Parallelizability of 3-manifolds

Question

Robert Bryant's answer here ( https://mathoverflow.net/a/149496/85500 ) states that any orientable 3-manifold is parallelizable.

Previously I was under the impression that only closed (compact & without boundary) orientable 3-manifolds were necessarily parallelizable.

I have also heard somewhere (unfortunately I don't remember the source) that most noncompact 3-manifolds are parallelizable, but even if this statement is correct in some sense, I have no idea whether this "most" was meant informally or if this statement can be interpreted in a technical sense.

In either case, my knowledge of differential/algebraic topology is nowhere near my knowledge of local differential geometry so I do not know any proofs and would probably have difficulty understanding one anyways.

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Primary motivation for asking this question is to try to understand how restrictive is demanding spacetimes in general relativity to be parallelizable. If we want to have a well-posed initial value problem, spacetime must be of the form $\mathbb R\times\Sigma$, where $\Sigma$ is a 3-manifold, so it is parallelizable if $\Sigma$ is parallelizable.

It is also known that a spacetime admits spin structures if and only if it is parallelizable. So if one wishes to include fermions on spacetime, one wants a parallelizable spacetime.

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Question: So what is the proper statement about the parallelizability of orientable 3-manifolds, especially in regards to non-compact ones?

Answer 1

Edit: the following argument is wrong. Please ignore this post.

I think that all orientable three-manifolds $M^3$ are parallelizable. Here is a proof in the smooth setting.

Assume we know that it always holds if $M^3$ is closed. $(*)$

Then it holds also when $M$ is compact and has some boundary $(**)$, because we can reduce to the closed case by "doubling" $M$, i.e. gluing two copies of $M$ along the boundary $\partial M$.

Now for the general case. We can assume that $M$ is connected and, again, that $\partial M=\emptyset$. Fix a Riemannian metric $g$ on $M$ and an exhaustion $M_1\subset M_2\subset\cdots$ of $M$ with compact connected manifolds (possibly with boundary), where $M_k$ is in the interior of $M_{k+1}$.

Choose a trivialization for $TM_1$ (using $(**)$), consisting of three (pointwise) linearly independent vector fields $X,Y,Z$. Assuming we have one for $TM_k$, here is how to extend it to $M_{k+1}$: say that $M_{k+1}'$ is obtained from $M_{k+1}$ by adding a small tubular neighborhood of $\partial M_{k+1}$. Since the tangent bundle $TM_{k+1}'$ is trivial, we can think it as $\mathbb R^3$ (using again $(**)$). Hence, the problem reduces to extending three continuous vector valued functions $X,Y,Z:M_k\to\mathbb R^3$ to the whole of $M_{k+1}\subset M_{k+1}'$, preserving pointwise linear independence.

To do this, temporarily set $X,Y,Z$ to vanish on $\partial M_{k+1}'$. Take their harmonic extensions $X',Y',Z'$ on the manifold $N:=M_{k+1}'\setminus\operatorname{int}(M_k)$ (whose boundary is $\partial M_{k+1}'\cup\partial M_k$). The restriction of these vector fields to $M_{k+1}$ works: any nontrivial linear combination $\lambda X'+\mu Y'+\nu Z'$ is either nonnegative or nonpositive on $\partial N$ (since $M_k$ is connected and $\lambda X+\mu Y+\nu Z\neq 0$ here), hence cannot vanish anywhere on the interior of $N$ (maximum principle), hence anywhere on $M_{k+1}$. $X',Y',Z'$ then identify with vector fields on $TM_{k+1}$ extending $X,Y,Z$.

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Remark. Other approaches are already given in other answers:

• via spin structures, see here
• by means of a result of Whitehead, which says that a noncompact (orientable) $M^3$ immerses into $\mathbb R^3$, see here

Answer 2

Suppose we have a non-compact $3$ dimensional manifold $M$ which is spinable. Fix a CW structure of the manifold, then being spinable is equivalent to the tangent bundle being trivial on the $2$-skeleton. There are two ways to proceed now. One can either use that every non-compact $n$-manifold is homotopy equivalent to a subset of its $n-1$-skeleton or one can use that $\pi_2(SO(3))$ vanishes and hence every trivialization on the two skeleton can be extended to the $3$-skeleton.

So all that is left to show is that an orientable non-compact $3$-manifold is spinable. Since $H^2(M;\mathbb{Z}/2)\cong \text{Hom}(H_2(M;\mathbb{Z}/2),\mathbb{Z}/2)$ the second Stiefel-Whitney class is non-zero if and only if there exists an element $\alpha$ in $H_2(M;\mathbb{Z}/2)$ such that $w_2(\alpha)\neq 0$. Note that $\alpha$ can be represented by an embedded compact surface $S$ i.e a subsurface such that $\alpha=[S]$ (Represent $\alpha$ by an immersed subsurface and then resolve the singularities). Since $S$ is compact we know that it is contained in a compact $3$-dimensional submanifold $N$. Since $i^*(w_2(M))=w_2(N)=0$ we see that this implies that $w_2(M)([S])=0$ and hence $w_2(M)$ is zero i.e. $M$ is spinable.