39
$\begingroup$

A parallelizable manifold $M$ is a smooth manifold such that there exist smooth vector fields $V_1,...,V_n$ where $n$ is the dimension of $M$, such that at any point $p\in M$, the tangent vectors $V_1(p),...,V_n(p)$ provide a basis for the tangent space at $p$. Equivalently, a manifold is parallelizable if its tangent bundle is trivial.

There is a theorem that states that any compact orientable 3-manifold is parallelizable, and there is a proof of this result which uses $spin^c$ structures and the Steifel-Whitney class.

I am wondering whether there exists a more elementary, perhaps more straightforward proof. Otherwise, I would be grateful for some intuition on why this is true. Also, it the theorem still true without the compactness assumption? If so, is there a relatively simple proof in that case?

$\endgroup$
7
  • $\begingroup$ Do you know the standard proof («$w_1=w_2=0$ implies parallelizable by elementary obstruction theory» + «$w_1=0$ implies $w_2=0$ by Wu's formulas»)? It's not that hard — and at least the first part is, in a sense, the most straightforward approach possible (but I indeed don't know any intuitive explanation of the second part). $\endgroup$
    – Grigory M
    Jan 17, 2015 at 11:26
  • 2
    $\begingroup$ I think you should add the condition "M is compact". $\endgroup$ Jan 17, 2015 at 12:02
  • $\begingroup$ @GeorgesElencwajg - Both according to the Wikipedia article and this article in nLab: ncatlab.org/nlab/show/framed+manifold this condition is not assumed... $\endgroup$
    – Pandora
    Jan 17, 2015 at 13:19
  • 1
    $\begingroup$ I have no particular trust in these sources, which don't give self-contained proofs. Milnor-Stasheff sketch a proof (Exercise 12-B) based on previous results which depend on $M$ being compact (by the way, you may like it because it doesn't use spin structures). Viro-Fuchs also assume $M$ compact. Of course this does in no way prevent Stiefel's theorem from being true for non compact $M$: maybe this will be settled in some future answer to your question $\endgroup$ Jan 17, 2015 at 13:54
  • 1
    $\begingroup$ Qiaochu's very complete answer indicates the exact role of compactness: it allows one to skip the condition $w_2=0$, which however is required for open manifolds. $\endgroup$ Jan 17, 2015 at 19:43

5 Answers 5

46
$\begingroup$

I don't know of a totally elementary proof of this result, but here is some context for it. More generally, an $n$-manifold $M$ (without boundary) has a classifying map $f : M \to BO(n)$ for its tangent bundle. Knowing when $M$ is parallelizable is equivalent to knowing when $f$ is null-homotopic. There is a general machine for doing this involving lifting $f$ higher and higher through the stages of the Whitehead tower of $BO(n)$, and it tells us that the complete set of obstructions to solving this problem are a set of cohomology classes in $H^k(M, \pi_k(BO(n)), k \le n$, each of which is well-defined provided that the previous one vanishes, such that $f$ is null-homotopic iff all of the classes vanish.

The construction of the first such class goes like this. Assume for simplicity that $M$ is connected. The first question is whether $f$ lifts to the universal cover of $BO(n)$, which is true iff $f$ induces the zero map on $\pi_1$ by standard covering space theory. Now, $\pi_1(BO(n)) \cong \mathbb{Z}_2$, and the induced map on $\pi_1$ gives a homomorphism $\pi_1(M) \to \mathbb{Z}_2$ which corresponds precisely to the first Stiefel-Whitney class $w_1$. This class vanishes iff $f$ lifts to the universal cover of $BO(n)$, which is $BSO(n)$, iff $M$ is orientable.

Now we want to try lifting $f$ to the $2$-connected cover of $BSO(n)$; this is analogous to the universal cover but involves killing $\pi_2(BSO(n)) \cong \mathbb{Z}_2$ (for $n \ge 3$) instead of $\pi_1$. Whether this is possible is controlled by the map

$$BSO(n) \to B^2 \mathbb{Z}_2$$

inducing an isomorphism on $\pi_2$. This is equivalently a universal characteristic class in $H^2(BSO(n), \mathbb{Z}_2)$ which turns out to be precisely the second Stiefel-Whitney class $w_2$. This class vanishes iff $f$ lifts to the $2$-connected cover of $BSO(n)$, which is $BSpin(n)$, iff $M$ has a spin structure.

The first surprise in this story is that (when $n \ge 3$) $BSpin(n)$ also turns out to be the $3$-connected cover; in other words, $\pi_3(BSpin(n)) = 0$, so the next step of this story involves $\pi_4$ and can be ignored for $3$-manifolds. If $M$ is a $3$-manifold, not necessarily compact, admitting both an orientation and a spin structure, then the classifying map of its tangent bundle lifts to a map $M \to BSpin(3)$, but since the latter is $3$-connected any such map is nullhomotopic. So:

A $3$-manifold, not necessarily closed, is parallelizable iff the first two Stiefel-Whitney classes $w_1, w_2$ vanish iff it admits an orientation and a spin structure.

To give an indication of the generality of this machinery, for $4$-manifolds the next step involves computing $\pi_4(BSpin(4)) \cong \mathbb{Z}^2$ and then lifting to the $4$-connected cover of $BSpin(4)$, which is called $BString(4)$. This says that the next obstruction to a $4$-manifold with an orientation and a spin structure being parallelizable is a pair of cohomology classes in $H^4(M, \mathbb{Z})$ which I believe turn out to be the Euler class $e$ and the first fractional Pontryagin class $\frac{p_1}{2}$ (a certain characteristic class of spin manifolds that when doubled gives the Pontryagin class $p_1$) respectively. Since $H^4(M, \mathbb{Z})$ is always torsion-free for a $4$-manifold, the conclusion is that

A $4$-manifold, not necessarily closed, is parallelizable iff the characteristic classes $w_1, w_2, e, p_1$ all vanish iff it admits an orientation, a spin structure, and a string structure.

And for $5$-manifolds and higher we really need to talk about $\frac{p_1}{2}$ and not just $p_1$.

The second surprise in this story is that for closed $3$-manifolds the condition that $w_2$ vanishes is redundant: it is implied by orientability by a standard computation with Wu classes. In other words, closed orientable $3$-manifolds automatically admit spin structures. I don't have a good intuitive explanation of this; it comes from a relationship between the Stiefel-Whitney classes, Steenrod operations, and Poincaré duality that I don't understand very well.

$\endgroup$
4
  • $\begingroup$ Great answer, especially the clarification of the role of compactness as a sufficient condition for avoiding the hypothesis $w_2=0$ in the theorem under discussion. What do you think of PVAL's suggestion for a proof which, even in the non-compact case, doesn't require the hypothesis $w_2=0$ ? $\endgroup$ Jan 17, 2015 at 19:54
  • 2
    $\begingroup$ «The second surprise in this story» — that's, IMHO, the most interesting part... $\endgroup$
    – Grigory M
    Jan 17, 2015 at 21:16
  • 4
    $\begingroup$ I was curious about a more geometric argument for the second surprise too (I'd only seen the computation with Wu classes), but I dug up some alternatives. First, use a (nontrivial) fact from 3-manifold topology that any closed, oriented $X^3$ admits a branched cover $X \to S^3$ branched outside an embedded $3$-ball $B$. But $TS^3$ is trivial, and the obstruction to extending that trivialization across $B$ lies in $\pi_2 SO(3) = 0$. $\endgroup$
    – anomaly
    Aug 27, 2015 at 19:33
  • 2
    $\begingroup$ The second approach is more complicated. To prove $w_2(X^3) = 0$, the idea is to construct a spin structure on $X$. Since $\chi(X^3) = 0$, we have $TX = \theta^1 \oplus \eta$ for some $\eta\to X$. Lift $w_2(\eta)$ to an integral class and take its Poincare dual $S$. We have a spin structure outside $S$; and on $S$, we can use some algebraic topology arguments (not involving the Wu class, though) to show the obstruction $w_2$ to constructing a spin structures vanishes. Then show that the two spin structures agree on their intersection. (It seems much easier to just use the Wu class.) $\endgroup$
    – anomaly
    Aug 27, 2015 at 19:41
13
$\begingroup$

Here is (in my estimate) the approach outlined in Milnor-Stasheff. Let $M$ be a closed, orientable 3-fold, $w_i$ denote the $i$th Whitney class, and $o_i$ the $i$th obstruction class as it is defined in Steenrod's Topology of Fibre Bundles . As $M$ is orientable, $w_1=0$ and a straightforward calculation with Wu's formula gives $w_2$ and $w_3=0$ as well.

From a remark in Ch.12 of Milnor-Stasheff, we have $o_2$ vanishes. So there exists a 2-frame over the 2-skeleton of $M$ (thinking of $M$ as a 3-dim CW-complex). As $\pi_2(V_2(\Bbb R^3))$ is trivial by some basic homotopy theory, the 2-frame extends over $M$ (one just takes a null-homotopy of the map on the boundary to be the map on the 3-cells).

Now we have a trivial 2-plane bundle inside of $TM$. By endowing $TM$ with a metric we can find a complement for said 2-plane bundle, which is orientable hence trivial. So $TM$ is the Whitney sum of trivial bundles hence trivial.

$\endgroup$
0
5
$\begingroup$

Benedetti and Lisca wrote a paper on this same question. They write “the answers given [on this question] until July 18, 2018 use the same tools employed in the proofs mentioned above”, and give three more-elementary answers. https://arxiv.org/abs/1806.04991

$\endgroup$
3
$\begingroup$

Kirby gives a nice proof that every orientable 3-manifold is parallelizable at https://math.berkeley.edu/~kirby/papers/Kirby%20-%20The%20topology%20of%204-manifolds%20-%20MR1001966.pdf . (See the unnumbered non-blank page between page 46 and page 47.)

$\endgroup$
2
$\begingroup$

Cf. "On some theorems of Geroch and Stiefel" by Phillip E. Parker, Journal of Mathematical Physics 25, 597 (1984); https://doi.org/10.1063/1.526209 Where "Modern proofs of two theorems of Geroch (spinor spacetimes and globally hyperbolic spacetimes are parallelizable) and a theorem of Stiefel (orientable 3-manifolds are parallelizable) are given, using the computationally efficient obstruction theory of algebraic topology. These techniques also easily show that, in fact, Geroch's second theorem is a corollary of Stiefel's theorem."

$\endgroup$

You must log in to answer this question.

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