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.

# Simple and Indecomposable Modules

First, let’s set some global standards: $G$ denotes a finite group, and $k$ denotes a field. All vector spaces will be assumed to be finite-dimensional, and all $kG$-modules will have finite $k$-dimension, i.e. be finitely generated. Last time, we defined what a representation was in two separate ways - via a group homomorphism, and via modules.

Definition (1): Let $G$ be a group and $V$ be a finite-dimensional vector space over a field $k$. A representation $\rho$ of $G$ over $k$ is a group homomorphism $\rho: G \to \text{Aut}(V).$

Definition (2): A representation of $G$ over $k$ is a (finitely generated) $kG$-module.

Though we will primarily refer to representations in the module-theoretic sense, we will do a bit of hopping back and forth - hopefully it will be clear which definition we are using from the context. Typically, I’ll just call representations in the sense of the 2nd definition “$kG$-modules.” I will also generally use “$M$” or “$N$” to refer to representations as $kG$-modules, and “$\rho$” or “$\chi$” to refer to representations as group homomorphisms. We’re first going to discuss some $R$-modules that can be viewed in some sense as the “fundamentally small” $R$-modules. We’ll introduce two different notions, one stronger than the other, of $R$-modules that build up all the “larger” $R$-modules out there.

Like groups have subgroups and vector spaces have vector subspaces, $R$-modules have $R$-submodules, which are closed both under the module addition and the $R$-multiplication. Recall simple groups, groups with no proper normal subgroups. These groups can be seen as the “fundamental building blocks” of all groups, via a composition series (if you haven’t seen composition series, don’t worry, I didn’t see them until grad school - but Wikipedia is your friend!). There is a similar notion for modules.

Definition: A module is simple if it has no proper submodules, that is, its only submodules are itself and the zero module, $\{0\}$.

In the same way, these modules can be considered as building blocks for all modules, via composition series of modules - however, we won’t get into details there. For any ring $R$, an interesting question to ask is, “what are all the simple $R$-modules?” In the case where $R = kG$, we call simple $kG$-modules simple or irreducible representations. Given any group $G$ and field $k$, it is one of the main goals of representation theory to classify all of the simple representations, i.e. simple $kG$-modules!

Now, you may recall that given two groups $G$ and $H$, you can form a new group out of the two, the direct product $G\times H$, whose elements are tuples $(g,h)$ with $g\in G, h\in H$, and with multiplication given by componentwise multiplication. We have a similar notion for modules.

Definition: Let $M$ and $N$ be two $R$-modules. The external direct sum of $M$ and $N$, written $M \oplus N$, is, as an abelian group, the direct product $M \times N$, with $R$-multiplication performed componentwise:

$r\cdot (m,n) = (r\cdot m, r\cdot n).$

(Why this is called a direct sum rather than a direct product is a matter that I (might) discuss in a few posts later in this series!) In my subjective view, we call this an external direct sum since $M$ and $N$ in some sense live “outside” of $M\oplus N$, though you can see they are naturally embedded inside via the submodules $M \oplus \{0\}$ and $\{0\} \oplus N$. There is a corresponding internal direct sum which only applies for submodules of a given module - this gives us the ability to describe how modules are built up from their submodules.

Definition: If $M$ has proper submodules $N_1, N_2 \subset M$ such that every $m \in M$ can be written uniquely as a sum $m = n_1 + n_2$ for $n_1\in N_1, n_2 \in N_2$, then we say $M$ is the internal direct sum of $N_1$ and $N_2$, written $M = N_1 \oplus N_2$.

Reusing the $\oplus$ notation and using “$=$” may seem like an abuse of notation, but here, we truly mean that $M$ can be broken down into the sum of its submodules in a canonical way. There is additional internal structure of how $N_1$ and $N_2$ are embedded inside $M$ that an external direct sum may not communicate. When we refer to submodules “breaking down” from here on out, we’ll always mean internal direct sums. However, internal and external direct sums are always canonically isomorphic - so using the same symbol to denote both is acceptable (at least if the context is clear).

Note that just because an $R$-module has proper submodules does not mean that it can be expressed as the direct sum of its submodules! For a simple example, let $R = \mathbb{Z}$ and take the $\mathbb{Z}$-module $\mathbb{Z}$ itself. For any $n = 2,3,4,\cdots$, $\mathbb{Z}$ has $n\mathbb{Z}$, i.e. the multiples of $n$, as a proper submodule, and in fact, these are all of its proper submodules. However, it is impossible to find any pair $n_1, n_2$ such that $\mathbb{Z} = n_1\mathbb{Z} \oplus n_2 \mathbb{Z}$! We have a name for modules that cannot be expressed as direct sums of proper submodules like this.

