Dear all, it seems any domain in Euclidean space, that is, connected open set will not make it. Am I right or could anyone give me a reference??
What should I do if I want to construct a space with fundamental group A5?
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1$\begingroup$ en.wikipedia.org/wiki/Classifying_space $\endgroup$– Gabriel C. Drummond-ColeCommented Oct 20, 2012 at 15:38
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1$\begingroup$ (for your second question) $\endgroup$– Gabriel C. Drummond-ColeCommented Oct 20, 2012 at 15:38
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4$\begingroup$ Every finitely presented group is the fundamental group of a domain in $\mathbb{R}^5$, see e.g. mathoverflow.net/questions/30238/… $\endgroup$– Alain ValetteCommented Oct 20, 2012 at 21:42
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3 Answers
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Your question has been answered, but I thought I'd add something. For any space $X$, $H_1(X,\mathbb{Z})$ is just the abelianization of $\pi_1(X,*)$. So, to find an example of a space with zero first homology but non-trivial fundamental group, you just need to find a space $X$ with $\pi_1$ equal to its commutator subgroup $[\pi_1,\pi_1]$ (such groups are called perfect groups). Then $H_1(X,\mathbb{Z})\simeq \pi_1^{ab} = \pi_1/[\pi_1,\pi_1] = 0$. These spaces are guaranteed to exist - see, for example, Gabriel's comment.
An example of such a space $X$, as has already been mentioned, is the homology 3-sphere. This is the quotient of $S^3$ (viewed as the group of unit quaternions) by the binary icosahedral group $2I$, a perfect (sub)group. In fact, this quotient map is a covering and exhibits $S^3$ as the universal cover of $X$. Hence $\pi_1\simeq 2I$ and $H_1 = 0$.
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For any group $G$ there is a simple explicit construction of a 2-dimensional (connected) CW-complex $X$ such that $\Pi_1(X)=G$. This can be found in [Hatcher: Algebraic Topology, Corollary 1.28].
As an example take the group $G = A_5$ mentioned in the question. Then proceed the following steps:
- Choose a presentation, e.g. $A_5 = \langle a,b\mid a^2=b^3=(ab)^5=1\rangle$
- Form the bouquet $S^1 \vee S^1$ and consider $a$ resp. $b$ as loop around the corresponding $S^1$.
For each relation attach a 2-cell with attaching map just the relation-loop, i.e. we attach
- a 2-cell $D_1$ with attaching map $a^2=a \ast a$ (where $\ast$ is composition)
- a 2-cell $D_2$ with attaching map $b^3$
- a 2-cell $D_3$ with attaching map $(ab)^5$
In summary we have $X=S^1_a \vee S^1_b \coprod_{a^2} D_1 \coprod_{b^3} D_2 \coprod_{(ab)^5} D_3$. It has one 0-cell, two 1-cells and three 2-cells.
In general each generator of $G$ corresponds to one copy of $S^1$ and each relation corresponds to one attached 2-cell.
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$\begingroup$ I think the OP also wants the first homology group to be 0 -- going by the title of the question rather than the text. $\endgroup$ Commented Oct 20, 2012 at 22:48
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1$\begingroup$ The first homology is the abelianization of the fundamental group, i.e. if the fundamental group is perfect (like $A_5$) then $H_1$ vanishes. $\endgroup$– RalphCommented Oct 20, 2012 at 23:15
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If you want to look for such a space, look at the Wikipedia entry for Homology Sphere. The conjecture in the first sentence is false (take any homology sphere, embed it into some $\mathbb{R}^n,$ then thicken it to be an open set).
answered Oct 20, 2012 at 15:10
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$\begingroup$ Igor, it really helps. Spasibo. $\endgroup$– WfpiggieCommented Oct 20, 2012 at 15:27