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There surely are many answers to this question... For me, one of the key reasons is that there are lots of situations where existing geometric and algebraic structures exhibit some kind of associativi …

Operads that are quotients of the associative operad (which I believe your updated question is aimed at) are, in characteristic zero, fairly extensively studied since the 1950s or so, under the somewhat … Feel free to e-mail me if you are interested in some particular operads, maybe I happen to know precisely what you want to find out. …

To record my answer in comments properly: brace algebras coming from operads satisfy one obvious constraint: for every $x$ and sufficiently large $n$ we have $x\{x_1,\ldots,x_n\}=0$, since we cannot plug …

A non-cofibrant dg operad whose homology is $\mathrm{Pois}$ and which looks like $\mathrm{Com}\circ\mathrm{Lie}_\infty$ appears in Section 4.1 of this paper of Anton Khoroshkin and Pedro Tamaroff, the …

For $K=\mathbb{R}$, the positive part of your operad (mentioned in Gabriel's comment), and its algebras have been discussed by Tom Leinster and others in connection to entropy. See, for example, http …

One possible answer is contained in the paper of Victor Ginzburg and Travis Schedler, "Free products, cyclic homology, and Gauss-Manin connection", https://arxiv.org/abs/0803.3655. You will be in part …

I don't know if you found an answer since you posted the question, but I will write this just in case: there is a "cute" (easy) definition in case of nonsymmetric operads which generalises the A-infinity … story rather trivially (derivation homotopy), while for symmetric operads the definition is more involved. …

Depending on whether you want it to agree with the symmetric structure or only with monoidal structure, this would be usually referred to, respectively, as twisted commutative algebras or twisted asso …

In other words, the (co)associativity for (co)operads has a sequential axiom and a parallel axiom, and these terms vanish because of the parallel axiom. …

(Alternatively, one can use the language of operads to discuss that; I prefer that latter language but choose to write a more "classical" definition here). …

Consider the unique formal power series $g(z)$ with $g(0)=0$ and $g'(0)=1$ satisfying the equation $$ g(z)-g(z)^8+g(z)^{15}=z, $$ that is the inverse of $$ z-z^8+z^{15} $$ in the group of formal …

This is a reference request. I came across the class of nonassociative algebras satisfying the following identity: $$ (a,b,c)+\omega(b,c,a)+\omega^2(c,a,b)=0. $$ Here: by an "algebra" I mean a vect …

Let us denote by $p\colon Y\to X$ and $i\colon X\to Y$ the maps of your SDR. Since $pi=\mathop{\mathrm{id}}\nolimits_X$, the map $i$ is injective, and is an isomorphism with its image. The map $\pi=i\ …

To add a bit to what Damien says, addressing your question on how to generalise the gauge approach (which is equivalent to the approach outlined by Damien, as proved by several people): You can view …

Well, for your question 1 you presumably may ask yourself first about a relationship between (shifted) Lie algebras and Lie 2-algebras. Lie 2-algebras of Hanlon and Wachs can be viewed as $L_\infty$-a …