have find its covering,once we can say the covering is $K(\pi,1)$,so is the space itself.and the covering is

$\mathbb{R}^4-M$ where $M=M_1\cup M_2\cup M_3\cup M_4$,

$M_1=\{(x,y,z,w)|x,y \in \mathbb R,z,w \in\mathbb Z\}$

$M_2=\{(x,y,z,w)|y,z \in \mathbb R,x,w \in\mathbb Z\}$

$M_3=\{(x,y,z,w)|x,w \in \mathbb R,y,z \in\mathbb Z\}$

$M_4=\{(x,y,z,w)|z,w \in \mathbb R,x,y \in\mathbb Z\}$

I guess $\mathbb{R}^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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Since you arrangement is so simple, there are must be some purely topological methods to check asphericity (or non-asphericity) of its complement. This is a well-developed subject I do not know much about.

I am more familiar with the geometric approach developed by Daniel Allcock (following predictions of Gromov) in this paper. Unfortunately, the result in Allcock's does not apply to your arrangement because it is not "normal" (as defined in the paper). Allcock has a sequel in works where he gives a definitive answer to the question of asphericity for a large class of arrangements including the one you care about; if everything fails you might want to ask him for a preliminary version.

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    $\begingroup$ I guess (from the question) that the initial problem concerns a subset in $T^4$ and suggest to think in terms of it - try to fiber it, retract... $\endgroup$
    – Petya
    Commented Mar 20, 2010 at 0:38
  • $\begingroup$ Petya, I agree, this might be easier. $\endgroup$
    – Igor Belegradek
    Commented Mar 20, 2010 at 16:07
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Here is a sketch of a solution. As Petya tells us, $\mathbf{R}^4\setminus M$ is invariant under the action of $\mathbf{Z}^4$ by integral translations, so it suffices to show that the image of $\mathbf{R}^4\setminus M$ in $T^4=\mathbf{R}^4/T^4$ is a $K(\pi,1)$. Denote this image by $N$. It is mapped to the 2-torus by $p:[x,y,z,w]\mapsto [x,y]$ where the square brackets stand for the images in the tori. $N$ is homotopy equivalent to $p^{-1}$ of the wedge $W$ of two circles given by $[x]=[1/2]$ and $[y]=[1/2]$. Set $X=S^1\times (S^1\vee S^1)$. The preimage $p^{-1}(W)$ is homotopy equivalent to two copies of $X$ joined along $pt\times (S^1\vee S^1)$, with a circle in each of the copies deleted; the deleted circles correspond to the two different circles of $S^1\vee S^1$, so $p^{-1}(W)$ is homotopy equivalent to the wedge of two 2-tori, and hence so is $N$.

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  • $\begingroup$ It is $M_1$, etc. that is homotopy equivalent to $p^{-1}(W)$, not all of $M$. Also, the preimage $p^{-1}(W)$ has cohomology in dimension 3, so cannot be homotopy equivalent to $X$ or any 2-complex. Anyway, the question is about the complement of $M$, not $M$ itself. For $M$, you can apply Gromov's CAT(1) link condition to see that $M$ is locally CAT(0), and thus aspherical. $\endgroup$
    – Tom Church
    Commented Mar 20, 2010 at 23:50
  • $\begingroup$ Tom, thanks, I misread the question and used $M$ for the complement. This has been fixed. Re "the preimage $p^{−1}(W)$ has cohomology in dimension 3": why? $\endgroup$
    – algori
    Commented Mar 21, 2010 at 0:09
  • $\begingroup$ Since you were talking about $M$, I had taken $p$ as being defined from $T^4\to T^2$, rather than from $T^4\setminus \bar{M}\to T^2$. Now it seems clear that you meant the latter, so I withdraw my objection. $\endgroup$
    – Tom Church
    Commented Mar 21, 2010 at 0:37
  • $\begingroup$ Hi,algori.Thank you for your answer.I am eager to get a clear picture of the solution.would you please talk about some details. ---student $\endgroup$
    – student
    Commented Mar 21, 2010 at 12:17

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