Hello and welcome! I’m Sam, a Ph.D. candidate studying modular representation theory - in particular, equivalences between representation spaces. If you don’t know what those words mean, fear not. This series of blog posts will attempt to introduce my field and more! My end goal is to build up to painting a picture of the question which guides my dissertation, getting into some, but not too many, technical details. The intended audience is an undergraduate student who’s taken a course on group and ring theory, with basic linear algebra assumed as well. Though if you’re not there yet, don’t worry! You still might get something out of these posts, and I might write an appendix-like primer for everything you need from those courses to get what I’m doing here.

Today, we’re finally going to define one of the two main objects I’m concerned with - a splendid Rickard complex. To get there, we’re going to define what chain complexes are, define the chain complex category, and the homotopy category, and quickly handwave the notions of duals and tensor product of complexes.

# Chain Complexes

Chain complexes are an essential tool in higher mathematics - in particular, they are necessary for computing “homology” or “cohomology” of various things. Often, the application of chain complexes relies on creating one somehow from a seemingly unrelated object, such as a topological space (singular homology), a group (group cohomology), a manifold (de Rham cohomology), or a scheme/variety (Etale cohomology). However, chain complexes can be mathematically interesting to study in their own right as well. In fact, they are the objects for some particularly interesting categories.

**Definition:** Let $R$ be a ring. A \textbf{chain complex} $C$ of $R$-modules consists of the following data:

- For every integer $i \in \mathbb{Z}$, an $R$-module $C_i$
- For every integer $i \in \mathbb{Z}$, an $R$-module homomorphism $d_i: C_i \to C_{i-1}$ with the property that $d_{i-1} \circ d_i = 0$.

We call the homomorphisms **differentials** or **boundary maps**, and while there are some names for the modules, I usually just call them **components**. We also say the chain complex is **bounded** if there are only a finite number of nonzero components.

You may be wondering why we need to specify that $d\circ d = 0$. I’ll give a historical reason first. To my knowledge (but I’m no math historian), chain complexes were first used in defining the singular homology of a topological space, which essentially describes “the $n$-dimensional holes in a topological space.” The idea of this construction is that essentially, you approximate your space using “simplicies,” or essentially, $n$-dimensional triangles. Then, your boundary maps represent sending a simplex to a formal sum of its boundary (so for example, a triangle would be sent to a formal sum of its 3 sides). Because of how the formal sum is signed, it follows that if you take this boundary operator on the result, everything cancels to 0! So for example, if you had a formal sum of these 3 sides of the triangle, taking the boundary operator on an individual side would give you a difference corresponding to the two ends of the side - but doing this 3 times would result in adding and subtracting each corner of the triangle exactly once, hence everything cancels. I don’t want to dwell on details but if this is interesting to you, take a course in algebraic topology!

The second reason the $d\circ d = 0$ condition is necessary is that this is required to define the **homology** of the chain complex. Note that because of the $d\circ d=0$ condition, for any $i \in \mathbb{Z}$, we must have that $\operatorname{im} d_{i+1} \subseteq \operatorname{ker} d_i$. Moreover, $\operatorname{im} d_{i+1}$ and $\operatorname{ker} d_i$ are both submodules of $C_i$, so we can take the quotient!

**Definition:** The **homology** $H_i(C)$ of the chain complex $C$ is the $R$-module $\operatorname{ker} d_i/\operatorname{im} d_{i+1}$. We say $C$ is **exact** if $H_i(C) = 0$ for all $i \in \mathbb{Z}$.

You may have heard of “short exact sequences” or “long exact sequences” before, and this is that same “exact” term. Short exact sequences are exact sequences of the form \(0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0\) which amounts to saying that $f$ is injective, $g$ is surjective, and $\operatorname{ker} g = \operatorname{im} f$. Moreover, one can think of $A$ as being a submodule of $B$ and $C$ being isomorphic to the quotient $B/A$ - essentially, one could replace this sequence with the isomorphic (what this means will be made precise soon) chain complex \(0 \to A \hookrightarrow B \xrightarrow{\pi} B/A \to 0\) where $\pi$ is the canonical surjection of $B$ onto the quotient $B/A$, i.e. sending $b \in B$ to the coset $bA$. One can also define short exact sequences for arbitrary groups (rather than abelian ones - remember a $\mathbb{Z}$-module is just an abelian group), and this gives us the notion of a group extension.

We are almost ready to define the category of chain complexes of $R$-modules, but remember, categories have both objects and morphisms. The category of chain complexes should obviously have chain complexes as its objects, but what are its morphisms? Well, since chain complexes have in each component a module, we should probably have maps between all of the components. However, like how module homomorphisms have to satisfy an $R$-linearity condition as well as an additive linearity condition, chain complex homomorphisms also have to satisfy an additional linearity condition that makes things work nicely.

