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Problem: Let $F$ be a finitely generated free group. Prove that there is an $n$ manifold, $M$, $n > 2$ with $\pi(M) = F$.

  1. Let $F = F_S$ s.t. $|S| \in \mathbb{N}$.

  2. If I could show that there exists $n$-manifolds $M_1, \ldots , M_k$ s.t.

    \begin{equation} \tag{*} F_S = \pi_1 (M_1) \ast \pi_1(M_2) \ast \ldots \ast \pi_1(M_k) \end{equation}

    then I could use the result from this post to assert that

    $$ F_S = \pi_1 (M_1) \ast \pi_1(M_2) \ast \ldots \ast \pi_1(M_k) = \pi_1(M_1 \# M_2 \# \ldots \# M_k) $$

    so that if $M_1 \# \ldots \# M_k = M$, then we would have that $F_S = \pi_1(M)$ as desired.

Unfortunately, I'm not sure how we could show $(*)$, or if this is the right track towards an answer.

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    $\begingroup$ Let' start by understanding the problem and let $F=\mathbb Z$ what is a manifold $M$ with dimension $>2$ such that $\pi_1(M)=\mathbb Z$? $\endgroup$ – Walid Taamallah May 9 '14 at 16:24
  • $\begingroup$ Well I know $M$ cannot be $S^n$ for $n > 1$, for we have that $\pi_1(S^n) \cong \{e\}$ for $n > 1$. $\endgroup$ – user1770201 May 9 '14 at 16:38
  • $\begingroup$ That's a good step, do you know any manifold if any dimension with fundamental group Z? $\endgroup$ – Moishe Kohan May 10 '14 at 3:11
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    $\begingroup$ Yes -- $\pi_1(S^1) = \mathbb{Z}$. $\endgroup$ – user1770201 May 10 '14 at 9:37
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    $\begingroup$ You need an $n$-manifold whose fundamental group is $\Bbb Z$. Once you have that, you're done - because $\pi_1(S^n)$ is trivial, Van Kampen's theorem will easily show that $\pi_1(M\# M) = F_2$. So the first step is all you have to do. $\endgroup$ – user98602 May 11 '14 at 16:05
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Let's start by constructing an $n$-manifold $M_n$ with $\pi_1(M_n) \cong \Bbb Z$. We know that $S^1$ is a $1$-manifold and $\pi_1(S^1) \cong \Bbb Z$. The product of two manifolds is again a manifold. Consider $M_n = S^1 \times \Bbb R^{n-1}$. This is an $n$-manifold. Since $\Bbb R^{n-1}$ is contractible, $\pi_1(M_n) \cong \pi_1(S^1) \times \pi_1(\Bbb R^{n-1}) \cong \Bbb Z$.

Now, we use the following result:

Suppose $n$ is a natural number that is at least equal to 3. Suppose $M_1$ and $M_2$ are (possibly homeomorphic, possibly not) $n$-dimensional connected manifolds and $M_1 \# M_2$ is their connected sum. We then have the following relation between the fundamental groups of the manifolds: $$\pi_1(M_1 \# M_2) = \pi_1(M_1) * \pi_1(M_2)$$

You can find a proof here.

What's left is to take the connected sum of $k$ copies of $M_n$, where $k$ is the number of generators of the given free group.

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  • $\begingroup$ I think you meant $M_n = S^1 \times \Bbb R^{n-1}$. $\endgroup$ – user98602 May 11 '14 at 17:26
  • $\begingroup$ @MikeMiller You're right. Fixed. Thanks! $\endgroup$ – Ayman Hourieh May 11 '14 at 17:27
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Note that a bouquet of $n$ circles has fundamental group the free group on $n$ generators $F_n$. Take your desired dimension $N>1$, embed a bouquet of $n$ circles in $\mathbb{R}^N$, and then take a small open $\epsilon$-neighborhood of the image. This neighborhood is an open manifold $M$ which deformation retracts onto the bouquet, hence has $\pi_1(M) = F_n$.

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  • $\begingroup$ Isn't the resulting manifold a manifold with boundary? I think the OP is looking for a manifold without boundary. $\endgroup$ – Ayman Hourieh May 11 '14 at 17:14
  • $\begingroup$ @AymanHourieh I meant "neighborhood" to indicate an open set -- edited to clarify. If one takes an open neighborhood, then the manifold has no boundary. $\endgroup$ – Neal May 11 '14 at 17:22
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    $\begingroup$ Now it's clear. :) $\endgroup$ – Ayman Hourieh May 11 '14 at 17:23
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Since both answers here give non-closed manifolds as their examples, the following realizes your question for closed $n$-manifolds.

Fix $n>2$. Then $\pi_1(S^{n-1})$ is trivial, and thus (since the fundamental group of a product of spaces is the product of their fundamental groups) $\pi_1(S^1 \times S^{n-1}) = \mathbb Z$. Write $X_1 = S^1 \times S^{n-1}$. As mentioned above, because $\pi_1(S^{n-1})$ is trivial, an application of Van Kampen's theorem shows that $\pi_1(A \# B) = \pi_1(A) \ast \pi_1(B)$ when $A$ and $B$ are manifolds of dimension greater than 2. As a result, inductively defining $X_k = X_1 \# X_{k-1}$, we see that $\pi_1(X_k)$ is the free group on $k$ generators, where $X_k$ is a closed (compact without boundary) $n$-manifold.

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    $\begingroup$ I like this answer. Constructing an open manifold to meet given criteria feels like cheating. $\endgroup$ – Neal May 12 '14 at 13:06

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