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If $O$ is a cyclic operad, it can be regared as a modular operad $P$ with $P(g,n)=0$, for $g >0$. So we have cobar dual $BO$ and Feynman transform $FP$(with trivial cocycle). Is there any relationship between them? I guess

$Cyc(FP) = BO$

where $Cyc$ is the restricting or forgetting operator from modular operads to cyclic operads by keeping only its $g=0$ part.

Section 5.9 of Ginzburg-Kapranov has a related statement but there is a shift I don't understand.

Another related question is: what is the relationship if we replace cobar duality with dg duality? I am using the conventions of Getzler-Kapranov and Ginzburg-Kapranov.

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Your statement, up to the appropriate twists, is (5.9) in Getzler–Kapranov.

But let me take this opportunity to advertise the framework in Ben Ward's paper Six operations formalism for generalized operads that demonstrates a slightly more general result as a coherent part of a more general theory of passing among different types of operad-like structures.

One part of Corollary 8.2 implies the statement that you made. This part of the Corollary (which is also Corollary 1.4) says that the modular envelope of $BO$ is naturally isomorphic to the Feynman transform of the ``extension by zero'' of $O$ to a modular operad (taking appropriate twists into account). Using Ward's notation, your $P$ is $L^!O$ and your $FP$ is $DL^!O$ while your $BO$ is $DO$. Ward's statement, $DL^!O\cong LDO$, is at the level of modular operads, but applying the truncation ($Cyc$ in your notation, $R$ in Ward's), yields $$ Cyc(FP)=RDL^!O\cong RLDO\cong DO=BO $$ because $RL$ is naturally isomorphic to the identity.

Ward also gives a statement that he considers more obvious and less interesting, that $DR\cong RD$. This statement means, in your notation, that if you start with the modular operad $P$ instead of starting with the cyclic operad, that $$ Cyc(FP)=RDP\cong DRP=B(Cyc(P)). $$

The notation for $D$ (including twists) is discussed in Appendix A.2, using the notation of this paper by Kaufmann and Ward.

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  • $\begingroup$ Thanks!actually I konw statment 5.9, but that is different from what I posed here, it has an additional suspension applying to BO, right? Is what I posed here wrong? I dont quite understand why 5.9 needs a suspension operation.I have edited the question and want to know the relation for dg duality as well.Can you answer this question as well? $\endgroup$
    – Hao Yu
    Commented Apr 8, 2017 at 16:09
  • $\begingroup$ It's true that 5.9 is elliptical. Ward's theorem works just fine in the dg category but I don't know what you mean by "dg duality." If you want the details about shifts and signs then you need to be clearer about what conventions you are using. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Apr 8, 2017 at 16:17
  • $\begingroup$ dg duality is in the paper " Koszul dyality for operads " . it is defined just after the definition of cobar duality. Again, is 5.9 right? Can you elaborate more on 5.9 because i dont quite understand what it means. I cant see why it needs a shift. Yes, I am in dg category. $\endgroup$
    – Hao Yu
    Commented Apr 8, 2017 at 16:26
  • $\begingroup$ I don't see a precise definition for the shift in 5.9 but a shift is definitely necessary given your conventions. The Feynman transform $FP$ is not a modular operad but a $\mathfrak{K}$-twisted modular operad and the forgetful functor $Cyc$, which Getzler–Kapranov use to forget to cyclic $\mathbb{S}$-modules, NOT cyclic operads, cannot be naively lifted to cyclic operads because the signs of the composition are wrong. (Continued) $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Apr 8, 2017 at 16:43
  • $\begingroup$ That these two things need some kind of shift can be seen easily by just looking at degrees of the underlying modules. If you look at 3.2.8 in Koszul duality for Operads, you see that the underlying complex of $BO(n)$ is $F(O^*[-1])(n)$, that is, there is a shift in the generators. On the other hand, according to the first sentences of (5.1) in Getzler–Kapranov, the underlying stable $\mathbb{S}$-module is $\mathbb{M}_{\mathfrak{D}^\vee}(\mathcal{A}^*)$ which has no shift on the operad $\mathcal{A}$. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented Apr 8, 2017 at 16:45

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