Let $\mathrm{Ass} $ denote the operad, whose algebras are associative unital algebras, considered as a dg-operad. Denote $\mathrm{Ch} $ the category of chain complexes over a commutative ring $\mathrm{R} $ and denote $\mathrm{Bialg}_{ } (\mathrm{Ch} )$ the category of cocommutative (counital and unital) bialgebras in $\mathrm{Ch}. $

Let $\mathcal{O}$ be a dg-operad and $\phi: \mathcal{O} \to \mathrm{Ass} $ a map of dg-operads. $\phi$ induces a free-forgetful adjunction $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \rightleftarrows \mathrm{Alg}_{ \mathrm{Ass} } ( \mathrm{Ch} ).$

When does the left adjoint $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Alg}_{ \mathrm{Ass} } (\mathrm{Ch} ) $ lift to a functor $ \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \to \mathrm{Bialg}_{ } (\mathrm{Ch} )? $

For $\mathcal{O}$ the Lie operad there is such a lift.

For $\mathcal{O}$ the Lie operad the left adjoint $\mathcal{U}$ is the enveloping algebra. For every Lie-algebra $\mathrm{X}$ the bialgebra structure on $\mathcal{U}(\mathrm{X})$ is encoded in a symmetric monoidal structure on the category $\mathrm{LMod}_{ \mathcal{U}(\mathrm{X}) } (\mathrm{Ch} )$ (being the category of representations of $\mathrm{X}$) lifting the symmetric monoidal structure on $\mathrm{Ch} .$

The category of representations of $\mathrm{X}$ can also be described as the category $\mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) $ of $\mathrm{X}$-modules over the Lie-operad.

This leads to my 2. question: For which dg-operads $ \mathcal{O}$ and $ \mathcal{O}$-algebras $\mathrm{X}$ does the symmetric monoidal structure on $\mathrm{Ch} $ lift to a symmetric monoidal structure on $\mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) ?$

More generally given a dg-operad $\mathcal{O}$ and a $\mathcal{O}$-algebra $\mathrm{X}$ there is a enveloping associative algebra $\mathcal{U}_{ \mathcal{O}}( \mathrm{X} ) $ with the property that we have a canonical equivalence $$\mathrm{LMod}_{\mathcal{U}_{ \mathcal{O}}( \mathrm{X} ) } (\mathrm{Ch} ) \simeq \mathrm{Mod}^{\mathcal{O}}_{ \mathrm{X} } (\mathrm{Ch} ) .$$

As a cocommutative bialgebra structure on an associative algebra corresponds to a symmetric monoidal structure on its category of left modules lifting the symmetric monoidal structure on $ \mathrm{Ch} $, my 2. question is equivalent to the following:

For which dg-operads $ \mathcal{O}$ does the enveloping associative algebra functor $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Alg}_{ \mathrm{Ass} } (\mathrm{Ch} ) $ lift to a functor $ \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch} ) \to \mathrm{Bialg}_{ } (\mathrm{Ch} )? $

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    $\begingroup$ Interesting. I think the second question is good, but the first is a red herring. Yes, you can form the universal enveloping algebra from the map Lie -> Ass, however this isn't the definition that extends to all operads (see section 12.3.10 of the book Algebraic Operads). Perhaps if you rewrote the question in this form you would get an answer, unfortunately I don't know how to answer it. $\endgroup$ Apr 9, 2018 at 10:25

1 Answer 1


I was going to post this as a comment but rambled on too long. I haven't check the details of the following.

To every operad $O$ there's an algebra $A(O)$ in the category of right $O$-modules created by marking one of the inputs as special. The algebra structure is given by composing two elements of $A(O)$ along this special input and the right $O$-module structure given by composing with an element of $O$ in any non-special input. If this algebra is a bialgebra in the category of right $O$-modules and furthermore $A(O)$ is a free right $O$-module then I believe the enveloping algebra functor extends to the category of bialgebras. This is because $A(O)\circ_O\mathfrak{g}$ is the enveloping algebra of an $O$-algebra $\mathfrak{g}$.

The reason for asking that $A(O)$ is free is so that the functor $(-)\circ_O \mathfrak{g}$ from right O-modules to vector spaces is monoidal. However on second thoughts the functor might be monoidal without this condition. I would however be (pleasantly) surprised if you could find an operad $O$ whose algebra $A(O)$ was a bialgebra and not free as a right $O$-module.

Having said that I'm now doubting that $A(Lie)$ is free as a right $Lie$-module if we work over the integers. It is free if you work over rationals or with shuffle collections, but perhaps not with symmetric collections if you can't make the symmetric group actions respect the generators.

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    $\begingroup$ $A(Lie)$ is not free as a right $Lie$-module in positive characteristic. What you call $A(O)$ is denoted $O[1]$ in Fresse's book Modules over Operads and Functors. In Proposition 10.2.7 he proves that $L[1]$ is free as a right Lie module in characteristic zero, but in positive characteristic it is stated that the failure comes from the failure of part of the PBW theorem (which is only an isomorphism $\operatorname{gr} T(X) \cong S(L(X))$). $\endgroup$ Apr 13, 2018 at 11:20
  • $\begingroup$ I am interested in the case of Lie algebras over the integers or more generally Lie algebras in spectra: $\endgroup$ Apr 14, 2018 at 19:15
  • $\begingroup$ Here I take $\mathcal{O}$ to be (a cofibrant replacement of) the spectral Lie operad, which gives the derivatives of the identity of the category of spectra an operad structure and is a shift of the Koszul dual of the cocommutative cooperad in spectra. Then I take the enveloping algebra functor $ \mathcal{U} : \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Alg}_{ \mathrm{Ass} } (\mathrm{Ch} )$ and ask if it lifts to a functor $ \mathrm{Alg}_{ \mathcal{O} } ( \mathrm{Ch}) \to \mathrm{Bialg} (\mathrm{Ch} ).$ $\endgroup$ Apr 14, 2018 at 19:15
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    $\begingroup$ Dear James, you may be aware of it but in here Anton Khoroshkin gives a few examples of operads $P$ such that $\partial P$ (what you write $A(P)$, which I write using the derivative functor on Joyal's species) is not a free right $P$-module. For example, this is true in case $P$ is the Poisson operad, the Permutative operad, and the Leibniz operad. The proof uses the "PBW framework" of Dotsenko and myself. $\endgroup$
    – Pedro
    May 22, 2021 at 20:20

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