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Oct 16, 2014 at 9:50 comment added Vladimir Dotsenko @SamueleGiraudo: it is a symmetric operad which is not regular, so knowing the free algebra on one generator is not equivalent. In any case, to the best of my knowledge the answer to your question is no, even though it may be quite feasible.
Oct 16, 2014 at 9:50 comment added Vladimir Dotsenko @MarianoSuárez-Alvarez: This operad is not Koszul. The Koszul dual has the Hilbert series $f(t)=t+t^2+t^3/2+t^4/24$ (and nothing else, this operad is nilpotent), and the compositional inverse of $-f(-t)$ has negative coefficients (e.g. at $t^{12}$), contradicting the Ginzburg--Kapranov functional equation.
Apr 25, 2013 at 16:05 comment added Samuele Giraudo Thanks. Is there a realization of this operad (a basis and an explicit definition for its composition maps $\circ_i$--or equivalently a description of the free algebra over this operad on one generator)?
Apr 24, 2013 at 8:55 answer added Carlo Beenakker timeline score: 3
Apr 23, 2013 at 23:19 comment added Gjergji Zaimi @Samuele, if you want a definition of flexible algebras that doesn't have several occurrences of a variable in the same monomial, notice that an algebra is flexible iff $(x,y,z)+(z,y,x)=0$. The parenthesis denote associators.
Apr 23, 2013 at 23:19 comment added Mariano Suárez-Álvarez The flexible identity is equivalent (over a field with more than two elements) to the multilinear identity $$(xy)z-x(yz)+(zy)x-z(yx)=0,$$ so you can express flexible algebras are algebras over an quadratic operad with a cubic relation. (Can someone see if this is Koszul?)
Apr 23, 2013 at 22:25 history edited user9072
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Apr 23, 2013 at 22:21 comment added Samuele Giraudo In spite of the fact that the flexibility relation involves two occurrences of $x$, is there an operad governing flexible algebras?
Apr 23, 2013 at 22:01 history asked Mariano Suárez-Álvarez CC BY-SA 3.0