2
$\begingroup$

Let $\mathrm{Sq}^i$ the Steenrod operations. These are the Adem relations:

when $ i<2j$ $$ \mathrm{Sq}^i \mathrm{Sq}^j=\sum_{k \in \mathbb{Z}}\binom{j-k-1}{i-2 k} \mathrm{Sq}^{i+j-k} \mathrm{Sq}^k,$$ with the understanding that $\binom{n}{m}=0$ if $m>n$ or $m<0$, and $\mathrm{Sq}^k=0$ if $k<0$ (so in practice $k$ ranges from $\max (0, i-j+1)$ to $\left.\left\lfloor\frac{i}{2}\right\rfloor\right)$.

In this link https://dec41.user.srcf.net/misc/adem I found this consequences: \begin{aligned} & \mathrm{Sq}^1 \mathrm{Sq}^1=0 \\ & \mathrm{Sq}^1 \mathrm{Sq}^2=\mathrm{Sq}^3 \\ & \mathrm{Sq}^1 \mathrm{Sq}^3=0 \\ & \mathrm{Sq}^2 \mathrm{Sq}^2=\mathrm{Sq}^3 \mathrm{Sq}^1 \\ & \mathrm{Sq}^1 \mathrm{Sq}^4=\mathrm{Sq}^5 \\ \end{aligned}

The problem is that if i try to compute $\mathrm{Sq}^1\mathrm{Sq}^3$ (that should be $0$ as suggested) this is what i get:
Since $\max (0, i-j+1)=\max(0,1-3+1)=0$ and since $\left\lfloor\frac{i}{2}\right\rfloor = \left\lfloor\frac{1}{2}\right\rfloor=0$ then $k=0$, so $\mathrm{Sq}^1\mathrm{Sq}^3=\binom{3-0-1}{1-2*0} \mathrm{Sq}^{1+3-0} \mathrm{Sq}^0=2\mathrm{Sq}^{4}$.
What is the problem? Why don't I get $0$?

I've got a similar problem trying to compute $\mathrm{Sq}^1\mathrm{Sq}^4$ (that should be $\mathrm{Sq}^5$ as suggested):
Since $\max (0, i-j+1)=\max(0,1-4+1)=0$ and since $\left\lfloor\frac{i}{2}\right\rfloor = \left\lfloor\frac{1}{2}\right\rfloor=0$ then $k=0$, so $\mathrm{Sq}^1\mathrm{Sq}^4=\binom{4-0-1}{1-2*0} \mathrm{Sq}^{1+4-0} \mathrm{Sq}^0=3\mathrm{Sq}^{5}$.
Why don't I get $\mathrm{Sq}^{5}$?

$\endgroup$

1 Answer 1

5
$\begingroup$

The Steenrod algebra is an algebra over $\mathbb{F}_2$, so in particular the coefficients are elements of $\mathbb{F}_2$. In other words, $$ 2 \mathrm{Sq}^4 = 0 $$ and $$ 3 \mathrm{Sq}^5 = \mathrm{Sq}^5 $$ and your problems disappear.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .