2.1. THE SERRE RELATIONS 11

submodule a. Since all off-diagonal matrix units are in the submodule, we have

that ad(Ei,i+1)Ei+1,i ∈ a for all i. We have proved that any non-zero submodule

must coincide with g.

Assertion 3

(1) The algebra U is a U-module via ad(x)(a) = xa − ax (x = ei,fi,hi).

(2) The space L =

∑n−1

i=1

Chi +

∑

i=j

CLij is stable under ad(L). In particular,

it is a U-submodule.

(3) Define a linear map ι : g → L by

Eij → Lij (i = j), Eii − Ei+1,i+1 → hi.

Then ι is an isomorphism of U-modules.

Let a, b, c be elements of U. Then the Jacobi identity [a, [b, c]] + [b, [c, a]] +

[c, [a, b]] = 0 holds and thus ad ([a, b]) = [ad(a), ad(b)]. In particular, the defining

relations of U imply that ad(ei),ad(fi) and ad(hi) all satisfy the same defining

relations. Hence (1) is proved. The Jacobi identity also implies the following

Leibniz rule.

ad(x)[a, b] = [ad(x)a, b] + [a, ad(x)b] (x, a, b ∈ U)

In order to prove (2) and (3), we establish explicit formulas for [x, Ljk] (x =

ei,fi,hi) in the following steps.

(step 1) [ hi, Ljk ] = (δij − δi+1,j − δik + δi+1,k)Ljk (j = k).

Exercise 2.3. Prove this formula.

(step 2) [ei, Lik] = 0 (k = i).

For k = i ± 1,i + 2, this is a direct consequence of the defining relations. For

other values of k, it is proved by induction.

(step 3) [ ei, Lj,i+1 ] = 0 (j = i + 1).

The proof is the same as (step 2).

(step 4) The following hold. In particular, we have [ei, Ljk] ∈ L for any j, k.

(i) [ei, Ljk] = 0 (j = i + 1, k = i, j = k).

(ii) [ei, Li+1,k] = Lik (k = i, i + 1).

(iii) [ei, Lji] = −Lj,i+1 (j = i + 1,i).

(iv) [ei, Li+1,i] = hi.

We start with (i). It is obvious for the cases j k, i j k and j k i.

The cases j = i k and j i = k − 1 k are nothing but (step 2) and (step 3).

In the remaining case j i k − 1, we know that [ei, Ljk] = 0 by the following

computation.

[ei, [Lj,i+1, Li+1,k]] = [Lj,i+1, [ei, Li+1,k]] = [Lj,i+1, Lik]

= [[Lji, ei], Lik] = [[Lji, Lik], ei]

= −[ei, Ljk]

Next we prove (ii). It is obvious for the case k ≥ i + 2. If k ≤ i − 1, it follows

from

[ei, [fi, Lik]] = [hi, Lik] = Lik.

The proof of (iii) is the same as (ii), and (iv) is obvious. Hence we have proved the

assertion.