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.

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    $\begingroup$ Your question has an easy answer in the affirmative. Replace $fD_n(1)$ with the subspace $SO(n)$. i.e. these are the automorphisms of a single disc, and we throw out all the proper endomorphisms of the disc. Keep all the other arities the same, i.e. retain $fD_n(k)$ for all $k \geq 2$. This is a topological operad equivalent to the framed discs operad. $\endgroup$ – Ryan Budney Jun 21 '19 at 19:26
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    $\begingroup$ @Budney Oh, you mean including the "deleted" framed little disks operad into the framed little disks operad, not projecting through the homotopy equivalence? -Silly me: it sounds great. Thank you. I'll check it. $\endgroup$ – Agustí Roig Jun 21 '19 at 19:34
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    $\begingroup$ Another one would be the framed Fulton–MacPherson operad. Let me also mention that your notion of "topological model" is rather strange. Typically one would require the maps to be (weak) homotopy equivalences. For simply connected spaces what you have defined a rational model, but $SO(n)$ is not simply connected. $\endgroup$ – Najib Idrissi Jun 21 '19 at 23:20
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    $\begingroup$ I agree with Najib the FM "compactification" (which in that case is not actually compact, the name's a bit misleading) of the framed configuration space should do the trick. For $n=2$ one can also look at the moduli space of genus 0 punctured Riemann surfaces with tangent vectors at the punctures. $\endgroup$ – Adrien Jun 22 '19 at 8:16
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    $\begingroup$ @Adrien I think it's compact, no? $FM_n(r)$ is a compact space, and $FM_n^{fr}(r) = FM_n(r) \times SO(n)^r$ is a product of compact spaces. $\endgroup$ – Najib Idrissi Jun 26 '19 at 14:49

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 $\mathsf{FM}_n^{fr}(1) = SO(n)$.
  • $\begingroup$ Thank you very much, Idrissi. Now, I have all I needed. :-) $\endgroup$ – Agustí Roig Jun 26 '19 at 15:07

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