There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called Steenrod squaring: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ 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 $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. 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 $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?


9 Answers 9


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. (Sorry, the link is now dead.)

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    $\begingroup$ 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/… $\endgroup$ Commented Sep 29, 2010 at 4:46
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    $\begingroup$ Here's a very elementary case. Take a 1 simplex of X and compose with the diagonal X --> XxX. Cellularly approximate by either the left+top or bottom + right. The 2-chain with this as boundary is exactly the two 2-simplices whose sum is the square, the product of the original 1-simplex with itself. This gives Sq^1(x) = x^2 when x is of degree 1. $\endgroup$ Commented Jan 29, 2015 at 15:04
  • $\begingroup$ Unfortunately the link is broken $\endgroup$
    – Thomas Rot
    Commented Mar 8, 2016 at 10:09
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    $\begingroup$ @Eric Wofsey: Is the link still working? $\endgroup$ Commented Jun 10, 2016 at 3:02
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    $\begingroup$ Here is a link to (I believe) the same document: isites.harvard.edu/fs/docs/icb.topic1525101.files/… $\endgroup$ Commented Feb 10, 2017 at 16:12

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.

  • $\begingroup$ that's really cool, do you have a reference? $\endgroup$ Commented Apr 14, 2010 at 12:03
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    $\begingroup$ 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 $\endgroup$
    – Mark Grant
    Commented Nov 1, 2010 at 11:03
  • $\begingroup$ 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 $\endgroup$
    – Tintin
    Commented Jun 14, 2011 at 14:10
  • $\begingroup$ $v$ is the normal bundle sorry ( i don't how to érase the comment) but still: what is $w_i$? thanks $\endgroup$
    – Tintin
    Commented Jun 14, 2011 at 14:15
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    $\begingroup$ $w_i$ is the ith Stiefel-Whitney class. $\endgroup$
    – Dev Sinha
    Commented Jun 17, 2011 at 23:20

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 $BG$.

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    $\begingroup$ 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? $\endgroup$ Commented Jul 18, 2013 at 6:49
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    $\begingroup$ I have the same question. $\endgroup$ Commented Nov 1, 2014 at 11:23
  • $\begingroup$ I believe this was Serre's proof that the Adem relations generate all the relations. Note also that one only needs Sq^1(x) = x^2, for x of degree 1, together with the Cartan formula, to determine all the Sq^i on these polynomial rings, and then, by Serre's observation that this is a faithful representation, one has the whole Steenrod algebra. To get the first fact, Sq^1(x) = x^2, see my comment following Wofsey's answer. $\endgroup$ Commented Jan 29, 2015 at 15:01
  • $\begingroup$ Is there a geometric (or whatever) motivation behind considering $x\mapsto x+x^2$ and its components? $\endgroup$
    – FShrike
    Commented Jul 12 at 9:02

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 $E^k(-)$ to $E^l(-)$ 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 $E^k(X)$.

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 $\mathbb{Z}/2\mathbb{Z}$ coefficients) there are the power operations: $x \to x^k$. These are additive if the coefficient ring has the right characteristic. In particular, squaring is additive in $\mathbb{Z}/2\mathbb{Z}$ cohomology.

Given an unstable cohomology operation $r: E^k(-) \to E^l(-)$ there is a way to manufacture a new operation $\Omega r: \tilde{E}^{k-1} \to \tilde{E}^{l-1}(-)$ using the suspension isomorphism (where the tilde denotes that these are reduced groups):

$E^{k-1}(X) \cong E^k(\Sigma X) \to E^l(\Sigma X) \cong E^{l-1}(X)$

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 $\Omega s = r$? And how many such are there? Most interesting is the question of when there is an infinite chain of operations, $(r_k)$ such that $\Omega r_k = r_{k-1}$. When this happens, we say that $r$ comes from a stable operation (there is a slight ambiguity here as to when the sequence $(r_k)$ 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 $\mathbb{Z}/2\mathbb{Z}$ 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 its $\mathbb{Z}/2\mathbb{Z}$-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.)

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    $\begingroup$ 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? $\endgroup$ Commented Jul 16, 2013 at 10:16
  • $\begingroup$ @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. $\endgroup$ Commented Jul 16, 2013 at 10:32

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 $H^n(-; G)\to H^m(-;H)$ as a map on two spaces, $K(G,n)\to K(H,m)$. [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. $K(\mathbb{Z},1)=S^1$ doesn't need any higher cells, but $\mathbb{RP}^\infty$ is the usual $K(\mathbb{Z}/2, 1$). 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 $\mathbb{RP}^\infty$ and do Steenrod squares on that, and to think about the Steenrod algebra.


I give an intuitive account of the origin of Steenrod operations in cohomology, Dyer-Lashof operations in homology of infinite loop spaces, and Steenrod operations in the cohomology of cocommutative Hopf algebras in Chapter 3 of my (unfinished) Adams Spectral Sequence Primer. Then in Chapter 4, I define the precise conditions needed and the resulting theorems, following Peter May's A General Algebraic Approach to Steenrod Operations. The role of symmetries in producing the operations in the first place, and then in producing their properties, is the focus. A key point that May makes is that all these operations have the same source, a category I call $Sym_\infty$ in my Primer. It is a category of DGAs with some additional structure encoding the symmetries.

One other beautiful way to see the Steenrod operations is through the dual of the Steenrod algebra. This is the Hopf algebra of coordinate functions on the automorphisms of the additive group. Precisely, let $F_2[[x]]$ be the power series ring in one variable $x$ with primitive coproduct $\psi(x) = 1 \otimes x + x \otimes 1$, corepresenting the additive group.

