algebraic-topology
I'm doing past papers for a first course in algebraic topology.
The question is:
Let $M$ be a 3-dimensional, closed, connected, non-orientable manifold. Show that $M$ has infinite fundamental group.
Is there any way of answering this question without simply quoting a classification theorem for 3-manifolds with finite fundamental group?
There are precisely two: $S^2 \times S^1$ and a twisted product $S^2 \widetilde \times S^1$.
Suppose $M$ is closed, connected, with fundamental group $\Bbb Z$. Then it must be prime (by Poincare and because $\Bbb Z$ is indecomposable under free product), but not necessarily irreducible (meaning that every embedded 2-sphere bounds a 3-ball).
Indeed, it can't possibly be irreducible: consider $\tilde M$. It has trivial fundamental group, and because $M$ is irreducible, has $\pi_2 M = 0$ by the sphere theorem (which states that if $M$ is orientable, some nontrivial element of $\pi_2 M$ is represented by an embedded sphere). (If $M$ here is not orientable, pass to the oriented double cover.) Because $\Bbb Z$ is infinite, $\tilde M$ is not compact, so has $\pi_3 = H_3 = 0$; the first equality by Hurewicz and the latter by noncompactness. Inductively by Hurewicz we see that $\pi_n = 0$ for all $n$. So $M$ is aspherical, and hence a $K(\Bbb Z,1)$. But this would imply that $$\Bbb Z/2 = H_3(M;\Bbb Z/2) = H_3(S^1;\Bbb Z/2) = 0!$$ So $M$ is prime but not irreducible. We now follow an argument from Hatcher's 3-manifolds notes.
Because $M$ is prime, any embedded sphere either bounds a ball or does not separate $M$. But because $M$ is not irreducible there is some embedded sphere that does not bound a ball. So there is an embedded, non-separating, 2-sphere in $M$. Because $S^2$ is simply connected, any line bundle (which are in bijection with $H^1(X;\Bbb Z/2)$) over it is trivial. In particular, the normal bundle to any $S^2$ in $M$ is trivial.
So embed this separating 2-sphere, which extends to an embedding of $S^2 \times I$. Pick a curve that goes from one side of the thickened $S^2$ to the other (such a thing exists by the non-separating assumption); thicken it to get an embedded $I \times D^2$, whose ends are glued to each side of the $S^2$. More precisely, we can glue these together by a map $\varphi: \{0,1\} \times D^2 \to \{0,1\} \times S^2$ with $\varphi(0,\cdot) \subset \{0\} \times S^2$ and similarly with $\varphi(1,\cdot)$. In different language, I'm attaching a 1-handle to $S^2 \times I$. The results depend only on whether or not $\varphi$ preserves orientation on both components. (That is, if it preserves orientation on both components, or reverses orientation on both components, you get one manifold; if it preserves orientation on one component but reverses orientation on the other, you get a second, distinct manifold). Call the results $X_0$ and $X_1$, where $X_0$ is the orientable one; $X_0$ is what you get if $M$ is orientable, and $X_1$ if it's not. The boundary of $X_i$ is $S^2$; because this $S^2$ does separate $M$, it bounds a ball on the other side. There is only one way up to homeomorphism to glue a ball onto a manifold with boundary $S^2$. So we get that $M$ is one of either $M_0$ or $M_1$, where these are the $X_i$ with that ball capped off.
To finish, we just need to find two manifolds with such a decomposition. $S^2 \times S^1$ is one, where here the embedded $S^2$ is just $S^2 \times \{*\}$; and $S^2 \widetilde \times I = (S^2 \times I)/(x,0) \sim (-x,1)$ is another (again, pick, say, $S^2 \times \{1/2\}$.) The former is orientable; the latter is non-orientable. So $M_0 = S^2 \times S^1$ and $M_1 = S^2 \widetilde \times S^1$ are the only closed 3-manifolds with fundamental group $\Bbb Z$.
Let $F$ be a path-connected topological space and $\phi : F \to F$ a homeomorphism. Denote the mapping torus of $\phi$ by $M$ so that we have a fiber bundle $F \to M \to S^1$. Explicitly, $M = F\times[0, 1]/\sim$ where $(x, 0) \sim (\phi(x), 1)$. From the long exact sequence in homotopy, there is a short exact sequence $$0 \to \pi_1(F) \to \pi_1(M) \to \pi_1(S^1) \to 0.$$ The covering space of $M$ corresponding to the image of $\pi_1(F)$ in $\pi_1(M)$ is homeomorphic to $F\times\mathbb{R}$. This follows from the fact that $M$ can be constructed as the quotient of $F\times\mathbb{R}$ by the $\mathbb{Z}$-action generated by $(x, t) \mapsto (\phi(x), t + 1)$. In particular, the universal cover of $M$ is homeomorphic to $\widetilde{F}\times\mathbb{R}$ where $\widetilde{F}$ is the universal cover of $F$.
Suppose now that $F$ is a closed $(n-1)$-manifold, so that $M$ is a closed $n$-manifold. If $\pi_1(M) \cong \mathbb{Z}$, then $\pi_1(F) = 0$ by the short exact sequence above. As suggested by Moishe Kohan, let $N = M \# X$ where $X$ is a closed simply connected $n$-manifold, then $\widetilde{N}$, the universal cover of $N$, is homeomorphic to the connected sum of $\widetilde{F}\times\mathbb{R} = F\times\mathbb{R}$ and countably many copies of $X$. If $H_k(X; \mathbb{Z}) \neq 0$ for some $0 < k < n$, then it follows from Mayer-Vietoris that $H_k(\widetilde{N}; \mathbb{Z}) \cong H_k(\widetilde{F}; \mathbb{Z})\oplus\bigoplus_{i\in\mathbb{Z}}H_k(X)$; in particular, $H_k(\widetilde{N}; \mathbb{Z})$ is not finitely generated. As $\pi_1(N) \cong \pi_1(M) \cong \mathbb{Z}$, if $N$ were a mapping torus, the fiber $F'$ would be a closed simply connected manifold, so $\widetilde{N}$ would be homeomorphic to $\widetilde{F'}\times\mathbb{R} = F'\times\mathbb{R}$ which has finitely generated homology. Therefore $N$ is not a mapping torus. Note, if $H_k(X; \mathbb{Z}) = 0$ for all $0 < k < n$, then $X$ is a simply connected homology sphere and hence homeomorphic to $S^n$, in which case $N$ is homeomorphic to $M$ (which is a mapping torus).
I believe that the hypothesis $\pi_1(M) \cong \mathbb{Z}$ is unnecessary, but the proof above only extends to the case where $\pi_1(M)$ is an extension of $\mathbb{Z}$ by a finite group, i.e. $F$ has finite fundamental group.
Best Answer
$\def\QQ{\mathbb Q}$If $\pi_1(M)$ is finite, $H_1(M;\QQ)=0$. If $M$ is non-orientable, $H_3(M;\QQ)=0$. So $\chi(M)=h_0(M;\QQ)-h_1(M;\QQ)+h_2(M;\QQ)-h_3(M;\QQ)=1+h_2(M;\QQ)>0$.
But by Poincaré duality, any odd-dimensional manifold has zero Euler characteristic.