Reminder: throughout this series, G is a finite group and K is a field. All K-vector spaces are assumed to be finite-dimensional over K.
G4. Maschke’s Theorem
If is a K[G]-submodule, it turns out V is isomorphic to the direct sum of W and some other submodule W’.
Maschke’s Theorem. Suppose char(K) does not divide #G (e.g. char(K) = 0). If is a K[G]-submodule, then there’s a submodule such that:
This gives .
As a consequence, if is a nontrivial submodule, then one can always decompose ; since V is finite-dimensional, this process must eventually terminate.
Definition. A K[G]-module V≠0 is said to be simple if it has no submodule other than 0 and itself. The corresponding representation is then said to be irreducible.
Thanks to Maschke’s theorem, every K[G]-module is a direct sum of simple modules.
Proof of Maschke’s Theorem.
Let X be a K-subspace of V such that and , which is always possible by linear algebra. Take the projection map Now define the linear map:
We claim: q is K[G]-linear; it suffices to prove that q(hv) = h·q(v) for all But
Next claim: if then q(w)=w. Indeed since W is a K[G]-submodule of V, so for every Hence q(w) is 1/|G| times a sum of |G| copies of w.
To complete the proof, we’ll show that W’ = ker(q) satisfies the conditions of the theorem:
- Since q is K[G]-linear, its kernel W’ is a K[G]-submodule of V.
- If , then q(v) = 0 since on the other hand, since we have q(v)=v so v=0. Thus,
- For any , write . Since we have also. So and we get Thus ♦
Consider acting on by permuting the coordinates (see example 3 in G1). The 1-dimensional subspace W spanned by (1, 1, 1) is clearly a G-invariant subspace. Applying the construction in the above proof gives us the complementary submodule
G5. Schur’s Lemma
Assume char(K) = 0. By Maschke’s theorem, every K[G]-module is a direct sum of simple modules. The following result is handy:
Schur’s Lemma. Let V and W be simple K[G]-modules.
- If f : V → W is a K[G]-linear map, then f=0 or f is an isomorphism.
- Suppose K = C. If f : V → V is a K[G]-linear map, then it’s a scalar multiple of the identity.
Suppose f : V → W is non-zero. Its kernel is a submodule of V so it’s either 0 or V; since f≠0, we have ker(f) = 0. Likewise, its image is a submodule of W so it’s either 0 or W; since f≠0, we have im(f) = W. Thus f is bijective.
Let λ be an eigenvalue of f with eigenvector v, i.e. f(v) = λv. Then ker(f – λ) is non-zero so it must be whole of V, i.e. f = λ·1V. ♦
Another way of looking at Schur’s lemma: for K = C and simple C[G]-modules V, W,
From now onwards, we let K = C.
G6. The Space of Fixed Vectors.
For any C[G]-module V, consider the space:
the space of all vectors fixed by G. The following result is critical.
Lemma. We have:
the average of the traces of acting on V.
Take the averaging map (which is C-linear):
We claim: for any , we have . Indeed, for any , we have
On the other hand, if then gw = w, which gives us p(w) = w. Just as in the proof of Maschke’s theorem, this shows and If we pick a basis of ker(p) and of VG, then the matrix for p is:
G7. Introducing Characters
Since the trace of g‘s action is important, let’s define:
Definition. Let be a representation of G. Its character is the function
Looking at V as a C[G]-module, we have
From G6, we see that Suppose V and W are C[G]-modules. Let’s consider the various constructions in G3.
- Direct sum : .
- Tensor product : .
- Dual : .
The first two equalities are easy. For the third, let’s pick a basis of V and consider the dual basis If M is the matrix for g : V → V with respect to B, the corresponding matrix for g : V* → V* with respect to B* is given by Hence:
Since each g : V → V has finite order, it is diagonalisable (via expressing g in its Jordan canonical form) and every eigenvalue is a root of unity. With respect to a basis of eigenvectors, g is diagonal with entries Thus g-1 is diagonal with entries and as desired.
- Hom :
Now consider HomC(V, W)G. A C-linear map f : V → W is fixed by G if and only if:
if and only if f is C[G]-linear. Thus and:
Together with Schur’s lemma, we have:
Corollary (Orthogonality of Characters).
If V and W are simple C[G]-modules, then:
G8. Class Functions
Since tr(AB) = tr(BA) for square matrices of the same size (in fact, it’s even true for rectangular matrices, as long as AB and BA are both square), we have
Definition. A function is called a class function if for any
Thus, the character of a representation is a class function. Define an inner product between class functions as:
From G7, the set of characters of simple C[G]-modules (or irreducible representations) forms an orthonormal set. In fact, we can say more:
Theorem. The set of irreducible characters forms an orthonormal basis for the space of all class functions.
It suffices to show that the irreducible characters span the space of class functions. Otherwise, there’s a class function f : G → C which is orthogonal to all irreducible characters. Take the element For any simple C[G]-module V, σ acts on V as a C-linear map. Now,
so it is in fact C[G]-linear and must be a scalar multiple of the identity. To compute this identity, we take the trace:
But f is orthogonal to each so c=0. Thus σ=0 on an irreducible representation V and hence on any representation as well. In particular, σ=0 on C[G] itself so 0 = σ·1 = σ. ♦
Corollary. The number of irreducible representations of G is the number of its conjugancy classes m.
The m × m table comprising of values is called a character table.
On an intuitive level, characters thus provide an excellent way to analyse non-abelian groups. Indeed, if G is “highly non-abelian” (e.g. simple), it has fewer conjugancy classes and thus a smaller character table. On the other hand, abelian groups have the largest possible character table since every conjugancy class has size 1.
G9. Degrees of Irreducible Characters
If V is a C[G]-module, its dimension as a C-space is called its degree.
Let’s take the obvious C[G]-module, i.e. C[G] itself (on the representation side, this gives rise to the regular representation we saw in example 3, G1). Writing C[G] as a direct sum of irreducible representations , the corresponding character gives:
, where each .
To compute each di, we take the inner product . But is easy to compute: it takes e to |G| and all other to zero (since if g≠e, the action of g on G via left-multiplication has no fixed point). Thus,
We have thus proven:
Theorem. If V is an irreducible representation of G, the number of times it occurs in the regular representation is dim(V).
Corollary. . [ Follows from ]
Corollary. If G is abelian, then each . [ This follows from: size of character table is |G| × |G|. ]
Example of a Character Table
Here’s the character table of the symmetric group S4. It’s 5 × 5 since the number of conjugancy classes of S4 is p(4) = 5. The size of each conjugancy class is written in square brackets [s].
In the next installation, we’ll see how this table is obtained.