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Let $F$ be a finite field, $Sq^i$ be the $i$-th Steenrod operation

$$ H^*(-;F) \to H^{*+i}(-;F).$$

By Yoneda lemma, such operation is a map $\phi_i: B^{*}F \to B^{*+i} F$, where $B$ denotes the delooping operator. By applying its inverse $\Omega$ many times, we get a map $\psi_i: \Omega^iF \to F$. I would like to understand Steenrod operation $Sq^i$ by $\psi_i$.

  1. Can you describe $\psi_i$ explicitly?
  2. What do Adem's relations translate to in this perspective? Can I see the combinatorics much clearly?
  3. Can we generalize this construction from $F$ to all abelian groups? It seems to me that the crux hides in the natural transformation

$$\Omega^i \to Id,$$

and does nothing with the coefficient system $F$.

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  • $\begingroup$ What is the loop space of a finite field? $\endgroup$
    – Greg Friedman
    Commented May 21, 2020 at 17:09

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You cannot extend the Steenrod squares to integer coefficients (as in have a set of cohomology operations satisfying the same axioms).

Consider the tangent space $TS^2$ of the 2-sphere. With a little persistence, we can identify the Thom space of this vector bundle with the two cell complex $S^2 \cup_{2\eta} e^4$ (here $\eta$ is as usual the Hopf map generating $\pi_3(S^2))$. This map $\eta$ has Hopf invariant 1, and so the cup product on this Thom space is $x \cup x = 2y$. Now the tangent bundle of the sphere is trivial after adding a trivial line bundle, so the suspension of this space is $S^3 \vee S^5$. By naturality, any cohomology operation applied to the 3-cell is trivial.

Hence, there is no stable transformation $Sq^2:H^*(-) \rightarrow H^{*+2} (-)$ so that for a 2 dimensional cohomology class x, $Sq^2 (x)=x \cup x$.

Of course, there other ways to see that the suspension of $2 \eta$ is trivial, but I like this one.

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  • $\begingroup$ Thanks for your answer! However, do you find the delooping viewpoint makes sense? If there's a good definition of $Sq^i: Id \to \Omega^i$, we don't need it to square at the right degree. $\endgroup$
    – Student
    Commented May 19, 2020 at 18:01
  • $\begingroup$ @Student Well in your post your map should be $F \rightarrow B^i F$ (you have the Yoneda lemma the wrong way around). You then need to find a delooping of this (nullhomotopic) map, and then a delooping of that map, etc. The issue with taking your delooping to be $B(F \rightarrow B^i F)$ is that this is always trivial. I would be surprised if you can get a good geometric description of any of the Steenrod operations. $\endgroup$
    – Connor Malin
    Commented May 19, 2020 at 18:13
  • $\begingroup$ If you are interested in more no go results, the only stable cohomology operations for rational cohomology is multiplication by a rational number. $\endgroup$
    – Connor Malin
    Commented May 19, 2020 at 18:16
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    $\begingroup$ @Student While what you've written is correct $\Omega^i F =0$ if $i>0$. The point is that this cohomology operation will always be 0 on 0th cohomology, but it can deloop to something nontrivial. So any additive way of delooping it will give the trivial operation. $\endgroup$
    – Connor Malin
    Commented May 19, 2020 at 18:22
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    $\begingroup$ @Student F is a discrete set of points. $\Omega^i F$ is the space of pointed maps from the i-sphere into $F$. $\endgroup$
    – Connor Malin
    Commented May 19, 2020 at 18:33

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