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Basics of group representation theory July 10, 2009

Posted by Akhil Mathew in algebra, representation theory.
Tags: , , ,

Today, I want to talk a bit about group representation theory. Many of us (such as myself) are interested in representation theory in general and will likely talk more about it in the future, so it will be useful to summarize the essential ideas here to refer back. But the basics are well known and have been discussed at length on other blogs (see, e.g. here, which is discussing the subject right now), so I am merely going to summarize these facts without proofs. The interested reader can read these notes for full details. Then, I’ll mention a property to be used later on.

What is a group representation?

Start with a group {G}. At least for now, we’re essentially going to be constructed with finite groups, but many of these constructions generalize. A representation of {G} is essentially an action of {G} on a finite-dimensional complex vector space {V}.

Formally, we write:

Definition 1 A representation of the group {G} is a finite-dimensional complex vector space {V} and a group-homomorphism {G \rightarrow Aut(G)}. In other words, it is a group homomorphism {G \rightarrow GL_n(V)}, where {n = \dim \ V}, and {GL_n} is the group of invertible {n}-by-{n} matrices.

An easy example is just the unit representation, sending each {g \in G} to the identity matrix.

Alternatively, we can form the group ring {{\mathbb C} [G]} of {G}; this consists of formal sums {\sum_{g \in G} c_g g} for coefficients {c_g \in {\mathbb C}}. It is made into a ring by pointwise multiplication: we demand that multiplication in the group ring match multiplication in the group, and be linear.

Phrased in this language, a representation of {G} is just a {{\mathbb C}[G]}-module.

Maschke’s Theorem

One of the nice facts about group representation theory is that we can split a representation of a group {G} into “atoms” which contain no smaller elements. Formally, we write:

Definition 2 A nonzero representation {V} of the group {G} is said to be simple or irreducible if {V} (which is a {{\mathbb C}[G]} module) contains no proper {{\mathbb C}[G]}-modules other than zero.

The following is Maschke’s theorem, which we state without proof:

Theorem 3 A representation {V} of {G} can be written as a direct sum,

\displaystyle V = \oplus_i n_i V_i

where the {V_i}‘s are simple representations, and the {n_i} are integers.

The decomposition is also unique up to isomorphism. (Note that the direct sum of representations is a representation itself.)

This is equivalent to saying that the group ring {{\mathbb C}[G]} is semisimple.

[Edit, 6/14- for a proof of Maschke’s theorem, see this post, which appeared after this one..]

The Regular Representation

Any ring {R} is automatically a (left) {R}-module. In particular, the group ring {{\mathbb C}[G]} is itself a {{\mathbb C}[G]}-module, which is also a finite-dimensional vector space, if {G} is a finite group.

Thus the underlying vector space {V_G} of {{\mathbb C}[G]} is a representation of {G}. This is called the regular representation.

The following is the fundamental property of the regular representation:

Theorem 4 When decomposed into simple components, the regular representation of a finite group {G} contains each simple component with multiplicity equal to the dimension of that component.

So, in particular, there are only finitely many such (non-isomorphic) simple representations, a fact which can be proved more generally for semisimple algebras.

Characters and Class Functions

Given a representation {\rho: G \rightarrow Aut(V)} for a vector space, we define the corresponding function {\chi_V: G \rightarrow {\mathbb C}} by {\chi_V(g) = Tr(\rho(g))}.

Definition 5 {\chi_V} is called the character of {V}.

From the basic properties of traces, characters are additive on direct sums, i.e. {\chi_{V \oplus W} = \chi_V + \chi_W}. Moreover, characters are multiplicative on tensor products.

From the properties of traces (specifically {Tr(AB)=Tr(BA)} for matrices {A,B}), we see that the characters {\chi} are class functions; that is, {\chi_V( ugu^{-1}) = \chi(g)} for all {u,g \in G}. The characters are invariant under inner automorphisms.

Theorem 6 The irreducible characters of {G} form a basis of the vector space of class functions on {G}.

In particular, this implies the number of irreducible representations of {G}, up to isomorphism, is given by the number of conjugacy classes.

This basis is actually orthonormal with respect to the Hermitian inner product on the space of class functions defined as follows:

\displaystyle (f,f') = \frac{1}{Card \ G} \sum_{g \in G} f(g) \overline{ f'(g) }

because of the following:

Theorem 7 For irreducible representations {V, V'}, we have {(\chi_V, \chi_{V'}) = 1} if {V, V'} are isomorphic, and {(\chi_V, \chi_{V'})=0} otherwise.

In particular, putting Maschke’s theorem and this together, we see that to find the multiplicties with which an irreducible {A} occurs in {V}, one just takes the inner product {(\chi_V, \chi_A)}.

I want to end this post with a random remark that will become useful later on.

First of all, our matrices {\rho(G)} for {\rho: G \rightarrow Aut(V)} defining a representation are actually diagonalizable. We know from linear algebra that we can put such a matrix in a triangular form, but we also know that some power of them is the identity: {G^{Card \ G} = 1}. So fix {M}, a triangular matrix, which satisfies {M^n = I} for some {n}. I claim {M} is diagonal. Indeed, the condition means {M^n-I = 0}, or {M} satisfies {X^n-1 = 0}. In particular, the minimal polynomial of {M} divides {X^n-1}. Since {X^n-1} has no repeated roots, the minimal polynomial also has none. By a general theorem of linear algebra, this implies {M} is diagonalizable. Moreover, let the diagonal entries be {\zeta_1, \dots, \zeta_d}. Then each {\zeta_j} satisfies {\zeta_j^{n} = 1}.

In this case, the trace {Tr M} is just the sum {\sum_j \zeta_j}. It follows that the character {\chi_V} is always a sum of {Card \ G}-th roots of unity and is in particular an algebraic integer.

This fact will be useful to us later, as we talk about applications of group representation theory. It will also enable us to show that the dimensions of group representations divide the group order. For now though, I’ll leave it here.


1. Qiaochu Yuan - July 12, 2009

The Princeton Companion article on representation theory has an interesting snippet about why representation theory works out so much more nicely than the theory of group actions. Offhand, the two are superficially very similar: just as a representation is a homomorphism G \to \text{Aut}(V) where the latter lies in the category of finite vector spaces, a group action is a homomorphism G \to \text{Aut}(S) where the latter lies in the category of finite sets and bijections.

The problem, according to the article, is that understanding the “irreducible” (transitive) group actions requires understanding the entire subgroup lattice of a group, which can be exponentially larger than the group itself. However, the nice categorical properties of \text{FinVect} somehow force us to restrict our attention to conjugacy structure, with the result that there are as many irreducible representations as conjugacy classes. Of course there are fewer of these than group elements.

The issue appears to be that \text{FinSet} isn’t additive. Decomposing a given group action with respect to disjoint union is easy (they’re given by the orbits), but decomposing with respect to Cartesian product appears to be hard (at least I’ve never heard anyone talk about it), and in \text{FinVect} we get to do both at the same time.

2. Akhil Mathew - July 12, 2009

Thanks for the comment! That is, indeed, an interesting observation, that the simple imposition of linearity conditions on a group action can drastically change the structure of irreducible objects. This also seems to relate to the reason characteristic zero (or more generally a characteristic prime to the group order) is necessary for the nice parts of character theory to work; in this case, the additive structure still exists, but averaging doesn’t work (just as it doesn’t for finite sets).

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