Let $n>0$. At the prime $p=2$ the Snaith splitting of $\Omega^kS^{n+k}$ realises the hight/weight filtration on the homology $H_*(\Omega^kS^{n+k};\mathbb{Z}/2)$. If I choose only the pieces $D_{2^r}(\mathbb{R}^k,S^n)$ with $r\geqslant 0$ then homologically, just for the algebra structure, it seems I can recover $H_*(\Omega^kS^{n+k};\mathbb{Z}/2)$ from the free commutative algebra that homology of these pieces generate. Now, the space $\bigvee_{r=0}^{+\infty}D_{2^r}(\mathbb{R}^k,S^n)$ fails to be a monoid under the obvious pairings $D_r\wedge D_s\to D_{r+s}$ as it is not closed really!

I wonder if one can put a multiplicative monoid structure on this wedge so that its group-completion is weak equivalent to $\Omega^kS^{n+k}$.


Your initial statement is not correct. It is easiest to describe what happens by thinking of $H_*(\Omega^kS^{n+k})$ as a bigraded ring with $H_iD_j$ in bidegree $(i,j)$. (All my homology is with coefficients $\mathbb{Z}/2$.) Then $H_*(\Omega^2S^3)$ is polynomial with generators $x_i$ of bidegree $(2^{i+1}-1,2^i)$ for $i\geq 0$. So your putative space of generators contains all monomials whose second degree is a power of two, such as $x_i^{2^j}$ or $x_5^2x_6x_7$. Clearly these do not generate the whole ring freely.

  • $\begingroup$ You think it is the free part of my statement that is not correct, right! Since, for a given monomial $H_*\Omega^kS^{n+k}$ I can decompose it to a product of elements of height/weight $2^i$, but not in a unique way, right! $\endgroup$ – user51223 Jul 9 '18 at 7:06
  • $\begingroup$ Would it possible to correct this and make something out of these $2^i$-adic constructions on $S^n$ so that we approximate $\Omega^kS^{n+k}$ when localised at $2$? My own guess is that probably NO! But, I also feel that maybe one can use some transfer maps to relate these pieces to each other?! $\endgroup$ – user51223 Jul 9 '18 at 7:32

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