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# TRIALITY CONSTRUCTION OF EXCEPTIONAL LIE ALGEBR A S

De
10 pages
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.

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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 :
18ϕ([v,ψ]w) =K([w,ϕ], [v,ψ]) =K([v,ψ], [w,ϕ]) = 18ψ([w,ϕ]v).
Proposition 4
∗With the previous notations, the bracket [·] makes g := Ŋl ⊕ W ⊕W into a semisimple Lie algebra, for3
which the Dynkin diagram is (G ). So it is isomorphic to g .2 2
Demonstration :
• We have just seen that g is a Lie algebra. Now we must prove that g is semisimple and its root system
is the one of g .2
• Let us choose a CSA of Ŋl , written h = hh :=Diag(1,−1,0), h :=Diag(0,1,−1)i. By construction,3 1 2
the elements of h act in a diagonalizable manner on g. Let us consider the weights of h in the h-module g :
in fact g is a Ŋl -module, so a h-module since h ⊂ Ŋl . Let r ⊂ g be a resoluble ideal. In particular,3 3
∗[Ŋl , r] ⊂ r and [h, r]⊂ r. But the Ŋl -submodules of g are Ŋl ,W,W and the sums of such modules. The3 3 3
4reason for this is that the three submodules are simple and not isomorphic. Let us show that r={0} :
⋄ Ŋl * r, because Ŋl is simple and r is resoluble.3 3
∗⋄ W* r. Indeed, suppose W ⊂ r. Since [.]| = 0, [W,W] is a nontrivial Ŋl -submodule of W whichW×W 3
∗ ∗ ∗is a Ŋl -module simple, so [W,W] =W and W ⊂ r. But similarly [.]| ∗ = 0, so [W,W ] = Ŋl , and3 W×W 3
Ŋl ⊂ r, what is absurd.3
∗⋄ W * r : we proceed likewise.
Consequently r={0}, so g is semisimple.
• We note that h is its own normalizer, because by construction, the space of weight 0 has dimension
2=dim h. So h is a CSA of g. And the root system of g is (G ).2
2.3 Constructing the other exceptional Lie algebras
We will generalize proposition 4 :
∗Let g be a semisimple Lie algebra, W a representation of g , and W its dual representation. Let us0 0
deﬁne g by :
∗g = g ⊕ W ⊕W .0
′Suppose there exists T and T , two trilinear forms :
3 ′ 3 ∗T : Λ W → C and T : Λ W → C.
Now let us consider the maps :
2 ∗ 2 ∗∧ : Λ W → W and ∧ : Λ W → W,
deﬁned by :
′(u∧v)(w) =T(u,v,w) and χ(ϕ∧ψ) =T (ϕ,ψ,χ).
Then we deﬁne a bracket on g, by :
[X,Y] = [X,Y], [X,v] =Xv, [X,ϕ] =Xϕ, [v,w] =a(v∧w), [ϕ,ψ] =b(ϕ∧ψ), [v,ϕ] =c(v∗ϕ),
where v∗ϕ is the only element in g verifying : ∀ Z ∈ g , K(v∗ϕ, Z) =ϕ(Zv), where K is the Killing0 0
form on g , and where a,b, c are scalars to be determined.0
By deﬁnition, [.] is bilinear and skew-symmetric. Moreover [.] satisﬁes the Jacobi identity. Indeed, the
veriﬁcations are the same as these made for g , except in the two last cases :2
∗⋄ If v, w ∈W and ϕ∈W , Jacobi is : [[v,w], ϕ]+[[w,ϕ], v]+[ [ϕ,v] , w] = 0, ie
|{z} |{z} |{z}
a(v∧w) c(w∗ϕ) −c(v∗ϕ)
ab(v∧w)∧ϕ =c(v∗ϕ)w−c(w∗ϕ)v. (4)
∗⋄ If v ∈W and ϕ, ψ ∈W , similarly Jacobi is :
ab(ϕ∧ψ)∧v =c(v∗ψ)ϕ−c(v∗ϕ)ψ. (5)
Remark 5
Both identities (4) and (5) are equivalent.
So, if the relations (4) and (5) are satisﬁed, then g is a Lie algebra.
Theorem 6 (Freudenthal)
′Let W be a representation of a semisimple Lie algebra g , and two trilinear forms T and T inducing0
maps
2 ∗ 2 ∗∧ : Λ W → W and ∧ : Λ W → W,
such that relations (4) and (5) are satisﬁed for certain scalars a, b, c.
