TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...

1.2. Nilpotent orbits
Lie algebra g. We start with a review of the basic results give in [Hum72]. Recall that sl(2) is
spanned by
1 0 0 1 0 0
H = , X = , Y =
0 −1 0 0 1 0
and these satisfy the following relations:
[H,X] = 2X, [H,Y ] = −2Y and [X,Y ] = H.
Let V be a finite-dimensional sl(2)-module. Since H is semisimple then H acts diagonally
on V . Therefore, we can decompose V as a direct sum of eigenspaces V λ = {v ∈ V | H.v =
λv}, λ ∈ C where the notation H.v denotes H acting on v by the representation.
Definition 1.2.2. If V λ = {0} then we call λ a weight of H in V and we call V λ a weight space.
Lemma 1.2.3. If v ∈ V λ then X.v ∈ V λ+2 and Y.v ∈ V λ−2.
Proof. Since [H,X] = 2X one has H.(X.v) = X.(H.v) + 2X.v = (λ + 2)X.v. So X.v ∈ Vλ+2. And this is done similarly for Y .
Since V =
Vλ and dim(V ) is finite then there must exist Vλ = {0} such that Vλ+2 = {0}.
λ
In this case, each non-zero x ∈ V λ is called a maximal vector of weight λ (note that X.v = 0 if
v is a maximal vector).
Now, we assume that V is an irreducible sl(2)-module. Choose a maximal vector, say
v0 ∈ V λ . Set v−1 = 0, vi = 1 i! Y i .v0 (i ≥ 0). Then one has the following lemma.
Lemma 1.2.4.
(1) H.vi = (λ − 2i)vi,
(2) Y.vi = (i + 1)vi+1,
(3) X.vi = (λ − i + 1)vi−1, (i ≥ 0).
Proof.
(1) One has H.vi = 1 i! HY i .v0 = 1 i! (Y H − 2Y )Y i−1 .v0. Since H.v0. = λv0, Y.vi−1 = ivi and by
induction on i, we get H.vi = (λ − 2i)vi.
(2) It follows from the definition of vi.
(3) We prove (3) by induction on i. If i = 0, it is clear since vi−1 = 0 and X.v0 = 0. If i > 0,
one has
iX.vi = XY.vi−1 = [X,Y ].vi−1 +Y X.vi−1 = H.vi−1 +Y X.vi−1.
By (1), (2) and induction, we obtain
iX.vi = (λ − 2(i − 1))vi−1 + (λ − i + 2)Y.vi−2 = i(λ − i + 1)vi−1.
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