There is no shortage of sign conventions in homological algebra. And once these conventions are set out, there is no shortage of diagrams where an obvious commutative diagram on the underlying abelian groups has to be corrected by an appropriate sign. Many people take a fairly relaxed attitude towards these.

Unfortunately, the signs don't always "just work themselves out". Conrad's book "Grothendieck duality and base change" (see his webpage) gives an extensive list of diagrams, and several signs that are likely opaque for most students of the subject initially. (For example, I find it unlikely that most graduate students would successfully identify the two signs going with canonical isomorphisms $(A[p]) \otimes (B[q]) \cong (A \otimes B)[p+q]$, and the reason they're different.) Poincare duality and the cap product give another offender.

A number of the signs going with these diagrams have conceptual explanations going with them, but most of the references I know generally present sign conventions (and the signs going into the appropriate diagrams) as dropped from the sky. I'd like to know if there is a place in where I can direct students who might want to understand this, rather than having to explain myself.

Do there exist references in the mathematical literature giving a conceptual basis from which these signs might be derived?

(By no means am I claiming that I know most, or all, of the relevant literature.)

  • 3
    $\begingroup$ I sadly doubt it :/ $\endgroup$ Jan 26, 2011 at 4:26
  • 2
    $\begingroup$ Dear Tyler, perhaps you could ask your students to e-mail sign questions to you, and that way, when you type up the answers, you could save them and become the author of such a document (assuming that Mariano is right and such a document does not exist)! I mean, if you're going to be answering the questions anyway (the hard part), you might as well compile the answers (the easy part!), no? $\endgroup$ Jan 26, 2011 at 5:17
  • 1
    $\begingroup$ By the way, Tyler, is there anything to be said for the phrase "keep it simplicial, stupid"? Oftentimes, we can completely avoid sign annoyances by keeping things simplicial until we need to compute (granted, this is only a viable option for left or right bounded complexes, but it seems like it definitely allows us to avoid playing with signs a lot of the time). $\endgroup$ Jan 26, 2011 at 9:10
  • 2
    $\begingroup$ @Harry: I'm already behind on my writing! So far as keeping things simplicial, yes, it often helps to translate problems into combinatorial versions. But sometimes the pure chain version is far more compact (see the Koszul complex). And at the end of the day, there are always computations that need to be done and chains tend to be more amenable to those. $\endgroup$ Jan 26, 2011 at 14:55
  • 1
    $\begingroup$ If one accepts the simplicial basis for signs, the question breaks up into two natural pieces: (1) signs needed if one works in simplicial modules; and (2) how signs for chain complexes are forced by simplicial modules. Once one has answered the first question, the second seems likely to have a single right answer. Also, one might model general chain complexes as simplicial-cosimplicial-modules and ask how the two diagrams interact. This might be more conceptual than choosing signs in unbounded complexes by analogy with positive complexes, which are forced by simplicial objects. $\endgroup$ Jan 26, 2011 at 17:51

4 Answers 4


This is much more of a comment than an answer, but I ran out of room in the comment box:

Tyler, my new standard reference for sign issues is YOUR paper, which I post here for the benefit of others http://www.math.umn.edu/~tlawson/papers/signs.pdf

Jim McClure and I have done a lot of wrestling with signs lately, and it seems that your intuition in your paper is the right one: start with the axiom that evaluation should be a degree 0 chain map $C^\ast\otimes C_*\to R$, and most of the rest of the standard conventions will follow. By the way, Jim and I have also come to the conclusion that the Poincare duality map should not be a cap product but a signed cap product $\alpha\to (-1)^{|\alpha|\dim(M)}\cap [M]$ in order to make it a degree $-\dim(M)$ chain map (with cohomological indexing) when it acts on the left (like all good maps should). This is assuming the cap product as defined in Dold, which Jim and I have concluded is a good definition (after playing with some other possible definitions and having them backfire for one reason or another).

Other than this issue, while he doesn't really provide much in the way of explanation, Dold's classic textbook tends to be at least accurate and consistent most of the way through. The exception is in his definition of the transfer (though he notes about three pages later that he should really have a sign there as well - this sign issue seems to correspond to the issue of making the duality map a properly signed chain map). This slightly messes with the signs in the remainder of the book, for example in the definition of the intersection product, although this issue doesn't occur until near the end so there's not much damage.

Sorry to ramble without providing much of an answer, but this is an issue close to my heart. At one point I decided that I want $(-1)^?$ chiseled into my tombstone.

  • 5
    $\begingroup$ Couldn't help but give +1 for $(-1)^?$. $\endgroup$ Jan 26, 2011 at 7:15
  • 21
    $\begingroup$ In general we should give this answer $(-1)^{?}$. $\endgroup$ Jan 26, 2011 at 8:16
  • 2
    $\begingroup$ I note that the sign that Tyler uses in his note for the cochain complex dual to a given chain complexes, $\partial f=(-1)^{|f|+1}f\partial$, is the same one Bredon takes pain to use in much of his Topology and Geometry textbook (see Ch. VI.2), especially when he discusses duality. $\endgroup$ Jan 26, 2011 at 15:17

There is a short classical paper that has not been mentioned:

Boardman, J. M. The principle of signs. Enseignement Math. (2) 12 1966 191–194.

