It has been known since Milnor’s famous paper that two smooth manifolds can be homeomorphic without being diffeomorphic. Milnor showed that certain sphere bundles over ${S^4}$ were homeomorphic but not diffeomorphic to the 7-sphere ${S^7}$. In later papers, Milnor constructed a number of additional examples of exotic spheres.

In this post, I’d like to give a detailed presentation of the argument in Milnor’s first paper.

1. Distinguishing homeomorphic manifolds

Suppose you have a ${4k-1}$-dimensional manifold ${M}$ which is known to be homeomorphic to the sphere ${S^{4k-1}}$. There are a number of criteria for this: for instance, ${M}$ could admit a cover by two charts, or ${M}$ could admit a function with only two critical points. The goal is to prove then that ${M}$ is not diffeomorphic to ${S^{4k-1}}$. Obviously the standard invariants in topology see only homotopy type and are useless at telling apart ${M}$ and ${S^{4k-1}}$. One needs to define an invariant that relies on the smooth structure of ${M}$ in some way.

It can be shown that any such ${M}$ is an oriented boundary, ${M = \partial B}$, for a ${4k}$-manifold ${B}$. This is a deep fact, but in practice, the manifolds ${M}$ come with explicit ${B}$‘s already, and one might as well define the invariant below on boundaries. Milnor’s strategy is to define an invariant in terms of ${B}$ (which will depend very much on the smooth structure on ${M}$), in such a way that it will turn out to not depend on the choice of ${B}$.

What can we say about ${B}$? It is a ${4k}$-dimensional manifold with boundary. We know that if ${B}$ were aclosed manifold, then we could compute the signature

$\displaystyle \sigma(B) = L(p_1, \dots, p_k)[B] \ \ \ \ \ (1)$

by computing the Pontryagin classes of the tangent bundle ${TB}$ and evaluating on the fundamental class. Here ${L}$ is a somewhat complicated polynomial; the last term ${p_k}$ occurs to degree one as a linear term ${c_k p_k}$. In particular, if we write

$\displaystyle L(p_1, \dots, p_k) = L'(p_1, \dots, p_{k-1}) + c_k p_k,$

we have

$\displaystyle \frac{1}{c_k} \left( \sigma(B) - L' (p_1, \dots, p_{k-1})[B] \right) \in \mathbb{Z}. \ \ \ \ \ (2)$

This is a fairly nontrivial divisibility relation.

However, since ${B}$ is not a closed manifold, the signature formula (1) is not true for ${B}$, and we cannot assert (2). The existence of the boundary ${M}$ may introduce some fractional terms. Nonetheless, the (nontrivial) claim is that the quantity ${\frac{1}{c_k} \left( \sigma(B) - L' (p_1, \dots, p_{k-1})[B] \right)}$, while not necessarily integral, has residue in ${\mathbb{Q}/\mathbb{Z}}$ which is independent of the choice of manifold ${B}$.

Definition 1 The Milnor invariant ${\lambda(M)}$ is defined as

$\displaystyle \lambda(M) = \frac{1}{c_k} \left( \sigma(B) - L' (p_1, \dots, p_{k-1})[B] \right) \in \mathbb{Q}/\mathbb{Z};$

this is defined for any ${4k-1}$-manifold ${M}$ which is a homology sphere and the boundary of some ${B}$.

Here ${\sigma(B)}$ is defined as the signature of the quadratic form (cup product) on ${H^{2k}(B, W; \mathbb{Q})}$ and the fundamental class of ${B}$ lives in relative cohomology ${H^{4k}(B, W)}$. The Pontryagin classes ${p_1, \dots, p_{k-1}}$ of ${B}$ are pulled back to ${H^\bullet(B, W)}$ because ${H^\bullet(B, W) \simeq H^\bullet(B)}$ in dimensions below the top dimension; this is because ${M}$ is a homology sphere.

It is important to use cohomology relative to the boundary since ordinary Poincaré duality does not hold for ${B}$ (i.e., we need to take homology and cohomology relative to the boundary).

Milnor’s observation is:

Proposition 2 ${\lambda(M)}$ is actually well-defined; it does not depend on the choice of the bounding manifold ${B}$.

Proof: To see this, suppose ${B, \widetilde{B}}$ were two manifolds both with boundary ${M}$. Then we could form the sum ${C = B \sqcup_M \widetilde{B}}$; this becomes a compact, oriented manifold (without boundary) if we orient ${\widetilde{B}}$ the opposite way. Then the cohomology ring of ${C}$ is the sum of the cohomology rings of ${H^*(B, M)}$ and ${H^*(\widetilde{B}, M)}$ except in the top dimension. The fundamental class of ${C}$ is the “difference” of the fundamental classes of ${B, \widetilde{B}}$ because of our choice of orientation.

