Today I would like to take a break from the index theorem, and blog about a result of Wu, that the Stiefel-Whitney classes of a compact manifold (i.e. those of the tangent bundle) are homotopy invariant. It is not even a priori obvious that the Stiefel-Whitney classes are homeomorphism invariant; note that “homeomorphic” is a strictly weaker relation than “diffeomorphic” for compact manifolds, a result first due to Milnor. But in fact the argument shows even that the Stiefel-Whitney classes (of the tangent bundle) can be worked out solely in terms of the structure of the cohomology ring as a module over the Steenrod algebra.

Here is the idea. When $A \subset M$ is a closed submanifold of a manifold, there is a lower shriek (Gysin) homomorphism from the cohomology of $A$ to that of $M$; this is Poincaré dual to the restriction map in the other direction. We will see that the “fundamental class” of $A$ (that is,  the image of 1 under this lower shriek map) corresponds to the mod 2 Euler (or top Stiefel-Whitney) class of the normal bundle. In the case of $M \subset M \times M$, the corresponding normal bundle is just the tangent bundle of $M$. But by other means we’ll be able to work out the Gysin map easily. Once we have this, the Steenrod operations determine the rest of the Stiefel-Whitney classes.

The Gysin map

Let ${M}$ be a compact manifold. All cohomology will be with ${\mathbb{Z}/2}$-coefficients.

Given a closed submanifold ${A}$ with an inclusion ${i: A \hookrightarrow M}$, there is a Gysin map ${i_!: H^*(A) \rightarrow H^{* + c}(M)}$ where ${c}$ is the codimension. This can be defined as the Poincaré dual to the map ${H^*(M) \rightarrow H^*(A)}$ given by restriction.

Definition 1 The fundamental class of ${A}$ is the ${i_!(1) \in H^c(M)}$.

Alternatively, ${i_!(1)}$ is Poincaré dual to the image ${i_* [A]}$ of the “fundamental class” ${[A] \in H_{\dim A}(A)}$.

We are going to determine the pull-back ${i^* i_!(1)}$ of the fundamental class. The claim is that it is going to be the top Stiefel-Whitney class of the normal bundle ${N \rightarrow A}$ of ${A \subset M}$. When we apply this to the case of a manifold sitting inside its square (via the diagonal embedding), we will get a formula that will imply Wu’s theorem.

Namely, let us start by considering ${H^*(M, M-A)}$, which maps to ${H^*(M)}$. We know that there is a neighborhood of ${A}$ in ${M}$ isomorphic to the normal bundle ${N}$, so by excision we have an isomorphism

$\displaystyle H^*(M, M-A) \simeq H^*(N, N-A)$

where ${A}$ is imbedded in ${N}$ via the zero-section. However, by the Thom isomorphism, we also know that ${H^*(N, N-A)}$ (which is in fact the reduced cohomology of the Thom space of ${N}$), is a free module over ${H^*(A)}$, generated by a class ${e \in H^c(N, N-A)}$ (as ${c = \dim N}$). If we were working over ${\mathbb{Z}}$ with oriented bundles, this would be the Euler class.

Proposition 2 The class ${e \in H^c(N, N-A)}$ maps to the class ${i_!(1) \in H^c(M)}$ under restriction ${H^*(N, N-A) \simeq H^*(M, M-A) \rightarrow H^*(M)}$.

To prove this, we will work out what the Gysin map should be for an imbedding of a manifold in the zero section of a vector bundle.

There is still a Gysin map for an imbedding

$\displaystyle j: A \hookrightarrow Y$

where ${Y}$ is noncompact, except here we have to use compactly supported cohomology to apply Poincaré duality. The map goes ${H^*_{cp}(A) \rightarrow H^{*+c}_{cp}(Y)}$ where ${c}$ is the codimension and ${cp}$ indicates that compact supports are used. Suppose now that ${Y}$ is the total space of a vector bundle over ${A}$, and that in particular implies ${H^{*+c}_{cp}(Y) = H^*( Y, Y - A)}$. Then the claim is that the Gysin map is just multiplication by the class ${e(Y)}$ (the top Stiefel-Whitney or “Euler mod 2” class). To be precise, we mean that there is a class in ${H^c(Y, Y-A)}$, given by the Thom isomorphism theorem, such that multiplication by it is the Gysin map; this class becomes the “Euler” class when restricted to ${H^*(Y) = H^*(A)}$.

