# Signs in the tensor product and internal hom of chain complexes

Let $R$ be a commutative ring and $\text{Ch}(R)$ the category of chain complexes of $R$-modules. $\text{Ch}(R)$ is first of all an abelian category, but it can also be equipped with the structure of a (symmetric?) closed monoidal category: it has

• a tensor product $A \otimes B$ whose component in degree $n$ is $\bigoplus_{i+j=n} A_i \otimes_R B_j$ and where the differential is defined by $d(a \otimes b) = da \otimes b + (-1)^{|a|} a \otimes db$, and
• an internal hom $\text{hom}(A, B)$ whose component in degree $n$ is $\prod_i \text{Hom}_R(A_i, B_{i+n})$ and where the differential is defined by $(df)(a) = d(fa) - (-1)^{|f|} f(da)$

which are related via the adjunction $\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{hom}(B, C))$.

How can I motivate the sign convention in either of these definitions?

I can sort of answer that question: both sign conventions are just the graded Leibniz rule. As Theo Johnson-Freyd explains on MO, this is because $\text{Ch}(R)$ is precisely the category of modules over a one-dimensional graded-commutative Lie algebra over $R$ concentrated in degree $1$. But this is unsatisfying because I don't know the answer to this question:

What do graded Lie algebras have to do with the topological motivation for studying chain complexes?

By "the topological motivation" I mean the idea, made precise by the Dold-Kan correspondence, that a chain complex is, roughly speaking, a linearization of a combinatorialization of a topological space. I think that the sign convention for the tensor product can be motivated by thinking about simplicial chains, but this doesn't have quite the air of inevitability that I would like out of so fundamental a definition.

I have the feeling that everything goes back to Euler characteristic. A bounded complex can be sent to an alternating sum of elements of $K_0(R)$, and this sum is invariant under chain homotopy and additive in short exact sequences and possibly satisfies a universal property of some kind, and it clearly shows there is a real distinction to be made between the even and odd elements of a chain complex. But I don't know the conceptual route from this idea to the graded Leibniz rule.

• See also this MO-thread. – t.b. May 21 '11 at 12:41
• @Theo: I've seen that thread, but it seems to be discussing more confusing sign conventions than the one I am trying to motivate. In Tyler Lawson's linked notes he claims that the sign convention above follows directly from the symmetric monoidal structure, but I still don't understand why $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$ is a natural convention in this context. (I could convince myself that this is a good idea using the de Rham complex or simplicial chains, but I feel like there has to be a more fundamental reason than that.) – Qiaochu Yuan May 21 '11 at 13:09
• I think Boardman's paper Peter May mentions in that thread motivates this to a certain extent (in case you're wondering: you can download the paper as a pdf by clicking on the icon next to the title -- it took me a while to figure this out on that page). Of course, you're still left with wondering why the involution $-1$ should be the right one to look at and, more importantly, why you should want to consider dimension as grading modulo $2$, but still it seems like a step in the right direction. – t.b. May 21 '11 at 13:30
• About the $\mod 2$ thing: one can also consider other values of $2$! Searching arXiv for «$N$-complex» will bring up information about complexes where $d^N$ is zero and not the square: in that context you of course (?) use other roots of unity and not $-1$. – Mariano Suárez-Álvarez May 21 '11 at 16:10

## 1 Answer

One algebraic way to motivate this is to observe that the signs in the differential for the Hom are precisely what is needed for 0-cycles in the $\hom(A,B)$ complex to be the set of morphisms of complexes $A\to B$ (and also, that the 0th homology group $H_0(\hom(A,B))$ is the set of homotopy classes of morphisms $A\to B$). This is quite great.

Once you decide you want this, all the other signs you mention follow because you need various things to hold. For example, you want the adjuntion between $\hom$ and $\otimes$ to hold for the internal versions, so this forces you to add signs to the $\otimes$, and so on.

A topological, indirect explanation for the appearence of most signs is that in the long exact sequence of bases maps corresponding to a map $f:X\to Y$, which looks like $$X\to Y\to C(f)\to S(X) \to S(Y) \to S(C(f)) \to \cdots$$ there is a "sign" which you cannot get rid of. This sign reproduces itself in every algebraic version.

• Of course I agree but I have a very silly question: Do you have a good explanation why I would want to decide I want this? In other words: How do you motivate the leap from morphisms to $0$-cycles or homotopy classes? – t.b. May 21 '11 at 12:46
• @Theo: When the complexes are projective resolutions of modules over a ring, you want the homology in degree zero to be $Ext^0$, which is the $\hom$ between the modules. One can probaby flesh this into a motivation :) – Mariano Suárez-Álvarez May 21 '11 at 12:53
• @Theo: if one thinks of chain complexes as algebraic analogues of topological spaces and internal hom as an algebraic analogue of the internal hom in a nice category of topological spaces, then $0$-cycles are analogues of points and $1$-cycles are analogues of homotopies between points, so a $0$-cycle in the internal hom is a continuous map and a $0$-cycle up to boundaries is a homotopy class of continuous maps. – Qiaochu Yuan May 21 '11 at 13:01
• this is precisely my reason for caring about the internal hom, but it feels like putting the cart before the horse. I feel like there should be a natural reason this all falls out of the definition of the category of chain complexes and I don't know what that reason is. Can you explain that last paragraph a little more? I'm not sure I know what the notation in it means. – Qiaochu Yuan May 21 '11 at 13:04
• @eric, the sequence is the so called Puppe sequence. If you look up its construction, you'll see that one of the maps involves a reversal of orientation (you construct the suspension of something in an indirect way, which gives you the space "upside down") – Mariano Suárez-Álvarez Jun 22 '16 at 18:55