I’ve been trying to fix the (many) gaps in my knowledge of classical algebraic topology as of late, and will probably do a few posts in the near future on vector bundles, K-theory, and characteristic classes.

Let ${B}$ be a base space, and let ${p: E \rightarrow B}$ be a real vector bundle. There are numerous constructions for the characteristic classes of ${B}$. Recall that these are elements in the cohomology ring ${H^*(B; R)}$ (for ${R}$ some ring) that measure, in some sense, the twisting or nontriviality of the bundle ${B}$.

Over a smooth manifold ${B}$, with ${E}$ a smooth vector bundle, a construction can be made in de Rham cohomology. Namely, one chooses a connection ${\nabla}$ on ${E}$, computes the curvature tensor of ${E}$ (which is an ${\hom(E,E)}$-valued 2-form ${\Theta}$ on ${B}$), and then applies a suitable polynomial from matrices to polynomials to the curvature ${\Theta}$. One can show that this gives closed forms, whose de Rham cohomology class does not depend on the choice of connection. This is the subject of Chern-Weil theory, and it applies more generally to principal ${G}$-bundles on a manifold for ${G}$ a Lie group.

But there is something that this approach misses: torsion. By working with de Rham cohomology (or equivalently, cohomology with ${\mathbb{R}}$-coefficients), the very interesting torsion phenomena that algebraic topologists care about is lost. For the purposes of this post, we’re interested in cohomology classes where the ground ring is ${R = \mathbb{Z}/2}$, and so de Rham cohomology is out. However, in return, we have cohomology operations. We can use them instead.

OK, so we want to construct the Stiefel-Whitney classes. I explained slightly less than a year ago what the properties we want of them are. Let me repeat them. We want a class ${\mathrm{Sw}(E) \in H^*(B, \mathbb{Z}/2)}$ for every vector bundle ${E \rightarrow B}$, which is multiplicative:

$\displaystyle \mathrm{Sw}(E \oplus E') = \mathrm{Sw}(E) \mathrm{Sw}(E').$

We also want a dimension vanishing condition: ${\mathrm{Sw}(E)}$ should vanish in degrees greater than the dimension of ${E}$; we next want that the zero-coefficient of ${\mathrm{Sw}(E)}$ should be one. Last, but not least, we want ${\mathrm{Sw}(E)}$ to be one on a trivial bundle, but ${\mathrm{Sw}}$ should not be identically one!

Let’s review the Thom isomorphism, discussed earlier. If ${p: E \rightarrow B}$ is a vector bundle of dimension ${n}$, then of course the cohomology of ${E}$ is the same as that of ${B}$. The cohomology of the space ${E_0 = E - B}$ (where ${B}$ is imbedded in ${E}$ via the zero section), or of the pair ${(E, E_0)}$ is more interesting.

The main theorem is that this is also just the cohomology of ${B}$, but shifted. When ${B = \ast}$, this is clear, because the pair is then ${(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$.

More generally:

Theorem 1 There is a unique fundamental class ${\Phi \in H^n(E, E_0; \mathbb{Z}/2)}$ which restricts to a generator of each fiber ${(\mathbb{R}^n, \mathbb{R}^{n} \setminus \left\{0\right\})}$. Then the map

$\displaystyle H^*(B; \mathbb{Z}/2) \rightarrow H^{* + n} (E, E_0; \mathbb{Z}/2)$

given by ${x \mapsto p^* x \cup \Phi}$, is an isomorphism.

I described the proof of this result earlier, but the idea is to prove it for a product ${E = B \times \mathbb{R}^n}$, and then use a Mayer-Vietoris argument to get in general.

The idea is that ${\Phi}$ is going to determine the Stiefel-Whitney classes of ${E}$. Note that this is a reasonable idea, because ${\Phi}$ isnatural. If ${B' \rightarrow B}$ is a map, then the fundamental class of ${E \times_B B' \rightarrow B'}$ is the pull-back of ${\Phi}$. This follows by the characterization that ${\Phi}$ restrict to a generator in each fiber.

Note also that instead of considering relative cohomology, we could have chosen a metric on ${E}$, and considered the disk bundle ${D(E) \subset E}$ and the sphere bundle ${S(E)}$ (consisting of vectors of norm at most one, and norm one). Since ${E}$ deformation retracts onto ${D(E)}$ and ${E_0}$ onto ${S(E)}$, we find that

$\displaystyle H^*(E, E_0; \mathbb{Z}/2) \simeq H^*(D(E), S(E); \mathbb{Z}/2) \simeq H^*(D(E)/S(E); \mathbb{Z}/2).$

The space ${D(E)/S(E)}$, which can be thought of as the one-point compactification of ${E}$ when ${B}$ is compact, is called the Thom space of ${E}$. The above theorem, equivalently, states that the cohomology of the Thom space is determined in terms of the space.

1. The (unoriented) Euler class

We can start by describing one invariant constructed from the vector bundle. Let’s just take the fundamental class ${\Phi \in H^n(E, E_0; \mathbb{Z}/2)}$, restrict to ${E}$, and pull back to ${B}$.

