Problem Set 5: Linear Representations

The geometrician teaches me how to work out the size of my estates—rather than how to work out how much a man needs in order to have enough. What use is it to me to be able to divide a piece of land into equal areas if I’m unable to divide it with a brother? The geometrician teaches me how I may avoid losing any fraction of my estates, but what I really want to learn is how to lose the lot and still keep smiling.

Solutions must be typeset in LaTeX and written in complete, grammatical sentences. You must list all your collaborators explicitly. Feel free to DM on Discord if you have any questions. Problems marked (Challenge) are optional but encouraged.

Up to this point we have studied permutation representations of a group $G$. A simple summary is that we have studied and classified all the homomorphisms from a (finite) group $G$ into symmetric groups. Next we begin studying linear representations, which may be similarly summarized as the study of homomorphisms from $G$ into general linear groups $\gl_n(K)$ for some field $K$. First we need to review some linear algebra.

5.1Modules

Linear algebra is fundamentally the theory of systems of linear equations. It is likely that your first exposure to linear algebra was fairly concrete and practical, before you learned how to write proofs or to think about abstract structures. Now that you're all grown up, we can revisit linear algebra through the lens of modern algebra. From this perspective, linear algebra is the study of vector spaces and the linear maps between them.

In fact, $\endo(V)$ is a non-commutative ring with the addition operation as defined above and multiplication given by composition. The additive identity $0$ is the constant zero function $0v = 0$ for all $v$. The multiplicative identity $1$ is given by the identity function $1v = v$ for all $v \in V$. We call $\endo(V)$ the endomorphism ring of $V$.

Now suppose that $(V, \vp)$ is an $R$-module. If $a \in R$, then we write $\vp_a \in \endo(V)$ for the endomorphism of $V$ associated to $a$. This succinct definition packs a lot of information. For example, if $a, b \in R$ and $v \in V$ we have

The first equality follows from the additivity of $\vp$, the second equality is the definition of addition in $\endo(V)$. Notice that this is sort of like saying $v$ distributes over the sum $a + b$ on the right.

On the other hand, if $a \in R$ and $v, w \in V$, then we also have a distributive-like property

but here this just follows from $\vp_a$ being an endomorphism of $V$.

We can think of an $R$-module as a representation of a ring $R$: it is an abelian group on which $R$ acts. The extra operations in a ring mean that the simplest kinds of objects on which it makes sense for them to act on/be represented by are abelian groups.

Just like with group actions, the homomorphism $\vp : R \to \endo(V)$ is often understood in context and suppressed. This leads to the following elegant notation. Given $a \in R$ and $v \in V$ we write \[ av := \vp_a v. \] That is, we think of the elements of $V$ as vectors and elements of $R$ as scalars; the action of the endomorphism $\vp_a$ on $v$ is then simply scalar multiplication of $v$ by $a$. For example, in this notation (5.0.1) becomes simply \[ (a + b)v = av + bv, \] and (5.0.2) becomes \[ a(v + w) = av + aw. \] Another example is the calculation \[ \vp_{ab}v = (\vp_a\vp_b)v = \vp_a(\vp_b v), \] which uses the multiplicativity of $\vp$ for the first identity and the definition of multiplication in $\endo(V)$ as composition of functions for the second identity. This becomes simply an associative-like identity, \[ (ab)v = a(bv). \] Hence the way we usually think about $R$-modules is that they are abelian groups $V$ which have a notion of scalar multiplication by the ring $R$.

Just as we can learn a lot about a group $G$ by studying its representations, we can learn a lot about a ring $R$ by studying its $R$-modules. Commutative algebra is essentially the branch of mathematics devoted to the study of $R$-modules when $R$ is a commutative ring.

You have already studied some important classes of $R$-modules, though probably not expressed in that language. For example, if $V$ is any additive abelian group and $n \in \ZZ$, then there is already a default way to interpret $n v$ for any vector $v \in V$: if $n$ is positive, then $nv$ is the same as adding $v$ to itself $n$-times; if $n$ is negative, $nv$ is the sum of the inverse $-v$ to itself $|n|$ times. In other words, any abelian group may be viewed as a $\ZZ$-module in exactly one way. ¹ Thus abelian groups are the same thing as $\ZZ$-modules.

Here is the other essential example:

Indeed, a $K$-vector space is an additive abelian group $V$ for which we can scale the vectors $v \in V$ by elements of $K$. When you took linear algebra, you may have only studied vector spaces over $\RR$ or maybe over $\CC$, but the definition and the basic theory works verbatim if we replace those with any arbitrary field $K$.

Vector spaces tend to feel more familiar than modules, and so it is common to think of modules as vector spaces which allow scaling by a ring instead of a field. However, vector spaces are the simplest kind of module; modules over rings that are not fields can be much more complicated than the kinds of vector spaces you may be familiar with.

Later in the course we will return to study more kinds of modules–including for a special class of noncommutative rings–but for now we turn our focus to vector spaces.

5.2Vector spaces

Let's recall some important definitions.

The following could be called the Fundamental Theorem of Linear Algebra.

So, up to $K$-linear isomorphism, there is only one vector space of each dimension. A simple model for the $K$-vector space of dimension $n$ is $K^n$, $n$-tuples of elements of $K$. Other examples of $K$-vector spaces show up in nature, but they will all be isomorphic to some $K^n$.

