Recall: We present an operad (with $S_n$-action) in $R-Mod$ for any commutative ring $R$ not of characteristic 2 generated by a single element in degree $2$ satisfying the following identities:

  • $\theta+\theta\tau=0$
  • $\theta(1,\theta)+\theta(1,\theta)\sigma + \theta(1,\theta)\sigma^2$.

where $\tau$ and $\sigma$ are 2-cycles and 3-cycles respectively.

However, in characteristic $2$, this fails to characterize Lie algebras in the obvious way, since the first equation says that $\theta$ is skew-symmetric (and hence symmetric in characteristic 2).

The proper axiom to include is that $[x,x]=0$, i.e. that $[-,-]$ is alternating rather than skew-symmetric. Can we present this relation operadically? It seems like on the face of it, we can't, but I'd be happy to be surprised.

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    $\begingroup$ Operads only describe algebraic structures given by multilinear operations and equalities between their composites. Your suggested structure is generated by multilinear operations but has a quadratic relation which can not be expressed in multilinear terms. $\endgroup$ Commented Oct 17, 2010 at 15:18
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    $\begingroup$ I don't see what goes wrong if you define a "better" version of the Lie operad by taking the standard Lie operad over the integers, and modding out by symmetric trees of the form [x,x], as Gindi seems to be suggesting. Incidentally, the skew-symmetry axiom defines what's called a quasi-Lie algebra, so the standard Lie operad should really be called the quasi-Lie operad. $\endgroup$
    – Jim Conant
    Commented Oct 17, 2010 at 15:53
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    $\begingroup$ You don't have trees of the form $[x,x]$, trees will always accept any input and you cannot introduce a relation only when the two arguments are equal. Well you can but then you get outside of operads. $\endgroup$ Commented Oct 17, 2010 at 16:51
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    $\begingroup$ Harry, I don't know the answer to this question, but I'm inclined to agree with your intuition. There is a certain amount of work on which monads come from operads; in the case where the base category is $Set$, see for example appendix C.2 in Tom Leinster's book, where operads are described as equivalent to strongly regular finitary algebraic theories. There is also relevant information in the Appendix of Joyal's article in SLNM 1234; I don't recall whether his arguments handle a base category like vector spaces over Z mod 2, but if I were investigating this, these are places where I'd start. $\endgroup$
    – Todd Trimble
    Commented Oct 17, 2010 at 21:31
  • $\begingroup$ @Torsten: I see what you're saying now. $\endgroup$
    – Jim Conant
    Commented Oct 18, 2010 at 11:53

1 Answer 1


In fact, several monads can naturally be associated to an operad $P$ and this might be used to answer your question.

In the usual setting, one considers a generalized symmetric algebra $S(P,X) = \bigoplus_n (P(n)\otimes X^{\otimes n})_{\Sigma_n}$ where we form coinvariants under the action of the symmetric groups $\Sigma_n$. But we can also take invariants instead of coinvariants and form another functor $\Gamma(P,X) = \bigoplus_n (P(n)\otimes X^{\otimes n})^{\Sigma_n}$ associated to $P$. The image of the norm map from coinvariants to invariants still gives another functor $\Lambda(P,X)$ associated to $P$.

Under the assumption $P(0) = 0$, we have a monad structure on $\Lambda(P): X\mapsto\Lambda(P,X)$ and $\Gamma(P): X\mapsto\Gamma(P,X)$ inherited from the operadic composition structure of $P$. See (1.2.12-1.2.17) in http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf (ref.: http://www.ams.org/mathscinet-getitem?mr=2005g:18015)

For the operad $P = Lie$, the algebra category associated to $\Gamma(Lie)$ can be identified with the category of $p$-restricted Lie algebras (where $p$ is the cateristic of the ground ring), while the algebra category associated to $\Lambda(Lie)$ can be identified with the category of Lie algebras equipped with an alternating Lie bracket.

  • $\begingroup$ Thanks for the great references! After a first look at the paper, I haven't found the statement about $\Lambda(P)$, though. (Your paper references "On the Homotopy of Simplicial Algebras over an Operad", where I couldn't find the fact. Could you enlighten me? $\endgroup$ Commented Dec 9, 2011 at 15:11

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