∗If the weight spaces of W are one dimensional, and if the weights of W, W and the roots of g are all0
distinct, with abc = 0, then g is semisimple, and has the same Cartan subalgebra as g .0
Example 7
∗9 9f = Ŋo ⊕C ⊕C .4 9
∗3 9 3 9e = Ŋl ⊕ ΛC ⊕ΛC . And e , e are Lie subalgebras of e .8 9 6 7 8
5
6663 Third construction of g : derivations of octonion algebra2
Another way to construct an exceptional Lie algebra is to realize it as the Lie algebra of derivations of
an algebra, not necessarily associative.
3.1 Properties of derivations
Deﬁnition 8
Let A be an algebra, not necessarily associative. We call derivation of A an endomorphism D of A such
that ∀ a, b∈A, D(ab) =D(a)b+aD(b). We write der(A) the set of derivations of A.
Proposition 9
LetA be an algebra, not necessarily associative. Then gl(A) is a Lie algebra for the bracket [u,v] =uv−vu,
and der(A) is a Lie subalgebra of gl(A).
Theorem 10
Let A be a ﬁnite-dimensional algebra on the ﬁeldK =R orC, not necessarily associative. Let Aut(A) be
the group of automorphisms of the algebraA. It is a closed subgroup ofGl(A), so a Lie group. Moreover,
Lie(Aut(A)) = der(A).
Demonstration :
• Since Gl(A) is metric, the closeness of Aut(A) is obvious, by taking a sequence (v ) in Aut(A)n n∈N
which converges on v ∈Gl(A) and by applying it to ab.
• Let G =Aut(A). We will show that Lie(G) = der(A) :
⊂ Let g ∈Lie(G) and a, b∈A.
1Let us show that g(ab) =g(a)b+ag(b) : by hypothesis, g ∈T G, so there exists an arc C in G, writtenI
′γ, such that γ(0) =I and γ (0) =g. Since γ(t)∈Aut(A), we have γ(t)(ab) =γ(t)a.γ(t)b,
and by diﬀerentiating, we obtain :
′ ′ ′γ (t)(ab) =γ (t)(a)γ(t)(b)+γ(t)(a)γ (t)(b),
so for t = 0, g(ab) =g(a)b+ag(b).
⊃ Let D∈ der(A). We want to show that D∈ Lie(G). We identify D with an element of M C, wheren
tDn =dim(A). Let t∈R, and let us show that e ∈G : indeed,
P P Pj j∞ ∞ jtD t j t l l j−le (ab) = D (ab) = C D (a)D (b), andj=0 j=0 l=0 jj! j!
p∞ ∞ ∞m nX X X Xt t 1 1tD tD m n p m p−m tD(e (a))(e (b)) = ( D (a))( D (b)) = t D (a)D (b) =e (ab).
m! n! m!(p−m)!
m=0 n=0 p=0 m=0| {z }
mCp
p!
tDSo, e ∈G, and D∈Lie(G).
Example 11
LetH be the real quaternion algebra. We have Aut(H) =SO , so der(H) = Ŋo .3 3
Demonstration :
3We haveH =R⊕R , equipped with the product (a,u)(b,v) = (ab−uv, av +bu+u∧v). Let ϕ be an
3 2automorphism of the algebraH. ϕ preserves the center ofH which isR. Besides, q ∈R ⇔ q ∈R ,−
3 2 2 2 3therefore if q ∈ R , then ϕ(q) = ϕ(q ) = q ∈ R , so ϕ induces an automorphism of R . Since−
3 2 2∀ q ∈R , kqk =−q , ϕ preserves the norm. Last, ϕ| 3 preserves the cross product so the orientation.R
So, the map
Aut(H) → SO3
3ϕ →ϕ|R
is an injective morphism. It is also surjective since if ψ is in SO , then id ⊕ψ is an automorphism of3 R
H : indeed, ψ preserves the cross product, so id ⊕ψ preserves the product ofH.R
6Corollary 12
Let A be a ﬁnite-dimensional algebra on the ﬁeldK =R orC, not necessarily associative. Then g :=
der(A) is a Lie algebra. Let G be the only connected, simply connected Lie group such that Lie(G) = g.
Then Aut (A) is a quotient of G.0
Demonstration :
According to the preceding theorem, we have : Lie(Aut(A)) = der(A), and Aut(A) is a Lie group.
Moreover the other connected groups H such that Lie(H) = g are quotients of G.
