Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There is a function on Z/2Z-cohomology called Steenrod squaring: Sqi:H^k(X,Z/2Z) --> Hk+i(X,Z/2Z). (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if a∈Hk(X), then Sqk(a)=a∪a∈H2k(X) (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle E, the ith Stiefel-Whitney class is given by wi(E)=φ-1∘Sqi∘φ(1), where φ is the Thom isomorphism.

I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that Sq1 corresponds to the "Bockstein homomorphism" of the exact sequence 0-->Z/2Z-->Z/4Z-->Z/2Z-->0. Explicitly, if we denote by C the chain group of the space X, we apply the exact covariant functor Hom(C,-) to this short exact sequence, take cohomology, then the connecting homomorphisms Hi(X)-->Hi(X) are exactly Sq1. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?

share|improve this question
add comment

7 Answers

up vote 13 down vote accepted

Here's one way to understand them. The external cup square $a \otimes a \in H^{2n}(X \times X)$ of $a \in H^n(X)$ induces a map $f:X \times X \to K(Z_2, 2n)$. It can be show that this map factors through a map $g:(X \times X) \times_{Z_2} EZ_2 \to K(2n)$, where $Z_2$ acts on the product by permuting the factors and $EZ_2$ can be taken to just be $S^\infty$. If you unravel what this means, it says that our original map $f$ was homotopic to the map obtained by first switching the coordinates and then applying $f$. It also says that this homotopy, when applied twice to get a homotopy from $f$ to itself, is homotopic to the identity homotopy, and we similarly have a whole series higher "coherence" homotopies. Now $X \times BZ_2$ maps to $(X \times X) \times_{Z_2} EZ_2$ as the diagonal, so we get a map $X \times BZ2 \to K(2n)$. But $BZ_2$'s cohomology is just $Z_2[t]$, so this gives a cohomology class $Sq(a) \in H^*(X)[t]$ of degree $2n$. If we write $Sq(a)=\sum s(i) t^i$, it can be shown that $s(i)=Sq^{n-i}a$.

What does this mean? Well, if our map $f$ actually was invariant under switching the factors (which you might think it ought to be, given that it appears to be defined symmetrically in the two factors), we could take $g$ to be just the projection onto $X \times X$ followed by $f$. This would mean that $Sq(a)$ comes from just projecting away the $BZ_2$ and then using $a^2$, i.e. $Sq^n(a)=a^2$ and $Sq^i(a)=0$ for all other $i$. Thus the nonvanishing of the lower Steenrod squares somehow measures how the cup product, while homotopy-commutative (in terms of the induced maps to Eilenberg-MacLane spaces), cannot be straightened to be actually commutative. Indeed, in the universal example $X=K(Z_2,n)$, the map $f$ is exactly the universal map representing the cup product of two cohomology classes of degree $n$.

Some somewhat terse notes on this can be found here; see particularly part III.

share|improve this answer
3  
This is great. I return to it every few months and I find that I understand more each time. For everyone's convenience, I've texed this up (with minimal changes), and the result is available at math.berkeley.edu/~aaron/xkcd/… –  Aaron Mazel-Gee Sep 29 '10 at 4:46
add comment

Here's how I explain Steenrod squares to geometers. First, if $X$ is a manifold of dimension $d$ then one can produce classes in $H^n(X)$ by proper maps $f: V \to X$ where $V$ is a manifold of dimension $d-n$ through many possible formalisms - eg. intersection theory (the value on a transverse $i$-cycle is the count of intersection points), or using the fundamental class in locally finite homology and duality, or Thom classes, or as the pushforward $ f_*(1) $ where $1$ is the unit class in $H^0(V)$. Taking this last approach, suppose $f$ is an immersion and thus has a normal bundle $\nu$. If $x = f_*(1) \in H^n(X)$ then $Sq^i(x) = f_*(w_i(\nu))$. This is essentially the Wu formula.

