k!

span, and are eigenspaces of h of weight » ’ 2k. For any » ∈ C we can construct

such a module as follows: Let b+ denote the subalgebra of sl(2) generated by

h and e. Then U (b+ ), the universal enveloping algebra of b+ can be regarded

as a subalgebra of U (sl(2)). We can make C into a b+ module, and hence a

U (b+ ) module by

h · 1 := », e · 1 := 0.

Then the space

U (sl(2)) —U (b+ ) C

with e acting on C as 0 and h acting via multiplication by » is a cyclic module

with cyclic vector v» = 1 — 1 which satis¬es (2.4). It is a “universal” such

module in the sense that any other cyclic module with cyclic vector satisfying

(2.4) is a homomorphic image of the one we just constructed.

This space U (sl(2)) —U (b+ ) C is in¬nite dimensional. It is irreducible unless

1

there is some k! f k v» with

1k

e f v» = 0

k!

where k is an integer ≥ 1. Indeed, any non-zero vector w in the space is a ¬nite

1

linear combination of the basis elements k! f k v» ; choose k to be the largest

integer so that the coe¬cient of the corrresponding element does not vanish.

Then successive application of the element e (k-times) will yield a multiple of

2.3. THE CASIMIR ELEMENT. 39

v» , and if this multiple is non-zero, then U (sl(2))w = U (sl(2))v» is the whole

space.

But

1k 1k 1

f k’1 v» .

v» = (1 ’ k + »)

e f v» = e, f

(k ’ 1)!

k! k!

This vanishes only if » is an integer and k = » + 1, in which case there is a

unique ¬nite dimensional quotient of dimension k + 1. QED

The ¬nite dimensional irreducible representations having zero as a weight

are all odd dimensional and have only even weights. We will call them “even”.

They are called “integer spin” representations by the physicists. The others are

“odd” or “half spin” representations.

2.3 The Casimir element.

In U (sl(2)) consider the element

12

C := h + ef + f e (2.5)

2

called the Casimir element or simply the “Casimir” of sl(2).

Since ef = f e + [e, f ] = f e + h in U (sl(2)) we also can write

12

C= h + h + 2f e. (2.6)

2

This implies that if v is a “highest weight vector” in a sl(2) module satisfying

ev = 0, hv = »v then

1

Cv = »(» + 2)v. (2.7)

2

Now in U (sl(2)) we have

[h, C] = 2([h, f ]e + f [h, e])

= 2(’2f e + 2f e)

=0

and

1

· 2(eh + he) + 2e ’ 2he

[C, e] =

2

= eh ’ he + 2e

= ’[h, e] + 2e

= 0.

Similarly

[C, f ] = 0.

40 CHAPTER 2. SL(2) AND ITS REPRESENTATIONS.

In other words, C lies in the center of the universal enveloping algebra of sl(2),

i.e. it commutes with all elements. If V is a module which possesses a “highest

weight vector” v» as above, and if V has the property that v» is a cyclic vector,

meaning that V = U (L)v» then C takes on the constant value

»(» + 2)

C= Id

2

since C is central and v» is cyclic.

2.4 sl(2) is simple.

An ideal I in a Lie algebra g is a subspace of g which is invariant under the

adjoint representation. In other words, I is an ideal if [g, I] ‚ I. If a Lie

algebra g has the property that its only ideals are 0 and g itself, and if g is not

commutative, we say that g is simple. Let us prove that sl(2) is simple. Since

sl(2) is not commutative, we must prove that the only ideals are 0 and sl(2)

itself. We do this by introducing some notation which will allow us to generalize

the proof in the next chapter. Let

g = sl(2)

and set

g’1 := Cf, g0 := Ch, g1 := Ce

so that g, as a vector space, is the direct sum of the three one dimensional

spaces

g = g’1 • g0 • g1 .

Correspondingly, write any x ∈ g as

x = x’1 + x0 + x1 .

If we let

1

d := h

2

then we have

x = x’1 + x0 + x1 ,

[d, x] = ’x’1 + 0 + x1 , and

[d, [d, x]] = x’1 + 0 + x1 .

Since the matrix «

1 11

’1 0 1

1 01

is invertible, we see that we can solve for the “components” x’1 , x0 and x1 in

terms of x, [d, x], [d, [d, x]]. This means that if I is an ideal, then

I = I1 • I0 • I1

2.5. COMPLETE REDUCIBILITY. 41

where

I’1 := I © g’1 , I0 := I © g0 , I1 := I © g1 .

Now if I0 = 0 then d = 1 h ∈ I, and hence e = [d, e] and f = ’[d, f ] also belong

2

to I so I = sl(2). If I’1 = 0 so that f ∈ I, then h = [e, f ] ∈ I so I = sl(2).

Similarly, if I1 = 0 so that e ∈ I then h = [e, f ] ∈ I so I = sl(2).

Thus if I = 0 then I = sl(2) and we have proved that sl(2) is simple.

2.5 Complete reducibility.

We will use the Casimir element C to prove that every ¬nite dimensional rep-

resentation W of sl(2) is completely reducible, which means that if W is an

invariant subspace there exists a complementary invariant subspace W so that

W = W • W . Indeed we will prove:

Theorem 2 1. Every ¬nite dimensional representation of sl(2) is completely

reducible.

2. Each irreducible subspace is a cyclic highest weight module with highest

weight n where n is a non-negative integer.

3. When the representation is decomposed into a direct sum of irreducible

components, the number of components with even highest weight is the

multiplicity of 0 as an an eigenvector of h and

4. the number of components with odd highest weight is the multiplicity of 1