# Showing $\pi_1(M) = F$ ($F$ a finitely generated free group and $M$ an $n$-manifold of dimension greater than $2$)

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.

$$\tag{*} F_S = \pi_1 (M_1) \ast \pi_1(M_2) \ast \ldots \ast \pi_1(M_k)$$

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.

• Let' start by understanding the problem and let $F=\mathbb Z$ what is a manifold $M$ with dimension $>2$ such that $\pi_1(M)=\mathbb Z$? – Walid Taamallah May 9 '14 at 16:24
• Well I know $M$ cannot be $S^n$ for $n > 1$, for we have that $\pi_1(S^n) \cong \{e\}$ for $n > 1$. – user1770201 May 9 '14 at 16:38
• That's a good step, do you know any manifold if any dimension with fundamental group Z? – Moishe Kohan May 10 '14 at 3:11
• Yes -- $\pi_1(S^1) = \mathbb{Z}$. – user1770201 May 10 '14 at 9:37
• 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. – user98602 May 11 '14 at 16:05

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.

• I think you meant $M_n = S^1 \times \Bbb R^{n-1}$. – user98602 May 11 '14 at 17:26
• @MikeMiller You're right. Fixed. Thanks! – Ayman Hourieh May 11 '14 at 17:27

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 $\pi_1(M) = F_n$.

• Isn't the resulting manifold a manifold with boundary? I think the OP is looking for a manifold without boundary. – Ayman Hourieh May 11 '14 at 17:14
• @AymanHourieh I meant "neighborhood" to indicate an open set -- edited to clarify. If one takes an open neighborhood, then the manifold has no boundary. – Neal May 11 '14 at 17:22
• Now it's clear. :) – Ayman Hourieh May 11 '14 at 17:23

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.

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