10 2. THE SERRE RELATIONS

Relations:

eifj − fjei =δijhi, hihj = hjhi,

hiej − ejhi =

⎧

⎪2ej

⎨

⎪

⎩

(i = j)

−ej (i − j = ±1)

0 (otherwise)

,

hifj − fjhi =

⎧

⎪−2fj

⎨

⎪

⎩

(i = j)

fj (i − j = ±1)

0 (otherwise)

,

ei

2ej

− 2eiejei + ejei

2

= 0 (i − j = ±1), eiej = ejei (otherwise),

fi2fj

− 2fifjfi +

fjfi2

= 0 (i − j = ±1), fifj = fjfi (otherwise).

Proof. We define elements Lij ∈ U (i = j) by induction on |j − i|:

Li,i+1 = ei, Lij = [Lik, Lkj] (i k j),

Li+1,i = fi, Lij = [Lik, Lkj] (i k j).

For a, b ∈ U, [a, b] is ab − ba by definition.

Assertion 1 The Lij are well-defined.

We prove this by induction on N = |j −i|. Since the proof for i j is the same

as the proof for i j, we prove it only for i j. Assume that it is already proved

up to N − 1. We shall show for 1 ≤ l ≤ N − 2 that

[Li,i+l+1, Li+l+1,i+N ] = [Li,i+l, Li+l,i+N ] . (2.1)

Since we have Li,i+l+1 = [Li,i+l, ei+l] by the induction hypothesis, the left hand

side of (2.1) equals [[Li,i+l, ei+l] , Li+l+1,i+N ], which is the same as

− [ei+l, [Li,i+l, Li+l+1,i+N ]] + [Li,i+l, [ei+l, Li+l+1,i+N ]] .

We now notice that Li,i+l and Li+l+1,i+N are non-commutative polynomials in

ei,...,ei+l−1 and ei+l+1,...,ei+N−1 respectively. Hence the first term is 0. The

second term equals the right hand side of (2.1) by the induction hypothesis.

Assertion 2 The Lie algebra g is an irreducible U-module via the following action:

ei → ad(Ei,i+1), fi → ad(Ei+1,i), hi → ad(Eii − Ei+1,i+1),

for 1 ≤ i ≤ n − 1. (Recall that the Eij are matrix units.)

To prove that this defines an action of U, we check that ad(Ei,i+1), ad(Ei+1,i)

and ad(Eii −Ei+1,i+1), for 1 ≤ i ≤ n−1, satisfy the defining relations of U. By the

Jacobi identity, it is enough to see that Ei,i+1,Ei+1,i and Eii − Ei+1,i+1 satisfy the

relations, but this is easily checked by direct computation. We remark here that

we also have seen that Lij → ad(Eij).

Next we show that g is irreducible as a U-module. Let a be a non-zero U-

submodule of g. It is easy to see that a is not contained in h and that a has

a simultaneous eigenspace decomposition with respect to ad(h). So a contains a

matrix unit Ekl (k = l). We shall show that a contains all Eij (i = j). If i = k and

j = l, then Eij = ad(Eik)Ekl ∈ a. If i = k and j = l, then Eij = −ad(Elj)Ekl ∈ a.

If i = k and j = l, we consider ad(Eik)ad(Elj)Ekl and we have that Eij is in the