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1 vote

Poisson and homotopy Poisson operads

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 …
Vladimir Dotsenko's user avatar
3 votes
Accepted

Are there examples of brace algebras that are not operads?

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 …
Vladimir Dotsenko's user avatar
10 votes
2 answers
214 views

Degree 8 multilinear operations on Jordan algebras

I am interested in the dimension, or, even better, in the $S_8$-module structure of the space of degree 8 multilinear operations on Jordan algebras. Recall that a Jordan algebra is a commutative but n …
1 vote

Degree 8 multilinear operations on Jordan algebras

I managed to run Albert on a very powerful computer at work, and the computation of the desired dimension converged: it seems equal to 19089. I would very much like to confirm that it is correct (I am …
Vladimir Dotsenko's user avatar
8 votes
0 answers
111 views

Identity for the associator involving a third root of unity

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 …
3 votes
Accepted

Differential of the Twisted complex for algebraic operads

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. …
Vladimir Dotsenko's user avatar
5 votes
Accepted

Infinity-homotopies

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. …
Vladimir Dotsenko's user avatar
4 votes
Accepted

What is the relation between 2-Gerstenhaber, CohFT, and Gerstenhaber geometrically?

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 …
Vladimir Dotsenko's user avatar
2 votes

Analogy of Gerstenhaber algebra

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 …
Vladimir Dotsenko's user avatar
5 votes

Affine spaces as algebras for an operad?

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 …
Vladimir Dotsenko's user avatar
9 votes
Accepted

How to define the equivalence of Maurer-Cartan elements in an $L_{\infty}$-algebra?

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 …
agt's user avatar
  • 4,306
8 votes
Accepted

Does this notion related to species/operads/FI-modules have a name?

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 …
Vladimir Dotsenko's user avatar
17 votes
1 answer
583 views

Positivity of coefficients of the inverse of a certain power series

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 …
3 votes
Accepted

Homotopy transfer in the opposite direction

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\ …
Vladimir Dotsenko's user avatar
17 votes
Accepted

Is the Amitsur-Levitzki identity essentially unique?

(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). …
Vladimir Dotsenko's user avatar

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