In fact, what we have to do is to check that multiplication by ${e}$, which induces a map

$\displaystyle H^*(A) \rightarrow H^{*+c}(Y, Y-A) = H^{*+c}_{cp}(Y)$

is Poincaré dual to restriction ${H^*(Y) \rightarrow H^*(A)}$. The last map is an isomorphism. So we have to check the following. Let ${\pi: Y \rightarrow A}$ be the projection and ${i: A \rightarrow Y}$ the inclusion of the zero section. We have to check that if ${\iota \in H^*(A)}$ is a cohomology class, then for any ${y \in H^*(A)}$, we have a correspondence between

$\displaystyle \iota y \in H^{\dim A}(A) , \quad \text{and} \quad \pi^* \iota e y \in H^{\dim A + c}_{cp}(Y).$

where to say that they “correspond” just means that they are either both nonzero or both zero (by Poincaré duality, we know that the top cohomology is ${\mathbb{Z}/2}$). But this is just the Thom isomorphism theorem, which states that multiplication by ${e}$ induces an isomorphism

$\displaystyle H^{\dim A}(A) \simeq H^{\dim A + c}_{cp}(Y).$

This is the result.

In other words, we have also proved:

Proposition 3 The Gysin map for the inclusion ${A \hookrightarrow Y}$ of a manifold inside the zero section of a vector bundle ${\pi: Y \rightarrow A}$ is given by multiplication by the mod 2 Euler class or top Stiefel-Whitney class of the normal bundle. In particular, it is an isomorphism.

We can also use the above result, together with the tubular neighborhood theorem, to get something weaker for the inclusion ${A \hookrightarrow X}$ for a compact manifold ${X}$. Let ${U = X-A}$, so we can write ${X = A \mathrm{Sq}cup U}$.

Corollary 4 (Gysin sequence) There is a long exact sequence

$\displaystyle \dots \rightarrow H^*(M) \rightarrow H^*(U) \rightarrow H^{*-c}(A) \rightarrow H^{*+1}(M) \rightarrow \dots$

In fact, we need only consider the long exact sequence

$\displaystyle \dots \rightarrow H^*(M) \rightarrow H^*(M-A) \rightarrow H^{*+1}(M, M-A) \rightarrow H^{*+1}(M) \rightarrow \dots$

The first map is restriction, and the second map is the coboundary. If we use the isomorphism between ${H^{* - c}(A)}$ and ${H^*(M, M-A)}$ (the Thom isomorphism), then we are done.

In any event, we see that if ${i: A \hookrightarrow X}$ is a closed imbedding, then ${i^* i_!(1)}$ is the mod 2 Euler class of the normal bundle.

The diagonal map

Let ${M}$ be a compact manifold of dimension ${n}$, and consider the inclusion

$\displaystyle \Delta: M \rightarrow M \times M.$

We are interested in understanding what ${\Delta_!(1)}$ is. On the one hand, we know that the normal bundle of ${M}$ imbedded in ${M \times M}$ is just the tangent bundle ${TM}$ itself. So we can say that ${\Delta^*\Delta_!(1)}$ is going to have to correspond to the mod 2 Euler class of ${TM}$.

But we will now step away momentarily from characteristic classes. The goal is:

Proposition 5 Let ${\left\{b_i\right\}}$ be a basis for ${H^*(M)}$ and let ${\left\{\beta_i\right\}}$ be the dual basis. Then we have:

$\displaystyle \Delta_!(1) = \sum b_i \times \beta_i \in H^*(M \times M) = H^*(M) \otimes H^*(M).$

In fact, we know that ${\Delta_!(1)}$ has the following property. For any cohomology class ${\gamma \times \delta \in H^*(M \times M) = H^*(M) \otimes H^*(M)}$, we have that

$\displaystyle (\gamma \times \delta \cup \Delta_!(1), [M] \times [M]) = (\Delta^*(\gamma \times \delta), [M])$

where ${[M]}$ is the fundamental homology class. But this is ${(\gamma \delta, [M])}$. This is the definition of ${\Delta_!(1)}$ as being Poincaré dual to the map ${H^*(M \times M) \rightarrow \mathbb{Z}/2}$ given by pulling back to ${M}$ (via ${\Delta^*}$ and) “integrating.”

Now, by assumption, we can express ${\Delta_!(1) = \sum c_{ij} b_i \times \beta_j}$ for some ${c_{ij} \in \mathbb{Z}/2}$. We just need to compute the ${c_{ij}}$ as the Kronecker delta. In fact, fix ${i_0, j_0}$, and consider ${\gamma = \beta_{i_0}, \delta = b_{j_0}}$. Then

$\displaystyle \gamma \times \delta \cup \Delta_!(1) = \sum_{ij} c_{ij} (\beta_{i_0} \times b_{j_0}) \cup ( b_{i} \cup \beta_j).$

If we evaluate this on the homology fundamental class ${[M] \times [M]}$, we get ${\sum c_{ij} \delta_{i_0 i} \delta_{j_0 j} = c_{i_0 j_0}}$. On the contrary, if we compute ${(\gamma \delta, [M])}$, we get ${\delta_{i_0 j_0}}$. As a result, we find as claimed

$\displaystyle c_{i_0 j_0} = \delta_{i_0 j_0}.$

This proves the claim about ${\Delta_!(1)}$.