**Definition:** Let $C$ and $D$ be chain complexes of $R$-modules, with components $C_i$ and $D_i$ and boundary maps $c_i$ and $d_i$ respectively. A **chain complex homomorphism** $f: C\to D$ is a family of $R$-module homomorphisms ${f_i: C_i \to D_i}_{i \in \mathbb{Z}}$ which satisfy, for all $i \in \mathbb{Z}$, $f_{i-1}\circ c_i = d_i\circ f_i: C_i \to D_{i-1}$.

Drawing a picture helps here, and normally I’d include one, but unfortunately I’m not sure how to make tikzcd work in these blogs, so I’ll describe the picture instead and leave it to you to complete it! Thank of the homomorphism as an infinite ladder lying on the ground, with the top and bottom corresponding to $C$ and $D$, and the rungs corresponding to the $f_i$. Then, the compatibility condition amounts to saying “it doesn’t matter if first, you go down then right, or right then down.” In technical terms, it’s saying that “each square in the diagram commutes.”

This gives us the ability now to define the category of chain complexes of $R$-modules, $\text{Ch}({}_R\textbf{Mod})$. The objects are chain complexes of $R$-modules and the morphisms are chain complex homomorphisms!

# Contractible Complexes and the Homotopy Category

Our goal now is to define the homotopy category of $R$-modules, which is similar to the chain complex category (same objects), but with more isomorphisms. I’m going to try to do this by not stating the formal definition of what a homotopy is, because in my opinion it’s not very enlightening. Rather, I’m going to describe what objects are isomorphic in this category and do a bit of handwaving.

First, let’s note that we can take direct sums of complexes, $C \oplus D$, like we can with modules, vector spaces, or whatever. However here, we have to add the maps as well - it’s a good exercise to show that what results is still a chain complex, i.e. that $d\circ d = 0$. So for a basic example, the direct sum of the complex $0 \to M \xrightarrow{\text{id}} M \to 0$, with the nonzero terms in degrees 1 and 0, with the direct sum of the same complex, but with the nonzero terms in degrees 2 and 1, is: $0 \to M \xrightarrow{(\text{id}, 0)} M \oplus M \xrightarrow{(0, \text{id})} M \to 0$. With this in mind, we can define what a contractible complex is, and we can define what an isomorphism looks like in the homotopy category.

**Definition:** A complex $C$ is **contractible** if it is isomorphic (as chain complexes) to a direct sum of complexes of the form $0 \to M \to M \to 0$ for some varying $R$-modules $M$. Say two chain complexes $C, D$ are **homotopy equivalent** if there exist contractible complexes $C’, D’$ for which $C \oplus C’ \cong D \oplus D’$. We express this with $C \simeq D$.

This is not exactly the standard definition of either of these terms, but it’s the simplest one for me to give, at least with the context of what I’m trying to build up to. You may recognize the term “homotopy equivalent” from studying topology, or at least from pop-math and the joke “a mathematician can’t tell the difference between a coffee mug and a donut.” Two topological spaces are homotopy equivalent if loosely, one can be “continuously deformed” into the other (see for example a coffee mug and donut - they both have one hole). I mentioned earlier that there is a way of associating a chain complex to a topological space through the theory of singular homology. As it turns out, two topological spaces are homotopy equivalent if and only if their corresponding chain complexes are homotopy equivalent! This is, I believe, where the chain complex-theoretic definition first came from.

We’re now ready to very informally define the homotopy category.

**Definition:** The homotopy category of $R$-modules, $K({}_R\textbf{Mod})$ is given by taking the chain complex category $\text{Ch}({}_R\textbf{Mod})$ and insisting all homotopy equivalences are isomorphisms.

Formally, we define the category by defining an equivalence relation on chain complex homomorphisms (saying $f$ and $g$ are homotopic, written $f\sim g$ if some technical condition is satisfied), and then we define the homotopy category by taking the chain complex category and modding out the morphism classes by this equivalence relation. However, I don’t want to get too detailed there - the above informal definition in my opinion paints a clear enough picture!

# Splendid Rickard Complexes

We are now ready to define the first of two equivalences that are fundamental to my dissertation question! Jeremy Rickard proved that if $R$ is a symmetric algebra (don’t worry about what that means, but two examples are group algebras $kG$ and block algebras $B$), then equivalences of homotopy categories arise in a similar way as Morita’s theorem (recall from part 4) stated - they arise by tensoring an “invertible” chain complex of bimodules.

