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.

  • 1
    $\begingroup$ See this MathOverflow question $\endgroup$
    – Jim Belk
    May 9 '14 at 19:08
  • $\begingroup$ I modified my post to incorporate a proof sketch found elsewhere (similar to the MathOverflow post) with targeted questions in bold. $\endgroup$ May 9 '14 at 21:34
  • 1
    $\begingroup$ 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.) $\endgroup$
    – user98602
    May 12 '14 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*}

  • $\begingroup$ Does Seifert-van Kampen yield that $\pi(X) \cong \left\langle a_1, \ldots , a_{|S|} \right\rangle$ since $\pi(S^3)= \{e\}$? $\endgroup$ May 13 '14 at 14:48
  • $\begingroup$ 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? $\endgroup$ May 13 '14 at 18:08
  • $\begingroup$ @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 $N_j$. $\endgroup$ May 17 '14 at 19:28
  • $\begingroup$ Just checking, the manifolds constructed this way are all orientable right? $\endgroup$ Aug 21 '17 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.