4
$\begingroup$

Question. Let $p$ be a prime. Let $q$ be a power of $p$. Let $P^0$, $P^1$, $P^2$, ... be elements of some associative $\mathbb F_q$-algebra $A$. (Here, $P^i$ does not mean $P$ to the $i$-th power; instead, $i$ is an upper index.) Asume that the power series $\sum\limits_{a,k} \left(tu\right)^a P^a P^k$ and $\sum\limits_{c,j} u^c t^{qj}P^cP^j$ are equal, where $t$ is the indeterminate of our power series and $u=\left(1-t\right)^{q-1}=1+t+t^2+...+t^{q-1}$. Prove that any nonnegative integers $a$ and $b$ such that $a < qb$ satisfy

$\displaystyle P^aP^b = \sum\limits_j \left(-1\right)^{a-qj} \binom{\left(b-j\right)\left(q-1\right)-1}{a-qj} P^{a+b-j}P^j$.

(If you don't use the same conventions about negative binomial coefficients as I do, think of this sum as going from $j=0$ to $j=\left\lfloor a/q\right\rfloor$.)

Motivation. This question is equivalent to deriving the Adem-Wu relations in the Steenrod algebra (without the Bockstein) from the Bullett-Macdonald formula. I am working in the invariant-theoretical setting, so I want a proof which does not refer to the topological interpretation of the Steenrod algebra (at least not unless it shows that this is equivalent to the invariant-theoretical one).

Larry Smith, An algebraic introduction to the Steenrod algebra, arXiv:0903.4997 gives a proof using complex integration, but (the indexing mistakes put aside) I do not really believe it. It seems to work over $\mathbb Z$ first (which allows for integration) and then project onto $\mathbb F_q$, which is okay, but I think the condition that $\sum\limits_{a,k} \left(tu\right)^a P^a P^k = \sum\limits_{b\geq j} u^{b-j}t^{qj}P^{b-j}P^j$ cannot be "lifted" to $\mathbb Z$ in a straightforward way, to begin with, which puts the whole complex-analysis approach under question. Probably it works with the right incantations being said, but I was not able to come up with these incantations (and way too confused in this topic). A pedestrian algebraic proof would be preferred.

$\endgroup$

2 Answers 2

7
$\begingroup$

Larry Smith is not really using complex integration. Instead, he is using the residue map, which can be defined algebraically by the rule $$ \text{res}\left(\sum_{k=-N}^\infty a_k z^k dz\right)=a_{-1} $$ and the fundamental transformation property that $$ \text{res}\left(\sum_{k=-N}^\infty a_k \;f(z)^k\;f'(z) dz\right) = a_{-1} $$ under appropriate conditions on $f$. For example, this works if everything is happening over a commutative ring $R$, and $f(z)=\sum_{k=0}^\infty b_kz^k$ with $b_0$ nilpotent and $b_1$ invertible. The basic idea is old and well-known, perhaps due to Cartier. One possible set of technical details is explained in Section 5.4 of my paper "Formal groups and formal schemes": http://arxiv.org/abs/math/0011121

$\endgroup$
2
  • $\begingroup$ Thanks a lot. Since this is way more complicated than I thought, I cannot check it at the moment, but I certainly will. $\endgroup$ May 15, 2011 at 19:15
  • $\begingroup$ I see now. I was as far as formulating your Corollary 5.36 and proving it in characteristic $0$ myself, but I failed to see that it is a polynomial identities in the coefficients of $f$ and $a_k$, so it would immediately project on rings of arbitrary characteristic. $\endgroup$ May 15, 2011 at 22:39
2
$\begingroup$

Bullet and Macdonald gave an algebraic proof of this in their paper On the Adem relations

$\endgroup$
1
  • $\begingroup$ Interesting, because Smith claims that the proof they give is the same as he gives (and even that he gives more detail). Do you have the Bullett-Macdonald paper in electronic form? $\endgroup$ May 15, 2011 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.