Representation theory with fewer characters

Representation theory, in its most classical setting, is the study of groups acting on vector spaces by linear transformations. In the case of finite groups, when the vector space is finite dimensional and the underlying field of the vector space has characteristic prime to the order of the group, in particular when the characteristic is 0, the theory is especially nice and essentially everything you could hope to be true is true, which allows one to prove important structural results about such groups. A discussion of this, in particular for infinite groups, can be found here.

A group $G$ acting on a vector space $V$ over a field $k$ of dimension $n$ is equivalently a homomorphism $f: G \to GL_n(k)$, or equivalently a $kG$ module structure on $V$, where $kG$ is the group algebra over $k$. 

The fundamental objects of study in representation theory are the characters. Given a homomorphism $f: G \to GL_n(k)$, the character is the composition of $f$ with the trace function. This gives a map $\chi: G \to {k}$ which is constant along conjugacy classes of $G$ since the traces of conjugate matrices are equal.

For concreteness, the reader can choose to pretend that we work over $\mathbb{C}$.

For finite groups, all finite dimensional representations are direct sums of irreducible representations, which are the analogue of prime numbers in this setting. This fact is known as Maschke's theorem and is sometimes repackaged as the sentence $kG$ is semisimple. We will see that the characters satisfy a lot of nice properties. However, the proofs given of these results are more often than not direct computations of characters, which I find conceptually unsatisfying. The goal of this post is therefore to record what I think of as better proofs.

First, we recall Schur's lemma and a consequence. The proof of Maschke's theorem and Schur's lemma are standard and can be found in just about any introductory text on representation theory. I promise that I haven't hidden any character theory under the rug but the reader is of course welcome to verify this.

Recall that a module is simple if it has no proper submodules and a division ring/algebra is a (not necessarily commutative) ring with 1 where any non-zero element has a multiplicative inverse.

Schur's Lemma Let $M_1, M_2$ be simple right $A$-modules. Then either $M_1 \cong M_2$, or $Hom_A(M_1, M_2) = 0$. If $M$ is a simple $A$-module, then $End_A(M)$ is a division algebra.

this has the following consequence in the setting of group representations:

Let V and W be vector spaces; and let ${\displaystyle {\rho }_{V }}$$\rho_V$ and ${\displaystyle {\rho }_{W }}$$\rho_W$ be irreducible representations of G on V and W respectively.

1. If ${\displaystyle V}$V $V$ and ${\displaystyle W}$$W$ $W$ are not isomorphic, then there are no nontrivial G-linear maps between them.

2. If $V=W$ is ${\displaystyle V=W}$$V=W$  finite-dimensional over an algebraically closed field ${\displaystyle \mathbb{} }$and if ${\displaystyle {\rho }_V={\rho }_{W }}$$\rho_V=\rho_W$ , then the only nontrivial G-linear maps are the identity, and scalar multiples of the identity. (A scalar multiple of the identity is sometimes called a homothety.)

Lemma 1 Let \( V \) and \( W \) be representations of \( G \). Assume that \( W \) is irreducible and that
\[ V = V_1 \oplus \cdots \oplus V_k \]
with each \( V_i \) irreducible. Then the dimension of \( \text{Hom}_G(V, W) \) equals the number of \( V_i \) factors that are isomorphic to \( W \).



Proof An element of \( \text{Hom}_G(V, W) \) is determined by its restriction to each \( V_i \), and Schur's Lemma implies that
\[ \text{Hom}_G(V_i, W) = \mathbb{C} \text{if } V_i \cong W \] and is 0 otherwise $\blacksquare$

Now we can exhibit character-free proofs of foundational results. From now on all groups are finite and all fields are algebraically closed of characteristic prime to $|G|$.

Theorem 2 The group ring $kG$ decomposes into a direct sum of $dim(V)$ copies of
every irreducible representation $V$ of $G$.
 

Proof Let \( V \) be an irreducible representation of \( G \). Letting \( e \in G \) be the identity, a \( G \)-equivariant map \( \phi: \mathbb{C}[G] \to V \) is completely determined by \( \phi(e) \in V \), and any vector in \( V \) occurs as \( \phi(e) \) for some \( G \)-equivariant map \( \phi: \mathbb{C}[G] \to V \). It follows that \( \text{Hom}_G(\mathbb{C}[G], V) \) is \( \dim(V) \)-dimensional. The theorem now follows from Lemma 1. $\blacksquare$

With very little more effort, one can deduce that for a group $G$ with irreducible characters $\chi_1, \dots \chi_n$, $|G|= \sum_{i=1}^n \chi_1^2$.

Theorem 3 The number of irreducible representations of a group $G$ equals the number of conjugacy classes in $G$.

Proof We know every irreducible representation appears as a submodule of the group algebra, so we look for a submodule of \( k[G] \) whose codimension is both the number of irreducibles and the number of conjugacy classes.

Let \( T \) be the subspace of \( A = k[G] \) generated by commutators
\[ [x, y] = xy - yx. \]
We claim that this consists precisely of things of the form
\[ \sum a_g g \]
where the sum of \( a_g \) over every conjugacy class vanishes, which implies the codimension is equal to the number of conjugacy classes. \( T \) is spanned by things of the form \( [g, h] = gh - hg \), and replacing \( g \) by \( gh^{-1} \) we see that it’s spanned by things of the form \( g - hgh^{-1} \), so the characterization follows.

