TRIALITY CONSTRUCTION OF EXCEPTIONAL LIE ALGEBR A S

Frédéric Butin

Description

Niveau: Supérieur, Licence, Bac+3
TRIALITY, CONSTRUCTION OF EXCEPTIONAL LIE ALGEBR A S Frédéric BUTIN 20th February 2006 In the classification of the root systems, the Dynkin diagram (G2) (Diagram 1) appears. According to Serre's theorem, there exists an only complex semisimple Lie algebra that admits this root system. We will give three constructions of this Lie algebra : • A very concrete version by calculating its multiplication table. • A construction from the representations of ?l3. • A construction from the octonion algebra. 1 First construction of g2 1.1 Using the Dynkin diagram The classification of the root systems shows the sys- tem given by Diagram 2, for which the Dynkin dia- gram is Diagram 1. Diagram 1 Diagram 2 Theorem 1 (Serre) Let ∆ be a root system, let ? be a basis of ∆, and (n(?, ?))?,??? its Cartan matrix. Then the Lie algebra presented by generators E?, F?, H?, ? ? ∆ and Weyl-Serre relations is a finite-dimensional semisimple Lie algebra, and its root system is ∆. According to Serre's theorem, there exists so a complex Lie algebra which has this root system. But we want to verify it by hand. Suppose that such a Lie algebra g2 exists. Since it has 12 roots, and since dimC h = dimR h?R = 2, this algebra is necessary of dimension 14=12+2.

• ?w ?


• lie algebra


• c9 ?


• now


• ?3w ?


• w?w ?