For products in homology and cohomology theories, there is a systematic explanation of the products on the level of the symmetric monoidal (homotopy) category of spectra that dictates the signs. See Section 9 of the following (also on my web page: http://www.math.uchicago.edu/~may/)

May, J. P. The additivity of traces in triangulated categories. Adv. Math. 163 (2001), no. 1, 34–73.

The literature is strewn with mistakes of signs, which can have serious calculational effects. I once spent months tracking down a contradiction of signs involving the Steenrod and Dyer-Lashof operations. The mistake occurred in the unpublished errata of Steenrod and Epstein, which was meant to correct signs in the original classic text.

  • 5
    $\begingroup$ Welcome to MathOverflow! $\endgroup$ Feb 3, 2011 at 2:13

I don't have an answer to the question --- again, this is not an answer. It is, though, an explanation of some sign conventions that appear in dg vector spaces. My thesis is that if you thoroughly understand the signs for dg vector spaces, then you understand the signs for the rest of homological algebra. (But this thesis is not supported in this answer, and I don't intend to provide evidence; I am not a homological algebraist.) To reemphasize, I do not have a reference to the mathematical literature.

There is a very good category of functors from $\mathbb Z$, thought of as a category with only identity morphisms, to $\text{Vect}$ (or $\text{AbGp}$, but I'm going to pretend we're over a field). The addition map $+ : \mathbb Z \times \mathbb Z \to \mathbb Z$ leads to a tensor product on $\operatorname{Functors}(\mathbb Z \to \text{Vect})$ by push-forward. The resulting monoidal category is in some sense "the free monoidal [...] category on an invertible object", and so maybe deserves to be called $\text{Vect}[X,X^{-1}]$, where $X$ is the functor $\mathbb Z \to \text{Vect}$ that assigns the one-dimensional vector space $k$ to $1\in \mathbb Z$ and the zero-dimensional vector space $0$ to all other terms. So to give a braiding on this category it suffices to give the data of how to braid $X$ past itself, i.e. the braidings are parameterized by $\operatorname{Aut}(X^{\otimes 2}) = k^\times$. The symmetries are precisely the square-one elements of $k^\times$, and so if $k$ is not of characteristic $2$, there are two of these, $\pm 1$. The symmetric monoidal category $\text{GVect}$ of graded vector spaces is $\operatorname{Functors}(\mathbb Z \to \text{Vect})$, with $\otimes = $ push-forward along $+$, and with the symmetry given by the choice of braiding $-1 \in \operatorname{Aut}(X^{\otimes 2})$.

A symmetric monoidal linear category is enough data to talk about Lie algebra objects, and there is a distinguished such object in this category. Namely, the object $X$ has a unique Lie algebra structure, the abelian one. Let $\mathfrak X$ denote this Lie algbera (the abelian Lie algebra on $X$). The symmetric monoidal category $\text{DGVect}$ of dg vector spaces is the symmetric monoidal category of modules (in $\text{GVect}$) of the Lie algebra $\mathfrak X$.

I think this point of view does a good job of explaining what are the correct signs to include when working out tensor products and inner homs of dg vector spaces. In particular, if $A,B$ are dg vector spaces, then $\mathfrak X$ acts on $A\otimes B$ the way a Lie algebra ought to act on a tensor product. Namely, if $a\in A$ and $b\in B$, and writing $x\in \mathfrak X$ for the basis element of $\mathfrak X$, then the action $X \otimes A \otimes B \to A\otimes B$ is given by $$ x \cdot (a\otimes b) = (x\otimes 1 + 1 \otimes x)(a\otimes b) = (x\cdot a) \otimes b + \operatorname{flip}(x,a) \cdot b.$$ Here $\operatorname{flip}(x,a)$ is the element of $A\otimes X$ corresponding to $x\otimes a$ under the braiding. When $a\in A$ is a homogeneous element of degree $|a|$ for the $\mathbb Z$-grading, we have $\operatorname{flip}(x,a) = (-1)^{|a|}a\otimes x$. Then the "$\cdot b$" denotes the action of $X$ on $B$. (Of course, the whole point is that $x\in \mathfrak X$ implements the differential.) So this is the algorithm for computing signs in tensor products of dg vector spaces.

There are some particular distinguished objects of $\text{DGVect}$. Namely, for each $p\in \mathbb Z$, the object $X^{\otimes p}: \mathbb Z \to \text{Vect}$ is the functor that assigns $k$ to $p\in \mathbb Z$ and $0$ to all other elements. So $X^{\otimes p}$ is an object of $\text{GVect}$, but actually it has a unique structure as an object of $\text{DGVect}$ (namely, the structure where the Lie algebra $\mathfrak X$ acts trivially). As an object of a monoidal category, it determines an endofunctor of $\text{DGVect}$ (which in your notation acts "from the right"), namely the functor $[p] = \otimes X^{p}$.

