

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Are the Milnor's seven dimensional exotic spheres parallelizable?

share|cite|improve this question
A normal $S^7$ is parallelizable and admits a smooth global frame $e$. $S$ an exotic $S^7$ is homeomorphic to $S^7$. If we view the homeomorphism as a topological embedding, then we can think of $S \subset S^7$ and the smooth global frame $e$ almost works for $S$. But, the problem is $e$ is no longer smooth with respect to the smooth structure on $S$. Thus, $e$ doesn't make $S$ parallelizable. – Kelly Davis Mar 11 '11 at 8:25
As a related post see… – Ali Taghavi Aug 23 '15 at 21:18

A much more general result is true.

Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma $ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$.

It says that exotic spheres cannot be distinguished by looking at the tangent bundle. This result is one of the hidden gems of the golden age of topology and the proof invokes the whole plethora of topology of the 1950s.

The argument can be recollected from the old literature, but I do not know a coherent reference.

To start with, there are several invariants of the tangent bundles that do not depend on the smooth structure. Let $\Sigma$ be a homotopy sphere. Then:

  1. the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (Gauss-Bonnet, relatively easy).
  2. $T \Sigma^n \oplus \mathbb{R}$ is trivial. This is a deep result by Kervaire and Milnor (not in the Annals paper, but a small note published before). $T \Sigma \oplus \mathbb{R}$ is given by an element of $\pi_{n-1} (O)$, which is known, by Bott periodicity, to be either $Z$, $0$ or $Z/2$. The $Z$ groups are detected by the Pontrjagin class, which has to vanish by Hirzebruchs signature formula because the sphere evidently has signature $0$. In the $Z/2$ case, the argument is more delicate. Essentially, the normal spherical fibration of a manifold does not depend on the smooth structure. Since the normal fibration of the standard sphere is trivial, so is that of $T \Sigma$. The process of associating to a vector bundle its spherical fibration is the J-homomorphism which is injective in these dimensions by Adams' $J(X)$ paper.

Now look at the homotopy sequence of the fibration $O(n)\to O(n+1) \to S^n$, i.e. the piece

$$ \mathbb{Z}=\pi_n (S^n) \to \pi_{n-1} (O(n)) \to \pi_{n-1} (O(n+1)) = \pi_{n-1} (O). $$

It is known that $TS^n$ (for the standard smooth structure) is the image of the generator of $\pi_n (S^n)$ (not hard, see Steenrods book). By the above deep result, $T \Sigma$ lies in the kernel of $\pi_{n-1}(O(n)) \to \pi_{n-1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n) $. The image of $Z \to \pi_{n-1} (O(n)$ can be computed. It is $Z$ if $k$ is even (using the Euler class), it is $0$ if $n=1,3,7$ (follows directly from Adams' result on the parallelizability of the standard spheres) and it is $Z/2$ in the remaining cases. You can find the (not so hard, but clever) argument for the last assertion in Levine's lectures on homotopy spheres (which can be viewed as the sequel to the Kervaire-Milnor paper).

How to proceed? If $n=1,3,7$, it follows that $T \Sigma$ is trivial, as $TS^n$. If $n$ is even, then the kernel of $\pi_{n-1}(O(n))\to \pi_{n-1}(S^{n+1})$ is detected by the Euler class, and by Gauss-Bonnet, this characterizes $T \Sigma$.

It remains the case of odd $n$ apart from the "Adams dimensions". One has to argue that in these dimensions, $T \Sigma$ is nontrivial. In the introduction to his Hopf invariant paper, Adams attributes to Dold the result ''$T \Sigma$ parallel implies that $\Sigma$ (and hence $S^n$) is an H-space''. But he (Adams) proved that is not the case $n=1,3,7$. Adams does not give a reference for Dolds result, but in his answer to this question (and the subsequent comments), John Klein sketches a proof that looks like a 1950s argument.

EDIT: I supervised a Master's thesis (written by Julia Heller), who worked out the details of this argument. Here is the argument for the fact that if a homotopy sphere $\Sigma^n$ is parallelizable, then $n=0,1,3,7$.

Consider the diagonal $\Sigma \subset \Sigma \times \Sigma$. Its normal bundle $N$ is isomorphic to $T \Sigma$,

hence trivial. We have the Pontrjagin-Thom collapse $c:\Sigma \times \Sigma \to Th (N) $. Any trivialization $N \cong \Sigma \times \mathbb{R}^n$ induces a map $a: Th (N) \to S^n$. Let $$ \mu: \Sigma \times \Sigma \to Th (N) \to S^n $$ be the composition of the maps just explained. By counting intersections, it can be seen that the restriction of $\mu$ to the submanifolds $\{x\} \times \Sigma$ and $\Sigma \times \{x\}$ has degree $\pm 1$, hence is a homotopy equivalence. By composing with $(h,k):S^n \times S^n \to \Sigma \times \Sigma$ for suitably chosen homotopy equivalences, we get a map $$ \mu' : S^n \times S^n \to S^n $$ which restricts to degree 1 maps $S^n \times \{x\} \to S^n$ and $\{x\} \times S^n \to S^n$. Since maps of degree 1 are homotopic, it follows that $\mu'$ gives $S^n $ the structure of an $H$-space. By Adams' theorem, $n=0,1,3,7$.

Note: it is not important for Adams' theorem that the $H$-space structure is homotopy associative, but homotopy unitality is essential.

share|cite|improve this answer
This is an awesome answer. Thanks for posting! – Akhil Mathew Mar 6 '12 at 13:56
"The argument can be recollected from the old literature, but I do not know a coherent reference": Maybe you should give this to a masters student. The masters thesis (assuming it's well written) could then become the missing "coherent reference". – André Henriques Mar 4 '13 at 23:31
Is smoothness part of your definition of homotopy sphere, or is it known that a topological manifold with the same homotopy groups as a sphere necessarily admits a smooth structure? – Michael Albanese Apr 11 at 19:33
For anyone interested in my question in the previous comment, I asked about the situation here. – Michael Albanese Apr 13 at 20:57

The tangent bundle to a smooth structure on $S^7$ is classified by a map $S^7 \to G_7(R^{\infty})$. By the exact sequence for a fibration for the fiber bundle $O(7)\to V_7(R^\infty)\to G_7(R^\infty)$, we see that $\pi_7(G_7(R^\infty)) = \pi_6(O(7))$. But $\pi_6(O(7))=0$ (I found a table A1.1.3.2 of homotopy groups of orthogonal groups here(pdf), since this isn't in the stable range of Bott periodicity), so the tangent bundle is trivial, i.e. parallelizable.

share|cite|improve this answer

Here's another way to answer the original question. There is a theorem of Bredon and Kosinski (Annals, 1966) which says that if a manifold $M^n$ is stably parallelizable, then either $M^n$ is parallelizable or the maximum number of linearly independent vector fields on $M^n$ is the same as on $S^n$. Since $S^7$ is parallelizable, this implies that exotic 7-spheres are parallelizable (since they are stably parallelizable).

share|cite|improve this answer
This Bredon Kosinski theorem can be deduced from Smale-Hirsch immersion theory. (May be BK did this way, I do not know.) So immersion theory implies that a stably parallelizable manifold immerses into Euclidean space with codimension one. Then using the Gauss map one can pull back the vectorfields on the sphere to the immersed manifold. Hence a stably parallelizable manifold admits at least as many vector fields as the sphere of the same dimension. – András Szűcs Jul 28 '13 at 15:22
@AllenHatcher I am sorry if my question is elementary: Why exotic spheres are stably parallelizable)? – Ali Taghavi Aug 23 '15 at 20:56
@AllenHatcher I am sorry if I send this message again. May you answer to my previous comment question? – Ali Taghavi Aug 26 '15 at 14:21
The stable parallelizability of exotic spheres is Theorem 3.1 of the famous Kervaire-Milnor paper "Groups of homotopy spheres I" in the 1963 Annals. The proof is short but uses several big theorems from the previous decade such as Bott periodicity, the Hirzebruch signature theorem, and Adams' work on the J-homomorphism. – Allen Hatcher Aug 26 '15 at 19:46

The following is just an expansion of Johannes' last paragraph.

I went to Adams' paper where he attributes to Dold the statement that $S^n$ parallelizable implies $S^n$ is an $H$-space. No reference is given, so I had to think of why the statement is true (and also its converse).

Here's is one possible argument (that avoids Hopf invariant one considerations).

Consider the diagram (whose horizontal sequences are fibrations) $$ O_n \to O_{n+1} \to S^n $$ $$ \downarrow \qquad \qquad \downarrow \qquad\qquad \downarrow $$ $$ F_n \to G_{n+1} \to S^n $$ $$ \downarrow \qquad \qquad \downarrow \qquad\qquad \downarrow $$ $$ \quad \quad F_n \to F_{n+1} \to F_{n+1}/F_n $$ where $G_n$ is the unbased self homotopy equivalences of $S^{n-1}$ and $F_n$ is the based homotopy equivalences of $S^n$. The map $O_n \to G_n$ is given by restricting an isometry to its unit sphere and the map $G_n \to F_n$ is given by unreduced suspension.

Then $S^n$ is parallelizable iff the top fibration has a section which implies that the middle fibration has a section $S^n \to G_{n+1}$. We can assume without loss in generality that this map sends the base point of $S^n$ to the identity.

The adjoint of this section is of the form $S^n \times S^n \to S^n$ which is an $H$-space structure.

Conversely, if there's an $H$-space structure, then the middle fibration has a section. The map $S^n \to F_{n+1}/F_n$ is approximately $2n$-connected (this is a consequence of the EHP sequence). Consequently, there's a section of the bottom fibration up to around the $2n$-skeleton of the basespace $F_{n+1}/F_n$.

But, the square

$$ O_n \to O_{n+1} $$

$$ \downarrow \qquad \qquad \downarrow $$

$$ F_n \to F_{n+1} $$

is about $2n$-cartesian.

This implies that $O_{n+1} \to S^n$ has a section iff and only if the pullback of $F_{n+1} \to F_{n+1}/F_n$ to $S^n$ has one and that's if and only if $G_{n+1} \to S^n$ has a section.

Is this a correct argument?

share|cite|improve this answer
It seems to be valid, but not sufficient. According to Adams, Dold showed that if $S^n$ is parallelizable with respect to some smooth structure, then it is an H-space. You would have to create a map from the frame bundle of the exotic structure to $G_{n+1}$ and I do not see this map. – Johannes Ebert Mar 11 '11 at 17:10
That sounds more convincing. – Johannes Ebert Mar 11 '11 at 19:15
Let $P \to M$ be the frame bundle of a Riemannian manifold $M$: at $x\in M$ its fiber is given by linear isometries $\phi: \Bbb R^n \to M$. Let $Q \to M$ be the fibration whose fiber at $x\in M$ is given by the space of homotopy equivalences $f: S^{n-1} \to S_x$, where $S-x$ is the image of a small sphere under the exponential map. Then when $M$ is an exotic sphere we get $Q \simeq G_{n+1}$. The map $P \to Q$ is given by restricting the isometry $\phi$ to a small sphere at the origin, and then applying the exponential map. How does that sound to you? – John Klein Mar 11 '11 at 20:18
A direct construction of the H-space map is also given as thm 10.5.7 in Aguilar-Gitler-Prieto (algebraic topology from a homotopical viewpoint) – Dylan Wilson Mar 5 '13 at 1:20

Your Answer


