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](http://web.math.ku.dk/~wahl/SalvatoreWahl.pdf).


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](https://bookstore.ams.org/surv-217/) to encompass it: see [Extended](https://www.degruyter.com/abstract/j/gmj.2018.25.issue-4/gmj-2018-0061/gmj-2018-0061.xml)


**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.



A model for the framed little disks operad $f{\cal D}_n$ with arity one *equal* to $SO(n)$?
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 $\mathsf{FM}\_n^{fr}(1) = SO(n)$.