3.2 Deﬁnition of complex octonions
Complex octonion algebraO =C⊗ O can be deﬁned this way :C R
3Let V = (v , v , v ) be the canonical basis ofC .1 2 3
3 3ThenO is identiﬁed as vector space withC ⊕C ⊕C⊕C. We write the elements of this space asC
a v 3matrices , where a, b∈C and v, w∈C . We equip this vector space with the product deﬁned
w b
by :
′ ′ ′ ′ ′ ′ ′a v a v aa −v·w av +bv +w∧w
= ,′ ′ ′ ′ ′ ′ ′w b w b aw +bw +v∧v bb −w·v
3where · is the scalar product onC and ∧ the cross product. It makes this 8-dimensionalC-vector space
into a Lie algebra. A basis ofO is given by :C

1 0 0 0 0 e 0 0iC = c = , c = , c = , c = , i = 1...3 .1 2 2+i 5+i0 0 0 1 0 0 e 0i
Then we obtain the following multiplication table (see Table 5) and we remark that

1 0
C :=c +c = is the unity of the algebraO .1 1 2 C0 1
LetusalsowriteC :=c −c andC :=c forj = 3...8. ThesubspaceO spannedbyallcommutators2 1 2 j j C0
′ ′cc −cc has codimension one with basis (C ,...,C ), complementary toC(C ).2 8 1
Table 5
Remark 13
•O is the subspace of all complex octonions having “trace” 0, ie such that a+b = 0.C0
•C.C is in the kernel of any derivation ofO .1 C
73.3 Looking for the derivations ofOC
Now we look for all derivations ofO . Since the set of derivations of an algebra is a Lie algebra, andC
because of the exceptional character of the octonion algebra,O is supposed to be an exceptional LieC
algebra.
We use Maple.
We ﬁnd that the set of derivations ofO is a 14-dimensional subspace of gl which can be identiﬁed withC 8
a subspace of gl (for any derivation d, we have d(O )⊂O ).7 C C0 0
Letuswrite (E ) thecanonicalbasisof gl . Thenthe14followingvectorsformabasisof der(O ):i,j i,j=1...7 7 C
h = −E +E +E −E h = −E +E +E −E1 2,2 4,4 5,5 7,7 2 3,3 4,4 6,6 7,7
g = −E −E +E −2E g = −E +E1 1,2 3,7 4,6 5,1 7 2,3 6,5
g = E −E +E +2E g = −E +E2 1,3 2,7 4,5 6,1 8 2,4 7,5
g = E +E −E +2E g = −E +E3 1,4 2,6 3,5 7,1 9 3,2 5,6
g = −E −2E +E −E g = −E +E4 1,5 2,1 6,4 7,3 10 3,4 7,6
g = E +2E +E −E g = −E +E5 1,6 3,1 5,4 7,2 11 4,2 5,7
g = E +2E −E +E g = −E +E6 1,7 4,1 5,3 6,2 12 4,3 6,7
Lemma 14
Let V be a ﬁnite-dimensionalC-vector space, and g ⊂ Ŋl(V) a Lie algebra acting irreducibly on V. Then
g is semisimple.
Demonstration :
Let r = rad(g). According to Lie’s theorem, there exists a ﬂag which is invariant under the action of
r. In particular, we can ﬁnd v ∈ V, and λ a linear form on r such that for all r ∈ r, rv = λ(r)v.
Let W = {v ∈ V ; ∀ r ∈ r, rv = λ(r)v}. According to the proof of Lie’s lemma, W is a nontrivial
g-submodule of V. Since V is simple, W =V. So, ∀ v ∈V, ∀ r ∈ r, rv =λ(r)v, ie ∀ r ∈ r, r =λ(r)id.
As r ⊂ g ⊂ Ŋl(V), we havedim(V)λ(r) =tr(r) = 0, hencer = 0. Therefore r is null. So, g is semisimple.
