Characteristic classes of sphere bundles over spheres in terms of clutching functions

Question

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?

Answer 1

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.)