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So let $M$ be a smooth closed orientable real manifold such that $M$ is parallelizable, i.e., the tangent space $TM$ of $M$ is trivial. From the triviality of $TM$ we get that the Stiefel-Whitney and Pontryagin classes are trivial and therefore the Stiefel-Whitney and Pontryagin numbers are all equal to $0$. We have the following amazing theorem of C.T.C. Wall:

(Wall's theorem) A closed orientable real manifold $M$ is the boundary of a compact oriented manifold (with boundary) iff its Stiefel-Whitney and Pontryagin numbers are trivial.

Q: Using the stronger additional assumption that $M$ is parallelizable, is it possible to give a simple proof that $M$ is the boundary of an oriented manifold?

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  • $\begingroup$ Kirby's lecture notes "The topology of four-manifolds", Ch. VII gives a proof that every orientable $3$-manifold is an orientable boundary. The first observation is that since $M^3$ is parallelizable, it immerses in $R^5$ with a trivial normal bundle. I'm not sure if this is a general consequence of parallelizability. $\endgroup$
    – JHM
    Commented Mar 8, 2014 at 17:06
  • $\begingroup$ How about starting off with the apparently simpler problem of M a compact Lie group? Lie groups are parallelizable. Why does every compact Lie group bound? SU(2): check. How about SU(3)? $\endgroup$
    – Richard Montgomery
    Commented Mar 12, 2014 at 4:52
  • $\begingroup$ Yes indeed, once one deals with this case, then one may look at a quotient of $G$ by a discrete subgroup $\Gamma\subseteq G$. Then any frame on $G$ may be pushed down to a frame on $G/\Gamma$ $\endgroup$
    – Hugo Chapdelaine
    Commented Mar 12, 2014 at 17:11
  • $\begingroup$ Martel. If an n-manifold is stably parallelizable, then it can be immersed in n+1-dimensional Euclidean space. This is a trivial consequence of Hirsch's immersion theorem. $\endgroup$
    – András Szűcs
    Commented Mar 29, 2018 at 9:59

2 Answers 2

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I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $\mathbb{RP}^q$, and the boundary of $N'$ is mapped into $\mathbb{RP}^{m-1} \subset \mathbb{RP}^q$. Take an $\mathbb{RP}^{q-m+1}$ in $\mathbb{RP}^q$ that intersects $\mathbb{RP}^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $\mathbb{RP}^{m-1}$ (this can be supposed) then $f^{-1}(\mathbb{RP}^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

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  • $\begingroup$ If m is odd, then this gives even oriented null-cobordism (because if we choose q to be odd, then both RP^q and RP^{q-m+1} are orientable). $\endgroup$
    – András Szűcs
    Commented Oct 14, 2014 at 22:01
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    $\begingroup$ Where did you use that $M$ is parallelizable? $\endgroup$
    – André Henriques
    Commented Oct 14, 2014 at 22:49
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    $\begingroup$ @AndréHenriques: the boundary of $N$ is morally $S(TM)$; that is only the same as $M\times S^{m-1}$ because $M$ is parallelizable. $\endgroup$
    – Neil Strickland
    Commented Oct 14, 2014 at 23:38
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Rene Thom put the same question to Sullivan during Sullivan's talk at IHES in 1980: "Why Lie groups are null-cobordant?" Sullivan answered: "It follows from Thom's theorem" (A manifold is null-cobordant precisely when each of its characteristic numbers vanish.)

Thom: "OK, but can you prove this without this heavy tool?"

After few minutes Sullivan said:

"Take a circle subgroup in the Lie group. Take all the cosets. This gives a fiber bundle with total space G (the Lie group) and fiber S^1. Take the associated disc bundle. Its boundary is G." Q.E.D.

An elementary proof that paralellizable manifold is null-cobordant can be found in a paper by Buoncristiano- Hacon. I am not sure it gives oriented null-cobordism.

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  • $\begingroup$ Dear Andras, what do you mean by "take all the posets"? $\endgroup$
    – Hugo Chapdelaine
    Commented Oct 14, 2014 at 20:56
  • $\begingroup$ Sorry , I should have written cosets (instead of posets) $\endgroup$
    – András Szűcs
    Commented Oct 14, 2014 at 21:08
  • $\begingroup$ OK thanks! This is a very nice argument! $\endgroup$
    – Hugo Chapdelaine
    Commented Oct 15, 2014 at 0:21
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    $\begingroup$ Sullivan invented it on the spot. $\endgroup$
    – András Szűcs
    Commented Oct 15, 2014 at 11:44
  • $\begingroup$ Yes indeed, the anecdote behind your first answer is quite inspiring! $\endgroup$
    – Hugo Chapdelaine
    Commented Oct 15, 2014 at 14:37

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