Stiefel-Whitney cycles as intersections

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 \xi is an \mathbb{R}^n bundle over a base space B, the Stiefel-Whitney classes w_i(\xi) \in H^i(B;\mathbb{Z}/2) for i=0,1,\cdots are the unique classes which satisfy

  1. w_0(\xi) = 1, and w_m(\xi)=0 for m>n;
  2. the w_i are natural; i.e. if f:B \to B' and \xi is an \mathbb{R}^n bundle over B', then if f^*\xi denotes the pullback bundle over B (i.e. the bundle whose fiber over each b \in B is equal to the fiber of \xi over f(b)) then f^*w_i(\xi) = w_i(f^*\xi) for each i;
  3. the w_i(\xi) satisfy the Whitney Product Formula: w(\xi \oplus \eta) = w(\xi)w(\eta) for bundles \xi and \eta over the same space B, where w(\xi):=\sum_i w_i(\xi) and where the product in taken in the ring H^*(B;\mathbb{Z}/2); and
  4. if \gamma^1 is the twisted \mathbb{R} bundle over the circle P^1 then w_1(\gamma^1) is nontrivial.

(For convenience, I’m going to suppress \mathbb{Z}/2 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 E denotes the total space of an \mathbb{R}^n bundle \xi, and if \dot{E} denotes the complement of the zero section, there is a unique Thom class u \in H^n(E,\dot{E}) which restricts to the generator of H^n(F,\dot{F}) for each fiber F, and the Stiefel-Whitney classes w_i(\xi) are the unique classes in H^i(B) satisfying

\pi^*w_i(\xi) \cup u = Sq^i u

where \pi:E \to B is projection to the base, and Sq^i is the ith 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 \mathbb{Z}/2 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 X be a space and let a \in H^n(X) be a class. This class is represented by a unique homotopy class of map X \to K(\mathbb{Z}/2,n). The external cross product class a \times a \in H^{2n}(X \times X) is likewise represented by a unique homotopy class of map

f:X \times X \to K(\mathbb{Z}/2,2n)

by pulling back a tautological class \iota_{2n} \in H^{2n}(K(\mathbb{Z}/2,2n);\mathbb{Z}/2).

If T:X \times X \to X \times X acts by switching the two factors, the maps f and fT both pull back \iota_{2n} to the same class, so they are homotopic. So there is a map g_1: X\times X \times D^1 \to K(\mathbb{Z}/2,2n) which gives a homotopy from f to fT. Likewise, we can define g_1 T where T switches the two factors, which gives a homotopy from fT to f, and we can glue g_1 and g_1 T together to give a map f_1:X \times X \times S^1 \to K(\mathbb{Z}/2,2n) which factors through the \mathbb{Z}/2 action which switches the factors of X\times X, and acts as the antipodal map on S^1.

But by obstruction theory, the map f_1 fills in (canonically) to a map

g_2:X \times X \times D^2 \to K(\mathbb{Z}/2,2n)

and we can glue g_2 and g_2 T together to give

f_2:X \times X \times S^2 \to K(\mathbb{Z}/2,2n)

And so on by induction. In the end we obtain

f_\infty: X \times X \times S^\infty \to K(\mathbb{Z}/2,2n)

which factors through the \mathbb{Z}/2 action which switches the factors of X\times X and acts as the antipodal map on S^\infty. Let’s restrict to the diagonal \Delta(X) \subset X\times X and quotient out by \mathbb{Z}/2 to get a map

d:X \times \mathbb{R}P^\infty \to K(\mathbb{Z}/2,2n)

There is a ring isomorphism H^*(X\times \mathbb{R}P^\infty) = H^*(X)[t] where t has degree 1, and we can express the pullback of the class \iota_{2n} canonically as a polynomial in t with coefficients in H^*(X). This entire construction depended on the original class a, so the coefficients we obtain are functions of a, and these are exactly the Steenrod squares. I.e.:

d^*\iota_{2n} = \sum_{i=0}^n Sq^{n-i}(a)t^i

for canonical classes Sq^j(a) \in H^{n+j}(X).

This is a hell of a procedure to go through to get Stiefel-Whitney classes w_i(\xi). 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 M be a closed manifold of dimension m and let \xi be a (smooth) \mathbb{R}^n bundle over M with total space E. The total space E is a (noncompact) smooth manifold of dimension m+n. We identify M with a submanifold of E by taking it to be the zero section, and note that this inclusion is a homotopy equivalence. Thus \dot{E}=E-M. It’s not hard to see that H^*(E,\dot{E}) is isomorphic to the compactly supported cohomology of E, so that there is a Poincaré duality isomorphism between H^n(E,\dot{E}) and H_m(E)=H_m(M), and under this isomorphism, the Thom class u is seen to be dual to the class [M] represented by the zero section itself. Thus, cupping with u is dual to intersection with M. The Thom isomorphism H^*(M) \to H^{*+n}(E,\dot{E}) is the composition with Poincaré duality in M to identify H^*(M) with H_{m-*}(M) = H_{m-*}(E) with Poincaré duality in E to identify

H_{m-*}(E) = H^{n+*}_c(E) = H^{n+*}(E,\dot{E})

What is w_n(\xi)? It is Poincaré dual to a submanifold W_{m-n} of M which is Poincaré dual to the class Sq^n u \in H^{2n}(E,\dot{E}). But the top Steenrod square is just u \cup u (as can be seen from the construction above) so this is dual to the self-intersection M \cap M' where M' is obtained by perturbing M so the two copies are in general position. Geometrically, we can take a generic section s:M \to E and let W_{m-n} = M \cap s(M).

