Table Of ContentOutline of Basic Representation Theory
Henry Cohn
May 6, 2002
This is a quick outline of some basic representation theory. The proofs are only sketched, and many
other details are omitted, but I hope I’ve included enough to make them relatively easy to reconstruct.
We’llworkoverthefieldCofcomplexnumbers, sincetheargumentsthatfollowwillusethefactthat
it is algebraically closed and has characteristic 0. Let G be a finite group, and let n=|G|.
A representation of G is a complex vector space V with a homomorphism ρ : G → Aut(V). Here,
Aut(V)isthegroupofcomplexlinearoperatorsonV. Forus,allrepresentationswillbefinitedimensional
as well. We will typically write the action ρ(g)v of g ∈G on v ∈V as gv, and omit ρ from the notation
entirely when it seems convenient. (Typically, we will be thinking about only one representation at a
time on a given vector space, but if more than one come up, ρ will return to prominence.)
An invariant subspace of a representation is a subspace mapped into itself by G (so it too is a
representation). A representation is irreducible if it has no invariant subspaces other than itself and the
zero-dimensional subspace. It is completely reducible if it is a direct sum of irreducible representations.
Given two representations V and W, we can combine them in various ways. The most obvious are
V ⊕W and V ⊗W (all tensor products are over C). The group G acts on them in the obvious way:
g(v⊕w) = (gv)⊕(gw) and g(v⊗w) = (gv)⊗(gw) (with the latter action extended by linearity). We
make the dual space V∗ into a representation by (gf)(v) = f(g−1v) for g ∈ G, v ∈ V, and f : V → C.
The inverse is necessary for this to be a representation (otherwise associativity wouldn’t hold). It’s
natural from the point of view of translating graphs: to translate the graph of a function three units
to the right, we subtract three from its argument. We turn Hom(V,W) into a representation similarly:
(gf)(v)=g(f(g−1v))forf :V →W. Here,ontherighthandsidetheg−1 isactingintherepresentation
V,andtheoutergisactingintherepresentationW. Here’sacasewherenamingthehomomorphismsfrom
GtoAut(V)andAut(W)mightclarifythings,butitwouldmakethenotationmorecumbersome,andit’s
logically unambiguous as it stands. It’s worth checking that these operations on representations satisfy
allthepropertiesonewouldhope. Inparticular,thereisacanonicalisomorphismHom(V,W)=V∗⊗W.
Let Hom (V,W) denote the set of G-homomorphisms from V to W. In other words, linear trans-
G
formations f : V → W that respect the G action: f(gv) = g(f(v)) for all g ∈ G, v ∈ V. A useful
characterization is that Hom (V,W) is the set of elements of the representation Hom(V,W) that are
G
fixed by the action of every element of G.
Theorem 1 (Maschke’s Theorem). Every (finite-dimensional) representation of G is completely re-
ducible.
Proof. The proof uses a fundamental technique, namely averaging over the group.
Call the representation V, and let {,} be any Hermitian form on V. We can convert it to a new
Hermitian form (cid:104),(cid:105) as follows:
1 (cid:88)
(cid:104)v,w(cid:105)= {gv,gw}.
n
g∈G
This new form is G-invariant: (cid:104)gv,gw(cid:105)=(cid:104)v,w(cid:105) for all g ∈G, v,w ∈V.
Suppose V has a proper invariant subspace. Then its orthogonal complement (under the G-invariant
form) is also invariant, and V breaks down as a direct sum. Induction on the dimension completes the
proof.
1
NoticethattheproofreliesonthefinitenessofGfortheaveragingtomakesense. Onecangeneralize
tocompacttopologicalgroupsbyreplacingthesumwithanaverageoverthegroup(withrespecttoHaar
measure). However,itdoesnotholdforarbitrarygroups,evenrelativelyniceonesliketheadditivegroup
Rofrealnumbers. Considerthefollowingtwo-dimensionalrepresentationofR: forx∈Rand(a,b)∈C2,
letx(a,b)=(a+bx,b). Thereisaone-dimensionalinvariantsubspace, namely{(a,0):a∈C},butthere
is only one, so the whole representation is not a direct sum, yet not irreducible either. For non-compact
groups like R, the best representations to look at are unitary representations, which automatically come
with an invariant Hermitian form (and are Hilbert spaces with respect to that form).