• diagram


• weyl-serre relations


• let h1

Sujets

• Lie algebra
• Diagram

Informations

• Publié par: ondey
• Nombre de lectures: 35
• Langue: English

Extrait

Frédéric BUTIN
20th February 2006
In the classification of the root systems, the Dynkin diagram (G ) (Diagram 1) appears. According to2
Serre’s theorem, there exists an only complex semisimple Lie algebra that admits this root system.
We will give three constructions of this Lie algebra :
• A very concrete version by calculating its multiplication table.
• A construction from the representations of Ŋl .3
• A construction from the octonion algebra.
1 First construction of g2
1.1 Using the Dynkin diagram
The classification of the root systems shows the sys-
tem given by Diagram 2, for which the Dynkin dia-
gram is Diagram 1.
Diagram 1
Diagram 2
Theorem 1 (Serre)
Let Δ be a root system, let Π be a basis of Δ, and (n(α,β)) its Cartan matrix. Then the Lie algebraα,β∈Π
presented by generators E , F , H , α∈ Δ and Weyl-Serre relations is a finite-dimensional semisimpleα α α
Lie algebra, and its root system is Δ.
According to Serre’s theorem, there exists so a complex Lie algebra which has this root system.
But we want to verify it by hand.
∗Suppose that such a Lie algebra g exists. Since it has 12 roots, and since dim h = dim h = 2, this2 C R R
algebra is necessary of dimension 14=12+2.
We construct a basis of g this way :2
• Let us write α ,...,α the positive roots of g , and β their opposites.1 6 2 j
Let us take X ∈ g \{0} and X ∈ g \{0}. Similarly, let Y , Y be eigenvectors for β , β .1 α 2 α 1 2 1 21 2
• Let H = [X ,Y ] and H = [X ,Y ].1 1 1 2 2 2
We can choose Y and Y so that the elements H ∈ h verify α (H ) =α (H ) = 2.1 2 i 1 1 2 2
• Then we take X = [X ,X ], etc...3 1 2
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IFinally, we obtain for g the following basis :2
B = (H , H , X , Y , X , Y , X , Y , X , Y , X , Y , X , Y ).1 2 1 1 2 2 3 3 4 4 5 5 6 6
We calculate all the brackets of basis vectors, so we obtain, after having divided X and Y by 2,4 4
X , Y , X and Y by 6, and replaced X and Y by their opposites, the following multiplication table5 5 6 6 5 3
for this possible Lie algebra (see Table 3).
Table 3
Remark 2
The significance of the multiplication table is that it gives the uniqueness of the Lie algebra g subject to2
its existence.
1.2 Verifying the Jacobi identity
To assert that g is a Lie algebra and at the same time show the existence of a Lie algebra associated to2
the diagram (G ), there is still to verify the Jacobi identity :2
Therefore, we can calculate all the [[g , g ], g ]+[[g , g ], g ]+[[g , g ], g ] where the g are in B. It1 2 3 2 3 1 3 1 2 j
3necessitates C = 364 rather boring verifications !!!14
Luckily, the following constructions allow us to let us off this verifications !
2 Second construction of g : à la Freudenthal2
2.1 Watching the root diagram
We watch the root system for g : see Diagram 4.2
It consists of a regular hexagon and two equilateral
triangles. Now the regular hexagon is the root dia-
gramfor Ŋl , andthetwoequilateraltrianglesarethe3
weight diagrams for the standard representation W
∗of Ŋl and its dual W .3
All that suggests that we should construct a Lie
∗bracket on Ŋl ⊕ W ⊕W .3
Diagram 4
22.2 Constructing a Lie bracket
∗We define on Ŋl ⊕ W ⊕W the following bilinear map, by its restriction on the three subspaces :3
[·] is defined :
∗⋄ On Ŋl × W and Ŋl × W by the standard action of Ŋl ; on Ŋl × Ŋl by the Lie bracket of Ŋl .3 3 3 3 3 3
∗⋄ On W ×W, [·] by : W ×W → W ; (v,w) → −2v∧w.
