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184 5 Basic Representation Theory

5.7 More on Irreducibility

In the last section we introduced the notion of an irreducible representation, but we

did not prove that any of the representations we have met are irreducible. In this

section we will remedy that and also learn a bit more about irreducibility along the

way. In proving the irreducibility of a given representation, the following proposi-

tion about the irreps of matrix Lie groups and their Lie algebras is often useful. The

proposition just says that for a connected matrix Lie group, the irreps of the group

are the same as the irreps of the Lie algebra. This may seem unsurprising and per-

haps even trivial, but the conclusion does not hold when the group is disconnected.

We will have more to say about this later. The proof of this proposition is also a nice

exercise in using some of the machinery we have developed so far.

Proposition 5.3 A representation (, V ) of a connected matrix Lie group G is

irreducible if and only if the induced Lie algebra representation (π, V ) of g is irre-

ducible as well.

Proof First, assume that (, V ) is an irrep of G. Then consider the induced Lie

algebra representation (π, V ), and suppose that this representation has an invariant

subspace W . We will show that W must be an invariant subspace of (, V ) as well,

which by the irreducibility of (, V ) will imply that W is either V or {0}, which

will then show that (π, V ) is irreducible. For all X ∈g, w ∈W,wehave

π(X)w ∈W

⇒ e

π(X)

w ∈W

⇒ 





w ∈W. (5.73)

Now, using the fact that any element of G can be written as a product of exponentials

(cf. Proposition 5.2), we then have

(g)w =



···e



=









···





which must be in W by repeated application of (5.73). Thus W is also an invariant

subspace of (, V ), but we assumed (, V ) was irreducible, and so W must be

equal to V or 0. This then proves that (π, V ) is irreducible as well.

Conversely, assume that (π, V ) is irreducible, and let W be an invariant subspace

of (, V ). Then for all t ∈R, X ∈g, w ∈W we know that (e

)w ∈W , and hence

π(X)w =







t=0

= lim

t→0

(e

)w −w

must be in W as well, so W is invariant under π and hence must be equal to V

or {0}. Thus (, V ) is irreducible. 

5.7 More on Irreducibility 185

Note that the assumption of connectedness was crucial in the above proof, as

otherwise we could not have used Proposition 5.2. Furthermore, we will soon meet

irreducible representations of disconnected groups (like O(3, 1)) that yield Lie al-

gebra representations which do have non-trivial invariant subspaces, and are thus

not irreducible. The above proposition is still very useful, however, as we will see

in this next example.

Example 5.27 The SU(2) representation on P

), revisited

In this example we will prove that the su(2) representations (π

)) of Exam-

ple 5.7 are all irreducible. Proposition 5.3 will then tell us that the SU(2) repre-

sentations (

)) are all irreducible as well. Later on, we will see that these

representations are in fact all the ﬁnite-dimensional irreducible representations of

SU(2)!

Recall that the π

) are given by

) =



∂

∂z

∂

∂z



) =



∂

∂z

−z

∂

∂z



) =



∂

∂z

−z

∂

∂z



Deﬁne now the “raising” and “lowering” operators

≡π

) +iπ

) =−z

∂

∂z

≡−π

) +iπ

) =−z

∂

∂z

It is easy to see that Y

trades a factor of z

for a factor of z

, thereby raising the

) eigenvalue of a single term by i. Likewise, X

trades a z

for a z

and lowers

the eigenvalue by i.

Consider a nonzero invariant subspace W ⊂P

). If we can

show that W =P

), then we know that P

) is irreducible. Being nonzero, W

contains at least one element of the form

w =a

l−1

l−2

+···+a

where at least one of the a

is not zero. Let k

be the biggest nonzero value of k,so

that a

l−k

is the term in w with the highest power of z

. Then applying (X

)

You may object to the use of i in our deﬁnition of these operators; after all, su(2) is a real Lie

algebra, and so the expression S

+iS

has no meaning as an element of su(2), and so one cannot

say that, for instance, Y

= π(S

+iS

). Thus X

and Y

are not in the image of su(2) under π

This is a valid objection, and to deal with it one must introduce the notion of the complexiﬁcation

of a Lie algebra. A discussion of this here would lead us too far astray from our main goals of

applications in physics, however, so we relegate this material to the appendices which you can

consult at leisure.

