# Showing that every finitely presented group has a $4$-manifold with it as its fundamental group

Wikipedia: For any finitely presented group it is easy to construct a (smooth) compact 4-manifold with it as its fundamental group.

Question: How do we do this?

EDIT: Below is a proof sketch found elsewhere with some targeted questions of my own bolded:

1. Let $F = \left\langle S \mid R \right\rangle$ s.t. $|S| \in \mathbb{N}$.

2. Let $X$ be a $4$-manifold.

3. Represent each relation in $R$ by a loop in $X$.

What exactly does this mean?

4. Assume that the loop is a smooth, simple closed curve.

5. Replace the loop by a tubular neighborhood $N$, which is homeomorphic to $S^1 \times D^3$.

6. The boundary of $N$ is homeomorphic to $S^1 \times S^2$.

7. Delete the interior of $N$.

8. Now $S^1 \times S^2$ is also the boundary of $D^2 \times S^2$, so take a copy of $D^2 \times S^2$ and use it to fill in the whole by identifying on the boundary.

How is this identification procedure carried out?

9. Now use Seifert Van-Kampen to show that you have killed off the generator.

I'm not exactly sure what is meant here, but I think once I understand the above bolded this will make more sense.

• Commented May 9, 2014 at 19:08
• I modified my post to incorporate a proof sketch found elsewhere (similar to the MathOverflow post) with targeted questions in bold. Commented May 9, 2014 at 21:34
• I'm too lazy to write out the whole thing explicitly. For (3), relations are just words on $|S|$. You start with a 4-manifold $X$ whose fundamental group is the free group on $|S|$ generators, and pick an equivalence class of loops that represents the appropriate word; represent this class by a smooth simple loop in $X$. For (8), there's a homeomorphism between the boundaries of the tubular neighborhoods and the copies of $D^2 \times S^2$. Let $x \sim y$ if $f(x)=y$ under this homeomorphism, and pass to the quotient topology induced by this relation. (We're gluing along the boundaries.)
– user98602
Commented May 12, 2014 at 5:46

Start with the closed $4$-manifold $X = (S^3 \times S^1) \# \cdots \# (S^3 \times S^1)$, the connected sum of $|S|$ copies of $S^3 \times S^1$. Using Seifert-van Kampen, you can check that $\pi_1(X) \cong \langle a_1, \dots, a_{|S|}\rangle$, where $a_i$ is represented by the $S^1$-factor of the $i^\text{th}$ $S^3 \times S^1$ summand of $X$.

Each relation can be represented by a smooth loop in $X$. For example, the relation $a_2 a_1^2 a_3^{-1}$ is represented by the loop in $X$ that first goes around the $S^1$-factor in the $2^\text{nd}$ $S^3 \times S^1$-summand, then goes around the $S^1$-factor in the $1^\text{st}$ $S^3 \times S^1$-summand twice, and finally goes around the $S^1$-factor in the $3^\text{rd}$ $S^3 \times S^1$ factor in reverse (we need to have an orientation chosen for each $a_i$). Denote the loop in $X$ corresponding the the $j^\text{th}$ relation by $b_j$.

Each loop $b_j$ has a tubular neighborhood $N_j$, which is a copy of $S^1 \times D^3$ embedded in $X$. The boundary of $N_j$ is homeomorphic to $S^1 \times S^2$. Note that $D^2 \times S^2$ also has boundary $S^1 \times S^2$. Hence we can cut out $N_j$ from $X$ and insert a copy of $D^2 \times S^2$ in its place by attaching it to the "empty" $S^1 \times S^2$ left behind in $X \setminus \text{int}(N_j)$.

The only step left is to check that removing $N_j$ and replacing it with a copy of $D^2 \times S^2$ has the effect of killing $b_j$. Let $X_j$ denote the manifold obtained from $X$ by this process. We have that $$X_j = (X \setminus \text{int}(N_j)) \cup_{S^1 \times S^2} (D^2 \times S^2).$$ Now \begin{align*} \pi_1(X \setminus \text{int}(N_j)) & = \langle a_1, \dots, a_{|S|} \rangle, \\ \pi_1(D^2 \times S^2) & = 0, \\ \pi_1(S^1 \times S^2) & = \langle c \rangle, \end{align*} where $c$ is represented by $S^1 \times \{\text{pt}\}$ in $S^1 \times S^2$. Note that the image of $c$ in $X_j$ is precisely the curve $b_j$. Therefore by Seifert-van Kampen we have that $$\pi_1(X_j) = \langle a_1, \dots, a_{|S|} \mid b_j \rangle.$$

If $X'$ denotes the result of doing this surgery of all $b_j$'s, we have that \begin{align*} \pi_1(X') & = \langle a_1, \dots, a_{|S|} \mid b_1, \dots, b_{|R|} \rangle \\ & = \langle S \mid R \rangle. \end{align*}

• Does Seifert-van Kampen yield that $\pi(X) \cong \left\langle a_1, \ldots , a_{|S|} \right\rangle$ since $\pi(S^3)= \{e\}$? Commented May 13, 2014 at 14:48
• How do you know that the image of $c$ in $X_j$ is the curve $b_j$, and how is Van Kampen being used here to yield the result that comes after? Commented May 13, 2014 at 18:08
• @user1770201 SvK yields $\pi_1(X) \cong \langle a_1, \dots, a_{|S|} \rangle$ because $\pi_1(S^{\color{red} 2}) = 1$. You can check using SvK that $\pi_1(X \# Y) \cong \pi_1(X) \ast \pi_1(Y)$ when $X$ and $Y$ have dimension $\geq 3$. As for the image of $c$ in $X_j$, it's really homotopic to $b_j$. Note that $b_j = S^1 \times \{\mathrm{pt}\} \subset N_j$, so the image of $c$ is $S^1 \times \{\mathrm{pt}\} \subset S^1 \times S^2 = \partial N_j$, which is $b_j$ homotoped onto the boundary of $N_j$. Commented May 17, 2014 at 19:28
• Just checking, the manifolds constructed this way are all orientable right? Commented Aug 21, 2017 at 15:06