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Let $M$ and $N$ be two connected manifolds of the same dimension $n$. What is the fundamental group of their connected sum $M \# N$ in terms of $\pi_1(M)$ and $\pi_1(N)$ ?

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This suggests that "it depends" : topospaces.subwiki.org/wiki/… –  Patrick Da Silva Sep 20 '12 at 15:24

1 Answer 1

Consider the following open covers of $M,N,$ and $M\#N$:

$$M=U_1\cup U_2$$ where $U_1$ is $M$ without a point inside the small open ball we remove and $U_2$ is the small open ball which we delete to perform the connected sum operation;

$$N=V_1\cup V_2$$ where $V_1$ is $N$ without a point inside the small open ball we remove and $V_2$ is the small open ball which we delete to perform the connected sum operation;

$$M\#N=W_1\cup W_2$$ where $W_1$ is homotopy equivalent to $U_1$ and $W_2$ is homotopy equivalent to $V_1$, and each covers $U_1$ or $V_1$ respectively, plus a small collar neighborhood of the sphere we glue on which originally lived in $V_1$ or $U_1$ respectively.

Then, assuming we have that $M$ and $N$ are both of dimension 2 or higher, we can apply Seifert-Van Kampen to each of these situations (why the restriction? SVK needs the intersection of the covering sets to be path connected- this is never true for 1-manifolds, but always true for manifolds of higher dimensions. Also, you can casework the 1-manifold situation- the only 1-manifolds up to homeomorphism are $S^1, [0,1],[0,1),(0,1)$.). For each covering, the intersection of the open sets is homotopy equivalent to $S^{n-1}$, where $n$ is the dimension of $N$.

We have that $\pi_1(U_2)\cong\pi_1(V_2)\cong 0$, as they are contractible. The next calculation depends on the dimension $n$ of $N$ somewhat: In the case where $n>2$, $\pi_1(U_1\cap U_2)\cong\pi_1(V_1\cap V_2)\cong\pi_1(W_1\cap W_2)\cong 0$, and so $\pi_1(U_1)\cong\pi_1(M)$ and $\pi_1(V_1)\cong\pi_1(N)$. Knowing this, the pushout square given by Van Kampen turns into just the free product of the fundamental groups: $\pi_1(M) \ast \pi_1(N)$.

In the case where the dimension of $M$ and $N$ is 2, things are not so immediate. With the knowledge that $\pi_1(U_1\cap U_2)\cong\pi_1(V_1\cap V_2)\cong \pi_1(W_1\cap W_2)\cong \mathbb{Z}$, the pushout squares can get sort of messy- consider the example of the 2-torus: the fundamental group of the torus minus a small disc is the free group on 2 generators, while the fundamental group of the torus is the free abelian group on two generators.

But since we're in such a low dimension, we're saved by the constraints this places on the manifolds. By the classification of surfaces (2-manifolds), every 2-manifold (possibly with boundary) is homeomorphic to one and only one of the following: $$S^2\#^b D^2, \#^h \mathbb{R}P^2\#^b D^2, \#^g T\#^b D^2$$ where $T$ is the 2-torus. By using a presentation of each of these surfaces as an $n$-gon, it is possible to calculate the fundamental group explicitly. Unfortunately, this depends on knowing what manifolds you are connected-summing: recognizing a surface (possibly with boundary) from an arbitrary presentation of its fundamental group is a very hard problem. Even recognizing whether a group given by generators and relations is trivial or not is a problem which has been proven unsolvable.

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What happened when M, N are knot complements? –  John0417 Dec 13 '12 at 7:19
    
Do you believe that there are any steps that would change if $M$ and $N$ are knot complements? –  KReiser Dec 14 '12 at 1:47

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