Okay, this identification is well-known: the top class is the \mathbb{Z}/2 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 s takes values pointwise in vector spaces, it makes sense to define the antipodal section -s by (-s)(x)=-(s(x)) for each x \in M. The sections s(M) and -s(M) are not equal, but at least they have the property that M \cap s(M) and M \cap -s(M) are equal (ignoring signs, since we’re working over \mathbb{Z}/2). Now, of course any two sections of E are homotopic, so we can choose a generic path of sections s_1:M \times D^1 \to E from s to -s so that s_1(M\times D^1) intersects M in general position. Then define

W_{m-n+1} = M \cap s_1(M\times D^1)

Now, M \times D^1 is a manifold with boundary, so we should expect this boundary to contribute a boundary to W_{m-n+1}. But by construction, the two contributions to the boundary from the two points of \partial D^1 are both equal to W_{m-n}; so the boundary “glues up” and we get a closed manifold, representing a homology class in H_{m-n+1}(M) Poincaré dual to w_{n-1}(\xi).

And now proceed by induction. The two sections s_1 and -s_1 glue up to give a circle of sections M\times S^1 \to E which can be filled in to a disk of sections s_2: M\times D^2 \to E and W_{m-n+2} = M \cap s_2(M\times D^2). 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 s_n:M \times D^n \to E that by general position and symmetry each x \times D^n maps over the point 0, so that W_{m} = M which is Poincaré dual to w_0(\xi) = 1.

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 M is a smooth, triangulated manifold, then the (n-i)th Stiefel-Whitney class of the tangent bundle is Poincaré dual to the union of i simplices in the first barycentric subdivision of the triangulation. For i=0 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 W_{m-n+i} is exactly the union of the i-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 s_0 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 s_0 around the path which winds once around the base circle, it takes s_0 exactly to the section -s_0, so this is a perfectly legitimate choice of homotopy s_1. Under this homotopy, the zero section itself zips once around the circle, and sweeps out the fundamental class; said another way, w_1 is dual to a point (the zero of the original section s_0) and w_0 is dual to the entire circle (the “path” of zeros of the homotopy of sections from s_0 to -s_0).

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 M. It seems that there is, and it is probably worth spelling out.

If \xi is a (smooth) \mathbb{C}^n bundle over M with total space E, then we can also think of it as a real oriented \mathbb{R}^{2n} bundle \xi_{\mathbb{R}} (with the same total space). The image of a generic section s_0(M) is an oriented submanifold of the oriented total space E, so we can orient the intersection C_n:=M \cap s_0(M) and think of it as an integral homology class [C_n] \in H_{m-2n}(M;\mathbb{Z}) dual to the top Chern class c_n(\xi) \in H^{2n}(M;\mathbb{Z}). In fact, this is also the Euler class e_{2n}(\xi_{\mathbb{R}}) of the underlying oriented real bundle.

But now there is a natural S^1 action on the total space, coming from the natural multiplication of a vector in a complex vector space by the scalar e^{i\theta} for \theta \in [0,2\pi]. So the section s_0 determines a circle’s worth of sections \hat{s}_0:M \times S^1 \to E which can be filled in with a disk’s worth of sections s_1:M \times D^2 \to E. We can define C_{n-1}:=M \cap s_1(M\times D^2) and observe that the “boundary” of this manifold is just the zeros of \hat{s}_0:M \times S^1 \to E. But the S^1 action is trivial on the set of zeros of any section, so this is in turn just equal to M \cap s_0(M) = C_n. In other words, the image of the boundary has codimension 2, and therefore C_{n-1} represents a well-defined homology class [C_{n-1}] \in H_{m-2n+2}(M;\mathbb{Z}) dual to c_{n-1}(\xi) \in H^{2n-2}(M;\mathbb{Z}).

At the next stage we get \hat{s}_1:M \times D^2 \times S^1 \to E by multiplying s_1 by the S^1 action. But the restriction of this action to S^1 = \partial D^2 is just rotation (since that’s how we defined s_1 on the boundary) so it factors through \hat{s}_1:M \times (D^2 \times S^1/\sim) \to E where the equivalence relation \sim on \partial D^2 \times S^1 = S^1 \times S^1 quotients out the (1,-1) curves on this torus to points. But now observe (D^2 \times S^1/\sim) = S^3 and we actually have \hat{s}_1:M \times S^3 \to E which can be filled in generically to s_2:M \times D^4 \to E, and C_{n-2}:=M \cap s_2(M \times D^4) is a cycle (as before) dual to c_{n-2}(\xi) \in H^{2n-4}(M;\mathbb{Z}). 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 S^1 actions; i.e. (presumably) they are operations on S^1-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?

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7 Responses to Stiefel-Whitney cycles as intersections

  1. Anonymous says:

    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.

  2. ranicki says:

    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”

  3. ranicki says:

    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

      • ranicki says:

        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.

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