Definition: A nonzero $R$-module $M$ which cannot be expressed as a direct sum $M = M_1 \oplus M_2$ with $M_1, M_2\subset M$ proper submodules of $M$ is indecomposable.

In the case where $R = kG$, we say $M$ is an indecomposable representation. Any module can be broken down into a direct sum of indecomposable submodules (though not necessarily uniquely), so in this way, indecomposable modules can also be viewed as “building blocks” of modules! Like in the case of simple modules, given any ring $R$, an interesting question to ask is “what are all the indecomposable $R$-modules?” Well, for one, I’ll leave it as an exercise to the reader to prove that any simple module is indecomposable as well. On the other hand, in general most indecomposable modules are not simple. But perhaps there are some rings for which the indecomposable modules are precisely the simple modules…

One final note before we move on, you may be wondering how the notions of simple and indecomposable translate to our original definition of a representation, a group homomorphism $G \to \text{Aut}(V)$.

Definition: A representation $\rho: G\to \text{Aut}(V)$ is simple if no basis exists of $V$ such that after identifying $\text{Aut}(V) \cong GL_n(k)$, $\rho$ is of the form

$g \mapsto \begin{pmatrix} A(g) & B(g) \\\ 0 & C(g) \end{pmatrix}$

where $A, B, C$ are functions from $G$ to sets of matrices of fixed size. In other words, $V$ has no $\rho$-invariant proper subspace. Here, $C$ corresponds to a subspace of $V$ which would be $G$-invariant - this corresponds to the representation (in the module sense) having a submodule.

Definition: Given two representations $\rho_1, \rho_2$, their direct sum $\rho_1 \oplus \rho_2$ is given by

$(\rho_1 \oplus \rho_2)(g) := \begin{pmatrix} \rho_1(g) & 0 \\\ 0 & \rho_2(g) \end{pmatrix}.$

A representation $\rho$ is indecomposable if there do not exist subrepresentations $\rho_1, \rho_2$ for which $\rho = \rho_1 \oplus \rho_2$.

# Representation Theory over $\mathbb{C}$

With introducing indecomposability and irreducibility, we are now ready to survey some of the results of the representation theory of finite groups over the field $\mathbb{C}$. Some questions we wish to answer are:

1. How many irreducible representations are there?
2. How many indecomposable representations are there?
3. Which representations are projective, i.e. a direct summand of a finite direct sum $\mathbb{C} G \oplus \cdots \oplus \mathbb{C}G$?

We’ll begin answering these questions by stating a celebrated theorem of representation theory, Maschke’s theorem. It follows as a corollary of the following theorem:

Theorem: Let $k$ be any field, and assume $\mid G \mid$ is invertible in $k$. Suppose $M$ is a $kG$-module, with a $kG$-submodule $N_1 \subset M$. Then it has another submodule $N_2$ such that $N_1 \oplus N_2 = M$. In other words, $kG$ is semisimple.

Remember our example with $\mathbb{Z}$ - if one chooses a submodule $n\mathbb{Z}\subset \mathbb{Z}$, it is impossible to find another submodule for which $\mathbb{Z}$ decomposes into a direct sum of the two. But in the case of $\mathbb{C}G$, such a situation never arises! Any subrepresentation always belongs to a direct sum decomposition of the original representation. From this, it’s not too hard to prove the corollary:

Corollary: (Maschke’s Theorem) Let $k$ be any field, assume $\mid G\mid$ is invertible in $k$, and let $M$ be a $kG$-module. Then $M$ is isomorphic to a direct sum of irreducible representations. In particular, $M$ is indecomposable if and only if $M$ is irreducible, and this holds for $k = \mathbb{C}$.

So the irreducibles and the indecomposables coincide! So if we can find an answer to question (1.) or (2.) we’ve found an answer to both. But how will we go about answering either one? The answer can be found via character theory, which allows us, under $\mathbb{C}$, to simplify the objects we’re working with even further. Let’s first define what a character is.

