## Chain complexes as Graded C[epsilon]-modules, part 3: Bicomplexes and Superalgebras

< See Part 2

The last two posts in this topic were mostly exploring what homology looks like in the language of modules. This time, we will return to the task of shoving homological algebra into the theory of module categories, by considering another fundamental tool: bicomplexes. This is particularly fruitful, since it reveals that $\epsilon$ wants to anticommute with itself (which it technically does, because it squares to zero). To get the right framework for this, we will introduce the notion of a superalgebra.

Bicompexes and categorical products

A (graded) bicomplex M is a vector space (or module or whatever) with a $\mathbb{Z}^2$ -grading, together with two endomorphisms $\delta, \partial$ of degree $(-1,0)$ and $(0,-1)$ respectively, such that $\delta^2=0$, $\partial^2=0$ and $\delta\partial=-\partial\delta$.

There is a typical way of turning a bicomplex into a regular complex, called the total complex of the bicomplex. The homogeneous components are defined by $Tot(M)_k=\oplus_{i+j=k}M_{i,j}$, and the differential is defined as $\delta+\partial$. Notice that the bizarre sign that appeared in the identity $\delta\partial=-\partial\delta$ is absolutely necessary for this new differential to square to zero.

Why are these two notions useful? There are many diverse answers, but the one I like most is that it allows us to tensor two chain complexes together and get a new chain complex. Given two (graded) chain complexes $N_\bullet$ and $N'_\bullet$, I can build a bicomplex $N\otimes N'$, where

$(N\otimes N')_{i,j}=N_i\otimes N'_j$

and the two differentials are

$\delta_{N\otimes N'}(n\otimes n')=\delta_N(n)\otimes n'$, and

$\partial_{N\otimes N'}(n\otimes n')=(-1)^{\deg(n)}n\otimes\delta_{N'}(m)$.

Notice the pesky sign again! Anyway, by taking the total complex of this bicomplex, I get a chain complex that is a good notion of the tensor product of $N$ and $N'$.

What really happened is that we gave the category of chain complexes a monoidal structure. We can now multiply any two objects in this category to get a third, and this multiplication has some nice properties: it has an identity (the chain complex consisting of $\mathbb{C}$ concentrated in degree 0), it is associative up to natural equivalence, and it is bifunctorial (the left or right action of multiplying by a fixed object is a functor). Monoidal structures are useful for a bunch of reasons. First, we can now define ring-like objects in the category, which in the case of chain complexes will be the differential graded algebras. Also, we can now enrich other categories over this category, which is necessary if we want to enrich the category of chain complexes over itself (which we most certainly do).

However, before we go on, we really must figure out why that damn sign keeps popping up. It would be so much nicer if we could just tensor two complexes together and use the sum of their unadulterated differentials, but this won’t be a chain complex.

From the module viewpoint, what is the problem? The tensor product of two any two $R$-modules over $\mathbb{C}$ is naturally a module over $R\otimes_\mathbb{C} R$. The problem is that we would like to define a bicomplex as a (graded) $\mathbb{C}[\epsilon]\otimes_\mathbb{C}\mathbb{C}[\epsilon]=\mathbb{C}[\epsilon,\epsilon']$ module, but then the relation $\delta\partial=-\partial\delta$ won’t hold. We need to use the non-commutative algebra $\mathbb{C}\langle\epsilon,\epsilon'\rangle/(\epsilon\epsilon'+\epsilon'\epsilon)$; in other words, we need $\epsilon$ and $\epsilon'$ to anticommute. Therefore, to banish the annoying sign change, we need to find a way to make $\mathbb{C}[\epsilon]\otimes_\mathbb{C}\mathbb{C}[\epsilon]=\mathbb{C}\langle\epsilon,\epsilon'\rangle/(\epsilon\epsilon'+\epsilon'\epsilon)$.

How do we get an element of a commutative algebra to ‘know’ that it needs to anticommute with itself? To answer this question, we need to throw out ‘vector spaces’ and replace them with ‘super vector spaces’.

Super(things)

Next, we need to understand where commutativity comes from, categorically. When one says that an algebra $R$ is commutative, this is the same as saying that the multiplication map $m:R\otimes R\rightarrow R$ is invariant if I precompose it with the ‘swapping’ map $b:R\otimes R\rightarrow R\otimes R$ which swaps the left and right terms.

