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

Question

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-connected 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.

EDIT Idrissi kindly warns me that, in the presence of a non-simply connected $SO(n)$, this is the wrong notion of a topological model: I should have said weak homotopy equivalences instead of quasi-isomorphisms. (See the comments.)

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.

Answer 1

Let me summarize the comments. You have several possibilities:

• [Ryan's comment] You can consider the sub-operad $fD'_n \subset fD_n$ such that $fD'_n(r) = fD_n(r)$ for $r \ge 2$, and $fD'_n(1) = SO(n) \subset fD_n(1)$ is the subspace where the only little disk fills the whole ambient disk (i.e. is of radius $1$). Note that this only works if you consider the non-unital version of the framed little disk operad, i.e. with nothing in arity $0$; otherwise, this isn't a sub-operad.
• [Adrien's comment] For $n = 2$, you can consider $\mathcal{M}^{fr}_{0,\bullet+1}$, the moduli spaces of genus $0$ punctured Riemann surfaces with tangent vectors at the punctures.
• [my comment] Consider the Fulton–MacPherson compactification operad $\mathsf{FM}_n$. References include the original paper of Axelrod–Singer, the book of Lambrechts–Volić, or Sinha's paper Manifold-theoretic compactifications of configuration spaces. This operad admits a natural action of $SO(n)$, so you can take the semidirect product $\mathsf{FM}_n^fr = \mathsf{FM}_n \rtimes SO(n)$. Since $\mathsf{FM}_n(1)$ is a singleton, it follows that $\mathsf{FM}_n^{fr}(1) = SO(n)$.