Definition: Given a representation $\rho: G \to \text{Aut}(V) \cong GL_n(k)$, the $k$-character of $\rho$ is defined as the function

$\chi = \chi_\rho: G\xrightarrow{\rho} GL_n(k) \xrightarrow{\text{tr}} k,$

where $\text{tr}$ is the trace function, given by summing the diagonal of a square matrix. If $k = \mathbb{C}$, we call these complex characters, or just characters for short.

Recall that given any $n\times n$ matrix $A$ and invertible $n\times n$ matrix $B$,

$\text{tr}(A) = \text{tr}(BAB^{-1}),$

so the character $\chi$ of a representation of $G$ is truly independent of the choice of basis made to identify $\text{Aut}(V) \cong GL_n(k)$. Moreover by this property, if $g_1, g_2 \in G$ are conjugate, that is, $g_1 = gg_2g^{-1}$ for some $g \in G$, then $\chi(g_1) = \chi(g_2)$, which makes $\chi$ a complex class function on $G$, i.e. a function $G \to \mathbb{C}$ which is constant on conjugacy classes of $G$. Let’s denote the set of all class functions $G \to \mathbb{C}$ by $\text{CF}(G,\mathbb{C})$ - one can see that it forms a $\mathbb{C}$-vector space via function addition and $\mathbb{C}$-scalar multiplication, and has $\mathbb{C}$-dimension equal to the number of conjugacy classes of $G$.

We can make a few quick observations about characters. First, since any representation sends $1 \to \text{id}_n$, the $n\times n$ identity matrix, the corresponding character $\chi$ “knows” the degree of the representation - it is given by $\chi(1)$. Next, given two $\mathbb{C}G$-modules $M_1, M_2$ with corresponding characters $\chi_1$, $\chi_2$, what is the character of $M_1 \oplus M_2$? Well, since $(M_1 \oplus M_2)(g)$ (considered as a group homomorphism) looks like

$g\mapsto \begin{pmatrix} M_1(g) & 0 \\\ 0 & M_2(g) \end{pmatrix},$

we can see that the character $\chi_{M_1 \oplus M_2}$ is simply given by $\chi_1 + \chi_2$! Put simply, direct sums of modules correspond to sums of characters! We now state a few amazing results which one proves in a first course on representation theory:

Theorem: The function mapping a complex representation $\rho: G \to \text{Aut}(V)$ to its character $\chi_\rho: G\to \mathbb{C}$ is bijective. In other words, each character corresponds uniquely to a representation, so we have bijections (up to isomorphism):

$\{\mathbb{C}G\text{-modules}\} \leftrightarrow \{\text{Group homs } G\to \text{Aut}(V) \}\leftrightarrow \{\text{Complex characters of }G\}.$

Definition: On the $\mathbb{C}$-vector space $\text{CF}(G,\mathbb{C})$ of class functions from $G$ to $\mathbb{C}$, we introduce the Schur inner product

$(f_1, f_2) := \frac{1}{\mid G\mid } \sum_{g\in G} f_1(g) \overline{f_2(g)}.$

The Schur inner product is Hermitian and is symmetric on characters. Now, given an inner product $(-,-)$ on a vector space $V$, we can look for an orthonormal basis, that is, a basis ${b_1, \dots, b_n}$ of $V$ that satisfies $(b_i, b_i) = 1$ and $(b_i, b_j) = 0$ when $i \neq j$. As it turns out, an orthonormal basis of the Schur inner product isn’t too hard to find!

Theorem: The set of irreducible characters of $G$ (i.e. characters corresponding to irreducible representations of $G$) forms an orthonormal basis of $\text{CF}(G,\mathbb{C})$ with respect to the Schur inner product. In particular, the number of irreducible characters is finite and equal to the number of conjugacy classes of $G$, and for every class function $f \in \text{CF}(G,\mathbb{C})$, we have

$f = \sum_{i=1}^k(f,\chi_i) \chi_i,$

where ${\chi_1,\dots\chi_k}$ is the set of irreducible characters.

This answers questions (1.) and (2.) from the beginning of this section! To answer (3.), we don’t need much more work. Let’s consider the regular representation, the $\mathbb{C} G$-module $\mathbb{C}G$. I’m leaving an exercise for the reader to show that the corresponding character $\chi$ satisfies $\chi(1) = \mid G\mid$ and $\chi(g) = 0$ for any $g\neq 1$. With this, one can use the Schur inner product to show the following.