That is, if cohomology classes are represented by submanifolds, and for example cup product reflects intersection data, then Steenrod squares remember normal bundle data.

share|improve this answer
    
that's really cool, do you have a reference? –  Sean Tilson Apr 14 '10 at 12:03
1  
The original reference for this is René Thom's paper "Variétés plongées et i-carrés". Myself and Peter Eccles have proved a generalisation arxiv.org/abs/math/0509213 –  Mark Grant Nov 1 '10 at 11:03
    
please, could you define what is $w_i$ and $v$ in the formula $Sq^i(x)=f_*(w_i(v))$? Thanks a lot in advance –  Quetzalcube Jun 14 '11 at 14:10
    
$v$ is the normal bundle sorry ( i don't how to érase the comment) but still: what is $w_i$? thanks –  Quetzalcube Jun 14 '11 at 14:15
2  
$w_i$ is the ith Stiefel-Whitney class. –  Dev Sinha Jun 17 '11 at 23:20
add comment

You can understand the squares purely algebraically: let $R = F_2[x_1, x_2, \ldots, x_n]$ be a polynomial ring over the field $F_2$ of $2$ elements. A ring homomorphism $f$ on $R$ is completely determined by the values of $f(x_i)$, and those values are unrestricted, right? So define $f \colon R \to R$ by $f(x) = x + x^2$ for each $x = x_i$. (Since $\operatorname{char}(F)=2$, we actually have $f(x) = x + x^2$ for each $F$-linear combination of the $x_i$, too, i.e. this is a basis-free -- "gentlemanly"? -- definition.) Then for any homogeneous polynomial $p$ in $R$, we have $f(p) = p + p^2 + \text{other stuff}$; separating the components of $f(p)$ by degree, we can write $f(p) = \sum Sq^i ( p )$. E.g. $f(x_1 x_2) = (x_1 + x_1^2) \cdot (x_2 + x_2^2) = (x_1 x_2) + (x_1^2 x_1 + x_2^2 x_1) + (x_1 x_2)^2$, so $Sq^1( x_1 x_2 ) = x_1 x_2 (x_1 + x_2)$.

You can extend the definition of the $Sq^i$ to non-homogeneous polynomials by additivity if you like. This definition of the $Sq^i$ is consistent for polynomials in arbitrary numbers of variables (i.e. there are commutative diagrams involving the inclusions $F_2[x_1] \to F_2[x_1,x_2]$ etc.) So you can visualize the whole Steenrod algebra as an algebra of endomorphisms on an infinite polynomial ring.

This definition is sufficient to check the Adem relations.

Topologically, this mechanism defines $Sq^i$ on $H^\ast( (\mathbb{RP}^\infty)^n, Z/2Z)$. Then by naturality it defines the $Sq^i$ on cohomology classes that are pullbacks from those rings, i.e. you have a description of how $Sq^i$ acts on cohomology classes defined by vector bundles. This really is how I think of them when I play with classifying spaces $BG$.

share|improve this answer
    
I like this approach very much. And I understand it gives a way to define the Steenrod squares for SW classes of any vector bundle. But can one define in this way the Steenrod squares for any cohomology classes? –  András Szűcs Jul 18 '13 at 6:49
add comment

For the Steenrod squares, I highly recommend the first couple of chapters of the book "Cohomology operations and applications in homotopy theory" by Mosher and Tangora. It's beautifully written (and now available in a cheap Dover edition).

share|improve this answer
add comment

The Steenrod square is an example of a cohomology operation. Cohomology operations are natural transformations from the cohomology functor to itself. There are a few different types, but the most general is an unstable cohomology operation. This is simply a natural transformation from Ek(-) to El(-) for some fixed k and l. Here, one regards the graded cohomology functors as a family of set-valued functors so the functions induced by these unstable operations do not necessarily respect any of the structure of Ek(X).

Some do, however. In particular, there are additive cohomology operations. These are unstable operations which are homomorphisms of abelian groups.

In particular, for any multiplicative cohomology theory (in particular, ordinary cohomology or ordinary cohomology with Z/2 coefficients) there are the power operations: x → xk. These are additive if the coefficient ring has the right characteristic. In particular, squaring is additive in Z/2 cohomology.

