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 be a base space, and let be a real vector bundle. There are numerous constructions for the *characteristic classes* of . Recall that these are elements in the cohomology ring (for some ring) that measure, in some sense, the twisting or nontriviality of the bundle .

Over a smooth manifold , with a smooth vector bundle, a construction can be made in de Rham cohomology. Namely, one chooses a connection on , computes the curvature tensor of (which is an -valued 2-form on ), and then applies a suitable polynomial from matrices to polynomials to the curvature . 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 -bundles on a manifold for a Lie group.

But there is something that this approach misses: torsion. By working with de Rham cohomology (or equivalently, cohomology with -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 , 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 for every vector bundle , which is multiplicative:

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

Let’s review the Thom isomorphism, discussed earlier. If is a vector bundle of dimension , then of course the cohomology of is the same as that of . The cohomology of the space (where is imbedded in via the zero section), or of the pair is more interesting.

The main theorem is that this is also just the cohomology of , but shifted. When , this is clear, because the pair is then .

More generally:

Theorem 1There is a unique fundamental class which restricts to a generator of each fiber . Then the map

given by , is an isomorphism.

I described the proof of this result earlier, but the idea is to prove it for a product , and then use a Mayer-Vietoris argument to get in general.

The idea is that is going to determine the Stiefel-Whitney classes of . Note that this is a reasonable idea, because is*natural*. If is a map, then the fundamental class of is the pull-back of . This follows by the characterization that restrict to a generator in each fiber.

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

The space , which can be thought of as the one-point compactification of when is compact, is called the **Thom space** of . 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 , restrict to , and pull back to .

Definition 2The pull-back to of the fundamental class is called the(unoriented) Euler classof . It is an element of .

The usual Euler class is defined for an oriented vector bundle, and lives in ; 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 3For atrivialbundle, the Euler class is zero.

In fact, then , and we have under the Künneth isomorphism

The Künneth isomorphism is given by taking the cup-product. So is a projection, then we can clearly obtain the fundamental class of by pulling back the unique generator of . To get the Euler class, we pull this back to .

In other words, if is the inclusion of a zero section, and as above, then we pull back the element to via ; 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 4The Euler class for the tautological line bundle on is nonzero.

*Proof:* Here can be identified with a circle , and the tautological bundle is the Möbius band. Instead of considering the pair , we can consider the pair consisting of the {disk} bundle in and the {sphere bundle} in . The disk bundle is a Möbius band, and the sphere bundle is a double cover of , itself a copy of .

Now if you take and remove a small disk, you get a Möbius band. We find that the cohomology of the pair , by excision, is the reduced cohomology of , or that of . The fundamental class must be the unique nonzero class in , and the claim is that this restricted to is also nonzero.

But in the Thom isomorphism, whenever the fundamental class doesn’t square to zero, then the restriction to the base —the Euler class—is nonzero. Indeed, corresponds under the Thom isomorphism to the restriction , by unwinding the definition. So .

**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 be a vector bundle of dimension . Let be the fundamental class.

Definition 5TheStiefel-Whitney classof , , is the inverse to under the Thom isomorphism.

Here is the (total) Steenrod squaring operation . So to obtain the th Stiefel-Whitney class (i.e. the th graded piece of the total one) of , one takes the fundamental class , applies to get something in , and then “de-cups” with . We have (where by abuse of notation is identified with its image in )

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

Theorem 6 (Thom)The above defintion of satisfies the relevant properties.

*Proof:* There are various claims to check. First, we have to check that of a trivial bundle is zero, but we have seen that the fundamental class is not too interesting. Namely, if is trivial and a projection, and if a generator, then . In particular, . But on , the Steenrod operations (apart from the ) are trivial, and so we have

The defining equation (1) implies that .

Next, we have to check that the above construction is multiplicative. That is, if , then . We can see this by using the multiplicativity of fundamental classes. If are generators of , then is a generator of . This and the fiberwise criterion show that if are fundamental classes of , then we have the cross product identity

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

which becomes

But because the Thom isomorphism is an isomorphism, this gives

Third, we need to check that the ‘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 , and de-Thom-isomorphisming it gets to dimension in .

Finally, we need to check that of the canonical line bundle on 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 , we have

which shows that is the pull-back of the fundamental class to . So, we’re done.

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 7The smallest degree in which is nonzero (besides degree zero) is a power of two.

*Proof:* This follows because the *indecomposable* elements of the Steenrod algebras are the . Every for not a power of two can be written as a polynomial in for ; this is a consequence of the Adem relations.

This argument using indecomposable elements in the Steenrod algebra shows in fact that not all commutative graded -algebras can occur as cohomology rings of spaces. For instance, the algebra which is in dimension zero, in dimension , and in dimension , with a generator in dimension squaring to the generator in dimension , can’t occur for not a power of two. This again follows because the square from dimension to dimension , 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 of odd Hopf invariant except when 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.

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