Given a representation (V,ρ), we define the character χ :G→C of V by
V
χ (g)=trρ(g).
V
(Here, the ρ is convenient, since trg looks cryptic!) Notice that χ (1)=dim(V), χ (g−1)=χ (g), and
V V V
χ is a class function (i.e., it is constant on each conjugacy class of G). Clearly, χ depends only on the
V V
isomorphism class of V.
It’s also not hard to check that χ = χ +χ , χ = χ χ , and χ = χ . It follows that
V⊕W V W V⊗W V W V∗ V
χ =χ χ .
Hom(V,W) V W
Define a Hermitian form on functions from G to C by
1 (cid:88)
(cid:104)f ,f (cid:105)= f (g)f (g).
1 2 n 1 2
g∈G
Lemma 2 (Schur’s Lemma). Let V and W be irreducible representations of G. Then every G-
homomorphism from V to W is either zero or an isomorphism. Every homomorphism from V to itself
is multiplication by some scalar.
Proof. Let f : V → W be a G-homomorphism. Then the kernel and image of f are invariant subspaces
of V and W, respectively, so each must be trivial or the entire space. The only ways this can work out
are if f is zero or an isomorphism.
Now let f : V → V be a homomorphism. Since C is algebraically closed and f is a linear trans-
formation, f has an eigenvector v and eigenvalue λ. Consider the homomorphism f −λ, i.e., f minus
scalar multiplication by λ. We know that this map is either zero or an isomorphism, and it is not an
isomorphism because v is in its kernel. Thus, f must be multiplication by λ.
Theorem 3. If V and W are irreducible representations, than (cid:104)χ ,χ (cid:105)=1 if they are isomorphic, and
V W
(cid:104)χ ,χ (cid:105)=0 otherwise.
V W
Proof. Define an averaging map
π :Hom(V,W)→Hom (V,W)
G
by
1 (cid:88)
π(f)= gf.
n
g∈G
This map is a projection of Hom(V,W) onto its subspace Hom (V,W) (i.e., it fixes each element of
G
Hom (V,W)). Its trace is thus the dimension of its image, which is 1 or 0 according to whether V and
G
W are isomorphic (by Schur’s Lemma).
Ontheotherhand, thetraceofthemultiplicationbyg mapisχ (g), sobylinearitythetrace
Hom(V,W)
of π is
1 (cid:88)
χ (g),
n Hom(V,W)
g∈G
which equals
1 (cid:88)
χ (g)χ (g),
n V W
g∈G
as desired.
2
Thistheoremimplies,amongotherthings,thatthereareonlyfinitelymanyirreduciblerepresentations
of G (up to isomorphism, of course): their characters are linearly independent in the finite-dimensional
space of class functions.
It also implies that the isomorphism classes of the irreducible representations in a direct sum are
uniquely determined: the irreducible representation V occurs (cid:104)χ ,χ (cid:105) times in the representation W.
V W
Furthermore,(cid:104)χ ,χ (cid:105)=1iffV isirreducible(ascanbeseenbywritingV asadirectsumofirreducibles
V V
and expanding).
The regular representation of G is the representation on formal linear combinations of elements of G.
In other words, the elements are complex vectors indexed by elements of G, and G acts by permuting
the coordinates.
Theorem 4. Every irreducible representation V occurs dim(V) times in the regular representation.
Proof. Let χ be the character of the regular representation. Then
(cid:40)
n if g =1, and
χ(g)=
0 otherwise.
To see why, note that each group element acts by a permutation matrix, and the trace of a permutation
matrix is simply the number of fixed points of the permutation.
Thus,
1 1
(cid:104)χ ,χ(cid:105)= χ (1)χ(1)= dim(V)n=dim(V).
V n V n
Let d ,...,d be the dimensions of the irreducible representations of G. Then it follows from the
1 k
preceding theorem that
(cid:88)
n= d2.
i
i
The group algebra C[G] is a very useful object. Its elements are formal linear combinations
(cid:88)
c g
g
g∈G
of the group elements with complex coefficients c , added and multiplied in the obvious ways. A G-
g
representation is exactly the same thing as a C[G]-module. (This is one reason why module theory
over non-commutative rings is so important! However, C[G] has much more structure than a typical
non-commutative ring.)