Consequently, one sees that the cup square quadratic form on ${H^{2k}(C)}$ is the “sum” of the quadratic forms on ${H^{2k}(B; M)}$ and the opposite of the one on ${H^{2k}(\widetilde{B}; M)}$. Consequently, we have ${\sigma(C) = \sigma(B) - \sigma(B')}$.

By the same reasoning, the Pontryagin classes ${p_1, \dots, p_{k-1}}$ of ${C}$ are determined by their restrictions to ${H^*(B, M)}$ and ${H^*(\widetilde{B}, M)}$. These are just the Pontryagin classes of ${B}$ and ${\widetilde{B}}$, and we have to integrate them with respect to the fundamental class of ${C}$.

This means that

$\displaystyle \sigma(B) - L' (p_1, \dots, p_{k-1})[B] - \left( \sigma(\widetilde{B}) - L' (p_1, \dots, p_{k-1})[\widetilde{B}] \right) = \left( \sigma(C) - L' (p_1, \dots, p_{k-1})[C] \right) .$

The last thing is a multiple of ${c_k}$ because of the signature formula applied to ${C}$. This shows that when one divides by ${c_k}$, the element of ${\mathbb{Q}/\mathbb{Z}}$ that one gets is actually well-defined. $\Box$

Milnor was considering the situation where ${k = 2}$ in his original paper. In this case, the ${L}$-polynomial is

$\displaystyle L(p_1, p_2) = \frac{7p_2 - p_1^2}{45}.$

Consequently, we can define the invariant ${\lambda}$ as

$\displaystyle \lambda(M) = \frac{45}{7} ( \sigma(B) + \frac{1}{45} p_1^2[B]) \in \mathbb{Q}/\mathbb{Z}.$

It is convenient to define from this an invariant mod 7 by multiplying by ${14}$,

$\displaystyle \widetilde{\lambda}(M) = ( 90 \sigma(B) + 2 p_1^2 [B]) \equiv 2p_1^2[B] - \sigma(B) \mod 7.$

This is an invariant mod 7, well defined for any homology 7-sphere.

Clearly ${\widetilde{\lambda}(S^7) = 0}$ since we can take the 8-ball as the boundary. We will compute it on certain sphere bundles and show that it does not vanish.

2. Four-dimensional vector bundles over ${S^4}$

There is a Hopf fibration

$\displaystyle S^3 \rightarrow S^7 \rightarrow S^4 .$

Motivated by this, let’s consider the 7-manifolds that one might get by taking the sphere bundle in a four-dimensional vector bundle over ${S^4}$.

The classification of four-dimensional vector bundles of ${S^4}$ is equivalent to the determination of

$\displaystyle \pi_3( SO(4)).$

There is a double cover ${\mathrm{Spin}(4) \rightarrow SO(4)}$; here ${\mathrm{Spin}(4)}$ can be identified with the group ${\mathrm{Sp}(1) \times \mathrm{Sp}(1)}$, the group of unit quaternions squared.

More explicitly, we can construct the map

$\displaystyle \mathrm{Sp}(1) \times \mathrm{Sp}(1) \rightarrow SO(4)$

by sending a pair of unit quaternions ${(z,w)}$ to the map ${\mathbb{R}^4 \rightarrow \mathbb{R}^4}$, ${x \mapsto z x w}$ (using quaternion multiplication). Here we identify ${\mathbb{R}^4}$ with ${\mathbb{H} }$. The kernel of this map consists of ${(1, 1)}$ and ${(-1, -1)}$ since ${\mathbb{R}}$ is central in ${\mathbb{H}}$, and consequently (by a dimension count) we have constructed the universal cover of ${SO(4)}$.

It follows that we can determine ${\pi_3(SO(4))}$:

Proposition 3 ${\pi_3(SO(4)) = \mathbb{Z} \oplus \mathbb{Z}}$. We can choose an explicit identification as follows: given ${(i, j) \in \mathbb{Z}}$, we have a map ${\phi: S^3 \rightarrow SO(4)}$ which sends a unit quaternion ${u}$ to the map

$\displaystyle \phi(u): \mathbb{R}^4 \rightarrow \mathbb{R}^4, \ \phi(u)(x) = u^i x u^j.$

In fact, we have determined the structure of ${\mathrm{Spin}(1) = S^3 \times S^3}$, and clearly ${\pi_3}$ of this is ${\mathbb{Z} \oplus \mathbb{Z}}$. The explicit identification comes from the explicit map ${S^3 \times S^3 \rightarrow SO(4)}$.

3. The sphere bundles

Our goal now is to analyze the sphere bundles in these vector bundles over ${S^4}$. In particular, we want to show that under specific conditions, they will be homeomorphic to the sphere.

