## The Group Algebra: Another Perspective on Representations

In this post, we introduce the group algebra as another way to view representations and illustrate the usefulness of this approach by studying representations of cyclic groups by elementary ring theory. This is the second part of a series that started with this post. The numbering is consecutive, i.e. when I refer to some result 1.x, then this is written in that post.

## A Review of Modules in Linear Algebra

We will begin by reviewing what modules over the polynomial ring $K[X]$ mean in terms of linear algebra, as this will be helpful for motivating the module-theoretic perspective on representations.

Let $K$ be a field and $V$ be a (say finite-dimensional) vector space over $K$ and let $A$ be a $K$-linear endomorphism of $V$ (so after choosing a basis, we can think of $A$ as a square matrix.) Suppose we wish to understand $A$, e.g. find a basis such that $A$ has a particularly nice matrix representation with respect to that basis.

From the pair $(V,A)$, we can define a $K[X]$-module structure on $V$ by defining $K[X]$-scalar multiplication via $(\sum_{i=0}^n \lambda_i X^i)v= \sum_{i=0}^n \lambda_i A^i(v)$, where $A^i(v)$ means that we apply $A$ to $v$ $i$ times.

Conversely, given a $K[X]$-module $M$, we can think of it as a $K$-vector space $V=M$ by restricting the scalar multiplication to $K \subset K[X]$. We also get a $K$-linear endomorphism of $V$ given by multiplication with $X$.

These constructions are inverse to each other: Going from a pair $(V,A)$ to the associated $K[X]$-module, multiplication with $X$ is precisely the endomorphism we started with.
For a $K[X]$-module, because every polynomial is a linear combination of powers of $X$, we only need to know $K$-scalar multiplication and how $X$ acts to reconstruct the $K[X]$-scalar multiplication.

Thus, we can think of pairs $(V,A)$ of vector spaces equipped with an endomorphism as $K[X]$-modules and we can translate between notions for endomorphisms and notions for $K[X]$.

Here’s an excerpt of a possible dictionary one might use for translation:

Pairs $(V,A)$ of vector spaces and endomorphisms $K[X]$-modules $M$
Subspaces $W$ that are invariant under $A$, i.e. $A(W) \subset W$ $K[X]$-submodules
For pairs $(V,A)$, $(W,B)$, a $K$-linear map $f:V \to W$ such that $f \circ A = B \circ f$ $K[X]$-linear maps
For pairs $(V,A)$, $(W,B)$, a $K$-linear isomorphism $f:V \to W$ such that $A =f^{-1} \circ B\circ f$ $K[X]$-linear isomorphisms
Eigenspace of $A$ associated to $\lambda$ The submodule of $M$ consisting of all elements annihilated by $X-\lambda$
The minimal polynomial of $A$ The unique monic generator of the annihilator ideal associated to $M$, i.e. the unique monic polynomial $P \in K[X]$ of minimal possible degree such that $P(v)=0$ for all $v \in V$

One could add many more rows.

The important part for finding e.g. a nice basis for $A$ is the third row: If we can find a module that is isomorphic to the $K[X]$-module associated to $(V,A)$ such that we write down a basis such that we get a nice matrix representation for multiplication with $X$, then the third row tells us that this is also a possible matrix representation for $A$!

Now as $K[X]$ is a PID, finitely generated modules over $K[X]$ (in particular those modules that are finite-dimensional over $K$) are very well-understood. There’s a structure theorem that tells us that they are finite direct sums of modules of the form $K[X]/(f)$, where $f \in K[X]$. (One can also put some conditions of $f$ to make them unique.) From there, one can easily deduce existence and uniqueness of canonical forms such as the Jordan normal form, the Frobenius normal form, and also properties of the minimal polynomial such as Cayley-Hamilton.

## The Group Algebra

Let $G$ be a group and $V$ be a representation of $G$ over a field $K$. By abuse of notation, we denote the action of an element $g \in G$ on a vector $v$ by $gv$. Let $\lambda \in K$, $g \in G$ and $v \in V$, then there seems to be only one sensible way to define what we mean by $(\lambda+g)v$, clearly, this has to be $\lambda v + gv$ if any sensible rules hold.
But what is the expression $\lambda+g$ supposed to mean? After all, we can’t just add an element of a group and an element of a field.

