# Conceptual explanation for the sign in front of some binary operations

In several situations, I've seen that given a binary operation on a graded module $$m:A\otimes A\to A$$, a new operation $$M(x,y)=(-1)^{|x|}m(x,y)$$ is defined so that it satisfies some properties.

One example of this happens in Homotopy G-algebras and moduli spaces, where for a binary operation $$m\in\mathcal{O}(2)$$ such that $$m\circ m=0$$ for some operad $$\mathcal{O}$$, an associative product is defined by $$xy=(-1)^{|x|+1}m\{x,y\}$$, where the brace notation stands for the brace algebra structure on $$\mathcal{O}$$. In this case, the explanation I've been able to deduce is that this is necessary for the brace relation (equation (2) in the paper) to imply associativity of the product $$xy$$. In this case the sign $$(-1)^{|x|}$$ works for this purpose too.

Another more direct instance of this situations occurs in Cartan homotopy formulas and the Gauss-manian connection in cyclic homology, where given an $$A_\infty$$-algebra with $$m_i=0$$ for $$i>2$$, one gets a dg-algebra by defining again $$xy=(-1)^{|x|}m_2(x,y)$$. In this case this is because the author uses a convention for $$A_\infty$$-algebras in which the equations only have plus signs, so some extra sign is needed to produce the associativity relation and the Leibniz rule. So the reasons are very similar to the previous case even though the construction is simpler because there is no brace algebra here.

And another extra example for which I don't have any reference is in the case of Lie algebras. When one defines a generator of the operad of graded Lie algebras, often one takes $$l(x,y)=(-1)^{|x|}[x,y]$$ instead of directly defining $$l$$ as the bracket. If I remember correctly this was needed to obtain the Jacobi identity in purely operadic terms.

So it looks like it's very common to add that sign in order to make some relations hold. What I would like to know if there is a more conceptual explanation of why this holds systematically. Maybe it's just that it works when writing down the equations, but I'm looking for a more general intuition.

My motivation is generalizing this idea to maps of higher arity. More precisely, given an $$A_\infty$$-multiplication $$m\in\mathcal{O}$$ such that $$m\circ m=0$$, I want to define an $$A_\infty$$-structure $$M$$ on $$\mathcal{O}$$ that satisfies the sign convention

$$\sum_{n=r+s+t}(-1)^{rs+t}M_{r+1+t}(1^{\otimes r}\otimes M_s\otimes 1^{\otimes t})=0.$$

(There is also another possible convention where $$rs+t$$ is replaced by $$r+st$$)

So this is very similar to Getzler's paper where he defines $$M_j(x_1,\dots, x_j)=m\{x_1,\dots x_j\}$$, and this structure maps satisfy the relation $$M\circ M=0$$ but with all plus signs. So I need to modify these maps by some signs in a similar way as the associative case. Of course I can try to sit down and write the equations and find some necessary conditions for the signs and maybe find a pattern. But if there is a conceptual explanation for the associative case and the lie algebras, then maybe there is an easier way to find out what the signs I need are.

• I once heard Raghuram quote Harder as saying "when you are dealing with signs, you are doing very difficult mathematics." After a life of signs, that made me feel good about my work. It doesn't seem directly related to your question, but I reflectively link Lawson - In which I try to get the signs right for once in any such question. – LSpice Aug 19 '20 at 14:05
• @LSpice I think I've already seen your work before, maybe having a look will help me find some ideas, even if it's not for this particular question, probably for future questions, so thank you. – Javi Aug 19 '20 at 14:10
• In order to avoid some other signs, sometimes people define $A$-infinity algebras not using the $A$-infinity operad, but its operadic suspension. That causes those signs you ask about. – Fernando Muro Aug 19 '20 at 15:45
• (Re, just to be clear, the work I linked is by Tyler Lawson, not me. I also meant that I reflexively link it, not that I reflectively link it ….) – LSpice Aug 19 '20 at 16:31
• I think that your example is indeed the Koszul sign rule and it has to do with the conventions one chooses for dealing with a suspension. Signs like those ones arise when you compose two operations of the form s^{-1} m_i s^{\otimes i} and s^{-1} m_j s^{\otimes j} and compare them to s^{1} (m_i \circ_k m_j) s^{\otimes i+j-1}. Further differences arise because e.g. you might realize "s" as tensoring with something on the right while I realize it as tensoring on the left. – Gabriel C. Drummond-Cole Aug 20 '20 at 0:22

I find the question quite interesting (in the sense that similar questions related to sign factors appearing in various different algebraic structures with no apparent reason, have been going through my studies for quite some time in the past..)