186 5 Basic Representation Theory

to w lowers the z

degree by k

, killing all the terms except a

l−k

. In fact, one

can compute easily that



l−k



=(−1)

This is proportional to z

, and since W is invariant W must then contain z

.But

then we can successively apply the raising operator Y

to get monomials of the form

l−k

for 0 ≤ k ≤l, and so these must be in W as well. These, however, form a

basis for P

), hence W must equal P

), and so P

) is irreducible. 

Before moving on to our next examples, we need to state and prove Schur’s

lemma, one of the most basic and crucial facts about irreducible representations.

Roughly speaking, the upshot of it is that any linear operator on the representation

space of an irrep which commutes with the group action (i.e. is an intertwiner) must

be proportional to the identity. The precise statement is as follows:

Proposition 5.4 (Schur’s lemma) Let (

), i = 1, 2 be two irreducible repre-

sentations of a group or Lie algebra, and let φ :V

→V

be an intertwiner. Then ei-

ther φ =0 or φ is a vector space isomorphism. Furthermore, if (

) =(

)

and V

is a ﬁnite-dimensional complex vector space, then φ is a multiple of the iden-

tity, i.e. φ =cI for some c ∈C.

Proof We will prove this for group representations 

; the Lie algebra case follows

immediately with the obvious notational changes. Let K be the kernel or null space

of φ. Then K ⊂V

is an invariant subspace of 

, since for any v ∈K, g ∈G,



(g)v



= 

(g)



φ(v)



= 

(0)

= 0

⇒

(g)v ∈K.

However, since (

) is irreducible, the only invariant subspaces are 0 and V

,so

K must be one of those. If K =V

then φ =0, so henceforth we assume that K =0,

which means that φ is one-to-one, and so φ(V

) ⊂V

is isomorphic to V

. Further-

more, φ(V

) is an invariant subspace of (

), since for any φ(v) ∈φ(V

), g ∈G

we have



(g)



φ(v)



=φ



(g)v



∈φ(V

But (

) is also irreducible, so φ(V

) must equal 0 or V

. We already assumed

that φ(V

) =0, though, so we conclude that φ(V

) =V

and hence φ is an isomor-

phism.

Now assume that (

) = (

) ≡ (, V ) and that V is a ﬁnite-dimen-

sional complex vector space of dimension n (notice that we did not assume ﬁnite-

dimensionality at the outset of the proof). Then φ is a linear operator and the eigen-

value equation

5.7 More on Irreducibility 187

det(φ −λI )

is a nth degree complex polynomial in λ. By the fundamental theorem of algebra,

this polynomial has at least one root c ∈ C, hence φ −cI has determinant 0 and

is thus non-invertible. This means that φ − cI has a non-trivial kernel K. K is

invariant, though; for all v ∈K,

(φ −cI)



(g)v



= φ



(g)v



−c(g)v

= (g)



φ(v) −cv



= 0

⇒(g)v ∈K

and since we know K = 0, we then conclude by irreducibility of V that K = V .

This means that (φ −cI )(v) =0 ∀v ∈V , which means φ =cI, as desired. 

Exercise 5.33 Prove the following corollary of Schur’s lemma: If (

), i = 1, 2 are

two complex irreducible representations of a group or Lie algebra, and φ,ψ :V

→V

are

two intertwiners with φ =0,thenψ =cφ for some c ∈C.

As a ﬁrst application of Schur’s lemma, we have

Proposition 5.5 An irreducible ﬁnite-dimensional complex representation of an

abelian group or Lie algebra is one-dimensional.

