The intuitive idea is that the sphere connected the two manifolds is not contractible, which implies the (n-1)th homotopy group is not zero. Another argument, which I am not totally understand, uses the fact that the universal covering of an aspherical manifold has only one end. I am wondering if someone here could clarify these for me or give me a new argument.
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$\begingroup$ More generally, if $M$ and $N$ are closed manifolds of dimension $n > 2$, then $M\# N$ is not aspherical unless one of them is $S^n$ and the other is aspherical. $\endgroup$– Michael AlbaneseCommented Sep 2, 2020 at 22:26
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4 Answers
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If $N$ is an open contractible manifold of dimension at least two, then it has one end. For instance, you can compute the zeroth reduced homology of $N$ minus a closed ball $B$ by using the long exact sequence for $(N,B)$ by using excision and the fact that $N$ is contractible.
If $M$ is a closed aspherical manifold of dimension at least three and $M$ is a connected sum of $M_1$ and $M_2$, then the fundamental group of $M$ is the free product of the the fundamental groups of $M_1$ and $M_2$.
If $M_1$ and $M_2$ are closed aspherical manifolds of dimension at least three, then their fundamental groups are nontrivial, and so $\pi_{1}(M)$ would be a nontrivial free product. But the universal universal cover of $M$ is one ended, and one ended groups do not decompose as nontrivial free products.
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Here's a different way to see it. Let $M$ and $N$ be aspherical of dimension at least 3. Then the wedge $M \vee N$ is aspherical (but not a manifold). Let $M\sharp N$ be the connected sum. Then we get a collapse map $M\sharp N \to M \vee N$ given by pinching to a point the embedded $(n-1)$-sphere you used to form the connected sum. It is trivial to check that this map is $(n-1)$-connected, and therefore an isomorphism of fundamental groups.
If the connected sum were acyclic aspherical then it would have to be homotopy equivalent to
the wedge, since the latter is acyclic aspherical and has the same fundamental group. Moreover,
the collapse map would have to give such a homotopy equivalence. But this is ludicrous,
since that would violate the long exact homology sequence of the cofibration
$$
S^{n-1} \to M\sharp N \to M \vee N .
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$\begingroup$ By the way, this argument generalizes to connected sums of acyclic Poincare duality spaces. $\endgroup$ Commented Jan 13, 2011 at 4:17
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In fact if $M$ is a closed aspherical $n$-manifold, then $\pi_1(M)$ does not split as an amalgamated product or HNN extension over a subgroup $H$ of cohomological dimension $\le n-2$. For concreteness suppose $\pi_1(M)=A*_C B$. Look at Mayer-Vietoris sequences of the splitting in dimension $n$ with $\mathbb Z_2$-coefficients. The factors $A$, $B$ have infinite index in $\pi_1(M)$, so their $n$-dimensional cohomology vanish, and so does the $(n-1)$-cohomology of $C$. By exactness, $H^n(M;\mathbb Z_2)$ vanishes, so $M$ cannot be closed aspherical.
answered Nov 22, 2010 at 0:50
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1$\begingroup$ Kindly look at the 4-dimensional example I considered in this post (mathoverflow.net/questions/475255/…) where $S=T^2$, $M=S^2\times T^2$, and $N=T^4$, the torus. The resulting fiber sum is $T^4$, and the fundamental group of this sum, I believe, splits over $\mathbb{Z}^2$ which is a group of cohomological dimension 2. I’m confused: Your non-asphericity conclusion does not seems to follow here! $\endgroup$– JeremyCommented Jul 18 at 13:34
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$\begingroup$ Saying $S=T^2$ doesn't quite make sense because you need to specify how $S$ sits inside $M, N$, but I think you assume that $S$ is a direct factor in $M$ and $N=S\times S$. Then what happens is that you remove open tubular neighborhoods of $S$ in $M, N$ and glue the resulting manifolds along the boundary which is diffeomorphic to $T^3$. Thus the amalgamated subgroup $C\cong\mathbb Z^3$ has codimension $1$. My answer assumes codimension $\ge 2$. Also the boundary is not $\pi_1$-injective in $M$ with tubular neighborhood removed. $\endgroup$ Commented Jul 19 at 14:21
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$\begingroup$ I understand my mistake, thank you for clarifying! $\endgroup$– JeremyCommented Jul 19 at 14:50
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$\begingroup$ @TopologyStudent: by the way, there is a standard criterion of when the result of gluing is aspherical. A graph of spaces is aspherical if all vertex and edge spaces are aspherical and all vertex-to-edge inclusions are $\pi_1$-injective. $\endgroup$ Commented Jul 19 at 14:56
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$\begingroup$ Yes, I'm aware of this result but it is a sufficient condition. In my case, at least one edge/vertex is NOT aspherical. So, I'm unable to decide asphericity and therefore, I was looking for some necessary conditions. $\endgroup$– JeremyCommented Jul 19 at 15:01
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Darryl McCullough gives a very complete answer to your question in his "Connected Sums of Aspherical Manifolds" paper. Let $M$ be the connected sum of $g$ aspherical manifolds of dimension $d\geq 3$. Its universal cover $\tilde M$ is homotopy equivalent to $\bigvee_{\rho \in \pi_1(M)} \bigvee_{i=1}^{g-1}S^{d-1}$ (and in particular is not contractible).
I wonder if there is a short proof of this along the lines of Robert Bell's answer.
answered May 15, 2014 at 23:36