This quarter I’m teaching the “Differential Topology” first-year graduate class, and for a bit of fun, I decided to teach an introduction to characteristic classes, following the classic book of that name by Milnor and Stasheff. The book begins with a discussion of Stiefel-Whitney classes of real bundles, then talks about Euler classes, and then Chern classes of complex bundles, Pontriagin classes, the oriented and unoriented cobordism ring, and so on.

One often-lamented weakness of this otherwise excellent book is that Milnor does not really give much insight into the geometric “meaning” of the characteristic classes; for example, Stiefel-Whitney classes are introduced axiomatically, and then “constructed” by appealing to the axiomatic properties of Steenrod squares, applied to the Thom class. This makes it hard to get a geometric “feel” for these classes, especially in the important case of bundles over a manifold. So I thought it would be useful to give a “geometric” description of Stiefel-Whitney classes in this context (described via Poincaré duality as cycles in the manifold), which is at the same time elementary enough to give a feel, and at the same time is transparently related to the “geometric” definition of Steenrod squares, so that one can see how the two definitions compare.

Milnor’s treatment of Stiefel-Whitney classes is axiomatic. If is an bundle over a base space , the Stiefel-Whitney classes for are the unique classes which satisfy

- , and for ;
- the are natural; i.e. if and is an bundle over , then if denotes the pullback bundle over (i.e. the bundle whose fiber over each is equal to the fiber of over ) then for each ;
- the satisfy the
*Whitney Product Formula:*for bundles and over the same space , where and where the product in taken in the ring ; and - if is the twisted bundle over the circle then is nontrivial.

(For convenience, I’m going to suppress coefficients throughout the sequel.)

Uniqueness of classes satisfying these properties is established by a dimension count, after one shows that any natural characteristic class must be obtained by pulling back cohomology from a classifying map to a Grassmannian. Then Milnor shows existence via Thom’s formula involving Steenrod squares.

Explicitly, if denotes the total space of an bundle , and if denotes the complement of the zero section, there is a unique *Thom class* which restricts to the generator of for each fiber , and the Stiefel-Whitney classes are the unique classes in satisfying

where is projection to the base, and is the th Steenrod operation.

Well yes, exactly; clear enough if you are Thom or Milnor, but mysterious to the rest of us.

First of all, what are these Steenrod squares? They arise in a subtle way from the systematic failure of the commutativity of cup product on cohomology (with coefficients) to be represented by a commutative (and associative) product at the cochain level. Another way to say this is that they arise from the failure of cohomology classes to be represented by unique *maps* to classifying spaces, but rather to be represented only by unique *homotopy classes of maps*.

Let me explain. Let be a space and let be a class. This class is represented by a unique homotopy class of map . The external cross product class is likewise represented by a unique homotopy class of map

by pulling back a tautological class .

If acts by switching the two factors, the maps and both pull back to the same class, so they are homotopic. So there is a map which gives a homotopy from to . Likewise, we can define where switches the two factors, which gives a homotopy from to , and we can glue and together to give a map which factors through the action which switches the factors of , and acts as the antipodal map on .

But by obstruction theory, the map fills in (canonically) to a map

and we can glue and together to give

And so on by induction. In the end we obtain

which factors through the action which switches the factors of and acts as the antipodal map on . Let’s restrict to the diagonal and quotient out by to get a map

There is a ring isomorphism where has degree , and we can express the pullback of the class canonically as a polynomial in with coefficients in . This entire construction depended on the original class , so the coefficients we obtain are functions of , and these are exactly the Steenrod squares. I.e.:

for canonical classes .

This is a hell of a procedure to go through to get Stiefel-Whitney classes . So let me now explain how to “simulate” this construction geometrically to give a natural construction of Stiefel-Whitney cycles, at least in the case of a vector bundle over a manifold.

Let’s let be a closed manifold of dimension and let be a (smooth) bundle over with total space . The total space is a (noncompact) smooth manifold of dimension . We identify with a submanifold of by taking it to be the zero section, and note that this inclusion is a homotopy equivalence. Thus . It’s not hard to see that is isomorphic to the compactly supported cohomology of , so that there is a Poincaré duality isomorphism between and , and under this isomorphism, the Thom class is seen to be dual to the class represented by the zero section itself. Thus, cupping with is dual to intersection with . The Thom isomorphism is the composition with Poincaré duality in to identify with with Poincaré duality in to identify

What is ? It is Poincaré dual to a submanifold of which is Poincaré dual to the class . But the top Steenrod square is just (as can be seen from the construction above) so this is dual to the self-intersection where is obtained by perturbing so the two copies are in general position. Geometrically, we can take a generic section and let .

Okay, this identification is well-known: the top class is the obstruction to the existence of a nonzero section. Now what? We take our hint from the construction of Steenrod squares, and proceed as follows.

