A model for the framed little disks operad $f{\cal D}_n$ with arity one *equal* to $SO(n)$?

The framed little disks operad $$f{\cal D}_n$$ can be described as the semidirect product $${\cal D}_n \rtimes SO(n)$$, where $${\cal D}_n$$ is the little disks operad and $$SO(n)$$ is the special orthogonal group, or rotation group.

As a topological space, forgetting the operad structure, this simply means that, in each arity $$k$$, we have

$$f{\cal D}_n (k) = ({\cal D}_n \rtimes SO(n))(k) = {\cal D}_n (k) \times SO(n)^k \ .$$

See for instance, Salvatore-Wahl.

So, in arity one, we get

$$f{\cal D}_n(1) = {\cal D}_n(1) \times SO(n) \ ,$$

which is easily seen to be homotopically equivalent to $$SO(n)$$, since $${\cal D}_n(1)$$ is contractible:

$$f{\cal D}_n(1) \simeq SO(n) \ .$$

This is pretty exciting, because it’s an example of a non-unitary topological operad; that is, $$P(1) \neq *$$, and Fresse had to do a nice effort to adapt his Sullivan rational homotopy theory for topological operads Homotopy of operads to encompass it: see Extended

Now my question is the following: is there any other topological model $$P$$ of $$f{\cal D}_n$$ for which $$P(1)$$ is actually $$P(1) = SO(n)$$—not just homotopically equivalent?

By “topological model” I mean, any topological operad $$P$$ that can be joined with $$f{\cal D}_n$$ through a chain of quasi-isomorphisms (topological operad morphisms inducing isomorphisms in homology) like

$$P \stackrel{\sim}{\longleftarrow} \cdot \stackrel{\sim}{\longrightarrow} \cdot \dots \stackrel{\sim}{\longleftarrow}\cdot \stackrel{\sim}{\longrightarrow} f{\cal D}_n \ .$$

And when I say “homology”, I mean homology with coefficients in a zero characteristic field: over the real numbers, for instance, would be fine.

I guess a brute-force approach—just delete $${\cal D}_n(1)$$ from the definition, leave $$SO(n)$$ and compose with the homotopy equivalence—can not work because you would destroy the operad structure.

If $d_{n} = \frac{{\beta}_{n}}{10^n}$ where ${\beta}_{n}$ takes integer values between 0 and 9, does $\Sigma_{n=1}^{\infty}d_{n}$ converge?

The series looks like a converging geometric series: $$\Sigma_{n=1}^{\infty}\frac{1}{10^n} = \Sigma_{n=0}^{\infty}\frac{1}{10^n} – 1 = \frac{1}{9}$$ Where the terms are being arbitrarily multiplied by constants between $$0$$ and $$9$$. I’m not sure about how this affects the convergence of a series. I suspect that it converges because the constants eventually become small in comparison to the value of the geometric terms, but I’m not sure how I could prove it.

The “worst case” would be one where $$\beta_{n} = 9$$ for all $$n$$. In that case:

$$\Sigma_{n=1}^{\infty}d_{n} = 9\Sigma_{n=1}^{\infty}\frac{1}{10^n} = 1$$

So again I’m inclined to think it converges in any case.

Thanks.