Note that the natural representation of G on C[G] is the regular representation.
Any element x of C[G] defines a linear operator on any G-representation. This operator is a G-
homomorphism if x is in the center of C[G].
Lemma 5. Let ϕ:G→C be any function. Then
(cid:88)
ϕ(g)g.
g∈G
is in the center of C[G] iff ϕ is a class function.
Proof. This element is in the center iff it commutes with every group element. Because
(cid:88) (cid:88) (cid:88)
h ϕ(g)g = ϕ(g)(hgh−1)h= ϕ(h−1gh)gh,
g∈G g∈G g∈G
it commutes with h iff ϕ(h−1gh) = ϕ(g) for all g, so it commutes with everything iff ϕ is a class
function.
3
Theorem6. Thecharactersoftheirreduciblerepresentationsformabasisforthespaceofclassfunctions
on G (and thus the number of them equals the number of conjugacy classes).
Proof. Weknowthatthecharactersarelinearlyindependent(sincetheyareorthogonal),soweonlyneed
to show that they span the space. Suppose ϕ is any class function orthogonal to every character.
Let V be a representation of G, and consider the G-homomorphism
1 (cid:88)
f = ϕ(g)g
n
g∈G
from V to V (the fact that it is a G-homomorphism follows from Lemma 5). Its trace is (cid:104)ϕ,χ (cid:105) (by
V
linearity of the trace), which is 0 because ϕ is orthogonal to the characters. If V is irreducible, then
by Schur’s lemma f is constant, and thus 0 because its trace is 0. It follows that it is 0 for every
representation (by Maschke’s theorem). Let V be the regular representation. Then we compute f(1):
1 (cid:88) 1 (cid:88)
0=f(1)= ϕ(g)g·1= ϕ(g)g.
n n
g∈G g∈G
Thus, each coordinate ϕ(g) must vanish, so ϕ=0.
This proof might seem initially like an odd trick, but operators of this form play a fundamental role.
One key thing to understand is the structure of the group algebra C[G]. We will see that these operators
tell how to break it up as a product.
Writing a ring R as a direct product R ×···×R is equivalent to finding elements e ,...,e such
1 k 1 k
that they are in the center of the ring, e +···+e =1, e2 =e for all i, and e e =0 for i(cid:54)=j. Namely,
1 k i i i j
we can take R = Re . (Think of e as having 1 in the i-th coordinate and 0 elsewhere.) Note that
i i i
it is crucial that e ,...,e are in the center of R (and this is easy to forget). They are called central,
1 k
orthogonal idempotents.
Let χ ,...,χ be the characters of the irreducible representations V ,...,V of G (with dimensions
1 k 1 k
d ,...,d ), and define elements of the group algebra by
1 k
d (cid:88)
e = k χ (g)g.
i n i
g∈G
We will prove that they form a set of central, orthogonal idempotents, so the group algebra breaks up as
a direct product over irreducible representations.
Note the similarity to the sums we saw before. The intuition behind the last proof is that if the
characters didn’t span all the class functions, we could try to construct another idempotent.
By Lemma 5, e ,...,e are in the center of C[G], so we just need to show that they are orthogonal
1 k
idempotents that sum to 1.
Lemma7. Theelemente actsonV bymultiplicationby1or0,accordingasi=j ori(cid:54)=j,respectively.
i j
Proof. Since e is central, it acts by a G-homomorphism, so Schur’s lemma implies that it acts by a
i
constant. The trace of e acting on V is
i j
d (cid:88)
k χ (g)trg =d (cid:104)χ ,χ (cid:105).
n i k i j
g∈G
Thus, it acts by the constant (cid:104)χ ,χ (cid:105).
i j
Now view C[G] as the regular representation of G, so it breaks up as the sum over k of d copies of
k
V . The previous lemma implies that multiplication by e fixes the copies of V and kills everything else.
k i i
Thus,wehaveexplicitlylocatedthecopiesofV withintheregularrepresentation: theyarethemultiplies
i
4
of e in the group algebra. This is in itself a worthwhile result, but it furthermore implies that these
i
are orthogonal idempotents: since they act by 0 or 1 on each direct summand, they are idempotents;
since they act by 1 only on disjoint summands, they are orthogonal; since their sum acts by 1 on each
summand, it is 1. (This uses the fact that two elements of the group algebra are equal iff multiplication
by them gives the same linear transformation. That’s simple: look at what they do to the identity!)
