Multiplicative monoid structures?

Question

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 $\Omega^kS^{n+k}$.

Answer 1

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.