The automorphism which sends $x$ to a power series $f = \sum a_i x^i$ commutes with $\psi$ iff the only non-zero coefficients are those with $i$ a power of two, so we may write an automorphism $f$ as $\sum c_i x^{2^i}$ (with $c_0=1$ of course). Let $\xi_i(f) = c_i$. There are no relations between these coordinate functions $\xi_i$, so we get a polynomial ring $F_2[\xi_1, \xi_2,\ldots]$ of functions on the group of automorphisms.

Now consider $\xi_n(f\circ g)$. We easily calculate that $\xi_n(f \circ g) = \sum_{i+j=n} \xi_i(f) \xi_j(g)^{2^i}$, so that the coproduct on these coordinate functions is $\psi(\xi_n) = \sum \xi_i \otimes \xi_j^{2^i}$, which is the dual Steenrod algebra, as shown by Milnor in 1956. (Note that these $\xi_i%$ are the conjugates of Milnor's generators, so the coproduct is reversed.) Jack Morava showed me this derivation of the dual Steenrod algebra.


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).


If I interpret the request a bit differently, I would say that the Steenrod operations in the cohomology of a spectrum tell you about the attachments of the cells. If $Sq^1 x = y$, then a cell dual to $y$ is attached by a map of degree 2 mod 4 to a cell dual to $x$. Similarly, $Sq^2 x = y$ tells us the attaching map is $\eta$, $Sq^4$ detects $\nu$ and $Sq^8$ detects $\sigma$. This doesn't go very far, but may help with the need to 'get a real grip on what they're doing'.

Next, let's assume you're really interested in homotopy, not just (co)homology. A class dual to a homology class in the image of the Hurewicz homomorphism must be indecomposable under the action of the Steenrod algebra, by naturality w.r.t. the map $S^n \longrightarrow X$. This limits the homotopy of $X$ which can be detected by the homomorpism $\pi_* X \longrightarrow H_* X$: the homomorphism $H^* X \longrightarrow H^* S^n$ can only map indecomposables non-trivially, since all classes in degrees below $n$ must go to $0$.

Then there are the relations. The fact that $Sq^n$ is decomposable when n is not a power of two tells us that if $y = Sq^n x$, there must be other classes between $x$ and $y$. EG, $Sq^3 x = y \neq 0$ tells us that $Sq^2 x \neq 0$ also, since $Sq^3 = Sq^1 Sq^2$. So our spectrum can't have just two cells, dual to $x$ and $y$, but must have a three cell subquotient with top cell attached by 2 (mod 4) to a cell attached by $\eta$ to the bottom cell.

Or, if $Sq^2 Sq^2 x = y \neq 0$ then we must also have nonzero classes $Sq^1 x$ and $Sq^2 Sq^1 x$, since $Sq^2 Sq^2 = Sq^1 Sq^2 Sq^1$, and vice versa, if $Sq^1 Sq^2 Sq^1 x = y \neq 0$ then $Sq^2 x \neq 0$ as well. This leads to an easy proof that the mod 2 Moore spectrum $M$ isn't a ring spectrum, since $2 \pi_0M = 0$ but $\pi_2 M = Z/4$, by looking at the obstruction to attaching the top cell of a putative spectrum with nonzero cohomology spanned by $x$, $Sq^1 x$, $Sq^2 Sq^1 x$, and $Sq^1 Sq^2 Sq^1 x$. More, the fact that you can only add such a top cell if you also have a class $Sq^2 x$ so that the top cell can be attached by the sum of $Sq^1$ on $Sq^2 Sq^1 x$ and $Sq^2$ on $Sq^2 x$ shows that $\eta^2$ (corresponding to the path $Sq^2 Sq^2$ from bottom to top, must lie in the Toda bracket $\langle 2, \eta, 2\rangle$, corresponding to the path $Sq^1$, $Sq^2$, $Sq^1$ from bottom to top.

Similarly, $y = Sq^1 Sq^2 x$ tells us that homotopy supported on a cell dual to $x$ can be acted on by $\text{v}_1$ to get $y$, literally if we have a $ku$-module and multiply by $\text{v}_1 \in ku_2$, or as the Toda bracket $\langle 2, \eta, -\rangle$ more generally. The key fact here is that $\text{v}_1 \in ku_2$ is in $\langle 2, \eta, 1_{ku} \rangle$, where $1_{ku} : S \longrightarrow ku$ is the unit.

Likewise, $Sq^2 Sq^1 Sq^2 x = y$ corresponds to multiplication by the generator of $ko_4$, literally for $ko$-modules, or as a bracket $\langle \eta, 2, \eta, - \rangle$ more generally. Here you have to be in a situation where $2 \nu = 0$ to form the bracket, since $\langle \eta, 2, \eta \rangle = \{ 2\nu, 6 \nu\}$. This hints that the role of $\nu$ is non-trivial in real K-theory, despite going to $0$ under the homomorphism $\pi_* S \longrightarrow \pi_* ko$ and despite the cohomology of $ko$ being induced up from the subalgebra $A(1)$ generated by $Sq^1$ and $Sq^2$. The Adem relation $Sq^2 Sq^1 Sq^2 = Sq^1 Sq^4 + Sq^4 Sq^1$ shows that $Sq^4$ must act nontrivially if $Sq^2 Sq^1 Sq^2$ does. Also, the fact that $A(1)//A(0)$ is spanned by $1$, $Sq^2$, $Sq^1 Sq^2$, and $Sq^2Sq^1Sq^2$ tells us (with a bit more work) that we can build $HZ$ as a four cell $ko$-module.

A good way to organize all this information is the Adams spectral sequence, which tells you that the mod $p$ cohomology of $X$ gives a decent first approximation, $\text{Ext}_{A}(H^*X,F_p)$, to the homotopy of the $p$-completion of $X$.


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


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