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?

  • $\begingroup$ 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
    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)? $\endgroup$ 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 $G/\Gamma$ $\endgroup$ 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. $\endgroup$ Mar 29, 2018 at 9:59

2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$ Oct 14, 2014 at 22:01
  • 2
    $\begingroup$ Where did you use that $M$ is parallelizable? $\endgroup$ Oct 14, 2014 at 22:49
  • 3
    $\begingroup$ @AndréHenriques: the boundary of $N$ is morally $S(TM)$; that is only the same as $M\times S^{m-1}$ because $M$ is parallelizable. $\endgroup$ Oct 14, 2014 at 23:38

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.

  • $\begingroup$ Dear Andras, what do you mean by "take all the posets"? $\endgroup$ Oct 14, 2014 at 20:56
  • $\begingroup$ Sorry , I should have written cosets (instead of posets) $\endgroup$ Oct 14, 2014 at 21:08
  • $\begingroup$ OK thanks! This is a very nice argument! $\endgroup$ Oct 15, 2014 at 0:21
  • 1
    $\begingroup$ Sullivan invented it on the spot. $\endgroup$ Oct 15, 2014 at 11:44
  • $\begingroup$ Yes indeed, the anecdote behind your first answer is quite inspiring! $\endgroup$ Oct 15, 2014 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.