∗ ∗ ∗ ∗⋄ Similarly, on W ×W by : W ×W → W ; (ϕ,ψ) → 2ϕ∧ψ.
∗⋄ Last, on W ×W , by :
[v, ϕ](w) = 3ϕ(w)v−ϕ(v)w. (1)
Remark 3
∗i. The bracket [.]| :W ×W →W is well defined to within a scalar, and [.]| induces triality.W×W W×W
ii. In the last case, [v, ϕ] is the only element of Ŋl verifying the relation : ∀ Z ∈ Ŋl ,3 3
K([v, ϕ], Z) = 18ϕ(Zv), (2)
[v,ϕ]
where K is the Killing form on Ŋl . To simplify the notation, we write v∗ϕ := .3 18
Demonstration :
∗i. In fact [.]| ∈Hom(W ⊗W, W )≃Hom(W ⊗W ⊗W, C).W×W
Moreover, since [.] is skew-symmetric, [.]| ∈Hom(W ∧W ∧W, C)≃C (dimW = 3), and the mapW×W
3T : Λ W →C, u∧v∧w → [u,v](w) is a triality : indeed if we fix any two nonzero arguments, the linear
form induced on the other vector space is nonzero.
ii. The uniqueness results from the nondegeneracy of the Killing form on the semisimple Lie algebra Ŋl .3
For the existence, we have :
∗ ∗ ∗K([e ,e ], Z) = 6tr([e ,e ]◦Z) = 6tr((3E −δ I)◦Z) = 18tr(E ◦Z)−6tr(Z) = 18e (Ze ),i i i,j i,j i,j ij j j
| {z }
=0
what is sufficient.
The map [·] defined this way is bilinear and skew-symmetric. To show that it is a Lie bracket, there is
still to verify the Jacobi identity : all what we have to do is to check it on triples of arbitrary elements
of the three subspaces.
▽ If the 3 elements are in g = Ŋl : it is the Jacobi identity in Ŋl .0 3 3
▽ If 2 elements X, Y are in g :0
⋄ If v ∈W, then : [X, [Y,v]]+[Y, [v,X]]+[v, [X,Y]] =XYv−YXv−[X,Y]v = 0
|{z} |{z} | {z }
Yv −Xv −[X,Y]v
∗⋄ If 2 elements X, Y are in g and ϕ∈W , we have similarly :0
[X, [Y,ϕ]]+[Y, [ϕ,X]]+[ϕ, [X,Y]] =XYϕ−YXϕ−[X,Y]ϕ = 0
|{z} |{z} | {z }
Yϕ −Xϕ −[X,Y]ϕ
▽ If 1 element Z is in g :0
⋄ If v, w∈W :
[Z, [v,w]]+[v, [w,Z]]+[w, [Z,v]] = Z(−2v∧w)−2v∧(−Zw)−2w∧(Zv))
|{z} |{z} |{z}
−2v∧w −Zw Zv
= 2(−Z(v∧w)+v∧(Zw)+(Zv)∧w) = 0,
by definition of the action of a Lie algebra on the exterior product.
∗⋄ If ϕ, ψ ∈W , we proceed likewise .
∗ 1⋄ If v ∈W and ϕ∈W : {[Z, [v,ϕ]]+[v, [ϕ,Z]]+[ϕ, [Z,v]]} = [Z,v∗ϕ]−v∗(Zϕ)−(Zv)∗ϕ.
18
By nondegeneracy of K and because K(Y, [Z,X]) =K([Y,Z], X), this expression is null iff :
3∀ Y ∈ g , K(Y, [Z,v∗ϕ])−K(Y, v∗(Zϕ))−K(Y, (Zv)∗ϕ) = 0,0
ie ∀ Y ∈ g , −ϕ([Y,Z]v)+(Zϕ)(Yv)+ϕ(Y(Zv)) = 0.0
But ϕ([Y,Z]v) =ϕ(Y(Zv))−ϕ(Z(Yv)), therefore the Jacobi identity amounts to :
∀ Y ∈ g , (Zϕ)(Yv)+ϕ(Z(Yv)) = 0, ie (with w =Yv) ∀ w∈W, (Zϕ)(w)+ϕ(Zw) = 0,0
what is true by definition of the dual representation.
▽ If no element is in g :0
⋄ If u, v, w∈W : by nondegeneracy of the Killing form on g ,0
[u, [v,w]]+[v, [w,u]]+[w, [u,v]] = 0
|{z} |{z} |{z}
−2v∧w −2w∧u −2u∧v
iff ∀Z ∈ g , K(Z, [u,v∧w])+K(Z, [v,w∧u])+K(Z, [w,u∧v]) = 0, ie, according to (2),0
18((v∧w)(Zu)+(w∧u)(Zv)+(u∧v)(Zw)) = (v∧w∧(Zu)+w∧u∧(Zv)+u∧v∧(Zw)) = 18Z(u∧v∧w) = 0,
what is true.
⋄ If ϕ, χ, ψ ∈W, we proceed likewise.
∗⋄ If u, v ∈W and ϕ∈W :
We have : [[v,w],ϕ] =−2[v∧w,ϕ] =−4(v∧w)∧ϕ =−4(ϕ(v)w−ϕ(w)v), therefore Jacobi amounts to :
−4(ϕ(v)w−ϕ(w)v) =−[[w,ϕ], v]−[[ϕ,v], w],
ie −4(ϕ(v)w−ϕ(w)v) =−[w,ϕ](v)+[v,ϕ](w). (3)
Now, according to (1), we have −[w,ϕ](v)+[v,ϕ](w) =−3ϕ(v)w +ϕ(w)v +3ϕ(w)v−ϕ(v)w,
hence the wanted equality.
∗⋄ If u∈W and ϕ, ψ ∈W :
We must show that : [[ϕ,ψ], v] =−[[ψ,v], ϕ]−[[v,ϕ], ψ].
|{z} | {z }
−[v,ψ] [v,ϕ]ψ
Or [[ϕ,ψ], v] =−4(ϕ∧ψ)∧v =−4(ϕ(v)ψ−ψ(v)ϕ). Therefore the wanted equality amounts to :
|{z}
2ϕ∧ψ
−4(ϕ(v)ψ−ψ(v)ϕ) = [v,ψ]ϕ−[v,ϕ]ψ,
ie ∀ w∈W, −4(ϕ(v)ψ(w)−ψ(v)ϕ(w)) = [v,ψ]ϕ(w)−[v,ϕ]ψ(w) =−ϕ([v,ψ]w)+ψ([v,ϕ]w),
by definition of the dual representation.
Now, if we apply ψ to the identity (3), we find : −4(ϕ(v)ψ(w)−ϕ(w)ψ(v)) =−ψ([w,ϕ]v)+ψ([v,ϕ]w).
It is the formula we looked for, since by using the symmetry of the Killing form, we have :