Definition 2 The pull-back to ${B}$ of the fundamental class is called the (unoriented) Euler class of ${E}$. It is an element of ${H^n(B; \mathbb{Z}/2)}$.

The usual Euler class is defined for an oriented vector bundle, and lives in ${H^n(B; \mathbb{Z})}$; for an oriented bundle, this is the reduction mod 2 of the usual one.

Let’s note that the Euler class is an interesting invariant.

Proposition 3 For a trivial bundle, the Euler class is zero.

In fact, then ${E = B \times \mathbb{R}^n}$, and we have under the Künneth isomorphism

$\displaystyle H^*(E, E_0; \mathbb{Z}/2) \simeq H^*(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\}, \mathbb{Z}/2) \otimes H^*(B; \mathbb{Z}/2).$

The Künneth isomorphism is given by taking the cup-product. So ${q: E \rightarrow \mathbb{R}^n}$ is a projection, then we can clearly obtain the fundamental class ${\Phi}$ of ${E}$ by pulling back the unique generator of ${H^n(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$. To get the Euler class, we pull this back to ${B}$.

In other words, if ${s: B \rightarrow E}$ is the inclusion of a zero section, and ${q: E \rightarrow \mathbb{R}^n}$ as above, then we pull back the element ${\iota \in H^n(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$ to ${B}$ via ${q \circ s}$; this, however, is a constant map.

Well, we have shown that the Euler class vanishes for trivial bundles. Now we want to show that it’s nontrivial in general.

Proposition 4 The Euler class for the tautological line bundle on ${\mathbb{RP}^1}$ is nonzero.

Proof: Here ${\mathbb{RP}^1}$ can be identified with a circle ${S^1}$, and the tautological bundle ${E}$ is the Möbius band. Instead of considering the pair ${(E, E_0)}$, we can consider the pair ${(D(E), S(E))}$ consisting of the {disk} bundle in ${E}$ and the {sphere bundle} in ${E}$. The disk bundle is a Möbius band, and the sphere bundle is a double cover of ${S^1}$, itself a copy of ${S^1}$.

Now if you take ${\mathbb{RP}^2}$ and remove a small disk, you get a Möbius band. We find that the cohomology of the pair ${(D(E), S(E))}$, by excision, is the reduced cohomology of ${(\mathbb{RP}^2, D^2)}$, or that of ${\mathbb{RP}^2}$. The fundamental class ${\Phi}$ must be the unique nonzero class in ${H^1}$, and the claim is that this restricted to ${\mathbb{RP}^1}$ is also nonzero.

But in the Thom isomorphism, whenever the fundamental class ${\Phi}$ doesn’t square to zero, then the restriction to the base ${B}$—the Euler class—is nonzero. Indeed, ${\Phi^2}$ corresponds under the Thom isomorphism to the restriction ${\Phi|_B}$, by unwinding the definition. So ${\Phi|_B \neq 0}$. $\Box$

2. Constructing the Stiefel-Whitney classes

We’ve seen that we can construct an interesting invariant of a vector bundle, the Euler class, directly from the fundamental class. Now we want to get more.

Let ${p: E \rightarrow B}$ be a vector bundle of dimension ${n}$. Let ${\Phi \in H^n(E, E_0; \mathbb{Z}/2)}$ be the fundamental class.

Definition 5 The Stiefel-Whitney class of ${E}$, ${\mathrm{Sw} E}$, is the inverse to ${\mathrm{Sq}\Phi}$ under the Thom isomorphism.

Here ${\mathrm{Sq}\Phi}$ is the (total) Steenrod squaring operation ${\mathrm{Sq} = \sum_i \mathrm{Sq}^i}$. So to obtain the ${i}$th Stiefel-Whitney class (i.e. the ${i}$th graded piece of the total one) of ${E}$, one takes the fundamental class ${\Phi}$, applies ${\mathrm{Sq}^i}$ to get something in ${H^{n+i}(E, E_0; \mathbb{Z}/2)}$, and then “de-cups” with ${\Phi}$. We have (where by abuse of notation ${\mathrm{Sw} E}$ is identified with its image in ${H^*(E)}$)

$\displaystyle \mathrm{Sw} E \cup \Phi = \mathrm{Sq} \Phi. \ \ \ \ \ (1)$

Because the Steenrod operations and the Thom isomorphism are natural, this is a completely natural construction.

Theorem 6 (Thom) The above defintion of ${\mathrm{Sw} E}$ satisfies the relevant properties.

