With some classmates we are studying the Poincaré conjecture from Tao's Poincaré Legacies. In section 3.6, he prove the following lemma:

$\mathbf{Lemma~ 3.6.2}$ Let $M$ be a compact non-empty connected 3-manifold. Then it is not possible for $\pi_{1}(M), \pi_{2}(M)$, and $\pi_{3}(M)$ to simultaneously be trivial.

He used the following version of Hureciwz theorem:

$\mathbf{Proposition~ 3.6.3}$ (Baby Hurewicz theorem). Let $M$ be a triangulated connected compact manifold such that the fundamental groups $\pi_{1}(M),\ldots,\pi_{k}(M)$ all vanish for some $k \geq 1$. Then $M$ has trivial $j^{\mathrm{th}}$ homology group $H_{j}(M)$ for every $1 \leq j \leq k$.

Then, he prove the lemma 3.6.2 as follows

Suppose for contradiction that we have a non-empty connected compact 3-manifold $M$ with $\pi_{1}(M)$, $\pi_{2}(M)$ and $\pi_{3}(M)$ all trivial. Since $M$ is simply connected, it is orientable (as all loops are contractible, there can be no obstruction to extending an orientation at some point to the rest of the manifold). Also, it is a classical result of Moise that every 3-manifold can be triangulated. Using a consistent orientation on $M$, we may therefore build a $3-$cycle on $M$ consisting of the sum of oriented $3-$simplices with disjoint interiors that cover $M$ (i.e., a $fundamental~class$), thus the net multiplicity of this cycle at any point is odd. On the other hand, the net multiplicity of any $3-$boundary at any point can be seen to necessarily be even. Thus we have found a $3-$cycle which is not a $3-$boundary, which contradicts proposition 3.6.3.

The problem here is that I and my classmate do not know (singular) homology theory, so it is a little difficult understand some steps:

  1. First, we do not know what is the $net~multiplicity$ that he used. We searched in internet the definition but we did not find it.
  2. How can we have a $3-$boundary in a $3-$manifold? I mean, a $3-$boundary it is a boundary of a $4-$chain, and how can we have a $4-$chain in a $3-$manifold? (the problem here is with the dimensions).

Can you help us with this?.

In the other hand, with a friend we (think that we) find a proof of lemma 3.6.2. but when $M$ has no boundary:

Suppose for contradiction that $M^{3}$ is connected non-empty and that $\pi_{j}(M)=0$ for $1\leq j \leq 3$. By the resoult of Moise that Tao used, $M$ can be triangulated. Then by Baby Hurewicz theorem, $H_{j}(M)=0$ for $1 \leq j \leq 3$. Also, we have that $H_{j}(M)=0$ for all $j>3$ because $M$ has dimension $3$. If some homotopy group is non trivial, by (a corollary of?) Hurewicz theorem, we have that for the first index $i$ when $\pi_{i}(M) \neq 0$ (notice that $i>3$), we will have that that $\pi_{i}(M)\cong H_{i}(M)=0$, which is a contradiction. So we will have that $\pi_{j}(M)=0$ for all $j\geq 1$. Now, this implies that $M$ have the same cohomology type of a point (Here my friend says that there is a theorem that says that if we have the vanish of the homotopy groups, then $M$ is weakly homotopy equivalent to a point, and this fact implies what I wrote above). In particular, we have that $H^{3}(M)=0$. Now, as $M$ is simply connected (remember that $\pi_{1}(M)=0$), $M$ is orientable, so we have a $3-$form that nowhere vanish, in fact, we have a volume form, let us call it $\omega$. Now, it is clear that $\omega$ is closed because $d\omega$ is a $4-$form in a $3-$manifold, so it should be zero. In the other hand, if $\omega$ is exact, i.e., that there exists a $2-$form $\tau$ such that $\omega=d\tau$, using Stokes' theorem and the additional hypothesis that $\partial M=\emptyset$ we obtain that $$0<\int_{M} \omega=\int_{M}d\tau=\int_{\partial M}\tau=0,$$ so $\omega$ can not be exact. Then, $\omega \in H^{3}(M)\setminus \{0\}$ which it is a contradiction, and thus we should have that $M=\emptyset$.

PS: I apologize for my poor english.

$\mathbf{Edit}$ (3^{rd} september): In a footnote on page 358 (page 370 of the pdf) of Poincaré Legacies, Tao says that "Unless otherwise stated, all manifolds are assumed to be without boundary", so the proof that we give should work.

  • $\begingroup$ Is $M$ necessarily smooth? I haven't read all of Tao's notes, but it looks like he's just working with a topological $3$-manifold. It is true that $3$-manifolds are uniquely smoothable, but that's nontrivial, and I wouldn't expect Tao to use it without explicitly noting it. (Also, your English is totally fine.) $\endgroup$
    – anomaly
    Aug 31, 2019 at 15:32
  • $\begingroup$ In our context, Tao should be working with smooth manifold because he want to prove something for Riemannian manifolds that flow by a Ricci flow. In page 359 (371 of the pdf) Tao in remark 3.1.2. mentions that in the case of $3-$manifolds all smooth structures are diffeomorphic. $\endgroup$
    – Sebathon
    Sep 3, 2019 at 15:37

1 Answer 1


1) It's not a term of art. There's a set of (presumably embedded) oriented $3$-simplices $M_\alpha$ forming a locally finite cover of $M$ such that the finite sum \begin{align*} \sum_{x\in M_\alpha} \mu_x, \end{align*} where $\mu_x$ is $+1$ or $-1$ according to the orientation of $M_\alpha$ at $x$, is odd for all $x\in M$.

2) Tao's working with singular homology (presumably for concreteness, since it's in the context of $3$-manifolds), and it's not immediate that any cycle $\Delta_k \to M^n$ must be a boundary for $k > n$. This is clear in, for example, in simplicial homology, but working with that would involve getting into some issues of CW-complexes, covering certain approximation theorems, using abstract nonsense to prove that the usual versions of non-exotic homology coincide, or etc.

  • $\begingroup$ With point 1) and Tao's construction now I understand that the net multiplicity the $3-$cycle is odd. I think that I understand your point 2), but I still don't know how to see that the net multiplicity of any $3$−boundary at any point is even $\endgroup$
    – Sebathon
    Sep 3, 2019 at 15:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .