Are the Milnor's seven dimensional exotic spheres parallelizable?

1$\begingroup$ 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. $\endgroup$– Kelly DavisMar 11, 2011 at 8:25

1$\begingroup$ As a related post see math.stackexchange.com/questions/1407327/… $\endgroup$– Ali TaghaviAug 23, 2015 at 21:18
4 Answers
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:
 the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (GaussBonnet, relatively easy).
 $T \Sigma^n \oplus \mathbb{R}$ 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 $\pi_{n1} (O)$, 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 Jhomomorphism 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
$$ \mathbb{Z}=\pi_n (S^n) \to \pi_{n1} (O(n)) \to \pi_{n1} (O(n+1)) = \pi_{n1} (O). $$
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 $\pi_{n1}(O(n)) \to \pi_{n1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n) $. The image of $Z \to \pi_{n1} (O(n)$ 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 KervaireMilnor 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 $\pi_{n1}(O(n))\to \pi_{n1}(S^{n+1})$ is detected by the Euler class, and by GaussBonnet, 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 Hspace''. 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 PontrjaginThom 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.

3$\begingroup$ This is an awesome answer. Thanks for posting! $\endgroup$ Mar 6, 2012 at 13:56

15$\begingroup$ "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". $\endgroup$ Mar 4, 2013 at 23:31
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.
Addendum: From the fibration $O(7)\to O(8)\to S^7$, we have the fibration sequence $$\pi_7(O(7))\to \pi_7(O(8))\to \pi_7(S^7)\to \pi_6 O(7)\to \pi_6 O(8)\to \pi_6 S^7=0.$$
Since $S^7$ is parallelizable (as may be shown via the octonians for example), there is a splitting $\pi_7(S^7)\to \pi_7(O(8))$. Hence we have an isomorphism $\pi_6 O(7)\to \pi_6 O(8) = \pi_6 O(\infty)$, since this is in the stable range. By Bott periodicity, $\pi_6 O(\infty)=0$, so $\pi_6 O(7)=0$.
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 7spheres are parallelizable (since they are stably parallelizable).

2$\begingroup$ This Bredon Kosinski theorem can be deduced from SmaleHirsch 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. $\endgroup$ Jul 28, 2013 at 15:22

2$\begingroup$ @AllenHatcher I am sorry if my question is elementary: Why exotic spheres are stably parallelizable)? $\endgroup$ Aug 23, 2015 at 20:56

3$\begingroup$ The stable parallelizability of exotic spheres is Theorem 3.1 of the famous KervaireMilnor 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 Jhomomorphism. $\endgroup$ Aug 26, 2015 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) $\require{AMScd}$ $$ \begin{CD} O_n @>>> O_{n+1} @>>> S^n \\ @VVV @VVV @VVV\\ F_n @>>> G_{n+1} @>>> S^n\\ @VVV @VVV @VVV\\ F_n @>>> F_{n+1} @>>> F_{n+1}/F_n \end{CD} $$
where $G_n$ is the unbased self homotopy equivalences of $S^{n1}$ 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
$$ \begin{CD} O_n @>>> O_{n+1} \\ @VVV @VVV \\ F_n @>>> F_{n+1} \end{CD} $$
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?

$\begingroup$ 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 Hspace. 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. $\endgroup$ Mar 11, 2011 at 17:10

2$\begingroup$ 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 $f: S^{n1} \to S_x$, where $Sx$ 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? $\endgroup$ Mar 11, 2011 at 20:18

2$\begingroup$ A direct construction of the Hspace map is also given as thm 10.5.7 in AguilarGitlerPrieto (algebraic topology from a homotopical viewpoint) $\endgroup$ Mar 5, 2013 at 1:20

1$\begingroup$ Lemma 2.17 in Allen Hatcher's book "vector bundles and Ktheory" gives a proof that if $S^n$ is parallelizable, then it is an Hspace. pi.math.cornell.edu/~hatcher/VBKT/VBpage.html $\endgroup$– Ian AgolMar 2, 2019 at 6:05
