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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?

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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

1 Answer 1

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

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