How do we take tensor products of chain complexes? Well, I handwaved how we take tensor products of modules in the first place, but I can define how we take tensor products of chain complexes in terms of that handwave. We’ll phrase it in terms of bimodules for this case but the structure of a left module tensor is essentially the same.

**Definition:** Let $C$ be a chain complex of $(kG,kH)$-bimodules and $D$ be a chain complex of $(kH,kL)$-bimodules. Then $C\otimes_{kH} D$ is a chain complex of $(kG,kL)$-bimodules, where the $n$th component of $C\otimes_{kH}D$ is given by \(\bigoplus_{i + j = n} C_i \otimes_{kH} D_j\) and the transition maps are defined on the direct summands by the maps \(d_{i,j}: C_i \otimes_{kH} D_j: \to C_{i-1} \otimes_{kH} D_j \oplus C_i \otimes_{kH} D_{j-1},\) given by, for $m \in C_i, n\in D_j$ \(d_{i,j}(m\otimes n) = c_i(m)\otimes d + (-1)^{i} m \otimes d_j(n).\)

The $(-1)^i$ is necessary to make sure that the resulting structure is indeed a chain complex, that is, $d\circ d = 0$. One way you can visualize this is by picturing a 2-dimensional grid of $C_i \otimes_{kH} D_j$s, with the $i$s decrementing to the right, and the $j$s decrementing downwards. Then, each component of the tensor product is a diagonal direct sum of components, and the transition maps are given by the maps between each diagonal. Note that if we want the modules of the tensor product to be finitely generated, we need to insist that the chain complexes are bounded - otherwise we could have infinite direct sums. Also, one final thing to note is that while I gave the definition of the tensor product over the group algebra $kH$, you could replace this with any subalgebra of $kH$, such as a block of $kH$, or take this tensor over basically any algebra that works, i.e. if $C$ is a chain complex of $(R,S)$-bimodule and $D$ is a chain complex of $(S,T)$-bimodules, then $C\otimes_S D$ is a chain complex of $(R,T)$-bimodules.

There is one more thing we must note before we reach our conclusion, and that is a dual. I am assuming the audience reading this has seen the dual of a vector space, but if not, if we have a $k$-vector space $V$, then the dual $V^*$ is defined to be the vector space $\operatorname{Hom}_k(V,k)$. We can play this game over arbitrary rings $R$ rather than $k$, and we can extend this notion to $RG$-modules as well, since $RG$-modules when restricted to $R$ are simply $R$-modules! It turns out that if $V$ is a $RG$-module, then $V^*$ is a $RG$-module as well. In fact, taking duals becomes a functor for ${}_{RG}\textbf{Mod}$ with one caveat - the direction of the morphisms is reversed. A map $f: M \to N$ is sent to the map sending a functional $\delta: N \to R$ to the functional $\delta \circ f: M \to R$.

As a result, we can also take duals of chain complexes, but the same thing happens, the direction of the chain complex is reversed. One other note - if we instead consider the dual of a $(R,S)$-bimodule, the result is a $(S,R)$-bimodule, and hence the dual of a chain complex of $(R,S)$-bimodules is a chain complex of $(S,R)$-bimodules. We’ll leave things at that, I don’t want to get into the technicalities there.

We’re now finally ready to state what a splendid Rickard complex is!

**Definition:** Let $A,B$ be block algebras or group algebras over $\mathcal{O}$. A bounded chain complex of $(A,B)$-bimodules is a \textbf{splendid Rickard complex} if:

- $C \otimes_B C^*\simeq A$
- $C^* \otimes_A C \simeq B$
- Each component of $C$ is a perfect $p$-permutation bimodule.

Here, $A$ and $B$ are considered as chain complexes with $A$ or $B$ in the 0th degree. In this case, tensoring by $C$, $C \otimes_B -$, induces the equivalence of categories $K({}_B\textbf{mod}) \cong K({}_A\textbf{mod})$, and its “inverse” is instead tensoring by $C^*.$ So splendid Rickard complexes are in a sense, “invertible” chain complexes of bimodules!

You may wonder why the $p$-permutation stipulation is in there. I’m afraid I can’t dwell on this for too long, but there are two short answers. The first is that Rickard proved that these complexes are in bijection with complexes that share all the same properties but over $k$ instead of $\mathcal{O}$. The second, which is the real answer, is that $p$-permutation allows these complexes to induce isotypies, which are a bouquet of character level equivalences which has an extremely technical definition that I’m not qualified to speak about. I think we’ll wrap things up there - sorry for the delay between last post and this one, life got in the way for a while. Next time, we’ll be switching things up and introducing Grothendieck groups!