Now we have to compare the codimension to the number of irreducibles. To compute the codimension of \( T \), we again decompose
\[ A = \bigoplus_{V_i \text{ irreducible}} \text{End}(V_i). \]
Clearly the image of \( T \) in \( \text{End}(V_i) \cong \text{Mat}_{d_i}(k) \) is the subspace spanned by commutators, which is just the subspace with trace 0, which has codimension 1. By theorem 2 we know that every irreducible representation appears as a direct summand, so adding up gives the result. $\blacksquare$

We now work over $\mathbb{C}$. Let $n$ be the number of irreducible representations of a group $G$. In light of theorem 3 we can create an $n \times n$ grid where each row corresponds to the values of characters for a specific irreducible representation, and each column contains the characters of a specific conjugacy class as the irreducible representation varies. We will show that the rows are orthogonal to each other for an appropriate inner product, with a similar result for columns, using the following result.

Artin-Wedderburn theorem: A semisimple ring R is isomorphic to a product of finitely many n_i-by-n_i matrix rings over division rings D_i, for some integers n_i, both of which are uniquely determined up to permutation of the index i.

Proof: If the ring \( R \) is semisimple, the right \( R \)-module \( R_R \) is isomorphic to a finite direct sum of simple modules. Write this direct sum as
\[R_R \cong \bigoplus_{i=1}^{m} I_i^{\oplus n_i},\]
where the \( I_i \) are mutually nonisomorphic simple right \( R \)-modules, the \( i \)-th one appearing with multiplicity \( n_i \). By Schur's lemma, there are no \( R \)-maps from \( I_i \) to \( I_j \) if \( i \) and \( j \) are distinct. This gives an isomorphism of endomorphism rings
\[\mathrm{End}(R_R) \cong \bigoplus_{i=1}^{m} \mathrm{End}\big(I_i^{\oplus n_i}\big),\]
and we can identify \( \mathrm{End}\big(I_i^{\oplus n_i}\big) \) with a ring of matrices
\[\mathrm{End}\big(I_i^{\oplus n_i}\big) \cong M_{n_i}\big(\mathrm{End}(I_i)\big)\]
where the endomorphism ring \( \mathrm{End}(I_i) \) of \( I_i \) is a division ring by Schur's lemma, because \( I_i \) is simple. Since \( R \cong \mathrm{End}(R_R) \), we conclude
\[R \cong \bigoplus_{i=1}^{m} M_{n_i}\big(\mathrm{End}(I_i)\big). \quad \blacksquare\]

In particular, $\mathbb{C}[G] \equiv \bigoplus M_{dim(V)}(End(V))$.

The space of complex-valued class functions of a finite group G has a natural inner product: $\left\langle \alpha ,\beta \right\rangle :={\frac {1}{\left|G\right|}}\sum _{g\in G}\alpha (g){\overline {\beta (g)}}$.

${\displaystyle ⟨\alpha ,\beta ⟩:=\frac{1}{\left|G\right|}\sum _{g\in G}\alpha (g)\overline{\beta (g)}}$

where $\overline {\beta (g)}$${\displaystyle \overline{\beta (g)}}${\overline {\beta (g) denotes the complex conjugate of the value of ${\displaystyle \beta }$$\beta$ on g. Note that for any character $\chi$, $\overline{\chi(g)}=\chi(g^{-1})$. 

Row orthogonality: With respect to this inner product, the irreducible characters form an orthonormal basis for the space of class functions, and this yields the orthogonality relation for the rows of the character table: $\left\langle \chi_i ,\chi_j \right\rangle= \delta_{ij}$.

${\displaystyle ⟨{\chi }_i,{\chi }_j⟩=\{\begin{array}{ll}0& \text{ if }i\ne j,\\ 1& \text{ if }i=j.\end{array}}$

Proof: Fix an irreducible representation \( V \) and let \( e_V \in \mathbb{C}[G] \) correspond to \( \mathrm{Id}_V \in \mathrm{End}_{\mathbb{C}}(V) \) and \( 0 \) in all of the other coordinates. Write \( e_V = \sum c_g g \). To recover the coefficient of \( h \), consider
\[e_V h^{-1} = \sum c_g g h^{-1}\]
and apply \( \mathrm{tr}_{\mathbb{C}[G]} \) to both sides to get
\[ m(V) \chi_V(h^{-1}) = |G| c_h,\]
hence
\[e_V = \frac{\dim V}{|G|} \sum \chi_V(g^{-1}) g.\]
Now note that \( e_V e_W = \delta_{VW} e_V \), and the orthogonality relations follow from equating coefficients at the identity in \( \mathbb{C}[G] \). $\blacksquare$ C[G]\mathbb{C}[G] .

 

Column orthogonality  $    \sum_{i = 1}^k \overline{\chi_i (g_j)}\chi_i (g_\ell) = \delta_{j\ell} |C_G(g_\ell)|$.

This is analogous to the fact that if a matrix has orthonormal rows, then it also has orthonormal columns.

Proof Consider the character table $X = (\chi_i(g_j))$. We know

   $ \delta_{ij} = \bra \chi_i, \chi_j\ket = \sum_\ell \frac{1}{|C_G(g_\ell)|}\overline{\chi_i(g_\ell)} \chi_k(g_\ell)$.  Then

    $\bar{X} D^{-1} X^T = I_{k\times k}$

  where $D$ is the diagonal matrix whose $i$th entry is $|C_G(g_i)|$.
  Since $X$ is square, it follows that $D^{-1} \bar{X}^T$ is the inverse of $X$. So $\bar{X}^T X = D$, which is exactly the theorem. $\blacksquare$

There are more beautiful facts about characters, but I only know of proofs of these that do use characters, so in the interest of making the title of this post somewhat accurate, we will leave those for another time.