5.3Matrices

Linear maps $\vp : K^n \to K^m$ may be succinctly expressed as a matrix. Recall the standard basis of $K^n$ consisting of the vectors $e_1, e_2, \ldots, e_n$ where every entry of $e_i$ is 0 except the $i$th entry which is 1. A linear map $\vp : K^n \to K^m$ is completely determined by where it sends a basis, hence by the values \[ v_j := \vp e_j = \sum_{i=1}^m a_{ij}e_i. \] We may neatly package all this data into a matrix $[\vp] :=(a_{ij})$ where $a_{ij} \in K$, $1\leq i \leq m$, and $1 \leq j \leq n$. Composition of linear maps is given by the operation of matrix multiplication: If $\psi : K^r \to K^n$ has matrix $[\psi] = (b_{jk})$, then $\vp\psi : K^r \to K^m$ is $K$-linear and $[\vp\psi] = [\vp][\psi] = (c_{ik})$ where \[ c_{ik} := \sum_{j=1}^n a_{ij}b_{jk}. \] These formulas can be difficult to remember, especially the order of the indices. Here is the key thing to keep in mind: The columns of the matrix $[\vp]$ record the values of $\vp e_j$.

An automorphism of a vector space $V$ is a $K$-linear isomorphism $\vp : V \to V$. We write $\aut(V)$ to denote the group of all such automorphisms of $V$. In the case $V = K^n$ this group has a special name, the $n$th general linear group \[ \gl_n(K) := \aut(K^n). \] We can view the elements of $\gl_n(K)$ as invertible $n\times n$ matrices with entries in $K$. Recall that the simple way to check if a square matrix is invertible is to calculate its determinant.

The determinant is a homomorphism $\det : \gl_n(K) \to K^\times$ which may be calculated recursively using the cofactor expansion formula: If $A = (a_{ij})$, then for any $1 \leq i \leq n$ the $i$th row cofactor expansion of $\det(A)$ is \[ \det(A) = \sum_{j=1}^n (-1)^{i+j}a_{ij}\det(A_{ij}), \] and for any $1\leq j \leq n$ the $j$th column cofactor expansion of $\det(A)$ is \[ \det(A) = \sum_{i=1}^n (-1)^{i+j}a_{ij}\det(A_{ij}). \] In both cases $A_{ij}$ refers to the matrix we get by removing the $i$th row and $j$th column from $A$. Since we have been talking about symmetric groups, let me also share another formula for the complete expansion of the determinant using permutations, \[ \det(A) = \sum_{\sigma \in S_n}\sgn(\sigma) \prod_{i=1}^n a_{i\sigma(i)}. \] This one is sometimes useful for proving things but not very useful in practice (too many terms in the sum!) Determinants are challenging to calculate and to analyze—there are entire books filled with nothing but identities for various kinds of special determinants. The main thing to keep in mind for now is that a square matrix $A$ is invertible if and only if $\det(A) \neq 0$.

5.4Linear representations

Now we get to the point. Let $G$ be a group. A permutation representation of $G$ is an action of $G$ on a set $X$, or equivalently a homomorphism $\rho : G \to S_X = \aut(X)$.

When $\rho$ is understood, we follow the same convention as with $G$-sets and write $gv := \rho_g(v)$. What makes these actions linear is how they interact with the vector space operations. If $u, v \in V$ and $c\in K$, then for all $g \in G$ we have \[ g(u + v) = gu + gv \] and \[ g(cv) = c(gv). \] Note that if $V \cong K^n$ as $K$-vector spaces, then $\aut(V) \cong \gl_n(K)$ as groups. Hence a linear representation of $G$ is equivalent to a way of representing the elements of $G$ as matrices.

Up to this point we have let $K$ be an arbitrary field. While representations are interesting over all fields, they are simplest over the complex numbers. We will now switch from $K$ to $\CC$. The two features of $\CC$ which make it the simplest venue are

1. $\CC$ is algebraically closed, and
2. $\CC$ has characteristic zero.

Any field with these two properties will be just as good.

The simplest example of a linear representation is the trivial representation on a vector space $V$, where every element of $G$ acts as the identity: $gv = v$ for all $g \in G$ and $v \in V$. In other words, $\rho : G \to \aut(V)$ is the trivial homomorphism $\rho(g) = 1$. When we refer to the trivial representation without specifying $V$, we mean the trivial representation on the 1-dimensional space $\CC$.

Every permutation representation can be turned into a linear representation, giving us an immediate and ample supply of examples. Let $X$ be a finite set. For each $x \in X$ we create a formal vector $e_x$ (just a symbolic expression) and then let $\CC X$ be the $\CC$-vector space of all formal $\CC$-linear combinations of the vectors $e_x$, \[ \CC X := \Big\{\sum_{x \in X} c_x e_x : c_x \in \CC\Big\}. \] We call $\CC X$ the free $\CC$-vector space generated by $X$. The vectors $\{e_x : x \in X\}$ form a basis for $\CC X$ by fiat. Hence we have $\sum_{x \in X} c_x e_x = \sum_{x \in X} d_x e_x$ if and only if $c_x = d_x$ for all $x \in X$.

For example, if $X = \{1,2,3\}$, then \[ \CC X = \{c_1 e_1 + c_2 e_2 + c_3 e_3 : c_i \in \CC\} \cong \CC^3. \] More generally the dimension of $\CC X$ is $|X|$, hence \[ \CC X \cong \CC^{|X|}. \] If a group $G$ acts on a set $X$ by permutations, then we can extend this to a linear action of $G$ on $\CC X$ by defining \[ g\sum_{x\in X}c_x e_x := \sum_{x \in X} c_x e_{gx}. \] (Check that this really is a linear action of $G$ on $\CC X$!) The most important instance of this construction is when $X = G$ acting on itself by left multiplication. The resulting linear representation $\CC G$ is called the regular representation of $G$. It has dimension $|G|$, and $g \in G$ acts on the basis vectors by $g e_h = e_{gh}$. We will see later that the regular representation contains a copy of every representation of $G$ we could ever want.