The map $b$ here is part of a ‘braiding’ of the monoidal category $\mathbf{Vect}$ of vector spaces; that is, a natural transformation which sends $A\otimes B$ to $B\otimes A$ for every $A$ and $B$. This braiding has the property that it squares to the identity, such braidings are called ‘symmetric’.

A natural question to ask is, what new notions of commutative can I get if I replace the symmetric braiding above with a different one? At first, this question has a silly answer, since there are no other symmetric braidings on the category of vector spaces. The rough reason for this is that every point in every vector space can be in the image of $\mathbb{C}$ as a vector space, so every point must commute or anti-commute the same way as $\mathbb{C}$ (a coherence condition forces $\mathbb{C}$ to commute with itself). Hence, we are stuck with the old notion of commutativity.

To get more interesting answers, we need to look at a weirder category. Specifically, let’s look at $\mathbb{Z}_2-\mathbf{Vect}$, the category of vector spaces with an action of $\mathbb{Z}_2$, the group with two elements. Since this group is commutative, this action breaks apart the vector space into eigenspaces, a +1 eigenspace and a -1 eigenspace. Since the action can be reconstructed from this decomposition, we see that vector spaces with a $\mathbb{Z}_2$ action are the same as vector spaces with a $\mathbb{Z}_2$ grading. Given $V$, we will denote the +1 eigenspace as $V_0$ and the -1 eigenspace as $V_1$, considering the subscript to be the ‘degree’ of the component.

Now that the category is more interesting, there are more interesting symmetric braidings. The one we will use is $s:A\otimes B\rightarrow B\otimes A$ such that $s(a\otimes b)=(-1)^{\deg(a)\deg(b)}b\otimes a$. It is a fun computation to show that this is the only new symmetric braiding (up to sign), and also that if we look at $G-\mathbf{Vect}$ for any abelian $G$, we still don’t get any significantly different symmetric braidings.

We will call a $\mathbb{Z}_2$-graded vector space a super vector space when we have this braiding chosen in the back of our minds. It comes with a tensor product that looks identical to the old one, until we try to swap the left and the right terms. The next idea is that of a super algebra, which is a super vector space $R$ together with a multiplication map $m:R\otimes R\rightarrow R$ (that is unital, associative, etc) such that the multiplication is invariant under precomposition with this new braiding. One might also be tempted to call such an algebra a ‘graded commutative algebra’, but the physicists got there first and decided to make it a bit more grandiose.

The advantage of all of this is that if we think of $\mathbb{C}[\epsilon]$ as a super-algebra, with $\epsilon$ in the degree 1 component, then the (super) tensor product of the algebra with itself will be $\mathbb{C}\langle\epsilon,\epsilon'\rangle/(\epsilon\epsilon'+\epsilon'\epsilon)$. Excellent! This means that cooking up bicomplexes from two regular complexes is completely straight-forward: just (super) tensor them together, and they automatically have a $\mathbb{C}\langle\epsilon,\epsilon'\rangle/(\epsilon\epsilon'+\epsilon'\epsilon)$ action.

Thus, the category of chain complexes, thought of as $\mathbb{C}[\epsilon]$ modules, has a much more natural monoidal structure as long as we remember that we are playing with a super-algebra. I think this is a pretty useful statement, since a large portion of the sign errors in homological algebra can be traced to the sign correction in the definition of a bicomplex. Maybe all these sign errors are the mathematicians’ punishment for not being more comfortable with superalgebras?

Unfortunately, to get this to work we paid a terrible price. $\mathbb{C}[\epsilon]$ modules must themselves be super vector spaces, (ie, $\mathbb{Z}_2$-graded), and so we can no longer completely ignore the existance of a grading. Grumble, grumble…

Well, that post got long in a hurry. If I can think of enough interesting things to say, my fourth (and probably final) post on this subject will be how to use this monoidal structure to our advantage, by constructing commutative ring-like objects and enriched categories. If you want to learn more about braidings, check out some of John Armstrong’s excellent posts over at The Unapologetic Mathematician, he is going into the nitty-gritty of it all.