$k$-Disk algebras versus $E_k$ algebras Background:
The little $k$-cubes operad is the $(\infty,1)$-operad defined by embedding disjoint unions of $k$-dimensional open cubes rectilinearly into one another, that is using maps $(0,1)^k\rightarrow (0,1)^k$ of the form $(x_i)\mapsto (a_i i_k+b_i)$ for $x_i\in (0,1)$ and $a_i\geq 0$ for $i=1,...,k$. 2-morphisms are given by isotopies of embeddings, 3-morphisms are isotopies of isotopies etc. Call this operad $\square^k$. Disjoint union equips $\square^k$ with a symmetric monoidal structure.
The $(\infty,1)$-category of $E_k$-algebras with values in a symmetric monoidal $(\infty,1)$-category $(C,\otimes)$ is defined as the $(\infty,1)$-category of symmetric monoidal functors $\text{Fun}^{\otimes}(\square^k,C)$.
The little $k$-disks $(\infty,1)$-operad is similarly defined as the operad of framed embeddings of open(?) disks into one another, with (higher) isotopies as (higher) morphisms. Call this operad $\text{Disk}_k^{fr}$.
The $(\infty,1)$-category of $k$-disk algebras with values in a symmetric monoidal $(\infty,1)$-category $(C,\otimes)$ is again the $(\infty,1)$-category of symmetric monoidal functors $\text{Fun}^{\otimes} ( \text{Disk}_k^{fr},C)$
The question
On the nLab, it is written (see here and here) that the little $k$-cubes operad and the little $k$-disks operad are distinct objects, and the latter is a generalization of the former. The main difference I notice is that the little $k$-disks operad allows one to rotate disks when embedding them, while the little $k$-cubes operad does not. Still, rotations are homotopic to the identity, so it seems (to me) safe to assume that the operads are equivalent as $(\infty,1)$-categories. Am I mistaken?
Also, according to Ginot's notes (page 27, Example 12), the $(\infty,1)$-categories of algebras of $\text{Disk}_k^{fr}$ and $\square^k$ are equivalent, and this leads me to believe that the operads themselves should be equivalent.
I would not be suprised of the above reveals a severe lack of understanding on my part. I am just starting to try to understand these gadgets. Any help will be greatly appreciated.
 A: There is an unfortunate clash of terminologies here. Traditionally, the little discs operad comes in two variants:


*

*the "usual" $\mathtt{D}_n$: the space of arity $r$ operations consists of embeddings of  that do not allow rotations (with some other conditions). In other words, such embeddings $D^n \hookrightarrow D^n$ must preserve the framing.

*the "framed" version $\mathtt{fD}_n$: here the embeddings are allowed to rotate the disks, and do not necessarily preserve the framing. Basically it is $\mathtt{D}_n$ together with an action of $\mathrm{SO}(n)$ (in fact it's a semi-direct product $\mathtt{D}_n \rtimes \mathrm{SO}(n)$, see P. Salvatore and N. Wahl, Framed discs operads and Batalin-Vilkovisky algebras. Q. J. Math., 2003, 54, 213-231").


These two operads are not weakly equivalent, and their categories of algebras are different. To give you an idea, $H_*(\mathtt{D}_2) = \mathtt{Ger}$ is the operad of Gerstenhaber algebras, whereas $H_*(f\mathtt{D}_2) = \mathtt{BV}$ is the operad of BV-algebras -- morally we have a circle action in addition. More generally, $\mathtt{D}_n(1)$ is contractible, whereas $\mathtt{fD}_n(1) \simeq \mathrm{SO}(n)$ is non-contractible, so the operads cannot be weakly equivalent.
The first operad $\mathtt{D}_n$ is actually equivalent to $\mathrm{Disk}_n^{\mathrm{fr}}$. This makes perfect sense in this context: $\mathtt{D}_n$ is equivalent to the endomorphism operad of $\mathbb{R}^n$ in the category of framed manifolds and embeddings, and you can take the factorization homology of a $\mathtt{D}_n$-algebra only on a framed manifold.
On the other hand, $\mathtt{End}_{\mathbb{R}^n} = \mathrm{Disk}_n$ in the category of unoriented manifolds and embeddings is equivalent to an operad slightly larger than $\mathtt{fD}_n = \mathtt{D}_n \rtimes SO(n)$, I think it is $\mathtt{D}_n \rtimes O(n)$. Its endomorphism operad in the category of oriented manifolds $\mathtt{End}^{\mathrm{or}}_{\mathbb{R}^n}$ is weakly equivalent to $\mathtt{fD}_n$.
Unfortunately, as you can see, the two occurrences of "framed" refer to different things, and are applied in opposite manners. As far as I know, a recent trend in some circles is to do away with the terminology "framed little discs operad" altogether.

With all that being said, it is indeed true that $\mathtt{D}_n \simeq \mathrm{Disk}_n^{fr}$ is equivalent to the operad $\square^n$ of little $n$-cubes (where you don't allow rotations, of course). A few possible references:


*

*R. Steiner, A canonical operad pair, Math. Proc. Cambridge Philos. Soc. 86
(1979), 443–449.

*C. Berger, Opérades cellulaires et espaces de lacets itérés. Ann. Inst. Fourier
46 (1996), 1125–1157.


It's actually not so easy to prove. It's easy to see that they are arity-wise equivalent: in arity $r$, both spaces are equivalent to the configuration space $\operatorname{Conf}_r(\mathbb{R}^n)$ of $r$ ordered points in $\mathbb{R}^n$. It's finding an equivalence that respects the operad structure that is difficult (but possible).
