As alluded to at the end of the previous article, we shall consider the case where K is algebraically closed, i.e. every polynomial with coefficients in K factors as a product of linear polynomials. E.g. K = C is a common choice.

Having assumed this, we see that any division ring D ⊇ K which is finite-dimensional as a K-vector space must be K itself; indeed, if x ∈ D, then K(x) is a commutative division ring (i.e. a field) over K which is of finite dimension, so it must be K itself, and x ∈ K. Thus,

and for any simple K[G]-module V, we have End_K[G](V) = K. This also gives:

,

where V_i runs through all simple K[G]-modules up to isomorphism.

In particular, this mean for any simple K[G]-module V. Thus, their characters actually form an orthonormal set over K. The next theorem we want to show is:

Theorem. The set of , where V runs through all simple K[G]-modules, forms an orthonormal basis of the space of class functions.

Proof

We need to show the characters span the space of class functions. Let f : G → K be any class function; subtracting by a linear combination of characters, we may assume f is orthogonal to all Let

Since f is a class function, we get for any g in G. Thus α commutes with every element of K[G]. If V is a simple K[G]-module, let m : V → V be the multiplication-by-α map; since α commutes with every element of K[G], m is a K[G]-linear map. From above, we know that End_K[G](V) = K, so m is a scalar map represented by λ ∈ K.

To find this λ, we take the trace of m :

which is zero since f was picked to be orthogonal to all characters. Hence multiplication-by-α is the zero map for all simple K[G]-module V; since K[G] is semisimple, multiplication-by-α is also zero on it. But that map takes e to α so α=0. ♦

Summary. This is what we know so far:

• The number k of isomorphism classes of simple K[G]-modules is precisely the number of conjugancy classes of G (by the above theorem).
• Their dimensions satisfy
• Their characters form an orthonormal basis for the space of class functions G → K.

As runs through all characters of simple modules and g runs through conjugancy classes, we obtain a k × k tables for the values Here are the character tables for S₄ and S₅:

The number [n] in square brackets refers to the number of elements in each conjugancy class. Let us check that and are orthogonal:

For the details of computations, refer to an earlier article. Since both character tables comprise of solely integers, one suspects that in fact the modules are defined over Q, i.e. that the characters of simple Q[G]-modules are already orthonormal. This is, in fact, true for any symmetric group S_n – but we will have to revisit this another day.

Finally, we end this article with an example of a simple K[G]-module V where End(V) is a non-commutative division ring.

Division Ring Example

Let K = R, and consider the division ring of quaternions H. Let G = {±1, ±i, ±j, ±k} ⊂ H which is a non-abelian group of order 8.Let’s compute its character table. First note that G has 5 conjugancy classes: {+1}, {-1}, {±i}, {±j}, {±k}. Next:

• There’s the trivial representation, where all g → 1.
• Let N = {±1, ±i}, which is of index 2 in G and thus normal. The non-trivial representation of G/N ≅ C₂ → R* then pulls back to a representation of G which is irreducible since it has dimension 1.
• Do the same for {±1, ±j} and {±1, ±k}.

This gives us 4 irreducible representations of dimension 1. Let us compute the character table over C. Since the last character has dimension 2. The other entries are easy to compute:

So this is for C, but what about over R? Note from our description above that the 4 characters of dimension 1 are defined over R. Write

where the character τ of W is given by Is W simple?

Note that upon extension to C, we have where V is the simple C[G]-module with character χ = (2, -2, 0, 0, 0) above. Thus if W is not simple, we must have for some R[G]-module V, such that dim_RV = 2.

We shall prove that such a V does not exist; let ρ : G → GL(2, R) be the corresponding group homomorphism.

• We have ρ(+1) = I and ρ(-1) = -I (the latter because it has trace -2 from the character table).
• The other 6 elements all satisfy so and so ρ(g) has eigenvalues ±√-1. From the character table, its trace is zero so the eigenvalues are +√-1 and -√-1.
• Thus ρ(g) has det = 1 and trace = 0, corresponding to the product and sum of its eigenvalues.
• It remains to show that if A, B are 2×2 real matrices with det = 1 and trace = 0, then the same cannot hold fo AB.

For that, it is easy to show A and B are of the form:

 Upon expanding tr(AB) we obtain:

Over the reals, we must have b = d = 0, which is absurd. Thus, V does not exist so W is already simple as an R[G]-module. Hence the structure of the group ring R[G] is:

where D is a division ring of dimension 4 over R. [ In fact, it is the ring of quaternions. ]