Or can we?

Definition 2.1 Let $G$ be a group (we denote the neutral element by $1$) and $K$ be a field, then the group algebra $K[G]$ is defined as the vector space over $K$ freely generated by the elements of $G$. We denote the elements of the basis corresponding to elements in $G$ by the same symbols. Group multiplication $G \times G \to G$ defines a multiplication of basis elements that we extend linearly in each argument. This defines a multiplication $K[G] \times K[G] \to K[G]$ that extends the multiplication of $G$ and makes $K[G]$ into a $K$-algebra.

We leave the verification of the ring axioms to the reader. Intuitively, the group algebra $K[G]$ consists of finite formal linear combinations of elements in $G$, i.e. we can write them as $\sum_{g\in G} \lambda_g g$ where all but finitely many coefficients vanish. To compute a product of two such expressions, we expand using distributivity and then use the group multiplication to multiply the products of the basis vectors.

Lemma 2.2 For every representation $V$ of $G$ over $K$, there is a unique way to define a $K[G]$-module structure that extends the given group action and $K$-scalar multiplication, conversely, every $K[G]$-module gives rise to a representation in a canonical way. Using this identification, a morphism of representations corresponds precisely to a $K[G]$-linear map.

Proof Given a representation $V$, $v \in V$ and an element $\sum_{g \in G}\lambda_g g \in K[G]$ (this means the sum is finite), the only possible way to define $(\sum_{g \in G}\lambda_g g)v$ such that the module axioms hold and $K \subset K[G]$ acts via the given scalar multiplication and $G \subset K[G]$ acts via the group action is to set $(\sum_{g \in G}\lambda_g g)v=\sum_{g \in G} \lambda_g(g(v))$.
Note that the RHS is just defined in terms of the group action and the vector space structure. One checks that this defines a module structure, using the linearity of the group action.
Conversely, if we have a $K[G]$-module $V$, we can turn $V$ into a $K$-vector space by restricting scalars to the subring $K \subset K[G]$.
We can also restrict the scalar multiplication to the subset $G \subset K[G]$. Associativitiy and unitality in the axioms for a module imply that this defines a group action.
Finally, group action and vector space operations are compatible due the equality $g\lambda = \lambda g$ in $K[G]$ and associativity and distributivity of the scalar multiplication.
The correspondence of morphisms of representations and $K[G]$-linear maps follows by similar arguments: the idea is that every element in $K[G]$ is a $K$-linear combination of elements in $G$, so it’s enough that a map commutes with $K$-scalar multiplication and the $G$-action to see that it preverses $K[G]$-scalar multiplication.

The relation between the description of linear-algebraic objects as $K[X]$-modules and representations as $K[G]$-modules is as follows:
In the former, we looked at any endomorphism without any condition, but only at one endomorphism at a time, that’s why the $K$-algebra of choice to describe such an object is the polynomial algebra $K[X]$ which is generated freely by one element $X$, i.e. we don’t impose any relation.
For group representations, we consider many endomorphisms (actually automorphisms) at once, subject to all the relations that hold in the group $G$. That’s why $K[G]$ isn’t necessarily generated by one element and by inheriting the multiplication from $G$, $K[G]$ also inherits all the relations between elements in $G$.

With lemma 2.2, we have added another characterization of representations to our collection (cf. lemma 1.3).
If one is really careful with the constructions in the lemma, one sees that it defines an isomorphism of categories which is just a formalization of the inuition that representations and $K[G]$-modules are exactly the same, just with a different point of view.

Let’s also mention the universal property of the group algebra, which can be quite useful even if you’re not a category-aficionado and implies lemma 2.2 as a special case.

Lemma 2.3 Let $K$ be a field and $G$ be a group, then for any $K$-algebra $A$ and every group homomorphism $\varphi:G \to A^\times$ to the group of units, there is a unique $K$-algebra homomorphism $K[\varphi]:K[G] \to A$ that extends $\varphi$.

