Group Representations and Characters

Def. (Group Representation) Let G be a group, and let V be a finite dimensional vector space.  Then a group representation of G is a homomorphism

\displaystyle \rho : G \to \text{GL}(V)

where \text{GL}(V) = \{ \text{linear isomorphisms } T: V \to V \}.  The group structure on \text{GL}(V) is given by function composition.

Note: for most of our applications, V = \mathbb{R}^3 or, perhaps \mathbb{R}^n or \mathbb{C}^n.  In the first case we will have

\displaystyle \text{GL}\left(\mathbb{R}^3\right) = \{ 3\,\times\,3 \text{ invertible matrices} \}.

This topic requires us to better (and more quickly!) understand homomorphisms, vector spaces, invertible matrices / vector space isomorphisms, and some other topics, too.  To get started on these, let’s consider the following exercises.


Problem 1.  Let G_1 and G_2 be groups (each with their own group operations), and suppose \varphi : G_1 \to G_2 is a group homomorphism.

(a) Prove that the kernel of \varphi, which is defined as

\displaystyle \text{ker}\varphi = \left\{ g \in G_1 : \varphi(g) = e_2 \right\},

(where e_2 is the identity element in G_2) is a sub-group of G_1.  (Recall that we do not have to check that G_1‘s multiplication is associative for the proposed group \text{ker}\varphi, but, instead we need to check that e_1 \in \text{ker}\varphi, that this set is closed under products and that it is closed under taking inverses).

(b) Let V = \mathbb{R}^n.  Explain why the function

\rho(g) = I_n \,\,\,

(for all g \in G and I_n is the n \times n identity matrix) is a representation of G.  Moreover, explain why this is called the trivial representation.

(c ) Prove that the image of \varphi, which is defined as

\displaystyle \text{im}\varphi = \left\{ \varphi(g) : g \in G_1 \right \} = \left\{y \in G_2 : \, \exists \, g \in G_1, \, \varphi(g) = y \right\},

is a subgroup of G_2.

Problem 2.  Consider two 2\times 2 matrices A, B \in \text{GL}\left(\mathbb{R}^2\right).  Prove that the trace is invariant under conjugation; that is, prove that

\text{tr}\left(B^{-1}\,A\,B\right) = \text{tr} A.

Note: given the (relatively) small size of each matrix involved, this will likely be easiest to prove using an explicit formula for the inverse of a 2\times 2 matrix.


It is not too difficult to show that the property of traces for 2\times 2 matrices from Problem 2. also holds for traces of n \times n matrices — let us take this as a given for the time being.

Conjugating a given matrix, A, by a second matrix, B, to form the strange-looking product

B^{-1}AB

is, from the point of view of Linear Algebra, a very natural process.  The new matrix that results from this so-called conjugation is said to be similar to the original matrix A, and it encodes much of the same information that the matrix A does.

In Linear Algebra, we learn to view such a conjugation process as a “change of basis” process.  This process works as follows:  Initially, the invertible transformation T :\mathbb{R}^n \to \mathbb{R}^n is represented by matrix A in terms of how T transforms the standard basis vectors \{\vec{e}_1, \vec{e}_2, \cdots, \vec{e}_n\}.  In particular, we have that

T \text{ is encoded by } A = \left[\begin{array}{cccc} \uparrow & \uparrow &  & \uparrow \\ T(\vec{e}_1) & T(\vec{e}_2) & \cdots & T(\vec{e}_n) \\ \downarrow & \downarrow &  & \downarrow \end{array}\right].

Next, the invertible matrix B^{-1} is used to create a new (usually non-standard) basis for \mathbb{R}^n given by \left \{ B^{-1}\vec{e}_1, B^{-1}\vec{e}_2, \cdots, B^{-1}\vec{e}_n\right\} = \left\{\vec{w}_1, \vec{w}_2, \cdots, \vec{w}_n\right\}.  We then build a matrix that encodes how the original transformation acts on \mathbb{R}^n by writing down how it transforms each new basis vector:

T \text{ is encoded by } B^{-1}AB = \left[\begin{array}{cccc} \uparrow & \uparrow & & \uparrow \\ T(\vec{w}_1) & T(\vec{w}_2) & \cdots & T(\vec{w}_n) \\ \downarrow & \downarrow & & \downarrow \end{array}\right].

This, as it may seem, is a fairly significant topic in Linear Algebra, and there are details to be clarified (what, exactly, is a basis?  how are the columns exactly expressed as columns of numbers?), but we can nonetheless focus on the central conclusion: conjugation by invertible matrices is a natural process.

In fact, since the trace of a matrix is invariant under this process, this reveals to us that the trace of a matrix is a more natural and important concept than one may first think (after all, its defined via a cute-looking formula, nothing particularly deep or scary seeming).  If we want to regard the matrices A and B^{-1}A B as “essentially the same” or as “encoding the same basic information,” then the trace becomes our friend, something that does not detect “artificial difference” we are more interested in overlooking.


Problem 3.  Consider the set of invertible n \times n matrices \text{GL}(\mathbb{R}^n) equipped with the relation \sim where two matrices A, C \in \text{GL}(\mathbb{R}^n) are related

A \sim C \text{ means } \,\, \exists B \in \text{GL}(\mathbb{R}^n), \, \, B^{-1}AB = C.

Prove that \sim\, is an equivalence relation.  Note that this “breaks up” the original set \text{GL}(\mathbb{R}^n) into different equivalence classes, each class containing matrices that represent the same isomorphism or symmetry of \mathbb{R}^n.


We now return to our abstract group, G.  If we have a representation \rho : G \to \text{GL}(V), then we are able to use this representation to think of G‘s elements as matrices (isomorphically) transforming some vector space (in our case, \mathbb{R}^n or \mathbb{R}^3).

Because \rho is a group homomorphism, and because conjugating matrices in \text{GL}(V) is a natural or fundamental process, this means that conjugating the original elements of G should also be a natural process, too.  In particular, given g, h \in G the conjugate of g by h is the group element

h^{-1}\,g\, h.

Def.character of a group G is a function \chi : G \to \mathbb{R} given by a representation \rho according to the following rule:

\chi (g) = \text{tr} \rho(g)

For every element g \in G, the homomorphism \rho converts it into some matrix, and then we obtain a real number by computing the trace of that matrix.  Note that by Problem 2 (and its extension to n \times n matrices), a character takes the same value on an element g as it does on a conjugated element h^{-1} g h.  We rephrase this in Problems 4 and 5. below.

Problem 4.  Let G be a group.  Define a relation \sim on G by defining

g_1 \sim g_2 \text{ means } \, \exists h \in G, \, h^{-1} g_1 h = g_2.

Prove that \sim is an equivalence relation.

Note: the equivalence classes of \sim are called the conjugacy classes of G.

Problem 5.  Explain why / prove that the following statement is true:

Characters are constant on G‘s conjugacy classes.


We still need to discuss many more aspects of representations and characters, but for now the table is somewhat set for some big ideas.  In particular, we should be in a position to better understand and answer the following questions:

  1.  What is a character table?
  2. What is a reducible representation?
  3. What is an irreducible representation?
  4. Exactly how do the entries of a character table encode the group under consideration?
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