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

$w_0(\xi) = 1$ , and $w_m(\xi)=0$ for $m>n$ ;
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 $i$ ;
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
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 $i$ 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 $\mathbb{Z}/2$ 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?

9 Responses to Stiefel-Whitney cycles as intersections

Anonymous says:

February 25, 2016 at 1:48 am

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.

- ranicki says:
  
  February 25, 2016 at 2:00 am
  
  The previous reply was not meant to be anonymous! Andrew
  
  - Danny Calegari says:
    
    February 25, 2016 at 5:18 am
    
    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.
ranicki says:

February 25, 2016 at 2:02 am

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”

ranicki says:

February 25, 2016 at 3:17 pm

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

- Danny Calegari says:
  
  February 26, 2016 at 5:40 am
  
  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:
    
    February 26, 2016 at 6:07 am
    
    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.
Anonymous says:

July 6, 2021 at 8:42 am

I think that the Chern case is essentially the top of Quillen’s paper “Elementary Proofs of Some Results of Cobordism Theory Using Steenrod Operations”. It seems needs cobordism to have a sharp connection between operations and Chern classes, but Quillen’s approach is geometric.

Anonymous says:

July 6, 2021 at 8:43 am

(This was a comment from Dev Sinha.)

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