Representation Theory of Finite Groups - Character Theory

Character Theory

Main article: Character theory

There is a mapping from G to the complex numbers for each representation called the character given by the trace of the linear transformation upon the representation generated by the element of G in question

χ_ρ(g)=Tr.

All elements of G belonging to the same conjugacy class have the same character: in other words χ_ρ is a class function on G. This follows from

Tr=Tr=Tr

by the cyclic property of the trace of a matrix.

What are the characters of C? Using the property that gh−1 is only the same as g if h = e, χ_C(g) is |G| if g=e and 0 otherwise.

The character of a direct sum of representations is simply the sum of their individual characters.

Putting all of this together,

with the Kronecker delta on the right hand side.

Repeat this, working with characters of G×G instead of characters, of G which I'll call Δ. Then, Δ_C(g,h) is the number of elements k in G satisfying g k h−1 = k. This is equal to

where * denotes complex conjugation. After all, C is a unitary representation and any subrepresentation of a finite unitary representation is another unitary representation; and all irreducible representations are (equivalent to) a subrepresentation of C.

Consider

.

This is |G| times the number of elements which commute with g; which is |G|2 divided by the size of the conjugacy class of g, if g and k belong to the same conjugacy class, but zero otherwise. Therefore, for each conjugacy class C_i of size m_i, the characters are the same for each element of the conjugacy class and so we can just call χ_ρ(C_i) by an abuse of notation). Then,

.

Note that

is a self-intertwiner (or invariant). This linear transformation, when applied to C (as a representation of the second copy of G×G), would give as its image the 1-dimensional subrepresentation generated by

;

which is obviously the trivial representation.

Since we know C contains all irreducible representations up to equivalence and using Schur's lemma, we conclude that

for irreducible representations is zero if it's not the trivial irreducible representation; and it's of course |G|1 if the irreducible representation is trivial.

Given two irreducible representations V_i and V_j, we can construct a G-representation

,

this time not as a G×G representation but an ordinary G-representation. See direct product of representations. Then,

.

It can be shown that any irreducible representation can be turned into a unitary irreducible representation. So, the direct product of two irreducible representations can also be turned into a unitary representations and now, we have the neat orthogonality property allowing us to decompose the direct product into a direct sum of irreducible representations (we're also using the property that for finite dimensional representations, if you keep taking proper subrepresentations, you'll hit an irreducible representation eventually. There's no infinite strictly decreasing sequence of positive integers). See Maschke's theorem.

If i≠j, then this decomposition does not contain the trivial representation (Otherwise, we'd have a nonzero intertwiner from V_j to V_i contradicting Schur's lemma). If i=j, then it contains exactly one copy of the trivial representation (Schur's lemma states that if A and B are two intertwiners from V_i to itself, since they're both multiples of the identity, A and B are linearly dependent). Therefore,

Applying a result of linear algebra to both orthogonality relations (|C_i| is always positive), we find that the number of conjugacy classes is greater than or equal to the number of inequivalent irreducible representations; and also at the same time less than or equal to. The conclusion, then, is that the number of conjugacy classes of G is the same as the number of inequivalent irreducible representations of G.

Corollary. If two representations have the same characters, then they are equivalent.

Proof. Characters can be thought of as elements of a q-dimensional vector space where q is the number of conjugacy classes. Using the orthogonality relations derived above, we find that the q characters for the q inequivalent irreducible representations forms a basis set. Also, according to Maschke's theorem, both representations can be expressed as the direct sum of irreducible representations. Since the character of the direct sum of representations is the sum of their characters, from linear algebra, we see they are equivalent.

We know that any irreducible representation can be turned into a unitary representation. It turns out the Hilbert space norm is unique up to multiplication by a positive number. To see this, note that the conjugate representation of the irreducible representation is equivalent to the dual irreducible representation with the Hilbert space norm acting as the intertwiner. Using Schur's lemma, all possible Hilbert space norms can only be a multiple of each other.

Let ρ be an irreducible representation of a finite group G on a vector space V of (finite) dimension n with character χ. It is a fact that χ(g) = n if and only if ρ(g) = id (see for instance Exercise 6.7 from Serre's book below). A consequence of this is that if χ is a non-trivial irreducible character of G such that χ(g) = χ(1) for some g≠1 then G contains a proper non-trivial normal subgroup (the normal subgroup is the kernel of ρ). Conversely, if G contains a proper non-trivial normal subgroup N, then the composition of the natural surjective group homomorphism G → G/N with the regular representation of G/N produces a representation π of G which has kernel N. Taking χ to be the character of some non-trivial subrepresentation of π, we have a character satisfying the hypothesis in the direct statement above. Altogether, whether or not G is simple can be determined immediately by looking at the character table of G.