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Let $\mathcal{O}$ be a topological operad and $X$ an algebra over it. Let the base ring be $\mathbb{Z}_2$. If $C_*$ denotes the singular chain complex over $\mathbb{Z}_2$, the action of $\mathcal{O}$ gives us morphisms $$\mu:C_*(\mathcal{O}(r))\otimes_{r} C_*(X)^{\otimes r}\to C_*(X)$$ where $\otimes_{r}$ means the tensor product over the group algebra $\mathbb{Z}_2\mathfrak{S}_r$ of the $r$th symmetric group. Now consider the covering $\mathcal{O}(r)\to \mathcal{O}(r)/\mathfrak{S}_r$. For each simplex $s\in C_*(\mathcal{O}(r)/\mathfrak{S}_r)$ we choose one of the $r!$ lifts $\widetilde{s}$ and consider the map $$\Phi:C_*(\mathcal{O}(r)/\mathfrak{S}_r)\times C_*(X)\to C_*(\mathcal{O}(r))\otimes_r C_*(X)^{\otimes r}, (s,a)\mapsto \widetilde{s}\otimes_r a^{\otimes r}.$$ We see a few things immediately:

  1. This map is not a map of chain complexes, but linear in the first argument.
  2. As we do not have to care about signs, we have $\tau_*a^{\otimes r}=a^{\otimes r}$ and the map is independent of the choice of lifts.
  3. If $c\in C_*(\mathcal{O}(r)/\mathfrak{S}_r)$ and $a\in C_*(X)$ are cycles, then $\Phi(c,a)$ is a cycle.
  4. If $c\in C_*(\mathcal{O}(r)/\mathfrak{S}_r)$ is a boundary and $a\in C_*(X)$ is a cycle, then $\Phi(c,a)$ is a boundary.

Putting all this together, we get a well-defined map $$\Psi:H_*(\mathcal{O}(r)/\mathfrak{S}_r)\times C^{\text{cyc}}_*(X)\to H_*(C_*(\mathcal{O}(r))\otimes_r C_*(X)^{\otimes r})\stackrel{\mu}{\to}H_*(X).$$

Now if we have $r=2$ (so we only have a $2$-fold covering) we can even show:

  1. If $c\in C_*(\mathcal{O}(r)/\mathfrak{S}_r)$ is a cycle and $a+a'\in C_*(X)$ is a boundary, then $\Phi(c,a)+\Phi(c,a')$ is a boundary.

This finally gives us a map $$Q:H_*(\mathcal{O}(2)/\mathfrak{S}_2)\times H_*(X)\to H_*(X)$$ which is used to define Dyer–Lashof operations: If $\mathcal{O}=\mathcal{C}_{n+1}$ is the little $(n+1)$-cube operad, then $\mathcal{C}_{n+1}(2)/\mathfrak{S}_2\simeq \mathbb{R}P^n$ and each $e_i\in H_i(\mathbb{R}P^n)$ gives an operation $Q_i:H_p(X)\to H_{2p+i}(X)$.

I was wondering if the last ingredient, namely (5), is only true for $r=2$. I have not found a counterexample for $r=3$ (still, everything over $\mathbb{Z}_2$), but I also could not find a proof.

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    $\begingroup$ I think it is easier to look at $r$ a power of $2$, so the next case being $r=4$. I believe that one should be able to get iterated Dyer-Lashof (Kudo-Araki rather since you are working at 2) operations. $\endgroup$
    – user43326
    Apr 17, 2019 at 15:47
  • $\begingroup$ Thank you! So do you think the statement is wrong for $r$ not being a power of $2$? $\endgroup$
    – FKranhold
    Apr 20, 2019 at 14:46
  • $\begingroup$ Well I only mean that the case $r$ being a power of 2 is the easiest case to investigate. Presumably for other values of $r$, we would get something related to Kudo-Araki operations via the product. $\endgroup$
    – user43326
    Apr 20, 2019 at 18:05

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