Proof The same argument as in lemma 2.2 applies: The only way we can define an extension that is $K$-linear is by sending $\sum_{g \in G} \lambda_g g$ to $\sum_{g \in G} \lambda_g \varphi(g)$, this just follows from the fact that $G$ is a basis for $K[G]$. One checks that because $\varphi$ is a group homomorphism and the multiplication in $K[G]$ is inherited from $G$, this also respects the unit element and multiplication, so it is a $K$-algebra homomorphism.

To see how this implies one direction in lemma 2.2, note that for a vector space $V$, the endomorphism ring $\mathrm{End}_K(V)$ is a $K$-algebra (here multiplication is composition) and for the group of units, we get $\mathrm{End}_K(V)^\times=\mathrm{GL}(V)$.
Thus if we have a group homomorphism $\varphi:G \to \mathrm{GL}(V)$, the universal property tells us that there is a unique $K$-algebra homomorphism $K[\varphi]: K[G] \to \mathrm{End}_K(V)$. Now we can define a module structure by uncurrying:
Define for $r \in K[G]$ and $v \in V, r \cdot v:=K[\varphi](r) (v)$
The fact that $K[\varphi]$ is a ring homomorphism translate neatly into the module axioms and $K$-linearity gives us that the $K$-scalar multiplication on $V$ remains the same.

Having this perspective is quite useful, because there are a lot of constructions for modules that now carry over directly to representations: we can form direct sums and products of representations, quotients etc. and all the properties of those constructions that we know to hold for modules also hold in this case. For example, subrepresentations as defined in 1.18 are the same as $K[G]$-submodules.

## Representations of Cyclic Groups

We will use the accessible example of cyclic groups to show how the structure of the group algebra contains information about representations.

Lemma 2.4 If $G=\langle g \rangle$ is cyclic of order $n$, then $K[X]/(X^n-1) \cong K[G]$, where the isomorphism sends $X$ to $g$.

Proof One can take the map $K[X] \to K[G]$ that sends $X$ \to $g$ and compute the kernel. As $g$ generates $G$, so every element in $K[G]$ is a polynomial in $G$, which implies the surjectivity of that map.
Let’s instead show that both satisfy the same universal property:

• If $A$ is any $K$-algebra, then a $K$-algebra homomorphism $K[G] \to A$ corresponds to a group homomorphism $G \to A^\times$ by lemma 2.3.
Since $G=\langle g\rangle$ is generated by $g$, a group homomorphism is uniquely determined by where it sends $g$. As $g$ has order $n$, we can send it precisely to those elements $a \in A$ such that $a^n=1$ (This condition automatically give us that $a\in A^\times$). Thus $K[G]$ has the following universal property for this choice of $G$:
For any $K$-algebra $A$ and every element $a \in A$ such that $a^n=1$, there’s a unique $K$-algebra homomorphism $K[G] \to A$ that sends $g$ to $a$.
• If $A$ is still any $K$-algebra, then by the homomorphism theorem, a $K$-algebra homomorphism $K[X]/(x^n-1) \to A$ is the same as a $K$-algebra homomorphism $K[X] \to A$ that sends $x^n-1$ to $0$. A $K$-algebra homomorphism $K[X] \to A$ is uniquely determined by where it sends $X$ and we can send it to every element in $a \in A$, but due to the condition that $x^n-1$ must be sent to $0$, for homomorphisms from $K[x]/(X^n-1)$, we can send $X$ to precisely those elements $a \in A$ such that $a^n=1$.
Thus we have proved:
For any $K$-algebra $A$ and every element $a \in A$ such that $a^n=1$, there’s a unique $K$-algebra homomorphism $K[X]/(X^n-1) \to A$ that sends $X$ to $a$.

