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Why does one consider the dual of the Steenrod algebra?

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The dual of the Steenrod algebra is commutative while the original is not (though this means that the diagonal on the dual is fairly complicated). We have many more tools for dealing with commutative things than with noncommutative things. This is sort of a cheap answer, and certainly there are more reasons than just this, but imho it alone is enough. –  Eric Peterson Jun 13 '11 at 18:06
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Please read "how to ask". –  Todd Trimble Jun 13 '11 at 18:18
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Alternatively, the coproduct on the Steenrod algebra is much easier to work with than the product (in other words: the Adem relations are more complicated than the Cartan formula). On the other hand, working with coproducts are awkward... so we take the dual and then we have a nice product, and the coproduct is ugly but that's alright. –  Dylan Wilson Jun 13 '11 at 18:28
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Let me add that, if $A$ is the Steenrod algebra and $A^*$ it dual, then there is a map $\lambda : H^*(X)\to H^*(X)\otimes A^*$ (in fact you need to take a completion somewhere, doesn't really matter). You can recover the action of $A$ on $H^*(X)$ easily from this. The great thing about $\lambda$ though, is that it is a map of rings. There are situations when this makes horrendous computations trivial. –  Pierre Jun 13 '11 at 19:49

2 Answers 2

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.

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Hal Sadofsky's answer is great, and you should accept it.

If you want a more obscure point of view, one which is purely algebraic, then (at least at the prime 2), the dual of the Steenrod algebra represents the group of strict automorphisms of the additive formal group. That is, for any commutative $\mathbf{F}_2$-algebra $R$, the additive formal group for $R$ is just the group $(R,+)$, and an automorphism of this is determined by a power series $f(x)$ with coefficients in $R$ such that $f(x+y) = f(x) + f(y)$. The group operation is composition of power series. So such automorphisms are just power series of the form $$ r_0 x + r_1 x^2 + r_2 x^4 + \dots + r_{n} x^{2^n} + \dots, $$ where each $r_i$ is in $R$ and $r_1$ is in $R^\times$. Such an automorphism is called strict if $r_1=1$. Now if $A$ is the dual of the mod 2 Steenrod algebra, then it's easy to check that $$ \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) $$ is in bijection with the set of strict automorphisms, as above, where given a map $A \to R$, each coefficient $r_{n}$ of the power series is determined by the image of the polynomial generator $\xi_n$ (or maybe its antipode $\zeta_n$). It's fun to then verify that composition of power series is induced by the coproduct in $A$: the group operation $$ \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) \times \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) \to \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) $$ is induced by an $\mathbf{F}_2$-algebra map $$ A \to A \otimes A, $$ and it's precisely the coproduct of $A$, as described by Milnor.

At odd primes you can do this with just the polynomial part of the dual of the Steenrod algebra.


There is also a connection between $A$ and complete binary trees...

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There is also a statement for the whole odd primary Steenrod algebra, if you enlarge to thinking about a formal completion of an additive supergroup. –  Eric Peterson Jun 14 '11 at 8:31
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@Eric: Can you be more precise please. I guess the supergroup is $Spec(H^*(B\mathbb Z/p\mathbb Z;\mathbb F_p))$. But what replaces "strict"? Can it be phrased in purely (super-)geometric language? –  André Henriques Jun 18 '11 at 5:22

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