finite generated group realized as fundamental group of manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:31:15Z http://mathoverflow.net/feeds/question/15411 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds finite generated group realized as fundamental group of manifolds sara 2010-02-16T03:00:08Z 2010-02-17T00:57:12Z <p>This is discussed in the standard textbooks on algebraic topology. Pick a presentation of the group $G = \langle g_1,g_2,...,g_n|r_1,r_2,...r_m \rangle$ where $g_i$ are generators and $r_j$ are relations. Then we have a wedge of $n$ circles and attach two-cells to the wedge sum according to the relations $r_j$. Denote the final space $X$. Then van Kampen says $\pi_1(X)=G$. While usually $X$ is not a manifold, it is well-known that every finitely generated group $G$ can be realized as the fundamental group of some 4-manifold $X$. Can someone sketch the proof? Also, if $X$ could not be some manifold of dimension $&lt;4$, what is the obstruction?</p> http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds/15414#15414 Answer by HW for finite generated group realized as fundamental group of manifolds HW 2010-02-16T03:12:26Z 2010-02-16T03:12:26Z <p><strong>Theorem.</strong> Every finitely presentable group is the fundamental group of a closed 4-manifold.</p> <p><em>Sketch proof.</em> Let $\langle a_1,\ldots,a_m\mid r_1,\ldots, r_n\rangle$ be a presentation. By van Kampen, the connected sum of $m$ copies of $S^1\times S^3$ has fundamental group isomorphic to the free group on $a_1,\ldots, a_m$. Now we can quotient by each relation $r_j$ as follows. Realise $r_j$ as a simple loop. A tubular neighbourhood of this looks like $S^1\times D^3$. Do surgery and replace this tubular neighbourhood with $S^2\times D^2$. This kills $r_j$. <em>QED</em></p> <p>There are many restrictions on 3-manifold groups. One of the simplest arises from the existence of <a href="http://en.wikipedia.org/wiki/Heegaard%5Fsplitting" rel="nofollow">Heegaard splittings</a>. It follows easily that if $M$ is a closed 3-manifold then $\pi_1(M)$ has a <em>balanced</em> presentation, meaning that $n\leq m$.</p> <p>Other obstructions to being a 3-manifold group were discussed in <a href="http://mathoverflow.net/questions/6132/fundamental-group-of-3-manifold-with-boundary/6143#6143" rel="nofollow">this MO question</a>.</p> http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds/15421#15421 Answer by Somnath Basu for finite generated group realized as fundamental group of manifolds Somnath Basu 2010-02-16T03:59:40Z 2010-02-16T03:59:40Z <p>A slightly different way of proving the same is the following.</p> <p>Take a wedge of n circles, one each for the generators. Now attach a disc for each relation. Imagine this complex $X$ sitting inside $\mathbb{R}^5$. By general position and finitely presented nature of $G$, the discs have no intersections in the interior. Take a tubular neighbourhood of $X$ in $\mathbb{R}^5$ and then take its boundary. One can check that this is a $4$-manifold with the required property. </p> http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds/15524#15524 Answer by Paul for finite generated group realized as fundamental group of manifolds Paul 2010-02-17T00:57:12Z 2010-02-17T00:57:12Z <p>Yet another explanation of the same constructions given above is to add 1 and 2 handles to the 4 ball according t the given presentation, obtaining a 4 manifold $X$ with boundary. Now the boundary of $X\times I$ (i.e. the double of $X$) has the same fundamental group by Van Kampen and the fact that $\partial X\subset X$ induces a surjection on fundamental groups (turning $X$ upside down shows that $X$ is obtained from $\partial X$ by adding 2 and 3 handles).</p> <p>Since the first homology=abelianization of $\pi_1$ of closed 1 and 2-manifolds are known, it is easy to see most groups dont occur for $n=1$ or $2$. For $n=3$, another algebraic obstruction is to observe that if $\pi=\pi_1(M^3)$, then $H_2(M)\to H_2(\pi)$ is onto, and if $M$ is orientable, then $H_2(M)=H^1(M)=H^1(\pi)$. So if $H^1(\pi)$ is smaller than $H_2(\pi)$, it cannot occur (for an oriented 3-manifold, in any case).</p>