Proof: There are various claims to check. First, we have to check that ${\mathrm{Sw}}$ of a trivial bundle is zero, but we have seen that the fundamental class ${\Phi}$ is not too interesting. Namely, if ${E}$ is trivial and ${q: E \simeq \mathbb{R}^n}$ a projection, and if ${\iota \in H^n(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$ a generator, then ${\Phi = q^* \iota}$. In particular, ${\mathrm{Sq} \Phi = \mathrm{Sq} q^* \iota = q^* \mathrm{Sq} \Phi}$. But on ${(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$, the Steenrod operations (apart from the ${\mathrm{Sq}^0 = \mathrm{Id}}$) are trivial, and so we have

$\displaystyle \mathrm{Sq} \Phi = \Phi.$

The defining equation (1) implies that ${\mathrm{Sw} E = 1}$.

Next, we have to check that the above construction is multiplicative. That is, if ${E = E' \oplus E'}$, then ${\mathrm{Sw} E = \mathrm{Sw} E' \mathrm{Sw} E'}$. We can see this by using the multiplicativity of fundamental classes. If ${\iota_m, \iota_n}$ are generators of ${H^m(\mathbb{R}^m, \mathbb{R}^m \setminus \left\{0\right\}), H^n(\mathbb{R}^n, \mathbb{R}^n \setminus \left\{0\right\})}$, then ${\iota_m \times \iota_n}$ is a generator of ${H^{m+n}(\mathbb{R}^{m+n}, \mathbb{R}^{m+n} \setminus \left\{0\right\})}$. This and the fiberwise criterion show that if ${\Phi_{E'}, \Phi_{E''}}$ are fundamental classes of ${E', E''}$, then we have the cross product identity

$\displaystyle \Phi_E = \Phi_{E'} \times \Phi_{E''} .$

This implies, by the properties of the Steenrod operations, that

$\displaystyle \mathrm{Sq} \Phi_E = \mathrm{Sq}\Phi_{E'} \times \mathrm{Sq} \Phi_{E''} = (\mathrm{Sw} E' \cup \Phi_{E'}) \times (\mathrm{Sw} E'' \times \Phi_{E''})$

which becomes

$\displaystyle (\mathrm{Sw} E \cup \mathrm{Sw} E'') \cup ( \Phi_{E'} \times \Phi_{E''}) = (\mathrm{Sw} E' \cup \mathrm{Sw} E'') \cup \Phi_E.$

But because the Thom isomorphism is an isomorphism, this gives

$\displaystyle \mathrm{Sw} E = \mathrm{Sw} E' \cup \mathrm{Sw} E''.$

Third, we need to check that the ${\mathrm{Sw} E}$‘s thus defined have the above dimensional properties, i.e. only live in dimensions up to the dimension of the bundle; this follows from the properties of the Steenrod squares, again. The top Steenrod square is just the square, which goes up to dimension ${2n}$, and de-Thom-isomorphisming it gets to dimension ${n}$ in ${H^*(B; \mathbb{Z}/2)}$.

Finally, we need to check that ${\mathrm{Sw}}$ of the canonical line bundle on ${\mathbb{RP}^1}$ is nontrivial. But we’ve seen that the Euler class of that is nonzero, and let’s observe that the top Stiefel-Whitney class is the Euler class. This is because the top Steenrod operation is the usual cup square, so if we take the top component ${\mathrm{Sw}_n E}$, we have

$\displaystyle \mathrm{Sw}_n E \cup \Phi = \Phi \cup \Phi,$

which shows that ${\mathrm{Sw}_n E}$ is the pull-back of the fundamental class to ${B}$. So, we’re done. $\Box$

One of the nice things about this construction of the Stiefel-Whitney classes is that one can apply facts about the Steenrod algebra; in particular, the following property becomes transparent.

Proposition 7 The smallest degree in which ${\mathrm{Sw} (E)}$ is nonzero (besides degree zero) is a power of two.

Proof: This follows because the indecomposable elements of the Steenrod algebras are the ${\mathrm{Sq}^{2^k}}$. Every ${\mathrm{Sq}^i}$ for ${i}$ not a power of two can be written as a polynomial in ${\mathrm{Sq}^j}$ for ${j; this is a consequence of the Adem relations. $\Box$

This argument using indecomposable elements in the Steenrod algebra shows in fact that not all commutative graded ${\mathbb{Z}/2}$-algebras can occur as cohomology rings of spaces. For instance, the algebra which is ${\mathbb{Z}/2}$ in dimension zero, ${\mathbb{Z}/2}$ in dimension ${n}$, and ${\mathbb{Z}/2}$ in dimension ${2n}$, with a generator in dimension ${n}$ squaring to the generator in dimension ${2n}$, can’t occur for ${n}$ not a power of two. This again follows because the square from dimension ${n}$ to dimension ${2n}$, which is the top Steenrod square, would otherwise have to pass through middle cohomology degrees—but everything is zero there. This argument shows in particular that there cannot be a map ${S^{2n-1} \rightarrow S^n}$ of odd Hopf invariant except when ${n}$ is a power of two. The stronger result of Adams that specifies exactly which powers of two are allowable requires much more work, though.

Next time, I will try to expand further on the relations between the Steenrod algebra and the Stiefel-Whitney classes. In particular, there is a formula of Wu that shows how the Steenrod algebra acts on these classes.