There are two somewhat widely known theorems which say

  • if $A$ is a nonnegatively graded commutative algebra in char $0$, then forgetful map on operadic cohomology $H^*_{Harr}(A, A) \to H^*_{Hoch}(A, A)$ is injective

  • if $L$ is a Lie algebra in char $0$, then $H^*(L, L) \to H^*(L, UL)$ is injective

Are they related at all? Is it a part of more general phenomenon occuring for exact sequences of (probably Koszul) operads?

(Also, it's obvious that latter theorem does not care about grading at all, just because $L \to UL$ is split injection of modules. Is it true for former one? I gess not.)


The second result you mention is significantly easier than the first. Indeed from the PBW theorem we know that $L$ is a direct summand of $UL$ as an $L$-module, and the second result follows. But in fact there is a strong connection between the two results, since the first theorem may also be seen as a consequence of a sufficiently abstract version of the PBW theorem. This is much less well known and I haven't seen it written down anywhere.

(In particular the phenomenon is not something general for any sequence of Koszul operads - you really need a PBW theorem in the background.)

Here is how it goes. Via the map $\mathsf{Lie} \to \mathsf{Ass}$ we may consider $\mathsf{Ass}$ as a bimodule over the Lie operad. Recall that if $P$ is an operad, then a $P$-bimodule is the same thing as a $P$-algebra in the category of right $P$-modules, so we may think of $\mathsf{Lie} \to \mathsf{Ass}$ as a morphism of Lie algebras in a certain tensor category. In fact we may identify $\mathsf{Ass}$ with the universal enveloping algebra of $\mathsf{Lie}$ in this category, considered as a Lie algebra. By a general form of the PBW theorem (for Lie algebras in abstract symmetric tensor categories over a field of characteristic zero) it then follows that there exists a splitting $\phi \colon \mathsf{Ass} \to \mathsf{Lie}$ in the category of modules over the Lie operad, considered as an algebra over itself. Unraveling this, this means that $\phi$ is a map of infinitesimal bimodules.

Explicitly, the fact that $\phi$ is a morphism of infinitesimal bimodules means that we have two commutative pentagons. The first says that the composition $$ \mathsf{Lie} \circ_{(1)} \mathsf{Ass} \to \mathsf{Ass} \circ_{(1)} \mathsf{Ass} \to \mathsf{Ass} \stackrel\phi\to \mathsf{Lie}$$ coincides with $$ \mathsf{Lie} \circ_{(1)} \mathsf{Ass} \stackrel{\mathrm{id}\circ_{(1)} \phi}\to \mathsf{Lie} \circ_{(1)} \mathsf{Lie} \to \mathsf{Lie}.$$ The other says the same thing except with $\mathsf{Lie}$ acting on $\mathsf{Ass}$ on the right.

Now let $A$ be a $C_\infty$-algebra. The $C_\infty$-structure is given by a Maurer--Cartan element $\mu$ in the pre-Lie algebra $\mathfrak g := \mathrm{Hom}_{\mathbb S}(\mathsf{coLie},\mathsf{End}_A)$, which is essentially the Harrison chain complex. The pre-Lie product $f \star g$ of two Harrison chains is given by the composition $$ \mathsf{coLie} \to \mathsf{coLie} \circ_{(1)} \mathsf{coLie} \stackrel{f \circ g}\to \mathsf{End}_A \circ_{(1)} \mathsf{End}_A \to \mathsf{End}_A.$$ We also have the associative version $\mathfrak h := \mathrm{Hom}_{\mathbb S}(\mathsf{coAss},\mathsf{End}_A)$ with analogously defined pre-Lie product. Via the map $\mathsf{coAss} \to \mathsf{coLie}$ we can think of $\mathfrak g$ as a pre-Lie subalgebra of $\mathfrak h$. Now the dual of $\phi$ induces a map $\phi^\ast \colon \mathfrak h \to \mathfrak g$, which is not in general a morphism of pre-Lie algebras. However, the fact that $\phi$ is a map of infinitesimal bimodules is exactly the condition needed for $\phi^\ast$ to satisfy a "projection formula": for $f \in \mathfrak g \subset \mathfrak h$ and $g \in \mathfrak h$ we have $\phi^\ast(f \star g) = f \star \phi^\ast(g)$ and $\phi^\ast(g \star f) = \phi^\ast(g) \star f$. Indeed if we write out the definitions of the pre-Lie product then the conditions we need for the projection formula to hold become exactly that we have a map of infinitesimal bi-comodules $\mathsf{coLie} \to \mathsf{coAss}$.

In particular we have the Harrison differential on $\mathfrak g$ given by $df = f \star \mu - (-1)^{\vert f\vert}\mu \star f$ and the analogous Hochschild differential on $\mathfrak h$. The previous paragraph says in particular that the splitting $\mathfrak h \to \mathfrak g$ is compatible with these differentials, so that Harrison chains are a direct summand of Hochschild chains.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.