Proof Again we prove only the group case. Since G is abelian, each (g) com-

mutes with (h) for all h ∈ G, hence each (g) : V → V is an intertwiner! By

Schur’s lemma, this implies that every (g) is proportional to the identity (with

possibly varying proportionality constants), and so every subspace of V is an invari-

ant one. Thus the only way V could have no non-trivial invariant subspaces is to

have no non-trivial subspaces at all, which means it must be one-dimensional. 

Exercise 5.34 Show that the fundamental representation of SO(2) on R

is irreducible.

Prove this by contradiction, showing that if the fundamental representation were reducible

then the SO(2) generator

X =



0 −1



would be diagonalizable over the real numbers, which you should show it is not. This shows

that one really needs the hypothesis of a complex vector space in the above proposition.

Example 5.28 The irreducible representations of Z

Proposition 5.5 allows us to easily enumerate all the irreducible representations

of Z

. Since Z

is abelian any irreducible representation (

irr

,V) must be one-

dimensional (i.e. V =R or C), and 

irr

must also satisfy

See Herstein [9], for instance.

188 5 Basic Representation Theory



irr

(−1)



=

irr



(−1)



=

irr

(1) =1,

which means that 

irr

(−1) =±1, i.e. 

irr

is either the alternating representation or

the trivial representation! Furthermore, Z

is semi-simple (as you will show in Prob-

lem 5.8), and so any representation (, V ) of Z

is completely reducible and thus

decomposes into one-dimensional irreducible subspaces, on which (−1) equals

either 1 or −1. Thus (−1) is diagonalizable, with eigenvalues ±1 (cf. Exam-

ple 5.18 for examples of this). Let us say the Z

in question is Z

{I,P}⊂O(3),

and that we are working in a quantum mechanical context with a Hilbert space H.

Then there exists a basis for H consisting of eigenvectors of (P ); for a given

eigenvector ψ , its eigenvalue of ±1 is known as its parity. If the eigenvalue is +1

then ψ is said to have even parity, and if the eigenvalue is −1 then ψ is said to have

odd parity. 

5.8 The Irreducible Representations of su(2), SU(2) and SO(3)

In this section we will construct (up to equivalence) all the ﬁnite-dimensional ir-

reducible complex representations of su(2). Besides being of intrinsic interest, our

results will also allow us to classify all the irreducible representations of SO(3),

SU(2), and even the apparently unrelated representations of so(3, 1), SO(3, 1) and

SL(2, C). The construction we will give is more or less the same as that found in the

physics literature under the heading “theory of angular momentum”, except that we

are using different language and notation. Our strategy will be to use the commuta-

tion relations to deduce the possible structures of su(2) irreps, and then show that

we have already constructed representations which exhaust these possibilities, thus

yielding a complete classiﬁcation.

Let (π, V ) be a ﬁnite-dimensional complex irreducible representation of su(2).

It will be convenient to use the following shorthand, familiar from the physics liter-

ature:

≡iπ(S

)

≡iπ(S

) −π(S

) (5.74)

−

≡iπ(S

) +π(S

These “raising” and “lowering” operators obey the following commutation relations,

as you can check:

]=±J

−

]=2J

Now, as discussed in our proof of Schur’s lemma, the fact that V is complex means

that every operator on V has at least one eigenvector. In particular, this means that

has an eigenvector v with eigenvalue b. The above commutation relations then

imply that

5.8 The Irreducible Representations of su(2), SU(2) and SO(3) 189

v) =[J

]v +J

v) = (b ±1)J

so that if J

v is not zero (which it might be!), then it is another eigenvector of J

with eigenvalue b ±1. Now, we can repeatedly apply J

to v to yield more and more

eigenvectors of J

, but since V is ﬁnite-dimensional and eigenvectors with different

eigenvalues are linearly independent (see Exercise 5.35 below), this process must

end somewhere, say at N applications of J

.Letv

be this vector with the highest

eigenvalue (also known as the highest weight vector of the representation), so that

we have

=(J

)

=0.

