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We are trying to understand the definition of $A_\infty$-algebras. But we are puzzled by what appear to be two different sign conventions (and we cannot figure out how these two are equivalent)

We see that an $A_\infty$-algebra $A$ consists of a graded vector space $A=\underset{n\in \mathbb Z}{\bigoplus}A_n$ along with maps

$$ m_k:A^{\otimes k}\longrightarrow A$$

of degree $(2-k)$. But what is the relation that these $m_k$ satisfy?

(1) See, for instance, page 6 of Keller (https://arxiv.org/pdf/math/9910179.pdf)

$$ \sum (-1)^{pq+r} m_{p+1+r}(1^{\otimes p}\otimes m_q\otimes 1^{\otimes r})=0$$

where the sum is taken over all decompositions of each $n=p+q+r$

(2) See, for instance, page 25 of Kenji Lefevre-Hasegawa (https://arxiv.org/pdf/math/0310337.pdf) where we find the equation

$$\sum (-1)^{p+qr} m_{p+1+r}(1^{\otimes p}\otimes m_q\otimes 1^{\otimes r})=0$$

This difference in signs $(-1)^{pq+r}$ and $(-1)^{p+qr}$ seems to happen at many places in the literature. Is there some way to tinker with the indexing of the graded space, or put signs in front of the $m_k$'s to turn one definition into the other and vice-versa?

Thanks a lot for your time!

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    $\begingroup$ The right definition of a (strictly unital, augmented) $A_{\infty}$-algebra structure on a graded space $V \oplus 1$ is a square zero degree -1 coderivation on the tensor coalgebra $T(V[1])$. (One may argue that there are other equivalent definitions; but I'd say that you cannot make this one meaningless by one sign error, unlike those with sums over partitions). If you unwind all graded pieces of the equation $d^2 = 0$ carefully enough, Koszul rule will tell you everything you need about the signs. $\endgroup$
    – Denis T
    Commented Nov 24, 2023 at 3:02
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    $\begingroup$ To make unwinding this definition even more intuitive, the axioms satisfied by $d$ are equivalent to its adjoint $d^*$ being a square zero, degree +1 derivation on the tensor algebra $T(V^*[-1])$. I'm assuming here that the grading on $V^*$ is defined such that the natural duality paring between $T(V[1])$ and $T(V^*[-1])$ itself has degree zero. To give oneself extra peace of mind, one can work out all the formulas for the $m_k$ in the case when $V$ is finite dimensional (or at least degree-wise finite dimensional), and then declare them the correct axioms for $A_\infty$-algebra for any $V$. $\endgroup$ Commented Nov 24, 2023 at 11:20

1 Answer 1

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Alexander Polishchuk has written a Field Guide to $A_\infty$ Sign Conventions which I have found quite helpful (despite some typos and imprecisions). Interestingly, Raf Bocklandt gives this as the main reference for sign conventions in his recent book A Gentle Introduction to Homological Mirror Symmetry.

Btw, I think you mixed up indices in (1): The conventions in Keller and Lefèvre-Hasegawa agree. In Polishchuk's field guide, this is sign convention $\epsilon_1$.

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  • $\begingroup$ I am so happy to see people writing "field guides" like this, they're fantastically valuable for the community! $\endgroup$
    – Tim
    Commented Jun 21 at 10:29

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