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Is it true that fundamental group of a manifold with boundary cannot be simple? I think I read that during a hurried research run through the basement of Geisel library, but didn't mark the source.

If false please provide a counterexample. If true, is there a "simple" proof or intuitive reason?

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Any closed ball is a manifold with boundary that is also contractible. –  Neal May 9 '12 at 1:08
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If you have a manifold $X$ with simple fundamental group (like $SO(n), n\geq 3$ which has fundamental group of order $2$), what is to stop you to just consider $X\times [0,1]$? This new space has a boundary and shares its fundamental group with $X$. –  Olivier Bégassat May 9 '12 at 1:10
    
@OlivierBégassat, I don't know what's to stop me from considering the product, I'm an engineer not a mathematician. What's the fundamental group of [0,1] and what results relate the fundamental groups of the factors to that of the product? Would be grateful for any references/links. –  alancalvitti May 9 '12 at 1:55
    
@JasonDeVito, your comment seems to be saying (correct me if I'm misinterpreting) that the fundamental group of a boundary of a manifold can be simple, but does that imply the answer to my original Q? –  alancalvitti May 9 '12 at 1:58
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Well, the fundamental group of a product is the product of their fundakental groups and an interval is homotopy equivalent to a point so has trivial fundamental group. –  Olivier Bégassat May 9 '12 at 1:59
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1 Answer

up vote 11 down vote accepted

This is false.

In fact, every finitely generated group, simple or not, is the fundamental group of a compact 4-manifold $X$ with boundary $S^3$, a 3 dimensional sphere.

First, it is known that every finitely generated group $G$ is the fundamental group of a compact 4 dimensional manifold $M$. See for example http://mathoverflow.net/questions/15411/finite-generated-group-realized-as-fundamental-group-of-manifolds.

Now, let $B^4$ be a small open ball in $M$ and consider $X = M - B^4$. Then $X$ is a manifold with boundary $S^3$.

I claim that $\pi_1(X) = \pi_1(M) = G$.

To see this, write $M = X \cup B^4$ and apply Seifert-van Kampen. Using the fact that $X\cap B^4 = S^3$ is simply connected and that $B^4$ is simply connected, we learn that $\pi_1(M)\cong \pi_1(X)\ast_{\pi_1(S^3)} \pi_1(B^4) \cong \pi_1(X)$. Thus, $\pi_1(X) \cong \pi_1(M) = G$ as claimed.

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thank you, very nice. I can't follow all the details - particularly the subscripted factor in the last line (markup typo?) but seems reasonable and thanks for the refs... topology continues to amaze. –  alancalvitti May 9 '12 at 13:13
    
That subscripted factor refers to the almagamated product of 2 groups: en.wikipedia.org/wiki/…. The idea is you have two groups G and G' with a "common" subgroup H. Then you first form G*G' which consists of "words" with a "letter" from G follows by one of G', continually alternating. The group operation is concatenation followed by simplifciation. The subscript part is a quotienting out procedure where we move things in the H in G into the corresponding thing in the H in G'. –  Jason DeVito May 9 '12 at 13:27
    
Thanks again. I like geometry, and dislike algebra, and that's a major handicap. –  alancalvitti May 9 '12 at 14:20
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