## Why does one think to Steenrod squares and powers?

I'm studying Steenrod operations from Hatcher's book. Like homology, one can use them only knowing the axioms, without caring for the actual construction. But while there are plenty of intuitive reasons to introduce homology, I cannot find any for the Steenrod operations. I can follow the steps in the proofs given by Hatcher, but I don't understand why one introduces all these spaces like $\Lambda X$, $\Gamma X$ and so on (in Hatcher's notation, I don't know if it's universal). Does anyone know how to get an intuitive grasp of what's going on?

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Steenrod operations are an example of what's known as a power operation. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.

A cohomology class on $X$ amounts to a map $a: X\to R$, where $R = \prod_{n\geq0} K(F_2,n)$. So the cup product of $a$ and $b$ is given by $$X\times X \to R\times R \xrightarrow{\mu} R.$$ In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes addition of cohomology classes.)

You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/\Sigma_n$, where $\Sigma_n$ is the symmetric group, i.e., $$X^n \xrightarrow{a^n} R^n \rightarrow R$$ should factor through the quotient $X^n/\Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $\Sigma_n$, and a product map: $$\mu_n' : E(n)\times R^n\to R$$ which is $\Sigma_n$ invariant, so it factors through $(E(n)\times R^n)/\Sigma_n$. Thus, given $a: X\to R$, you get $$P'(a): (E(n)\times X^n)/\Sigma_n \to (E(n)\times R^n)/\Sigma_n \to R.$$ If you restrict to the diagonal copy of $X$ in $X^n$, you get a map $$P(a):E(n)/\Sigma_n \times X\to R.$$

If $n=2$, then $E(2)/\Sigma_2$ is what Hatcher seems to call $L^\infty$; it is the infinite real proj. space $RP^\infty$. So $P(a)$ represents an element in $H^* RP^\infty \times X \approx H^*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$.

Other cohomology theories have power operations (for K-theory, these are the Adams operations).

You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.

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Cool. Can one define the Adams operations the same way but with $R=BO(\infty)$? – roger123 Feb 4 2010 at 15:17
Yep. Though there are some subtleties related to understanding the K-theory of classifying spaces of finite groups. The cleanest construction of Adams ops as power operations uses equivariant K-theory: there is a nice paper by Atiyah on this ("Power operations in K-theory", QJ Math. Ox. (2) 17, reprinted as an appendix in some editions of his K-theory book.) – Charles Rezk Feb 4 2010 at 16:24
Is that written up somewhere non-equivariantly? – roger123 Feb 5 2010 at 11:52
Not that I know of. – Charles Rezk Feb 5 2010 at 14:37

Just think of it as category theory; it is not enough to define objects, one must also have morphisms. So you define cohomology groups, and those are objects. What you want is morphisms between cohomology theories. Such a morphism would in fact be a natural transformation

h(X) --> k(X)

for cohomology theories h and k. They should of course be additive, since cohomology theories land in abelian groups, and they should preserve suspension. They can be graded (raise or lower degree by some fixed amount).

Anytime you have two cohomology theories h and k you want to compute this graded abelian group of cohomology operations. But it can be a hard job. The simplest case is when k=h, in which case you will have a ring (since you can compose natural transformations). And the simplest case to compute is when h is Z/2-cohomology, when the answer is the Steenrod algebra.

This many not answer your local questions, but this is the global answer. The answer to your local questions is basically that the cup product is not (graded)-commutative on the cochain level, and if you look closely at this failure, you will find the obstructions to cochain-level commutativity are the Steenrod operations.

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 Said another way, cohomology operations are to cohomology as the homotopy groups of spheres are to homotopy groups. – Ryan Budney Nov 21 2009 at 21:50 I was not exactly looking for motivation to introduce Steenrod operations; rather what it was mysterious to me is how one actually comes to define them, since that seem to involve quite a lot of machinery. – Andrea Ferretti Nov 22 2009 at 1:04 If all you want is an intuitive definition, you can take the Steenrod operations to be the stable cohomology of Eilenberg-Maclane spaces. But it sounds like you want more than an intuitive definition -- you want it to be intuitive all the way down to an explicit chain-level definition? – Ryan Budney Nov 22 2009 at 4:53 Yes, I was rather trying to understand how does one explicitly produce such cohomology operations/cohomology classes on Eilenberg-MacLane spaces. – Andrea Ferretti Nov 23 2009 at 23:05

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 $n$-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.

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 This is a beautiful perspective on Steenrod operations. Thanks! – Tom Church Mar 29 2010 at 9:50 Thank you, a very interesting point of view! – Andrea Ferretti Mar 29 2010 at 9:51

The "chain level" structure that others alluded to is explained in Mosher and Tangora's book, which is quite nice. The idea is simple but it often gets tied up in the algebra involved.

As an analogy from category theory, if your category has a cartesian product then there are isomorphisms from Z x Z → Z x Z for two reasons: firstly because they are visibly the same object and so you have an identity morphism, and secondly because X x Y is always isomorphic to Y x X for any X and Y, using a natural twist map τ. This becomes a natural "twist" automorphism on Z x Z.

