Why does one consider the dual of the Steenrod algebra?
[Math] Why does one consider the dual of the Steenrod algebra
at.algebraic-topologysteenrod-algebra
Related Solutions
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.
Normally people think about Steenrod comodules as graded $\mathbb{Z}/2$-modules equipped with a graded coaction $\psi\colon M_*\to M_*[\xi_1,\xi_2,\dotsc]$. However, it is equivalent to consider ungraded modules with coaction $\psi\colon M\to M[\xi_0^{\pm 1},\xi_1,\xi_2,\dotsc]$; the grading is recovered by the formula $$ M_d = \{m : \psi(m)=\xi_0^dm \pmod{\xi_k:k>0}\}, $$ and the original coaction is recovered by setting $\xi_0=1$.
Now the action is unstable iff $\psi(M)\leq M[\xi_0,\xi_1,\dotsc]$ (so no negative powers of $\xi_0$ are involved).
If $M$ has a compatible product then we can reformulate things in terms of the scheme $X=\text{spec}(M)$. A Steenrod action is equivalent to an action of the group scheme $\text{Aut}(G_a)$ on $X$ (where $G_a$ is the additive formal group). The main instability condition (that certain Steenrod operations should vanish) is equivalent to saying that the action extends over the monoid $\text{End}(G_a)$. (It is easy to see that there is at most one extension.) In the multiplicative context one also wants to impose the condition $\text{Sq}^k(x)=x^2$ when $|x|=k$. This is equivalent to saying that the Frobenius endomorphism $F_{G_a}\in\text{End}(G_a)$ should act on $X$ as the Frobenius endomorphism $F_X$.
Best Answer
There are really two reasons, both of which are already alluded to in Eric Peterson's answer and in Pierre's answer.
Formally one can think about mod p cohomology as being a functor to the category of modules over the Steenrod Algebra. Similarly one can think about mod p homology as being a functor to the category of comodules over the dual to the Steenrod Algebra.
These seem formally the same, and the first seems preferable since most of us are happier thinking about modules than about comodules. But:
1) Certain kinds of formulas can be written down more easily for the dual of the Steenrod Algebra than for the Steenrod Algebra itself. In particular, the coproduct on the dual of the Steenrod Algebra has a formula that is simpler for some applications than understanding the non-commutative product on the Steenrod Algebra, via (for example) the Adem relations.
As an example, the Milnor basis of the Steenrod Algebra is defined by noting that the dual to the Steenrod Algebra is a polynomial algebra. The Milnor basis to the Steenrod Algebra is then the dual basis to the monomial basis of the dual algebra.
2) The Steenrod Algebra is the (Hopf) algebra of cohomology operations. By contrast, the dual to the Steenrod Algebra is the (Hopf) algebra of homology co-operations.
When constructing the Adams spectral sequence for other theories (other than mod p homology or cohomology) it is more convenient to use the homology version with the algebra of homology co-operations. This is because the cohomology of a product often involves completions, whereas the homology of a product does not.
Once one has incorporated this way of thinking about things it becomes convenient even for the ordinary Adams spectral sequence. It removes the confusion caused by contravariant functors, and it makes maps between Adams spectral sequences (for different homology theories) more transparent.