Are the Milnor's seven dimensional exotic spheres parallelizable?

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:
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. 


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. 


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). 


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 $S^{n1}$ 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 $$ O_n \to O_{n+1} $$ $$ \downarrow \qquad \qquad \downarrow $$ $$ F_n \to F_{n+1} $$ 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? 