Proposition 15
der(O ) is semisimple.C
Demonstration :
7All we have to do is to prove that g = der(O ) acts irreducibly onV =C . Indeed, letW be a nontrivialC
g-submodule of V. The element h =Diag(0, 1, 2, −3, −1, −2, 3) is in g, so h W ⊂W. Since all the0 0
eigenvalues of h are distinct, h is diagonalizable, so is its restriction on W. Consequently W contains0 0
at least an eigenvector for h . Since the eigenvectors for h are the elements of the canonical basis, W0 0
contains at least a vector of the canonical basis (v ,...,v ) of V. Now ∀ j = 2...7, g v = ±v ,1 7 j−1 j 1
therefore v ∈ W. Last, the relations g v = −2v , g v = 2v , g v = 2v , g v = −2v , g v =1 1 1 5 2 1 6 3 1 7 4 1 2 5 1
2v , g v = 2v show that : ∀ j = 1...7, v ∈W. Finally, W =V.3 6 1 4 j
We have the following relations : (∗)
[h ,g ] = g [h ,g ] = 0 [h ,g ] = −g [h ,g ] = g1 1 1 2 1 1 7 7 2 7 7
[h ,g ] = 0 [h ,g ] = g [h ,g ] = −2g [h ,g ] = −g1 2 2 2 2 1 8 8 2 8 8
[h ,g ] = −g [h ,g ] = −g [h ,g ] = g [h ,g ] = −g1 3 3 2 3 3 1 9 9 2 9 9
[h ,g ] = −g [h ,g ] = 0 [h ,g ] = −g [h ,g ] = −2g1 4 4 2 4 1 10 10 2 10 10
[h ,g ] = 0 [h ,g ] = −g [h ,g ] = 2g [h ,g ] = g1 5 2 5 5 1 11 11 2 11 11
[h ,g ] = g [h ,g ] = g [h ,g ] = g [h ,g ] = 2g .1 6 6 2 6 6 1 12 12 2 12 12
We deﬁne the set Δ by :
Δ ={L , L , −L −L , −L , −L , L +L , −L +L , −2L −L , L −L , −L −2L , 2L +L , L +2L }.1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
8Proposition 16
Let h = Ch ⊕Ch . Then h is a Cartan subalgebra of g = der(O ), and Δ is the root system of g.1 2 C
Demonstration : M
• According to the relations (∗), we have the following decomposition : g = h ⊕ g .α
α∈Δ
Besides, h is diagonalizable and h =N(h) : indeed, h is commutative, so h⊂Z(h)⊂N(h). The restriction
of a diagonalizable endomorphism on an invariant subspace is also diagonalizable. SinceN(h) is invariantM
′ ′under ad(h), we have N(h) = h ⊕ g for a Δ ⊂ Δ. Let us show that Δ is empty :α
′α∈Δ
according to (∗), for allj = 1...12, there existsn ∈{1,2} andl = 0 such that [h ,g ] =l g . All thisj j n j j jj
shows that for all root α∈ Δ, there exists h∈ h such that [h, g ] = g *h.α α
′Consequently, Δ =∅, so h = N(h), and h is a Cartan subalgebra of g.
• The relations (∗) show that the elements of Δ are roots of g. We have found 12 roots of g, so we have
found all the roots, since for all root α of g, g is one-dimensional.α
Theorem 17
der(O ) is a Lie algebra isomorphic to g .C 2
Demonstration :
• Calculation of the Killing form on g : the formula
X
′ ′′∀ h, h ∈ h, K(h,h ) = α(h)α(h )
α∈Δ
gives K(h ,h ) =K(h ,h ) = 16 and K(h ,h ) = 8.1 1 2 2 1 2
∗ f f fWe deduce the bilinear symmetric form h.i deﬁned on h by hλ,μi = K(h ,h ), where h is the onlyλ μ λ
felement of h such that ∀ h∈ h, K(h,h ) =λ(h).λ
1 1 1 1g gWe ﬁnd h = h − h and h =− h + h .L 1 2 L 1 21 212 24 24 12
1 1Hence hL ,L i =hL ,L i = and hL ,L i =−1 1 2 2 1 212 24
∗Finally h.i induces a scalar product on the two-dimensionalR-vector space : h =R⊗ QΔ.QR
hL ,L i1 2 1 2π\\Furthermore, cos(L ,L ) = =− , hence (L ,L ) = .1 2 1 2|L |.|L | 2 31 2
Then we can draw the root system of g : we obtain the root system of g .2
•C is an algebraically closed ﬁeld of characteristic 0. The Lie algebras der(O ) and g are semisimpleC 2
and have the same root system, so they are isomorphic.
3.4 Application to G2
According to corollary 12, the exceptional Lie group G can be realized as the group of automorphisms of2
the complex octonion algebraO . Indeed, the quotients of the only connected, simply connected groupC
which admits g for Lie algebra are in bijection with the subgroups of the quotient P/Q, where P is the
integer weight lattice and Q =ZΔ the root lattice. Moreover, [P : Q] = det(C) = 1, where C is the
Cartan matrix of g. ThusG =Aut (O ). And asAut(O ) is connected, we deduce thatG =Aut(O ).2 0 C C 2 C
References
[FH] Fulton-Harris (Graduate Texts in Mathematics 129) : lecture 22.
[JB] Web page of John Baez about the octonion algebra.
[HM] Humphreys : Introduction to Lie Algebras and Representation Theory.
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6Contents
1 First construction of g 12
1.1 Using the Dynkin diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Verifying the Jacobi identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Second construction of g : à la Freudenthal 22
2.1 Watching the root diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.2 Constructing a Lie bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.3 Constructing the other exceptional Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Third construction of g : derivations of octonion algebra 62
3.1 Properties of derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.2 Deﬁnition of complex octonions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.3 Looking for the derivations ofO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8C
3.4 Application to G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
10