Since the section takes values pointwise in vector spaces, it makes sense to define the *antipodal* section by for each . The sections and are not equal, but at least they have the property that and are equal (ignoring signs, since we’re working over ). Now, of course any two sections of are homotopic, so we can choose a *generic* path of sections from to so that intersects in general position. Then define

Now, is a manifold with boundary, so we should expect this boundary to contribute a boundary to . But by construction, the two contributions to the boundary from the two points of are both equal to ; so the boundary “glues up” and we get a *closed* manifold, representing a homology class in Poincaré dual to .

And now proceed by induction. The two sections and glue up to give a circle of sections which can be filled in to a disk of sections and . And so on. Each intersection is a cycle because the boundary terms all glue up by the symmetry of the construction.

Notice once we get to that by general position and symmetry each maps over the point , so that which is Poincaré dual to .

Personally I find that this construction bears a nice “family resemblance” to one of the standard constructions of Steenrod squares, and removes some of the mystery from Thom’s theorem.

One nice application of this geometric interpretation of Stiefel-Whitney cycles is that it gives an elementary proof of a theorem of Halperin-Toledo (originally conjectured by Stiefel), that if is a smooth, triangulated manifold, then the th Stiefel-Whitney class of the tangent bundle is Poincaré dual to the union of simplices in the first barycentric subdivision of the triangulation. For this reduces to Hopf’s observation that the Euler characteristic is equal mod 2 to the number of simplices (summed over all dimensions). To see this, build (in the usual way) a section of the tangent bundle over each simplex singular exactly at the vertices of the first barycentric subdivision. Then build inductively families of homotopies between these sections and their negatives in an obvious way so that they agree on the boundaries. Directly one sees that is exactly the union of the -simplices in the first barycentric subdivision.

**Update 2/18/2016:** Rob Kirby emailed me to point out the following nice “homework exercise”. Consider the real 1-dimensional bundle over the circle whose total space is a Mobius band. We can choose a section which is transverse to the zero section at exactly one point. Now, if we choose a (metric) connection on the bundle, then it makes sense to talk about “translating” a section by parallel transporting it around some path in the base. As we translate around the path which winds once around the base circle, it takes exactly to the section , so this is a perfectly legitimate choice of homotopy . Under this homotopy, the zero section itself zips once around the circle, and sweeps out the fundamental class; said another way, is dual to a point (the zero of the original section ) and is dual to the entire circle (the “path” of zeros of the homotopy of sections from to ).

**Another Update 2/18/2016:** It is natural to wonder whether there is an analog of this construction for Chern classes, at least for complex vector bundles over closed smooth oriented manifolds . It seems that there is, and it is probably worth spelling out.

If is a (smooth) bundle over with total space , then we can also think of it as a real *oriented* bundle (with the same total space). The image of a generic section is an oriented submanifold of the oriented total space , so we can orient the intersection and think of it as an integral homology class dual to the top Chern class . In fact, this is also the Euler class of the underlying oriented real bundle.

But now there is a natural action on the total space, coming from the natural multiplication of a vector in a complex vector space by the scalar for . So the section determines a circle’s worth of sections which can be filled in with a disk’s worth of sections . We can define and observe that the “boundary” of this manifold is just the zeros of . But the action is trivial on the set of zeros of any section, so this is in turn just equal to . In other words, the image of the boundary has codimension 2, and therefore represents a well-defined homology class dual to .

At the next stage we get by multiplying by the action. But the restriction of this action to is just rotation (since that’s how we defined on the boundary) so it factors through where the equivalence relation on quotients out the curves on this torus to points. But now observe and we actually have which can be filled in generically to , and is a cycle (as before) dual to . And so on.

**Question:** What is the analog in this context of the Steenrod squares? It seems they should be replaced by cohomology operations in *integral cohomology*, defined now not for arbitrary spaces but for spaces with actions; i.e. (presumably) they are operations on -equivariant cohomology groups. Probably such operations, and their relation to Chern classes, are classical and well known, but not by me. Can any readers fill me in?

The notes http://www.maths.ed.ac.uk/~aar/euler.pdf use Kreck’s derivation of the Chern and Stiefel-Whitney classes from the Euler class to prove the (well-known) identification of the mod 2 reduction of a complex vector bundle with the Stiefel-Whitney classes of the underlying real vector bundle.

The previous reply was not meant to be anonymous! Andrew

Thanks Andrew! The notes are very informative. In particular, the use of the external product with the tautological line bundle (to get Chern classes) is pretty close to the Steenrod squares construction.

Worse: there was a typo in the original reply. It should have been the “mod 2 reduction of the Chern classes of a complex vector bundle”

What do you think of combining your blog entry and my notes into a short expository paper?

Dear Andrew – thanks for the kind offer. But I think this is just the sort of thing I like blogs for.

best, D

Fair enough. At some later stage I may combine my notes and your blog – with credit of course, as I think the geometric properties of Steenrod squares are well worth writing up for graduate students of all ages! Best, A.