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.