Thus, C[G] = (C[G]e )×···×(C[G]e ). What can one say about the structure of the ring C[G]e ?
1 k i
It turns out to be isomorphic to a d ×d matrix ring over C, namely all linear operators on V (so it
i i i
has dimension d2, which makes sense). Now that we’ve gotten this far, that’s not hard to see: We just
i
need to work out the multiplicative structure of the ring. Because C[G]e is just the sum of d copies
i i
of V when viewed as a G-representation, the effect of multiplication by an element of C[G]e is just
i i
determined by its action on V as a linear transformation. Thus, C[G]e is isomorphic to some subring of
i i
the matrix ring. If, for any i, it were not the full ring, then there would be a shortfall in the dimension
(cid:80)
of C[G], since n = d2. (Note that while there is a unique way to break C[G] up as a direct product,
i i
the isomorphisms of the factors to matrix rings are not canonical.)
Besides the intrinsic interest of determining the structure of the group algebra, this will limit the
possible dimensions of the representations of G:
Proposition 8. For each i, d is a factor of n.
i
Proof. Let m be an exponent of G (i.e., gm =1 for all g ∈G). In the expansion
n (cid:88)
e = χ (g)g,
d i i
i
g∈G
all the coefficients on the right hand side are in Z[ζ], where ζ is a primitive m-th root of unity (because
the trace is the sum of the eigenvalues, and the eigenvalues are all m-th roots of unity).
LetM =Z[ζ]Ge betheintegerspanoftheelementsζjge inthegroupalgebra,withj ∈Zandg ∈G
i i
(this is a finite-dimensional, free Z-module). We see that multiplication by ne preserves M, because
the coefficients of ne are in Z[ζ]. Since e is a central idempotent, multiplydinigiby ne is the same as
di i i di i
multiplying by n/d , so that preserves M as well.
i
Thus, n/d is the root of a monic polynomial over Z, namely its characteristic polynomial as a linear
i
transformation on M. Every rational root of such a polynomial is an integer.
The V -isotypic component of a representation V is the sum of the copies of V occurring in V. The
i i
results so far imply that it is uniquely determined from V, because it is simply e V. Note however the
i
isotypic component’s decomposition into summands isomorphic to V need not be unique. For example,
i
if G acts trivially on a high-dimensional space, there are infinitely many ways to write it as a sum of
trivialone-dimensionalrepresentations. Sotheexactdetailsofhowtoexpressarepresentationasadirect
sum of irreducibles are not uniquely determined, but we have seen that the isomorphism classes of the
irreducibles are, as are their isotypic components.
Suppose G = H ×H , where H has order n . Given representations V and V of H and H ,
1 2 i i 1 2 1 2
respectively,wecanturnV ⊗V intoaG-representationintheobviousway: (h ,h )(v ⊗v )=(h v )⊗
1 2 1 1 1 2 1 1
(h v ). Then an easy computation shows that
2 2
(cid:104)χ ,χ (cid:105)=(cid:104)χ ,χ (cid:105)(cid:104)χ ,χ (cid:105).
V1⊗V2 V3⊗V4 V1 V3 V2 V4
In particular, if V ,...,V are irreducible, then so are V ⊗V and V ⊗V (using the characterization in
1 4 1 2 3 4
terms of having norm 1). Furthermore, it shows that V ⊗V ≈ V ⊗V iff V ≈ V and V ≈ V . This
1 2 3 4 1 3 2 4
gives us every irreducible representation of G (one can check completeness using the sum of squares of
the dimensions of the irreducibles).
Here’s an entertaining application of the results so far:
Proposition 9. Suppose V is a faithful representation (i.e., the homomorphism ρ : G → Aut(V) is
injective). Then each irreducible representation of G occurs in V⊗i for some i.
5
Proof. Let W be any irreducible representation. Consider the formal power series
(cid:88)
(cid:104)χ ,χ (cid:105)ti.
