Timeline for How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical?
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6 events
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| Jul 19 at 15:01 | comment | added | Jeremy | 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. | |
| Jul 19 at 14:56 | comment | added | Igor Belegradek | @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. | |
| Jul 19 at 14:50 | comment | added | Jeremy | I understand my mistake, thank you for clarifying! | |
| Jul 19 at 14:21 | comment | added | Igor Belegradek | 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. | |
| Jul 18 at 13:34 | comment | added | Jeremy | 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! | |
| Nov 22, 2010 at 0:50 | history | answered | Igor Belegradek | CC BY-SA 2.5 |