In particular, we note that ${\Delta_!(1)}$ is a homotopy invariant of ${M}$, in that it is determined in terms of the cohomology ring. If we compute ${\Delta^* \Delta_!(1) \in H^n(M)}$, we recall that this is the top Stiefel-Whitney class of the normal bundle of ${\Delta}$, or equivalently ${TM}$. Thus:

Proposition 6${w_n(TM)}$ is a homotopy invariant of ${M}$ (and in fact is determined solely in terms of the cohomology ring of ${M}$).

This is not terribly surprising, since $w_n$ is the mod 2 Euler characteristic.

The other Stiefel-Whitney classes

Next, the claim is that all the Stiefel-Whitney classes of ${M}$ are determined in terms of the cohomology ring of ${M}$, with the action of the Steenrod squares. In fact, let us recall how the Stiefel-Whitney classes can be constructed for the vector bundle ${TM \rightarrow M}$. Namely, consider the generator in ${H^n(TM, TM - 0)}$, and apply the total squaring operation ${\mathrm{Sq}}$ to it; then invert the Thom isomorphism to pull back to ${H^*(M)}$. This will be the total Stiefel-Whitney class.

So, we have that ${H^*(TM, TM - 0) \simeq H^*(M \times M, M \times M - \Delta)}$, and the claim is that the inverse Thom isomorphism

$\displaystyle H^*(TM, TM - 0) \simeq H^*(M)$

can be realized in another way.

Proposition 7 The inverse to the Thom isomorphism, which is a map ${H^*(TM, TM - 0) \simeq H^*(M)}$, is given by restricting to ${H^*(M \times M)}$ and then the slant product with ${[M] \in H_n(M)}$.

In fact, if we look at the description of ${\Delta_!(1)}$, we see that the slant product of this with ${[M]}$ is definitely ${1}$. Both are homomorphisms of ${H^*(M)}$ modules, so we are done.

In other words, if we want to compute the Stiefel-Whitney classes of ${TM}$, all we need to do is to take ${\Delta_!(1) \in H^n(M \times M)}$, apply ${\mathrm{Sq}}$ to that, and then slant product with ${[M]}$. We write this as:

Proposition 8 The Stiefel-Whitney class ${w(TM) = (\mathrm{Sq} \Delta_!(1)) / [M]}$.

In particular:

Corollary 9 (Wu) The Stiefel-Whitney class ${w( TM)}$ (and thus the Stiefel-Whitney numbers) is a homotopy invariant of ${M}$.

This is because we have seen ${\Delta_!(1)}$ is a homotopy invariant of ${M}$.

Incidentally, a deep result in algebraic topology due to Thom is that the Stiefel-Whitney numbers of a manifold determine the unoriented cobordism class. In particular, we find:

Corollary 10 Two manifolds which are homotopy equivalent are unoriented cobordant.

We can also get another formula for the class ${\mathrm{Sq} TM \in H^*(M)}$.

Definition 11 If ${M}$ is a compact manifold, consider the map ${H^*(M) \rightarrow \mathbb{Z}/2}$ given by ${s \mapsto (\mathrm{Sq} s, [M])}$. This defines by duality an element ${v \in H^*(M)}$, which is called the Wu class.

The main formula is:

Theorem 12 (Wu) The Stiefel-Whitney class ${w(TM)}$ is ${ \mathrm{Sq} v \in H^*(M)}$.

To see this, note that what we have is

$\displaystyle w(TM) = \mathrm{Sq} \Delta_!(1) / [M] = \sum (\mathrm{Sq} b_i \times \mathrm{Sq} \beta_i )/[M].$

This is equivalently ${\sum \mathrm{Sq} b_i ( \mathrm{Sq} \beta_i, [M])}$ where the last thing is the pairing. However, ${\sum b_i (\mathrm{Sq} \beta_i, [M])}$ is easily seen to be ${v}$, so we are done.

It is sort of surprising that things are very different when one works with Pontryagin classes. There, it is true that certain polynomials in the Pontryagin classes are homotopy invariant; these polynomials are those given by the Hirzebruch signature theorem. However, those are the only Pontryagin numbers which are oriented homotopy type invariants.