W V⊗i
i≥0
The coefficient of ti is the number of times W occurs in V⊗i, so we just need to show that this power
series is not identically zero. We can in fact explicitly compute it as follows:
(cid:88) (cid:88) 1 (cid:88)(cid:88)
(cid:104)χ ,χ (cid:105)ti = (cid:104)χ ,χi (cid:105)ti = |c|χ (c)χ (c)iti,
W V⊗i W V n W V
i≥0 i≥0 i≥0 c
wherecrunsoverallconjugacyclassesinG(andweviewthecharactersasfunctionsonconjugacyclasses,
rather than on G). This simplifies to
1 (cid:88) |c|χ (c)
W ,
n 1−tχ (c)
V
c
whichis a rationalfunction. Because V isfaithful, χ (c) isneverequal to dim(V)except onthe identity
V
conjugacy class. (Since the eigenvalues are roots of unity, the trace can be dim(V) iff the matrix is
the identity.) Thus, only one of the summands has a pole at 1/dim(V), so the rational function is not
identically zero.
We now turn to restriction and induction. Let H ⊆ G be a subgroup. It is easy to turn a G-
representation V into an H-representation ResG V, simply by forgetting G ever existed. This operation
H
is called restriction, and is useful but not exciting. What’s perhaps more interesting is its left-adjoint,
induction.
Suppose W is an H-representation, which we want to turn into a G-representation V =IndG W. We
H
want W to be contained in V, and W should be an H-invariant subspace. Consider the G-cosets of W
in V. Because W is H-invariant, these cosets can actually be indexed by cosets of H in G. The simplest
thing that can happen is for V to be the direct sum of them:
(cid:77)
V = gW.
g∈G/H
(This is a slight abuse of notation. I mean that g should run over some set of representatives for G/H.)
It’s not hard to check that such a V is uniquely determined, up to G-isomorphism, by W.
As group algebra modules, we can take
IndG W =C[G]⊗ W.
H C[H]
Another convenient representation is
IndG W ={f :G→W :hf(g)=f(hg) for all h∈H},
H
where g acts on f by (gf)(g(cid:48))=f(g(cid:48)g). Checking the equivalence is left as a exercise to the reader.
As two examples, the permutation representation of G on G/H is induced from the trivial represen-
tation of H, and the regular representation of G is induced from the regular representation of H.
Two fundamental properties of induction are that the induction of a direct sum is the direct sum of
the inductions, and that induction is transitive: if H ⊆H ⊆G, and W is an H -representation, then
1 2 1
IndG W ≈IndG IndH2W.
H1 H2 H1
Both are easy to prove from the characterization above.
The most fundamental property of induction is that it is the left adjoint of restriction:
6
Theorem 10. Let W be an H-representation, and V a G-representation. Then every H-homomorphism
from W to ResG V extends uniquely to a G-homomorphism from IndG W to V, i.e.,
H H
Hom (W,ResG V)=Hom (IndG W,V).
H H G H
Proof. Suppose ϕ:W →ResG V is an H-homomorphism. We wish to extend ϕ to a G-homomorphism
H
defined on (cid:77)
IndG W = gW.
H
g∈G/H
Clearly, the only possible way to do that is that on the gW piece, ϕ(gw) must be defined as gϕ(w). This
clearly defines a linear map ϕ:IndG W →V.
H
We just need to check that it is a G-map. Suppose g ∈G, and u∈IndG W. We can write u uniquely
H
in the form
(cid:88)j
g w ,
i i
i=1
where g ,...,g runs over representatives for G/H and w ∈W. We need to check that ϕ(gu)=gϕ(u).
1 j i
Multiplicationbyg permutesthecosetsofH bysomepermutationσ,withgg =g h forsomeh ∈H.
i σ(i) i i
Then
(cid:88)j
gu= g h w ,
σ(i) i i
i=1
so
(cid:88)j
ϕ(gu)= g h ϕ(w ),
σ(i) i i
i=1
so
(cid:88)j
ϕ(gu)= gg ϕ(w )=gϕ(u),
i i
i=1
as desired.
This adjointness is of fundamental importance. For example, it implies that induction is well-defined
up to a canonical isomorphism, not just any old isomorphism. More significantly, it immediately implies
Frobenius reciprocity by taking dimensions of both sides:
(cid:104)χ ,χ (cid:105)=(cid:104)χ ,χ (cid:105).
W ResGV IndGW V
H H
For example, if W and V are irreducible, then the number of times W occurs in ResG V equals the
H
number of times V occurs in IndG V. (This is why it is called a reciprocity law.)
H
7
