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 =
Chi +

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
[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
[ei, [fi, Lik]] = [hi, Lik] = Lik.
The proof of (iii) is the same as (ii), and (iv) is obvious. Hence we have proved the
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