At this point, to finish the proof, one can either mumble something about the Yoneda lemma with a smug expression or one can make the usual argument why two objects with the same universal property are isomorphic. (This should be familiar to anyone who has seen e.g. why the tensor product is unique)
Let’s do the latter: Because of the universal property of $K[X]/(X^n-1)$, we can find a unique $K$-algebra homomorphism $\varphi:K[X]/(X^n-1)$ that sends $X$ to $g$. This also works in the other direction: we get a unique $K$-algebra homomorphism $\theta: K[G] \to K[X]/(X^n-1)$. Then $\varphi \circ \theta$ is a $K$-algebra homomorphism $K[G] \to K[G]$ that sends $g$ to itself.
By the universal property, there can only be one such homomorphism, but we know that the identity is an example. Therefore $\varphi \circ \theta = \mathrm{id}$.
By the same argument, $\theta \circ \varphi = \mathrm{id}$.

Exercise Do a similar argument to determine the group algebra $K[G]$ where $G$ is a product of two cyclic groups as a quotient of the polynomial ring in two variables. Why can this approach not work in this form for nonabelian groups?

We can use this to describe the representations of cyclic groups by decomposing $K[X]/(X^n-1)$ with the Chinese remainder theorem. If we do that, we will end up with a product of rings, which is one of the reasons why it’s useful to think about modules over products of rings. If $R$ and $S$ are rings, then for every pair $(M, N)$ where $M$ is a $R$-module and $N$ is a $S$-module, we can make $M \oplus N$ into an $R \times S$-module by having $R$ act on the left factor and $S$ on the right factors. The following lemma tells us that every $R \times S$-module arises in such a way:

Lemma 2.5 If $R$ and $S$ are rings, then every $R\times S$-module is isomorphic to a direct sum $M \oplus N$ where $M$ is a $R$-module and $N$ is a $S$-module such that $R\times \{0\} \subset R \times S$ just acts on the first factor and $\{0\} \times S \subset R \times S$ just on the second one.
$M$ and $N$ are canonically determined.

Proof Let $X$ be a $T= R \times S$-module. Consider the central idempotents $e:=(1,0), f:=(0,1)$. Then $Te=eT=R \times \{0\}, Tf=fT=\{0\} \times S$ and $Te \cap Tf= \{0\}, Te+Tf=T$. Then set $M=eX$ and $N=fX$, we get that $eX \cap fX = 0, eX+fX=X$, so $X = eX \oplus fX = M \oplus N$. It’s clear that $S$ acts trivialy on $eX$ and $R$ acts trivially on $fX$ which shows the statement. We can use the same $e$ and $f$ for all modues, which makes this decomposition canonical.

(Note: The above construction can be enhanced into a category equivalence of $(R\times S)\textrm{-}\mathbf{Mod}$ and $R\textrm{-}\mathbf{Mod} \times S\textrm{-}\mathbf{Mod}$)

Now let’s finally describe the representations of cyclic groups over $\mathbb{C}$!

If $G$ is cyclic of order $n$, generated by $g$, then by Lemma 2.4, $\mathbb{C}[G] \cong \mathbb{C}[X]/(X^n-1)$, where the isomorphism sends $g$ to $X$. Let $\zeta_n = \exp(2\pi i/n)$, then we have the factorization $X^n-1=\prod_{k=0}^{n-1}(X-\zeta_n^k)$, so by the Chinese remainder theorem, we get
$\mathbb{C}[X]/(X^n-1) \cong \prod_{k=0}^{n-1} \Bbb{C}[X]/(X - \zeta_n^k)$
Note that this is an isomorphism both of rings, but also of $K[X]$-modules, which means that we send $X$ to $X$ in each component.

Therefore, by lemma 2.5, every $\mathbb C[G]$-module is a direct sum of $\Bbb{C}[X]/(X - \zeta_n^k)$-modules where $k$ varies. But for each $k$, we have $\Bbb{C}[X]/(X-\zeta_n^k) \cong \Bbb{C}$ via sending $X$ to $\zeta_n^k$. Modules over a field $F$ are easy to understand: they are just a (possibly infinite) sum of $F$. Thus we get that every $\Bbb{C}[X]/(X-\zeta_n^k)$-module is a direct sum of copies of $\Bbb{C}[X]/(X-\zeta_n^k) \cong \Bbb{C}$.