The same happens for the cup product on the level of cochains. The cup product (x∪y) of two cocycles always differs from (y∪x) by a coboundary, expressed by a natural operation called a cup-1 product (x∪1y). But then if z is a cocycle of degree n, (z∪1z) is always a cocycle because its coboundary is the difference between (z∪z) and itself. This natural operation gives you the top-but-one Steenrod operation Sqn-1 on mod-2 cohomology. The higher squares play the same game; there is a cup-2 product (x∪2y) expressing the difference between (x∪1y) and (y∪1x), and (z∪2z) represents Sqn-2.

This answers a very common question: Why is the formula for the cup product so goofy and asymmetric? The answer is that it has to be, because the Steenrod operations obstruct the possibility of doing better.

The use of the cohomology of real projective space mentioned in previous answers is very clever, and the cup-i products correspond in that procedure to certain cells of S, but I think some of the underlying story is hidden there.

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 Thank you, your point of view is very interesting. I will try to connect it to the other one mentioned above. – Andrea Ferretti Nov 23 2009 at 23:07 i think your indexing might be a little off, isnt (z∪iz)=:sq^n-i(z) where n=|z|?(just being a tidge pedantic) brayton gray's book is supposed to be a more modern treatment... i am told. – Sean Tilson Feb 23 2010 at 6:45 Hi Sean, welcome to Math Overflow. Yes, you're right, my indexing was incorrect, and I've edited accordingly. – Tyler Lawson Feb 23 2010 at 13:19

The thing that Hatcher does that is different from, say, Steenrod and Epstein as Charles mentioned, is that rather than construct anything on cohomology rings, he constructs an operation on Eilenberg-Maclane spaces. The $n$th cohomology group of any space $X$ with coefficients in $G$ is naturally isomorphic to the group of homotopy classes of maps from $X$ to an Eilenberg-Maclane space $K(G, n)$: this identification is nice and functorial because if you map $X\rightarrow Y$ you can map $X\rightarrow Y\rightarrow K(G, n)$ to get a pullback map on cohomology groups. Now if you want to make a cohomology operation, which can be defined naturally on cohomology groups for any $X$, all you need is an operation $K(G, n)\rightarrow K(H, m)$ and by composing with that you get a natural map $H^n(X; G)\rightarrow H^m(X;H)$.

So Hatcher's idea is to take a kind of (smash) product of $K(\mathbb Z/2, n)$ (which he calls $K_n$) with itself, and map that to $K_{2n}$ in a way that will be like the cup-product square in dimension $n$, since $Sq^n(\alpha)=\alpha^2$ when $\alpha$ is of dimeniosn $n$. More explicitly, since $H^n(X\wedge X)$ is isomorphic to $H^n(X)\otimes H^n(X)$ by the Kunneth formula, the point is to find an element of $H^{2n}(X\wedge X)$ which is the cup-product square. His notation is NOT standard.

Intuitively, the idea is to do this for $K_n$ so that it works on all $X$, and to put the copies of $K_n$ back together by quotienting by a $\mathbb Z/2$ action. But the $\mathbb Z/2$-action on $K_n\wedge K_n$ given by switching is not free, so to make it free he takes $S^\infty\times K_n\wedge K_n$ and uses the antipodal map on $S^\infty$ to map $(x, y, z)$ to $(-x, z, y)$ and give a free $\mathbb Z/2$ action. (The coordinates here are really in $S^\infty\times K_n\times K_n$). Then the whole rest of the construction is about dealing with the extra things that $\mathbb RP^\infty$ gives you, and from there he can calculate what happens to homology elements of $\wedge X$ when it is mapped to $K_{2n}$. That tells you the action of $Sq^{n-i}$ on $H^i(X)$.

Hope that helps.

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 Uhm, the idea that the product with $S^{\infty}$ is to make the action free is actually useful, thanks. – Andrea Ferretti Nov 22 2009 at 1:07

Let me say a few words on this: We do have the functor H^n, which are functors from HTop(the homotopy cat.) to Abelian gps satisfying some axioms. (or H^*, functors from HTop to graded rings) But there are some finer structures on it, which allows one to distinguish some spaces/morphisms that could not be detected by just cohomology "groups"/"rings". We want to find these structures.

Now here is the key point:

NATURAL TRANSFORMATIONS ADD SOME STRUCTURES ON FUNCTORS.

If our functor in question has several non-trivial endormorphism, then we have some more structures. [in down-to-earth terms, the commutative diagrams in the def. of natural transformations are not given for free, there are something non-trivial there.]

Specify to our case, the Steenrod sqs and powers are natural transformations of the functor H^*, and our finer structures, as people may have pointed out, are just module structures over the Steenrod algbra A. [BTW, some exercises in Whitehead's Elements of Homotopy theory (GTM 61), give examples of spaces with isomorphic H^* as graded rings, but not as modules over A.]

A good reference on the philosophy I mentioned here is a 50-pages booklet by Steenrod called "Cohomology Operations"(?) [not the one written down by Epstein.]

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 I was not exactly looking for motivation to introduce Steenrod operations; rather what it was mysterious to me is how one actually comes to define them, since that seem to involve quite a lot of machinery. – Andrea Ferretti Nov 22 2009 at 1:08