I'm trying to understand Milnor's proof of the existence of exotic 7-spheres.


Milnor finds his examples among $S^{3}$ bundles over $S^{4}$ (with structure group $SO(4)$ ). Such a bundle can be described as follows:


Given $M$, an $S^{3}$ bundle over $S^{4}$, if we restrict $M$ to the northern (or southern) hemisphere of $S^{4}$, it must trivialize since each hemisphere is contractible. Hence, we can build $M$ by specifying, for each point $p$ in $S^{3}$ = equator of $S^{4}$ = intersection of northern and southern hemispheres, an element of $SO(4)$ which glues $p\times S^{3}$ in the northern hemisphere to $p\times S^{3}$ in the southern hemisphere.


This defines a function $f:S^{3}\rightarrow SO(4)$, which is known as the clutching function for $M$. By usual fiber bundle theory, the isomorphism type of $M$ only depends on the homotopy class of $f$.


$SO(4)$ is double covered by $S^3\times S^3$, and hence $\pi\_3(SO(4)) = \mathbb{Z}\oplus\mathbb{Z}$. Thus, $f$ is really determined (at least, up to homotopy) by an ordered pair of integers (i,j).


Now, as the bundles have structure group $SO(4)$, it makes sense to talk about the Pontryagin classes of $M$. In Milnor's proof of the existence of exotic spheres, he needs to argue that $p\_1(M) = \pm 2(i-j)$. His first step in this argument is that "clearly $p\_1(M)$ is a linear function of $i$ and $j$."


It IS clear to me that the Pontragin classes associated to $(ni, nj)$ for $n\in \mathbb{Z}$ will depend linearly on $n$. For, if we let $N\_{i,j}$ denote the principal $SO(4)$ bundle over $S^{4}$ corresponding to $(i,j)$, then $N\_{ni,nj}$ is clearly obtained as the pullback of $N\_{i,j}$ via a degree $n$ map from $S^{4}$ to itself.


However, it's not clear to me why $p\_1(M)$ is additive in $(i,j)$. Am I missing something simple?


And while we're talking about it, is more true? That is, For any sphere bundle over a sphere, say, $S^{k}\rightarrow E\rightarrow S^{n}$, should any characteristic classes (Pontryagin, Stiefel-Whitney, Euler) be linear in terms of the clutching function?


For example, we can think of $p\_1$ as a map from $\pi\_{n-1}(SO(k+1))\rightarrow H^{4}(S^{n})$. Is this map a homomorphism? How about for the other characteristic classes?



Characteristic classes of sphere bundles over spheres in terms of clutching functions
There is a way to explain it that's similar to what you said about multiplication by $n$.


Let $G$ be a Lie group, and let $f\_1$ and $f\_2$ be any two clutching functions that describe $G$-bundles $E\_1$ and $E\_2$ on $S^n$. Suppose that $f\_1$ and $f\_2$ agree at a base point of $S^{n-1}$. Let $c$ be some characteristic class of $G$-bundles of degree $n$; it could even be a cup product of standard classes such as Chern or Pontryagin or whatever. Let $f\_3$ be the combination of $f\_1$ and $f\_2$ on the one-point union $S^{n-1} \vee S^{n-1}$. It is the clutching function of a bundle $E\_3$ on the suspension $\Sigma(S^{n-1} \vee S^{n-1})$, which is the union of two $n$-spheres along an interval and thus homotopy equivalent to $S^n \vee S^n$.


Whether you define $c$ the old-fashioned way by obstructions, or the more modern way by pullbacks from classifying spaces, it is easy to argue that $c(E\_3) = c(E\_1) \oplus c(E\_2)$. I.e., it's just the ordered pair of the characteristic classes of its parts. Now addition in $\pi\_n$ is modeled by a map $S^n \to S^n \vee S^n$, and the induced map on $H^n$ takes $a \oplus b$ to $a+b$. (Your generalized question about $H^k(S^n)$ is of course non-trivial only when $k=n$.)




---


A shorter, more modern summary of the same story is as follows. The $X$ is a space and $LX$ is its loop space, then $\pi\_{n-1}(LX) \cong \pi\_n(X)$. The loop space of the classifying space $B\_G$ is homotopy equivalent to $G$ itself, so $\pi\_{n-1}(G) \cong \pi\_n(B\_G)$. A characteristic class of degree $n$ is any cohomology class in $H^n(B\_G)$. The linearity that Milnor uses is the transposed form of the fact that the Hurewicz homomorphism $\pi\_n(B\_G) \to H\_n(B\_G)$ is linear. (If you expand this it out more explicitly, it isn't really different from what I say above.)


