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.

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 iff $\Sigma$ 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.

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

  • $\begingroup$ Side remark: the case of a compact orientable $M^3$ with boundary reduces to the closed case by gluing two copies of $M$ (with opposite orientations). $\endgroup$
    – Mizar
    Nov 26 '19 at 13:32
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    $\begingroup$ Also, if $\Sigma$ is parallelizable then clearly so is $\mathbb R\times\Sigma$, but the converse seems false in general dimension: e.g. $S^n$ is generally not parallelizable, but $\mathbb R\times S^n\cong\mathbb R^{n+1}\setminus\{0\}$ is. $\endgroup$
    – Mizar
    Nov 26 '19 at 13:34
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    $\begingroup$ Although I still haven't completely understood it, the answer to this question seems to be what you are looking for. $\endgroup$ Nov 26 '19 at 13:43

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 $X,Y,Z$.

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
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    $\begingroup$ Am I right that except at the very first step, you're not using dimension 3? If so it seems that it follows that a connected smooth $d$-manifold is parallelizable iff and only each $d$-dimensional submanifold (allowing boundary) is parallelizable. $\endgroup$
    – YCor
    Nov 26 '19 at 15:03
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    $\begingroup$ I can't access this paper by Whitehead at the moment, but I would be extremely surprised if parallelizability of open orientable three-manifolds is not an essential ingredient in his proof that they immerse into $\mathbb R^3$. So I strongly suspect that any argument of the form "$M$ immerses into $\mathbb R^3$ and must therefore be parallelizable" is circular. $\endgroup$ Nov 26 '19 at 15:41
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    $\begingroup$ @YCor If my argument is correct, then yes! This seems to avoid weird phenomena such as this: mathoverflow.net/a/140051/36952 $\endgroup$
    – Mizar
    Nov 26 '19 at 17:42
  • $\begingroup$ @JohnPardon good point... I don't know that paper myself honestly, but instead I just wanted to add a link to another answer. I guess that the author of that answer knows that the argument is not circular, as it seems that he read the paper. $\endgroup$
    – Mizar
    Nov 26 '19 at 17:46
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    $\begingroup$ @JohnPardon: On the 2nd page of Whitehead's article, he writes explicitly that the proof of this immersion result "leads to an alternative proof of the theorem that every orientable 3-manifold is parallizable". Also, outside of a short early paragraph which reduces smooth immersability to PL immersability, the rest of the paper takes place completely in the PL category. $\endgroup$
    – Lee Mosher
    Nov 26 '19 at 17:53

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.


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