There is no 3-dimensional manifold with fundamental group isomorphic to direct sum of four copies of $\Bbb Z$

Question

I want to show that there doesn't exist any $3$-dimensional manifold for which the fundamental group is isomorphic to the direct sum $\Bbb Z\oplus \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$ I have calculated the fundamental group of $K_{3,3}$ graph and $K_5$ graph and if i have done right it was the direct sum of four copies of $\Bbb Z$ for both of them...

Answer 1

Let us first suppose $M$ is compact.

Any compact orientable irreducible $3$-manifold $M$ with infinite $\pi_1(M)$ is a $K(\pi_1(M),1)$; this is a consequence of Papakyriakopoulos' sphere theorem; this is done in Hempel's book on $3$-manifolds.

So first, suppose $M$ is an irreducible orientable $3$-manifold with fundamental group $\Bbb Z^4$. Any two $K(G,1)$ spaces; see Eilenberg-Maclane spaces, $X$ and $Y$ are weak homotopy equivalent for any group $G$, thus they must have isomorphic cohomology groups $H^n(X;\Bbb Z)$ and $H^n(Y;\Bbb Z)$ for all $n$. As $M$ is just a $3$-manifold we have $H^4(M;\Bbb Z)=0$, however $\Bbb T^4$; the product of four $S^1$'s, is a $K(\Bbb Z^4,1)$ space, and also an orientable compact closed $4$-manifold, so $H^4(\Bbb T^4;\Bbb Z)\cong \Bbb Z$. Therefore such $M$ cannot exist.

Now if $M$ is just orientable, $M$ is a connected sum $N_1\#\cdots\# N_m$, where each $N_i$ is either an irreducible $3$-manifold or $S^1\times S^2$, hence $\pi_1(M)\cong \pi_1(N_1)\ast\cdots\ast \pi_1(N_m)$, thus as $\pi_1(M)$ is abelian we must have that all $\pi_1(N_i)$ are trivial except for one, say $\pi_1(N_1)$, and thus $\pi_1(N_1)\cong \pi_1(M)\cong \Bbb Z^4$, then $N_1$ is irreducible and orientable, and we already know this cannot happen.

Finally, if $M$ is non-orientable, let $\tilde M$ be its $2$-fold orientable covering space, then $\pi(\tilde M)$ is a subgroup of $\pi_1(M)\cong\Bbb Z^4$ of index $2$, so $\pi_1(\tilde M)$ must be isomorphic to $\Bbb Z^4$, and this is not possible by what we just saw.

Now if $M$ is not compact and $\pi_1(M)\cong \Bbb Z^4$, then as $\pi_1(M)$ is finitely generated, from $M$ we can build another compact $3$-manifold with the same fundamental group, but by what we showed above this is imposible.

Answer 2

Here is a proof for closed manifolds avoiding the sphere theorem. It is unclear to me if something like this can be made workable for manifolds with boundary.

Given a space $X$ with fundamental group $\pi$, the natural map $f: H_2(X) \to H_2(\pi)$, induced by the map $X \to B\pi$ classifying the universal cover, is surjective. This follows because one may add cells of rank 3 and higher to $X$ to make a space $X'$ with no higher homotopy groups, and thus factor the map up to homotopy as $X \to X' \xrightarrow{\simeq} B\pi$. In particular, adding cells of degree 3 and higher can only add relations to $H_2(X)$.

Now $H_2(\Bbb Z^4) = \Lambda^2 \Bbb Z^4 = \Bbb Z^6$, but of course for an oriented closed 3-manifold, $H_2(Y) = H^1(Y)$, and $H^1(Y) = \text{Hom}(\pi_1 Y, \Bbb Z)$. In your case, this is of course $\Bbb Z^4$. This is a contradiction.

If your 3-manifold wasn't oriented, pass to the double cover just as in Camilo's answer, which in this case has the same fundamental group.