The Koszul sign rule is a sign rule that arises from graded-commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded-commutative algebra generated by $n$ elements of respective degrees $\lvert x_i\rvert$. Then, the sign $\varepsilon(\sigma)$ of a permutation $\sigma$ on $(x_1,\dotsc, x_n)$ is given by $$x_1\wedge\dotsb\wedge x_n=\varepsilon(\sigma)x_{\sigma(1)}\wedge\dotsb\wedge x_{\sigma(n)},$$ which comes from the fact that in a graded-commutative algebra one has by definition $a\wedge b = (-1)^{\lvert a\rvert\lvert b\rvert}b\wedge a$.

There is also an antisymmetric Koszul sign rule which arises from graded-anticommutative algebras and it's just the previous sign times the sign of the permutation. Both signs are used for instance in Lada and Markl - Symmetric brace algebras.

However, I've been seeing the Koszul sign rule used in any graded context and even for operations that are not products in some algebra. For example, from Roitzheim and Whitehouse - Uniqueness of $A_\infty$-structures and Hochschild cohomology, given graded maps of graded algebras $f,g:A\to B$, if we want to evaluate $f\otimes g$ in an element $x\otimes y$, apparently we need to apply the sign rule to get $$(f\otimes g)(x\otimes y)=(-1)^{\lvert x\rvert\lvert g\rvert}f(x)\otimes g(y),$$ but I see no mathematical reason to do that, it just seems to be a convention.

A more complex example of application of the Koszul sign rule is in the definition of brace algebra (also in the Lada and Markl paper).

I could give many more examples. In some of them I can understand the reason. For instance, the differential of a tensor product of complexes $C$ and $D$ cannot simply be $d_C\otimes 1_D+ 1_C\otimes d_D$ (it can be defined this way if we use the sign rule when we apply it to elements, but in any case it needs the sign). But maps in general need not be differentials. In other cases, the signs appear in nature and one use this sign rule to justify them, as in $A_{\infty}$-algebras, but this feels too artificial for me and doesn't really explain why we should use that sign rule.

So, in the end, every time there is a sequence $(x_1,\dotsc, x_n)$ of graded objects of any kind and not necessarily all of them of the same kind (elements, maps, operations, …), and related in any way (they can be multiplied, or applied, or whatever), we use the Koszul sign rule to permute the sequence.

To me all of this seems more philosophical than mathematical, and as I said it feels to be just a convention. But, is there some general mathematical reason to use the sign rule in such an extensive way? And if it's just a convention, why should we use it? From my experience, it gets very messy when it comes to applying the sign rule to larger formulas, and in the end everything is just a $+$ or $-$ sign, so I see no advantage.

  • $\begingroup$ @NajibIdrissi The signs come from the sign rule applied to shifted maps, so if I understand correctly, without the sign rule there would be no signs at all. $\endgroup$
    – Javi
    May 11, 2020 at 11:05

2 Answers 2


A precise statement of the conventions (which Jesse is referring to) is that the authors are using the symmetric monoidal structure on graded vector spaces, where the braiding map,, $\tau: V \otimes W \to W \otimes V$, is given by $$v \otimes w \mapsto (-1)^{{\rm deg} ~w ~{\rm deg}~ v} w \otimes v.$$

Roughly, what it means to use this symmetric monoidal structure is that you have to make all of your definitions diagramatically, using only $\tau$ to exchange symbols.

For instance suppose we have two algebras $A, B$ and an $A$ module $M$ and a $B$ module $N$. Then if $A,B,M,N$ were ordinary vector spaces, we are used to the fact that $M \otimes N$ is an $A \otimes B$ module. In the graded context, under the Koszul conventions, we define the action $$A \otimes B \otimes M \otimes N \to A \otimes M \otimes B \otimes N \to M \otimes N,$$ where in the first step we have used $1 \otimes \tau \otimes 1.$ Something quite similar is happening in the example your mention.

So far, this is more of a unified answer to how rather than why people use this convention.

For the question of why, the main reason the Koszul convention is useful in homological algebra has to do with the origin of homological algebra---topology.

Consider $\mathbb R^{p +q}$ with its standard orientation. Then the switching map $$\tau: \mathbb R^p \times \mathbb R^q \to \mathbb R^{q} \times \mathbb R^p$$ multiplies this orientation by $(-1)^{p q}$. This fundamental fact manifests itself in several ways.

One is that homology functor $H_*(-, k)$ from topological spaces to graded vector spaces is symmetric monoidal, but only with respect to the Koszul sign rule. This means that if one has an algebraic structure on a topological space $X$, then $H_*(X)$ naturally carries the same algebraic structure, but with respect to the Koszul sign rule. For instance, $X$ is always a co-commutative coalgebra, so $H_*(X)$ becomes a graded co-commutative coalgebra (with sign conventions according to the Koszul rule).

Something similar happens with the $A_\infty$ operad. Namely, the $A_\infty$ operad is the $dg$ operad obtained by taking the cellular homology of a (cellular) operad in topological spaces. The orientations of the cells of this operad explain the signs which arise.

There is also the monoidal Dold Kan correspondence, which you can read about on the nLab.

At the end of the day, it is just a convention (and not always the right one) but the relation with topology explains why people like to use it systematically.

  • $\begingroup$ This was the kind of answer I was hoping for, thank you very much! $\endgroup$
    – Javi
    May 11, 2020 at 11:11

This is not a complete answer (none will be, really), but there is a definite reason for applying the specific sign convention you've described just when considering graded maps of graded vector spaces with the signed braiding $\tau_{A,B}:A\otimes B\to B\otimes A$.

With homogeneous maps $f : A_\bullet \to B_{\bullet-n} $ and $g : X_\bullet \to Y_{\bullet+m}$, there are two competing ways to try interchanging the roles of $f$ and $g$: one can consider together with $\tau_{A,X}(a\otimes x) = (-1)^{|a||x|}(x\otimes a)$ Alternatively, there's an evaluate-first ask-questions-later approach, $\tau_{B,Y} (f a)\otimes (g x) = (-1)^{(|f|+|a|)(|g|+|x|)} (g x)\otimes(f a) $.

Now, conceivably, one could push all the extra signums, the $|f||g|+|f||x|+|g||a|$ into just the interchange $f\otimes g \;\mathrm{vs}\; g\otimes f$, but on the whole it seems to be cleaner to remark that evaluating $(f\otimes g)(a\otimes x) = \mathrm{ev} (f\otimes g\otimes a\otimes x)$ already involves interchanging $g$ and $a$, and similarly evaluating $(g\otimes f)(x\otimes a) = \mathrm{ev} (g\otimes f \otimes x\otimes a) $ involves interchanging $f$ and $x$.

  • $\begingroup$ In your second paragraph, should "there are two competing ways …: one can consider … Alternatively …" be something like "there are two competing ways …: one can consider …; alternatively, …"? That is, is that all one sentence? $\endgroup$
    – LSpice
    Sep 9, 2021 at 1:44

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