Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S^d$ denote the $d$-sphere. The only non-trivial cohomology groups are $H^0(S^d;\mathbb Z_2)= \mathbb Z_2$ generated by $1$ and $H^d(S^d;\mathbb Z_2)= \mathbb Z_2$ generated by the fundamental class $u$. I want to write all the Steendod squares on the sphere, that is all the group homomorphisms $$Sq^i:H^n(S^d;\mathbb Z_2)\to H^{n+i}(S^d;\mathbb Z_2)$$ satisfying some axioms.

I claim that there are only three of them:

  • $Sq^0:H^0(S^d;\mathbb Z_2)\to H^0(S^d;\mathbb Z_2)$ which is the identity by the first axiom;

  • $Sq^d:H^0(S^d;\mathbb Z_2)\to H^d(S^d;\mathbb Z_2)$ which sends $1$ to $u$;

  • $Sq^0:H^d(S^d;\mathbb Z_2)\to H^d(S^d;\mathbb Z_2)$ which is the identity.

Is this correct?

share|cite|improve this question
One of the axioms is that «If $n>\dim(x)$ then $\mathrm{Sq}^n(x) = 0$». So your second claim is wrong. – Mariano Suárez-Alvarez Nov 27 '11 at 18:06
ok i see.. thanks.. so there are only two steenrod squares, the two identity group homomorphisms $Sq^0:H^0(S^d;\mathbb Z_2)\to H^0(S^d;\mathbb Z_2) $ and $Sq^0:H^d(S^d;\mathbb Z_2)\to H^d(S^d;\mathbb Z_2)$ – palio Nov 27 '11 at 18:23

So as Mariano pointed out, your second bullet is incorrect. Also, there are many squares, most of them happen to be zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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