Then v

has a J

eigenvalue of b +N ≡j (note that so far we haven’t proved any-

thing about b or j , but we will soon see that they must be integral or half-integral).

Starting with v

, then, we can repeatedly ‘lower’ with J

−

to get eigenvectors with

lower eigenvalues. In fact, we can deﬁne

≡(J

−

)

which has J

eigenvalue j −k. This chain must also end, though, so there must exist

an integer l such that

=(J

−

)

=0butv

l+1

=(J

−

)

l+1

=0.

How can we ﬁnd l? For this we will need the following formula, which you will

prove in Exercise 5.36 below:

) =



2jk −k(k −1)



k−1

. (5.75)

Applying this to v

l+1

=0gives

0 =J

l+1

) =



2j(l+1) −(l +1)l



and since v

=0 we conclude that



2j(l+1) −(l +1)l



=0 ⇐⇒ j =l/2.

Thus j is a nonnegative integer or half integer! (In fact, j is just the ‘spin’

of the representation.) We further conclude that V contains 2j + 1 vectors

|0 ≤k ≤2j} all of which are eigenvectors of J

with eigenvalue j −k. Further-

more, since

π(S

) =−

−

)

π(S

) =

−

−J

the action of S

and S

takeagivenv

into a linear combination of other v

,sothe

span of v

is a nonzero invariant subspace of V . We assumed V was irreducible,

however, so we must have V = Span{v

}, and since the v

are linearly independent,

they form a basis for V !

190 5 Basic Representation Theory

To summarize, any ﬁnite-dimensional irreducible complex representation of

su(2) has dimension 2j +1, 2j ∈Z and a basis {v

}

k=0−2j

which satisﬁes

) =0.

−

) =v

k+1

,k<2j

) =(j −k)v

(5.76)

−

) =0

) =



2jk −k(k −1)



k−1

,k=0.

What is more, we can actually use the above equations to deﬁne representations

(π

), where V

is a (2j + 1)-dimensional vector space with basis {v

}

k=0−2j

and the action of the operators π

) is deﬁned by (5.76). It is straightforward to

check that this deﬁnes a representation of su(2) (see exercise below), and one can

prove irreducibility in the same way that we did in Example 5.27. Furthermore, any

irrep (π, V ) of su(2) must be equivalent to (π

) for some j, since we can ﬁnd a

basis w

for V satisfying (5.76)forsomej and then deﬁne an intertwiner by

φ :V →V

→v

and extending linearly. We have thus proved the following:

Proposition 5.6 The su(2) representations (π

),2j ∈ Z deﬁned above are all

irreducible, and any other ﬁnite-dimensional complex irreducible representation of

su(2) is equivalent to (π

) for some j, 2j ∈Z.

In other words, the (π

) are, up to equivalence, all the ﬁnite-dimensional

complex irreducible representations of su(2). They are also all the ﬁnite-dimen-

sional complex irreducible representations of so(3), since su(2) so(3).

If you look back over our arguments you will see that we deduced (5.76)from

just the su(2) commutation relations, the ﬁnite-dimensionality of V , and the exis-

tence of a highest weight vector v

satisfying J

) =0 and J

) =jv

.Thus,if

we have an arbitrary (i.e. not necessarily irreducible) ﬁnite-dimensional su(2) rep-

resentation (π, V ) and can ﬁnd a highest weight vector v

for some j , we can lower

with J

−

to generate a basis {v

}

k=0−2j

satisfying (5.76) and conclude that V has an

invariant subspace equivalent to (π

). We can then repeat this until V is com-

pletely decomposed into irreps. If we know that (π, V ) is irreducible from the start

then we do not even have to ﬁnd the vector v

, we just use the fact that (π, V ) must

be equivalent to (π

) for some j and note that j is given by j =

(dimV −1).

These observations make it easy to identify which (π

) occur in any given su(2)

representation.

Note that if we have a ﬁnite-dimensional complex irreducible su(2) representa-

tion that is also unitary, then we could work in an orthonormal basis. In that case,

it turns out that the v

deﬁned above are not orthonormal, and are thus not ideal

basis vectors to work with. They are orthogonal, but are not normalized to have

5.8 The Irreducible Representations of su(2), SU(2) and SO(3) 191

unit length. In quantum mechanical contexts the su(2) representations usually are

unitary, and so in that setting one works with the orthonormal basis vectors |m,

−j ≤m ≤j . The vector |m is proportional to our v

j−m

, and is normalized whereas

j−m

is not. See Sakurai [14] for details on the normalization procedure.

Exercise 5.35 Let S ={v

}

i=1−k

be a set of eigenvectors of some linear operator T on a

vector space V . Show that if each of the v

have distinct eigenvalues, then S is a linearly

independent set. (Hint: One way to do this is by induction on k. Another is to argue by

assuming that S is linearly dependent and reaching a contradiction. In this case you may

assume without loss of generality that the v

,1≤i ≤k −1 are linearly independent, so that

is the vector that spoils the assumed linear independence.)

Exercise 5.36 Prove (5.75). Proceed by induction, i.e. ﬁrst prove the formula for k =1,

then assume it is true for k and show that it must be true for k +1.

Exercise 5.37 Show that (5.76) deﬁnes a representation of su(2). This consists of showing

that the operators J

, J

−

, J

satisfy the appropriate commutation relations. Then show that

this representation is irreducible, using an argument similar to the one from Example 5.27.

Example 5.29 P

), revisited again

In Example 5.7 we met the representations (π

)) of su(2) on the space of

degree l polynomials in two complex variables. In Example 5.27 we saw that these

representations are all irreducible, and so by setting dim P

) = l + 1 equal to

2j +1 we deduce that





(π

l/2

) (5.77)

and so the (π

)), l ∈ Z also yield all the complex ﬁnite-dimensional ir-

reps of su(2). What is more, this allows us to enumerate all the ﬁnite-dimensional

complex irreps of the associated group SU(2). Any irrep (, V ) of SU(2) yields

an irrep (π, V ) of su(2), by Proposition 5.3. This irrep must be equivalent to

(π

)) for some l ∈ Z, however, and so by Proposition 5.2 (, V ) is equiv-

alent to (

)). Thus, the representations (

)), l ∈ Z are (up to

equivalence) all the ﬁnite-dimensional complex irreducible representations of

SU(2)!

It is instructive to construct the equivalence (5.77) explicitly. Recall that the rais-

ing and lowering operators (which we called Y

and X

in Example 5.27)aregiven

−

=−z

∂

∂z

=−z

∂

∂z

and that

=iπ

) =



∂

∂z

−z

∂

∂z



It is easy to check that v

≡ z

is a highest weight vector with j = l/2, and so the

basis that satisﬁes (5.76) is given by

192 5 Basic Representation Theory

≡(J

−

)





=(−1)

(l −k)!

l−k

, 0 ≤k ≤ l. (5.78)

Exercise 5.38 Show by direct calculation that (π

)) satisﬁes (5.76)withbasis

vectors given by (5.78).

