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

1.2. Nilpotent orbits
We prove the Jacobson-Morozov theorem by induction on the dimension of g. If dim(g) = 3
then g is isomorphic to sl(2) and the result follows. Assume that dim(g) > 3. If X is in a
proper semisimple subalgebra of g then by induction, there is a sl(2)-triple with X its nilpositive
element. Hence we may assume that X is not in any proper semisimple subalgebra of g. By
Lemma 1.2.7 and [X,X] = 0, one has X ∈ (g X ) ⊥ = [g,X]. Therefore, there exists H ′ ∈ g such
that [H ′ ,X] = 2X.
Now, let H ′ = H ′ s + H ′ n be the Jordan decomposition of H ′ in g where H ′ s is semisimple and
H ′ n is nilpotent. Remark that any subspace which is stable by ad(H ′ ) is also stable by ad(H ′ s)
and ad(H ′ n). The nilpotency of the ad(H ′ n) action on the stable subspace CX gives [H ′ n,X] = 0
and therefore [H ′ s,X] = 2X. Set H = H ′ s.
If H ∈ [g,X] then there exists Y ∈ g such that H = [X,Y ]. Since ad(H) acts semisimply on
g then g is decomposed by
g = g λ1 ⊕ ··· ⊕ g λk ,
where g λ1 ,..., g λk are ad(H)-eigenspaces. Write Y = Y λ1 + ··· +Y λk with Yi ∈ g λi . Let Z ∈ g λi
then [H,Z] = λiZ. Hence, [X,[H,Z]] = λi[X,Z]. By the Jacobi identity, one has
[Z,[H,X]] + [H,[X,Z]] = λi[X,Z].
Then we get ad(H)([X,Z]) = (λi + 2)[X,Z]. It shows that [X,g λi ] ⊂ gλi+2,1 ≤ i ≤ k. Since
ad(H)(H) = 0, one has H ∈ g0. Moreover, H = [X,Y ] = ∑ i=1
k [X,Yi]. Therefore, there is some
Y ′ ∈ g−2 such that H = [X,Y ′ ]. If we replace Y by Y ′ then [H,X] = 2X, [H,Y ] = −2Y and
[X,Y ] = H. That means {H,X,Y } is a sl(2)-triple with X its nilpositive element.
From the above reason, it remains to prove H ∈ [g,X]. By contradiction, assume that H /∈
[g,X] then κ(H,g X ) = 0. According to the Jacobi identity
[X,[H,g X ]] + [g X ,[X,H]] + [H,[g X ,X]] = 0,
one get [X,[H,g X ]] = 0. It implies that ad(H)(g X ) ⊂ g X , i.e. g X is ad(H)-invariant. By acting
semisimply of ad(H) on g X , g X is decomposed into ad(H)-eigenspaces:
g X = g X τ1 ⊕ ··· ⊕ gXτi = gX0 ⊕ ∑ g
τi=0
X τi .
From the invariance of the Killing form, one has κ(H,[H,g X ]) = κ([H,H],g X ) = 0. Therefore
if Z is a non-zero element of g X τi with τi = 0 then
0 = κ(H,[H,Z]) = κ(H,τiZ) = τiκ(H,Z).
This shows that H ∈ (g X τi )⊥ . Since κ(H,g X ) = 0 there must exist Z ∈ g X 0 = {Y ∈ gX | [H,Y ] =
0} = (g X ) H such that κ(H,Z) = 0. If Z is nilpotent then κ(H,Z) = 0 as in the proof of Lemma
1.2.7. Therefore, Z is non-nilpotent. That means its semisimple component Zs = 0 and hence
g Zs is reductive. As a consequence, [g Zs ,g Zs ] is a semisimple subalgebra of g and it is a proper
subalgebra since g Zs = g only if Zs = 0.
On the other hand, since [X,Z] = [H,Z] = 0, apply the property of Jordan decomposition
one has [X,Zs] = [H,Zs] = 0. That means Zs ∈ (g X ) H and then X ∈ g Zs . Also, H ∈ g Zs so we
get 2X = [H,X] ∈ [g Zs ,g Zs ]. This is a contradiction then we obtain the result.
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