If you haven't found something strange: Steenrod Algebra, Lie Groups and Formal Series! Oh My!

http://math.berkeley.edu/~reb/courses/261/all.pdf
Page 24

Disclaimer: I don't know the difference between 1 and -1.

Recall the operations Sq which act on the cohomology as degree i maps and looking at the right domain $Sq^n \; H^n \to H^{2n}$ gives the cup square.

These operations have the ugly relations
$Sq^i Sq^j = \sum_{k=0}^{i/2} \binom{j-k-1}{i-2k} Sq^{i+j-k} Sq^k$

In addition there is a coproduct on this algebra

$\Delta Sq^n=\sum Sq^i \otimes Sq^j$

so the Steenrod algebra is a cocommutative Hopf algebra over the 2 element field.

But we know where cocommutative Hopf algebras come up a lot. Universal enveloping algebras of some Lie group.

This "lie group" will be like the automorphism group of something. What is that something?

Here is where we get to formal power series. Infinitesimal transformations of the real line fixing 0 are given by formal power series with 0 constant term and invertible linear term. Look at the composition of such power series and you end up finding a complicated product or in other words a complicated coproduct on $Z[a_0 , a_0^{-1},a_1,\cdots]$

Now we really want the additive homomorphisms. There aren't any besides multiplication by constants. But if we reduce mod p we are ok. We do the same process of formal power series and figure out the coproducts.

So our formal power series are
$a_0 x+ a_{p-1}x^p+a_{p^2-1}x^{p^2}+\cdots$

Try composing one with coefficients a's and one with b's. You get

$\sum a_{p^i-1} b_{p^j-1}^{p^i}x^{p^{i+j}}$

What are the coproducts from this?

$\Delta a_{p^n-1} = \sum_{i+j=n} a_{p^i-1} \otimes a_{p^j-1}^{p^i}$

Set p=2, treat the 0th term as 1 and dualize to get the relations we had before.

If you haven't found something strange

Sunday, March 11, 2012

Steenrod Algebra, Lie Groups and Formal Series! Oh My!

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