Through all the isomorphisms, we have kept track where the generator $g$ is sent: we send it first to $X$, then $X$ to $X$ (modulo something different in the CRT isomorphism), then $X$ to $\zeta_n^k$. This means that $g$ acts on the (one-dimensional) $\mathbb{C}[G]$-module corresponding to $\Bbb{C}[X]/(X-\zeta_n^k)$ by multiplication with $\zeta_n^k$. And in general, all modules are a direct sum of such modules (for different $k$), this means that $G$ really acts as a diagonal matrix where all the diagonal entries are $n$-th roots of unity. (Even for infinite-dimensional representations.)

We can also say something about a more general setting. Suppose that the characteristic of $K$ does not divide $n$. Then $X^n-1$ has distinct roots, so we can factorize $X^n-1=f_1 \cdot f_k$ where all $f_i$ are irreducible and pairwise distinct. Doing the same Chinese remainder theorem argument we get that:

$K[G] \cong \prod_{i=1}^k K[X]/(f_i)$. Now as the $f_i$ are irreducible, $K[X]/(f_i)$ will be a field and the dimension over $K$ will be equal to the degree of $f_i$, so we can again appeal to linear algebra and get the following result:

Lemma 2.6 Let $G$ be a cyclic group of order $n$ and let $K$ be a field such that the characteristic of $K$ does not divide $n$ and let $X^n-1=f_1 \cdot \ldots \cdot f_k$ be the factorization of $X^n-1$ into irreducibles. Then for every $i$, There is a $K[G]$-module corresponding to $f_i$ such that the dimension of the module is equal to the degree of $f_i$ and in general, every $K[G]$-module is a direct sum of such modules.

Example 2.7 For $K=\mathbb{R}$, the only factors that can occur for $X^n-1$ are $(X-1),(X+1)$ and quadratic factors of the form $X^2-2\cos(2\pi k/n)+1$. By choosing a clever basis for the modules corresponding to the quadratic factors, one obtains rotation representations as in example 1.6 (though the angle of rotation will be $2 \pi k/n$ instead of $2 \pi/n$ in the example.)

Example 2.8 For $K=\mathbb{Q}$, the factorization of $X^n-1$ is well-known, it factors as $X^n-1=\prod_{d \mid n} \Phi_d(X)$, where $\Phi_d(X)$ is the $d$-th cyclotomic polynomial. Thus the number of irreducible representations of $G$ over $\Bbb{Q}$ is equal to the number of divisors of $n$ and for each divisor $d$, there’s an irreducible representation of degree $\varphi(d)$.

We can also use this approach to say something about representations of cyclic groups in the case where the characteristic divides the group order. For simplicity, we just treat the case that $K$ has characteristic $2$ and $G$ is cyclic of order $2$. The factorization of $X^2-1$ is just $(X-1)^2$ and we get
$K[G] \cong K[X]/(X-1)^2$. Here the Chinese remainder theorem doesn’t help.
But one can apply the structure theorem for finitely generated modules over a PID mentioned in the first section, noting that every $K[X]/(X-1)^2$-module is also a $K[X]$-module to get that every finitely generated $K[X]/(X-1)^2$-module is a direct sum of copies of $K[X]/(X-1)$ and $K[X]/(X-1)^2$. If we look at the action of $X$ (which corresponds to the generator of $G$) on these modules, we see that it acts by multiplication with $1$ on $K[X]/(X-1)$, i.e. via the identity (we say “trivially”).
The action on $K[X]/(X-1)^2$ is more interesting: Using the basis given by (the residue classes of) $1$ and $X-1$, we see that the action of $X$ corresponds to a transvection action $g\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}a+b\\b\end{pmatrix}$ where $g$ is a generator of $G$ (cf. example 1.7)

We have seen how the algebraic structure of the group algebra can help to understand representations and in our example of cyclic groups, it turned out that when the conditions for Maschke’s theorem are satisfied, the group algebra is a product of fields.
We will investigate the structure of the group algebra in more detail in future posts and see that this was not a coindidence.