Example 5.30 S

) as irreps of su(2)

We suggested in Example 5.15 that the tensor product representation of

su(2) on S

), the totally symmetric (0, 2j) tensors on C

, is equivalent to

(π

)) (π

),2j ∈ Z. With the classiﬁcation of su(2) irreps in place,

we can now prove this fact. It is easy to verify that

≡e

⊗···⊗e

  

2j times

∈S





(5.79)

is a highest weight vector with eigenvalue j , so there is an irreducible invariant

subspace of S

) equivalent to V

. Since dim V

= dimS

) = 2j + 1(as

you can check), we conclude that V

is all of S

), and hence that



π,S





(π

) 





(5.80)

which also implies (S

,S

))  (

)). Thus, we see that every

irreducible representation of su(2) and SU(2) can be obtained by taking a sym-

metric tensor product of the fundamental representation. Thus, using nothing

more than the fundamental representation (which corresponds to j = 1/2, as ex-

pected) and the tensor product, we can generate all the representations of su(2)

and SU(2). We will see in the next section that the same is true for so(3, 1) and

SO(3, 1)

. 

Our results about the ﬁnite-dimensional irreps of su(2) and SU(2) are not

only interesting in their own right; they also allow us to determine all the ir-

reps of SO(3)! To see this, ﬁrst consider the degree l harmonic polynomial rep-

resentation (

, H

)) of SO(3). This induces a representation (π

, H

)) of

so(3) su(2). It is easy to check (see exercise below) that if we deﬁne, in analogy

to the su(2) case,

≡iπ

) =i



∂

∂x

−x

∂

∂y



≡iπ(L

) −π(L

) =i



∂

∂y

−y

∂

∂z



−



∂

∂z

−z

∂

∂x



−

≡iπ(L

) +π(L

) =i



∂

∂y

−y

∂

∂z





∂

∂z

−z

∂

∂x



then the vector f

≡ (x + iy)

is a highest weight vector with eigenvalue l, and

so H

) has an invariant subspace equivalent to (π

). We will argue in Prob-

lem 5.9 that dimH

) =2l +1, so we conclude that (π

, H

)) (π

), and

5.9 Real Representations and Complexiﬁcations 193

is hence an irrep of so(3). Proposition 5.3 then implies that (

, H

)) is an irrep

of SO(3)!

Are these all the irreps of SO(3)? To ﬁnd out, let (, V ) be an arbitrary ﬁnite-

dimensional complex irrep of SO(3). Then the induced so(3) su(2) representation

(π, V ) must be equivalent to (π

) for some integral or half-integral j .Wejust

saw that any integral j value is possible, by taking (, V ) =(

, H

)). What

about half-integral values of j ? In this case, we have (careful not to confuse the

number π with the representation π !)

2π·π(L

)

−i2πJ

−i2πj

since v

has eigenvalue j

=−v

since j half-integral.

However, we also have

2π·π(L

)



2π·L





(I)



by (4.86)

a contradiction. Thus, j cannot be half-integral, so (π, V ) must be equivalent to

(π

) for j integral. But this implies that (π, V )  (π

, H

)), and so by

Proposition 5.2, (, V ) is equivalent to (

, H

))! Thus, the representations

(

, H

)), j ∈Z are (up to equivalence) all the ﬁnite-dimensional complex

irreducible representations of SO(3).

An important lesson to take away form this is that for a matrix Lie group G with

Lie algebra g, not all representations of g necessarily come from representations

of G.IfG =SU(2) then there is a one-to-one correspondence between Lie algebra

representations and group representations, but in the case of SO(3) there are Lie

algebra representations (corresponding to half-integral values of j ) that do not come

from SO(3) representations. We will not say much more about this here, except to

note that this is connected to the fact that there are non-isomorphic matrix Lie groups

that have isomorphic Lie algebras, as is the case with SU(2) and SO(3). For a more

complete discussion, see Hall [8].

Exercise 5.39 Verify that

=lf

5.9 Real Representations and Complexiﬁcations

So far we have classiﬁed all the complex ﬁnite-dimensional irreps of su(2), SU(2)

and SO(3), but we haven’t said anything about real representations, despite the fact

that many of the most basic representations of these groups and Lie algebras (like

the fundamental of SO(3) and all its various tensor products) are real. Fortunately,

there is a way to turn every real representation into a complex representation, so