Given an unstable cohomology operation r: Ek(-) → El(-) there is a way to manufacture a new operation Ωr: Ek-1(-) → El-1(-) using the suspension isomorphism (these should be reduced groups, so but you'll have to just imagine the tildes on top):

Ek-1(X) ≅ Ek(ΣX) → El(ΣX) ≅ El-1(X)

This is quite straightforward and is a cheap way of producing more operations. When applied to the power operations it produces almost nothing since the ring structure on the cohomology of a suspension is trivial: apart from the inclusion of the coefficient ring all products are zero.

What is an interesting question is whether or not this looping can be reversed. Namely, if r is an unstable operation, when is there another operation s such that Ωs = r? And how many such are there? Most interesting is the question of when there is an infinite chain of operations, (rk) such that Ωrk = rk-1. When this happens, we say that r comes from a stable operation (there is a slight ambiguity here as to when the sequence (rk) is a stable operation or merely comes from a stable operation).

One necessary condition is that r be additive. This is not, in general, sufficient. For example, the Adams operations in K-theory are additive but all but two are not stable.

However, for ordinary cohomology with coefficients in a field, additive is sufficient for an operation to come from a stable operation. Moreover, there is a unique sequence for each additive operation. This means that the squaring operation in Z/2 cohomology has a sequence of "higher" operations which loop down to squaring. These are the Steenrod squares.

The sequence stops with the actual squaring (rather, becomes zero after that point) because, as remarked above, the power operations loop to zero.

One important feature of these operations is that they give necessary conditions for a spectrum to be a suspension spectrum of a space. If a spectrum is such a suspension spectrum then it's Z/2-cohomology must be a ring. That's not enough, however, it must also have the property that, in the right dimensions, the Steenrod operations act by squaring. (Of course, this is necessary but not sufficient.)

share|improve this answer
    
Could you give some reference for a fact that for singular cohomology with coefficients in the field, any additive cohomology operation uniquely extends to a stable one? –  Piotr Pstrągowski Jul 16 '13 at 10:16
    
@PiotrPstrągowski It's been a while since I wrote this! I suspect I meant "finite field" and I was thinking of the third paragraph of the Introduction to Boardman, Johnson, and Wilson's paper "Unstable Operations in Generalized Cohomology". Comparing the descriptions of the additive and stable cooperations in that paper and its companion would give you a more precise formulation. –  Andrew Stacey Jul 16 '13 at 10:32
add comment

I second the references to Hatcher and to Mosher & Tangora, though you can also find Steenrod's original paper. At least the first two of those start out by listing the various axioms of Steenrod squares and then construct them.

The reason to axiomatize the properties of Steenrod squares is that it is hard to understand the relation of different constructions to each other (I spent a semester struggling with that), and the construction is not the point. Steenrod's original construction, calculating on simplices, has the same properties as Hatcher's CW-complex construction (note above), but the former is hard to prove things with and the latter is hard to visualize. It's nice to get a grip on one construction or other but ultimately the axioms themselves and the proof that they uniquely determine the maps becomes more practical.

All that said, I like Hatcher's construction because it uses the fact that elements of H^n(X; G) are the same as homotopy classes of maps from X to the Eilenberg-Maclane space K(G, n), a connected CW-complex whose nth homotopy group is G and others are trivial. Then you can understand any cohomology operation H^n(-; G)-->H^m(-;H) as a map on two spaces, K(G,n)-->K(H,m). [You can make a K(G,n) starting with n-cells of trivial boundary, attaching n+1-cells to make the right relations, and then continually attaching cells to kill the higher homotopy groups. K(Z,1)=S^1 doesn't need any higher cells, but RP^\infty is the usual K(Z/2, 1). Sorry if you already know this.] This makes it easy to see why cohomology operations on the same group have to increase dimension, and it gives you an actual space to do Steenrod squares in.

I believe the following has been said above, but in brief: One reason we like Steenrod squares is that they commute with suspension, but suspension kills cup products (even though Sq^n is the cup product square on H^n(X), after suspension the cup product square would be Sq^{n+1} but Sq^n is still defined). It's also awesome that the sum Sq^* of the various maps forms a ring operation of the cohomology ring. It's a good idea to look at RP^\infty and do Steenrod squares on that, and to think about the Steenrod algebra.

share|improve this answer
add comment

Section 4L of Hatcher's Algebraic Topology is a downloadable reference.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.