In the previous section, we classified the four-dimensional vector bundles over ${S^4}$, and showed that they were in bijection with ${\mathbb{Z} \oplus \mathbb{Z}}$. Let’s write ${V_{ij}}$ for the vector bundle corresponding to the pair ${(i,j)}$.

Proposition 4 If ${i + j = 1}$, the sphere bundle in ${V_{ij}}$ is homeomorphic to ${S^7}$.

The main idea is to use the following criterion for a manifold to be homeomorphic to the sphere:

Theorem 5 (Reeb) If a compact manifold admits a function with only two critical points, it is homeomorphic to the sphere.

Any of the coordinate functions is such an example on the usual sphere.

In the present situation, we want to produce such a function on the sphere bundle in ${V_{ij}}$. Fortunately, ${V_{ij}}$ is given to us fairly explicitly in coordinates. Namely, we have two charts of the sphere with transition functions, and the vector bundle is given to us explicitly with transition functions.

More specifically, we have two charts ${U_+ \simeq \mathbb{R}^4, U_- \simeq \mathbb{R}^4}$ of the sphere ${S^4}$. The transition function over ${U_+ \cap U_- \simeq \mathbb{R}^4 \setminus \left\{0\right\}}$ is given by

$\displaystyle x_- = \frac{x_+}{\left\lvert x_+\right\rvert^2}.$

Here ${x_-}$ is the coordinate on ${U_-}$ and ${x_+}$ is the coordinate on ${U_+}$; we have specified the coordinate transformation relating the two.

For the total space of the vector bundle ${V_{ij}}$, we have two charts ${V_+ \simeq \mathbb{R}^4 \times \mathbb{R}^4 }$ and ${V_- \simeq \mathbb{R}^4 \times \mathbb{R}^4}$ corresponding to the trivializations of the vector bundle on the sphere minus the north and south poles. The intersection is isomorphic to ${\mathbb{R}^4 \setminus \left\{0\right\} \simeq \mathbb{R}^4}$ and in coordinates, we have

$\displaystyle (x_-, v_-) = \left( \frac{x_+}{\left\lvert x_+\right\rvert^2} , \frac{x_+^i v x_+^j }{\left\lvert x_+\right\rvert^{}} \right)$

Here ${x_{\pm}}$ is the coordinate in the base and ${v_{\pm}}$ is the coordinate for the fiber. We have actually extended the clutching function to ${U_{-} \cap U_+}$ rather than simply on the three-sphere; this is perfectly reasonable.

The sphere bundle of ${V_{ij}}$ clearly has two charts ${S_+}$ and ${S_-}$ isomorphic to ${\mathbb{R}^4 \times S^3}$, and the transition function is the same.

Now we need a function on the sphere bundle. This can be defined in coordinates, using the real part of a quaternion. The definition is:

$\displaystyle h_+(x_+, v_+) = \frac{\Re (v_+)}{(1 + \left\lvert x_+\right\rvert^2)^{1/2}}$

and

$\displaystyle h_-(x_-, v_-) = \frac{\Re( x_- v_-^{-1})}{(1 + \left\lvert x_{-}\right\rvert^2)^{1/2}} .$

We need to check that the function ${h}$ is actually defined everywhere and then that ${h}$ has only two critical points.

In fact, we can check that on the intersection, ${h_+(x_+, v_+) = h_-(x_-, v_-)}$ (using the fact that the real part of a quaternion is invariant under conjugation); consequently we get an honest function ${h}$ on the sphere bundle. It can be checked directly that there are no critical points for ${h_-}$ and ${h_+}$ has two (at ${(0, \pm 1)}$). In view of Reeb’s theorem, this completes the proof that the sphere bundles are homeomorphic to ${S^7}$ (for ${i + j = 1}$).

4. Computation of the Milnor invariant

Now the natural thing to do is to compute ${\widetilde{\lambda}}$ on these sphere bundles, which we’ll write as ${S(V_{ij}) \subset V_{ij}}$. The good news is that ${S(V_{ij})}$ is naturally the boundary of the ball bundle ${B(V_{ij}) \subset V_{ij}}$, and this is a fairly concrete manifold which we can get our hands on. There are two things to compute: the signature and the Pontryagin classes. Let’s start with the former. Throughout, we’ll assume ${i + j = 1}$, so that the sphere bundle is actually a topological sphere.