Although i am not really familiar with most of your examples, since you are also mentioning associative and Lie algebras, i will refer to a similar "phenomenon" from graded algebras: This has to do with the $$\mathbb{Z}_2$$-graded tensor product, between two associative superalgebras ($$\mathbb{Z}_2$$-graded algebras) $$A$$ and $$B$$. If $$b$$, $$c$$ are homogeneous elements of $$B$$ and $$A$$ respectively, then the so-called super tensor product algebra or $$\mathbb{Z}_2$$-graded tensor product algebra, of superalgebras, is the superalgebra $$A\underline{\otimes} B$$, whose multiplication is given by $$(a \otimes b)(c \otimes d) = (-1)^{|b| \cdot |c|}ac \otimes bd$$ with $$|b|, |c|\in\mathbb{Z}_2$$. Here the sign factor, reflects the braiding of the monoidal category of representations of the group hopf algebra $$\mathbb{CZ}_2$$: Recall that, superalgebras can be alternatively viewed as algebras in the the braided monoidal Category $${}_{\mathbb{CZ}_{2}}\mathcal{M}$$ (i.e. the Category of $$\mathbb{CZ}_{2}$$-modules) and that the above multiplication can be abstractly written as: $$m_{A\underline{\otimes} B}=(m_{A} \otimes m_{B})(Id \otimes \psi_{B,A} \otimes Id): A \otimes B \otimes A \otimes B \longrightarrow A \otimes B$$ Here, the braiding is given by the family of natural isomorphisms $$\psi_{V,W}: V\otimes W \cong W\otimes V$$ explicitly written: $$\psi_{V,W}(v\otimes w)=(-1)^{|v| \cdot |w|} w \otimes v$$ where $$V$$, $$W$$ are any two $$\mathbb{CZ}_2$$ modules.
Furthermore, this braiding is induced by the non-trivial quasitriangular structure of the group Hopf algebra $$\mathbb{CZ}_{2}$$, given by the $$R$$-matrix: $$$$R_{\mathbb{Z}_{2}} =\sum R_{\mathbb{Z}_{2}}^{(1)} \otimes R_{\mathbb{Z}_{2}}^{(2)}= \frac{1}{2}(1 \otimes 1 + 1 \otimes g + g \otimes 1 - g \otimes g)$$$$ through the relation: $$\psi_{V,W}(v \otimes w) = \sum (R_{\mathbb{Z}_{2}}^{(2)} \cdot w) \otimes (R_{\mathbb{Z}_{2}}^{(1)} \cdot v)=(-1)^{|v| \cdot |w|} w \otimes v$$.
For yet another point of view, the above mentioned $$R$$-matrix can be considered to be "generated" by the corresponding bicharacter (or: commutation factor) of the $$\mathbb{Z}_2$$ group.
There are bijections between $$R$$-matrices, braidings and bicharacters (which here are actually commutation factors) in the braided, graded setting for either assoc or Lie braided ("colored" is another name), graded algebras.

All these can be generalized for graded algebras, gradings and braidings, or $$R$$-matrices, or bicharacters of the corresponding groups, for any finite, abelian group. Also for $$\mathbb{G}$$-graded, $$\theta$$-colored Lie superalgebras, to produce more complicated bicharacters $$\theta:\mathbb{G}\times\mathbb{G}\to k$$ (which in the example above where $$\mathbb{G}=\mathbb{Z}_2$$ is exactly the sign factor of the $$\mathbb{Z}_2$$ abelian group).

To conlude: the sign factors here, are an "implicit" appearance of the corresponding group bicharacters. And they can also be viewed as braidings of the corresponding category of representations or as $$R$$-matrices for the corresponding quaitriangular group hopf algebras (of the fin, abelian, grading group).

If you are interested in these examples and you consider them relevant to your question, you can also take a look at the description in this answer: https://mathoverflow.net/a/261466/85967 and my linked paper therein.

• Thank you for your answer. In the case of that tensor product I think it's natural to obtain that sign factor, as well as in many other situation in which the braiding acts. But in the case of my question I don't see how the braiding could pop up, since no elements are being permuted. – Javi Aug 19 '20 at 19:05

As Gabriel C. Drummond-Co commented, it has to do with suspensions that are implicit. I'll do it with the example of Gerstenhaber and Voronov and the others should follow similarly. Let us denote $$M_2(x,y)=x\cdot y$$ the product that we want to define based on the brace $$m\{x,y\}$$. If we define it as a map $$(s\mathcal{O})^{\otimes 2}\to s\mathcal{O}$$ (suspension as graded vector spaces), then the natural thing to do is using the brace $$m\{-,-\}:\mathcal{O}^{\otimes 2}\to \mathcal{O}$$, but to do so one has to compose with suspensions and desuspensions. Namely, $$M_2(x,y)=s(m\{(s^{-1}x,s^{-1}y)\})$$. And it's applying $$(s^{-1})^{\otimes 2}(x,y)$$ what makes the sign $$(-1)^{|x|}$$ appear. If we use $$(s^{\otimes 2})^{-1}$$ instead then we get the original sign $$(-1)^{|x|+1}$$.