Theorem: Let $\rho$ be the regular character of $G$ and $\chi_1,\dots,\chi_k$ the irreducible characters of $G$. We have

$\rho = n_1\chi_1 + \cdots + n_k \chi_k,$

where $n_i = \chi_i(1)$, i.e. the dimension of the corresponding representation. In particular, we obtain the relation:

$\mid G\mid = n_1^2 + \cdots + n_k^2,$

and that if $M_1, \cdots, M_k$ are the corresponding simple $\mathbb{C}G$-modules, then

$\mathbb{C}G = M_1^{\oplus n_1} \oplus \cdots \oplus M_k^{\oplus n_k}.$

Corollary: All finitely generated $\mathbb{C}G$-modules are projective.

We’ll leave the results of classical representation theory of finite groups at that, though there are plenty more incredible results which I don’t have the time to talk about, including results which help one figure out precisely what those irreducible characters are - the orthogonality relations. I don’t know about you, but I find these structural theorems about representation theory over $\mathbb{C}$ quite beautiful - these are some of the results that led me towards studying representation theory in the first place. We’ll now move on from the beautiful and structured land of $\mathbb{C}G$ to a much more chaotic world. Just how chaotic, you ask? Well…

# Modular Representation Theory

First, let’s recall that the characteristic of a ring $R$, written $\text{char}(R)$, is the minimum positive integer $p$ such that $1 \in k$ added to itself $p$ times is $0$. If no such integer exists (such as in the case of $\mathbb{C}$), we define $\text{char}(R) = 0$. It is a standard ring theory exercise to prove that if $R$ is a field, $\text{char}(R)$ is either prime or 0.

The prototypical example of a field of nonzero characteristic is $\mathbb{Z}/p\mathbb{Z}$, the finite field of $p$ elements. In fact, all the finite fields have order $p^n$ for some prime $p$ and $n\in \mathbb{N}^+$, and if two finite fields have the same order, they are isomorphic! However, there are infinite fields of positive characteristic as well, such as the algebraic closure of any finite field.

Again, the questions we wish to ask regarding the representation theory of $kG$ with $\text{char}(k) > 0$, i.e. modular representation theory, are:

1. How many irreducible representations are there?
2. How many indecomposable representations are there?
3. Which representations are projective, i.e. a direct summand of a finite direct sum $kG \oplus \cdots \oplus kG$?

Now let’s recall the initial condition from Maschke’s Theorem:

Let $k$ be any field, and assume $\mid G\mid$ is invertible in $k$.

If $k$ is a field, then the only noninvertible element is 0. Thus, this statement is equivalently stating that $\text{char}(k)$ divides $\mid G\mid$. In fact, if $\text{char}(k)$ does not divide $\mid G\mid$, then the representation theory is mostly the same as before. However, if $\text{char}(k) \mid \mid G\mid$, the converse of Maschke’s theorem holds as well. Let me state a stronger version.

Theorem: (Maschke’s Theorem, Module-Theoretic Version) Let $k$ be a field. $kG$ is semisimple if and only if $\mid G\mid$ is invertible in $k$.

(You may notice that this is actually a stronger version of the theorem that leads to Maschke’s theorem. Sometimes this is referred to as Maschke’s theorem instead, as the corollary is fairly immediate.)

From here on out, let’s assume that $k$ is a field with characteristic dividing the order of $G$ unless stated otherwise, so the group algebra $kG$ is not semisimple. So while every $kG$-module still breaks down into a direct sum of indecomposable $kG$-modules, it is no longer either true that all the indecomposable $kG$-modules are irreducible, or that all $kG$-modules are direct sums of irreducible $kG$-modules. So just how many more indecomposable $kG$-modules are there?

Well, in studying modular representation theory, one of the earlier results that is proven is that the number of irreducible representations is given in the following way: let $p = \text{char}(k)$, and let $G_{p’}$ denote the subset of $G$ (as a set) consisting of elements which have order not divisible by $p$ - we call these elements $p’$ elements. One can check that if $g$ has order not divisible by $p$, than all of the elements in its conjugacy class satisfies the same condition - hence $G_{p’}$ still can be partitioned into conjugacy classes of elements.

Theorem: The number of irreducible representations of $G$ over $k$ is equal to the number of conjugacy classes in $G_{p’}$.

So question (1.) is answered, and in some sense, this theorem mirrors the $\mathbb{C}G$ case. In particular, if $p \nmid \mid G\mid$, then the number of irreducible representations is the same as before! Now, how many indecomposable representations are there? Let’s illustrate with an example.

Example: Let $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (as an additive group), the Klein 4-group, and let $k$ be the algebraic closure of the field $\mathbb{Z}/2\mathbb{Z}$, an infinite field of characteristic 2. Let $a,b \in k$ with at least one of $a,b$ nonzero. We define a representation in the following way:

$\rho_{(a,b)}: (1,0) \mapsto \begin{pmatrix} 1 & a \\\ 0 & 1 \end{pmatrix}, \quad (0,1) \mapsto \begin{pmatrix} 1 & b \\\ 0 & 1 \end{pmatrix}.$

Then let $M_{(a,b)}$ be the corresponding $kG$-module. One can show the following things: $M_{(a,b)}$ is indecomposable, each 2-dimensional indecomposable $kG$-module is isomorphic to $M_{(a,b)}$ for some $a,b \in k$, and that $M_{(a,b)} \cong M_{(c,d)}$ if and only if $(a,b) = \lambda(c,d)$ for some $\lambda \in k$. Thus, this determines an infinite set (up to isomorphism) of indecomposable $kG$-modules.

So in general, there can be infinitely many indecomposable $kG$-modules. On the flip side, when $\text{char}(k)=2$, $kV_4$ only has one simple module, the trivial representation. Yikes!

However, there is hope. What if we rephrase the question of “how many indecomposables are there” to “how many projective indecomposables” are there? In this case, the theory plays a bit nicer. With a bit of idempotent theory, one can show that the projective indecomposable modules (which we call principal indecomposable modules) are in canonical bijection with the irreducible modules.

Theorem: Let $P_1, \dots, P_l$ be the complete set of principal indecomposable $kG$-modules (up to isomorphism). Then each $P_i$ has a unique maximal submodule $M_i \subset P_i$, and the quotient modules $P_1/M_1,\dots, P_l/M_l$ form a complete set of the irreducible $kG$-modules.

This more or less answers question (3.) and provides a bit more detail to the answer of (2.) Perhaps this classification is, in a way, consistent with the classical representation theory. After all, in the $\mathbb{C}$G case, the principal indecomposables are in bijection with the irreducibles, since every principal indecomposable is in fact irreducible. However, I might be reaching a bit with that comparison.

Unfortunately, I think discussing the details of how these results are proven may be a bit too much of a detour from where I’m trying to go, and this post is long enough as it is! But I did discuss characters in the previous section, and would like to briefly touch upon them here as well. One can compute characters in the same way as before, by using the trace function, but surprisingly, these aren’t the usual characters we study in modular representation theory. In fact, the characters that we study aren’t even functions from $G$ to $k$. Wait, what?

There’s something I’ve omitted from this discussion which is crucial to modular representation theory - we don’t just study the module theory over $kG$. We work with what is called a $p$-modular system, a triple of base rings $(K,\mathcal{O},k)$. Here, $\mathcal{O}$ is a particular type of ring with a unique maximal ideal $\pi$ such that $\mathcal{O}/\pi \cong k$, and $K$ is the field of fractions of $\mathcal{O}$ (a construction that turns $\mathcal{O}$ into a field in the same way $\mathbb{Q}$ is constructed from $\mathbb{Z}$). We then simultaneously study $KG$-modules, $\mathcal{O}G$-modules, and $kG$-modules, and some of the above results are derived by studying the interplay between the three types of modules. In fact, the $KG$-module theory is the same as the classical $\mathbb{C}G$-module theory!

For computing characters of $kG$-modules, we study what are called Brauer characters, which are functions $G_{p’} \to K$ - these are computed by first lifting the terms of the trace sum to $\mathcal{O} \subset K$, and then summing the lifts. These Brauer characters prove to be the “correct” notion of a character for $kG$-modules. Similar to before, these characters form a basis for $\text{CL}(G_{p’}, K)$, which is how we derive the number of simple $kG$-modules! However, there are other results about $\mathbb{C}$-characters which do not have equivalents for Brauer characters, and vice versa for Brauer characters. In fact, nonisomorphic $kG$-modules can have the same Brauer character - this follows from $kG$ being non-semisimple.

Whew, this survey ended up being a lot longer than I expected, so that’s all for this post! Next time, we’ll switch things up and introduce category theory, the category of $kG$-modules, and some categorical constructions I (and my thesis research) care about. If there’s anything that’s unclear or anything that needs correcting, feel free to contact me!

### Bibliography

• Course notes
• Linkelmann, “The Block Theory of Finite Groups”
• Nagao, Tsushima, “Representations of Finite Groups”