From this point of view, your isomorphism $(A[p]) \otimes (B[q]) \cong (A\otimes B)[p+q]$ is clear. Spelling it out, it is the isomorphism $$ A \otimes X^{\otimes p} \otimes B \otimes X^{\otimes q} \cong A \otimes B \otimes X^{\otimes(p+q)} $$ and so the only natural choice is to use the flip map $\operatorname{flip}: X^{\otimes p} \otimes B \to B \otimes X^{\otimes p}$, which has some signs given the choice of braiding. (On the homogeneous part of $B$ of degree $r$, it acts by $(-1)^{pr}$.) I tend to like to write all functors on the left, but I'll note that in this case it makes the most sense to use "shift" functors from the right, because in my convention $\mathfrak X$ (= the differential) acts from the left, and so it is fewer signs if the shifts act from the right.

One final note is important to make. Fix $p,q$. Then there are isomorphism $X^{\otimes p} \otimes X^{\otimes q} \cong X^{\otimes (p+q)} = X^{\otimes (q+p)} \cong X^{\otimes q} \otimes X^{\otimes p}$. But this is not the correct isomorphism to take. The correct choice is the flip map, which charges a sign of $(-1)^{pq}$ for the intermediate "$=$". Remembering this helps to clarify sign errors that can arise when folks cavalierly try to write $A[p][q] = A[p+q] = A[q][p]$.

I hope this explains the signs. What I can't be bothered to care about is whether the differential should increase degree by $1$ or decrease it (corresponding in my conventions to whether $\text{DGVect}$ is the category of $X$-modules or of $X^{-1}$-modules), and whether the shift maps should shift in the degree specified or the other direction (maybe your notation is to write $[p] = \otimes X^{\otimes (-p)}$). If you write the shifts on the right, they you are simply wrong to define $A[p] = X^{\otimes p} \otimes A$, although that choice is isomorphic to better ones (for example, you would be very justified to write functors from the left and choose $[p]A = X^{\otimes p}\otimes A$, although then note that if $\mathfrak X$ also acts from the left, then you'll have to think about signs to get correctly the action $\mathfrak X \otimes [p]A \to [p]A$, since this map is really $\mathfrak X \otimes [p] \otimes A \overset{\text{flip}}\longrightarrow [p]\otimes \mathfrak X \otimes A \overset{\text{act}}\longrightarrow [p] \otimes A$; then again, maybe you want to say that the differential is a map $[1]A \to A$, in which case in my conventions you do want functors from the left). The only thing that is truly wrong is to come up with operations that act "from the middle": all the natural structure is buildable out of tensor products and the like, done from the left and from the right, and if you keep track of all this, the signs work out.

  • 1
    $\begingroup$ Did you mean square-one instead of square-zero? $\endgroup$ Jan 26, 2011 at 5:44
  • $\begingroup$ Hi Theo, thanks for your response. The standard conventions in homological algebra make $A[p]$ canonically isomorphic to $X^{\otimes p} \otimes A$ (which annoys me to no end). $\endgroup$ Jan 26, 2011 at 5:46
  • 4
    $\begingroup$ One minor point to make is that this notation requires you to be a little careful. You have to make a decision once and for all whether you canonically make $X^{-1} \otimes X$ or $X \otimes X^{-1}$ isomorphic to the unit object. Choosing one means that the other one requires a flip first, which introduces a sign. $\endgroup$ Jan 26, 2011 at 14:59
  • $\begingroup$ @Tyler: yes, that's correct, and I was a bit sloppy, in spite of trying to be careful. Or, rather, both $X^{-1}\otimes X$ and $X \otimes X^{-1}$ are canonically isomorphic to the unit object, but when you try to write down these isomorphisms in terms of underlying vector spaces, they differ by a sign. @Dmitri: yes, and I'll correct it. @Tyler: that's awful. $\endgroup$ Jan 26, 2011 at 18:05

I propose two references:

First, there is a conceptual description of why a multi-functor (as tensor or hom...) has to carry a sign and relating the sign to a cocycle representing a class in $H^2(\mathbb{Z}^n,\mathbb{Z}/2\mathbb{Z})$ in Verdier's thesis "Des Catégories Dérivées des Catégories Abéliennes", Asterisque 239 (1996) See Chap. I, sect. 1.6. You can download the relevant chapter here. The discussion is conceptual.

Second, there is a discussion of signs (the dependence of choices) in the appendix of Calmès and Hornbostel "Tensor-triangulated categories and dualities" TAC 22 (2009).

I hope that this is of help to you. I myself tend to be really puzzled always about this stuff. It is depressing how harmful the question turns to be, especially when you want to make explicit some canonical defined maps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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