We have

$\displaystyle H^n(B(V_{ij}), S(V_{ij})) = \begin{cases} \mathbb{Z} & \text{if } n = 0 , 4, 8 \\ 0 & \text{otherwise} \end{cases} .$

This follows easily because ${B(V_{ij})}$ deformation retracts onto ${S^4}$. Poincaré duality for manifolds with boundary shows that the quadratic form on ${H^4}$ must be plus or minus squaring; if we choose our orientations for ${B(V_{ij})}$ (and thus for ${S(V_{ij})}$!) right, we can assume

$\displaystyle \sigma(B(V_{ij})) = 1.$

So, in order to compute ${\widetilde{\lambda}}$, we are left with determining the Pontryagin classes of ${B(V_{ij})}$. The tangent bundle ${T B(V_{ij})}$ splits into the “vertical” piece (just the pull-back of ${V_{ij}}$) and the “horizontal” piece (the pull-back of ${TS^4}$). In other words, if ${\pi: B(V_{ij}) \rightarrow S^4}$ is projection, we have

$\displaystyle TB(V_{ij}) = \pi^* ( TS^4 \oplus V_{ij}). \ \ \ \ \ (3)$

The tangent bundle ${TS^4}$ does not contribute to the Pontryagin classes as it is stably trivial. So, finally, we are left with determining the Pontryagin classes of ${\pi^* V_{ij}}$ as a function of ${i,j}$; we may as well do it on ${S^4}$.

Proposition 6 The Pontryagin class ${p_1(V_{ij}) = \pm 2 (i - j)}$.

Proof: It is easy to see that ${p_1(V_{ij})}$ is linear in ${i,j}$. Actually, it wasn’t easy for me when I first tried to read this paper, so let me spell it out briefly: If vector bundles ${V}$ and ${V'}$ are defined by clutching functions ${\alpha, \beta: S^3 \rightarrow SO(4)}$, then ${V \oplus V'}$ is stably isomorphic to the vector bundle defined by ${\alpha \beta: S^3 \rightarrow SO(4)}$; this follows by a bit of matrix manipulation.

Now we need to claim that ${p_1(V_{ij})}$ is always divisible by two, and that ${2}$ is in the image. This follows from the fact that ${c_2}$ is surjective onto ${\mathbb{Z}}$ for complex vector bundles over ${S^4}$ (a special case of a theorem of Bott, and easily checkable via Bott periodicity), and the fact that complexification

$\displaystyle \widetilde{KO}(S^4) \rightarrow \widetilde{K}(S^4)$

has image precisely of index 2. (Both groups are ${\mathbb{Z}}$ by Bott periodicity.)

To complete the proof, we need only check that ${p_1(V_{ij}) = p_1(V_{-j, -i})}$. In fact, ${V_{ij}}$ is defined by the map

$\displaystyle \phi_{ij}: S^3 \rightarrow SO(4), \quad \phi(u)(x) = u^i x u^j.$

Let ${\mu: \mathbb{R}^4 \rightarrow \mathbb{R}^4}$ be quaterionic conjugation.

Then

$\displaystyle (\mu \phi(u) \mu)(x) = u^{-j} x u^{-i}$

because ${\mu}$ is an anti-involution of the quaternions. But this means that ${\phi}$ and ${\mu \phi \mu^{-1}}$ are homotopic as maps ${S^3 \rightarrow SO}$ into the infinite orthogonal group. In fact, ${\mu}$ is not orientation-preserving, but ${\mu \oplus (-1) \in SO(5)}$ is, and consequently conjugation by it acts trivially on ${\pi_3(SO(5))}$.

Consequently, the vector bundles defined by ${(\mu \phi \mu): S^3 \rightarrow SO(4)}$ and ${\phi}$ are stably isomorphic. This means that the Pontryagin classes are the same. This is the last step in the computation of ${p_1}$. $\Box$

Let’s now put everything together.

Theorem 7 We have ${\widetilde{\lambda}(S(V_{ij})) = (i - j)^2 - 1}$. If ${i + j = 1}$ but ${i - j \not\equiv 1 \mod 7}$, then ${S(V_{ij})}$ is thus an exotic sphere.

Proof: Recall that

$\displaystyle \widetilde{\lambda}(S(V_{ij})) = 2 p_1^2[B] -\sigma(B)\mod 7.$

Here, take ${B = B(V_{ij})}$ as the interior manifold. We saw that ${\sigma(B) = 1}$. The first Pontryagin class of ${V_{ij}}$ is ${\pm 2(i - j)}$ times the generator of ${S^4}$, and consequently by (3), we have

$\displaystyle p_1^2[B] = 4 (i - j)^2.$

Consequently,

$\displaystyle \widetilde{\lambda}(S(V_{ij})) = (i-j)^2 - 1 \mod 7.$

$\Box$

Amazingly, the collection of manifolds homeomorphic to an $n$-sphere (at least for $n \neq 4$) forms a group (under connected sum), and the structure of this group can be determined in some cases: it is always finite, at least. This was a consequence of later work of Kervaire and Milnor.