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In particular, it is not nilpotent


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• (a) Cartan subalgebras

• (b) Root subspaces

• (c) Abstract root systems

• (d) Examples

1.1. Cartan subalgebras. Henceforwardgis a (finite-dimensional) semisim- ple Lie algebra over K, assumed to be algebraically closed of characteristic zero. In particular, it is not nilpotent. So by Engel’s theorem, it contains an elementX such thatXs6= 0, and therefore it contains semisimple elements.

Definition 1.1. A Cartan subalgebra ofgis a maximal subalgebra consist- ing of semisimple elements.

Lemma 1.2. Any subalgebra of g all of whose elements are semisimple is abelian.

Proof. Let h be such a subalgebra. We have to show adh(X) = 0 for all X ∈ h. Since X = Xs, it is diagonalizable, and so it suffices to show that all of the eigenvalues of X are zero. If not, suppose Y ∈ h is an eigenvector, [X, Y] =aY witha6= 0. Thenad(Y)(X) = [Y, X] =−aY is an eigenvector ofad(Y) with eigenvalue 0. WriteX =P

jXj where eachXj is an eigenvector ofY (which is also semisimple), with eigenvaluesaj. Then

ad(Y)(X) = X



is a linear combination of eigenvectors for non-zero eigenvalues. This con-

tradicts the previous statement.

Thus a Cartan subalgebra hofg is abelian, and since every element ofh is diagonalizable underad, and they all commute, it follows that the adjoint representation of hong is diagonalizable. Write




where Cg(h) is the centralizer and in particular containshand α runs over non-zero characters of h; gα ⊂ g is the subspace on which any X ∈ h has eigenvalueα(X). We let Φ⊂h be the set of elements such that gα6= 0.

We also writeCg(h) =g0. We will show thatCg(h) =h and each gα has dimension 1. First, two elementary properties.



Proposition 1.3. Let α, β ∈h. Then [gα,gβ]⊂gα+β. If α 6= 0 then any X∈gα is ad-nilpotent. If α, β∈h and α+β 6= 0 then Bad(gα,gβ) = 0.

Proof. The first statement follows from the Jacobi identity. Indeed, for any X∈h,Y ∈gα,Z ∈gβ,

[X,[Y, Z]] =−[Y,[Z, X]]−[Z,[X, Y]] = [Y,[X, Z]] + [[X, Y], Z]

= [Y, β(X)Z] + [α(X)Y, Z] = (α(X) +β(X))[Y, Z].

It follows that, if X ∈ gα, then ad(X)n(gβ) ⊂ gβ+nα. If α 6= 0 then the set of nsuch that β+nα ∈Φ∪ {0} is bounded, because dimg is finite. It follows thatad(X)n(gβ) = 0 for n >>0, even whenβ = 0.

Finally, if X ∈ gα and Y ∈ gβ, then adX ◦adY(gγ) ⊂ gγ+α+β and thus does not contribute to the trace of adX ◦adY unless γ+α+β =γ, which

proves the orthogonality.

Corollary 1.4. The restriction of the Killing form of g to Cg(h) is non- degenerate and defines an isomorphism h−→h .

Proof. The Killing form of g is non-degenerate, so for any X ∈ h there is Y ∈ g such that Bad(X, Y) 6= 0. Writing Y = Y0+P

Yα with Y0 ∈ g0 and Yα ∈gα for α ∈ Φ, it follows from the proposition that Bad(X, Y) =

Bad(X, Y0).

Proposition 1.5. Let h⊂g be a Cartan subalgebra. Then h=Cg(h).

Proof. WriteC=Cg(h). In several steps

Step 1: For anyX ∈C,Xs∈C,Xn∈C. Indeed, C={X∈g |adX(h) = 0}.

But Xs and Xn are polynomials in X without constant term, so if adX satisfies this property then so doad(X)s=ad(Xs) and ad(Xn).

Step 2: All semisimple elements of C are in h. Indeed, if X =Xs ∈ C, then h+KX is an abelian subalgebra consisting of semisimple elements.

Since his maximal, X∈h.

Step 3: The restriction ofBAtohis nondegenerate. LetX ∈hbe in the kernel. There exists Y ∈C such that T r(adXadY)6= 0. If Y is semisimple then Y ∈h; so we may assume Y is nilpotent. But [X, Y] = 0 so adXadY is the product of two commuting elements, one of which is nilpotent, and is therefore nilpotent. So its trace must be 0.

Step 4: The Lie algebra C is nilpotent. Let X = Xs+Xn ∈ C. Both Xsand Xn are ad-nilpotent onC, and they commute, so their sum is again ad-nilpotent. By Engel’s theorem, C is nilpotent.

Step 5: SinceBad is invariant,

Bad([X, Y], Z) =Bad(X,[Y, Z]) = 0,∀X ∈h, Y, Z∈C

because [X, Y] = 0. Thushis orthogonal to [C, C] and sinceBad is nonde- generate,h∩[C, C] = 0.


Step 6: C is abelian. If not, [C, C]6= 0. ThenZ(C)∩[C, C]6= 0 sinceC is nilpotent. Suppose 0 6= Z ∈ Z(C)∩[C, C]. If Z is semisimple it lies in h, which contradicts Step 5. So Zn6= 0, Zn∈C, henceZn∈Z(C) because Z(C) consists of elements that takeCto 0 under the adjoint representation.

But thenadYadZn is nilpotent for anyY ∈Zn, hence has trace 0, henceZn is in the kernel of the Killing form. Thus Zn= 0.

Step 7: If C 6=hthen C contains a non-zero nilpotent element Z. Since C is abelian, adYadZ is nilpotent for all Y ∈ C, which contradicts the

nondegeneracy ofBad on C. ThusC=h.

1.2. Root subspaces. Let g be a semisimple Lie algebra over K. Fix a Cartan subalgebrah.

Definition 1.6. A root ofhing is anα6= 0 such thatgα6= 0.

The set of roots is denoted Φ = Φ(g,h).

Example 1.7. Let g=sl(n), hthe diagonal subalgebra.

Let αi :h→ K denote the ith component. Then the roots are αi −αj, i 6= j. This proves that every element of h is semisimple and that h is maximal, hence a Cartan subalgebra.

Lemma 1.8. Suppose α∈Φ. Then −α∈Φand the Killing form places gα and g−α in duality.

Proof. We have seen that if α+β 6= 0 then Bad(gα,gβ) = 0. Since Bad is nondegenerate, if α ∈ Φ then −α ∈Φ and the two root subspaes are in


Lemma 1.9. The set of roots Φgenerates h.

Proof. Suppose not. Then there is an element H ∈ hsuch that α(H) = 0 for all α ∈ Φ. In particular, [H, Y] = 0 for all Y ∈ gα, for all α ∈ Φ;

and of course H commutes with h. So H is in the center of g, but g is


Proposition 1.10. (i) Letα∈Φ. Then dimgα = 1.

(ii) Ifn∈N and nα∈Φ thenn=±1.

We writeB =Bad.

Proof. Let 06=Eα ∈ gα, so that [H, Eα] =α(H)Eα for all H ∈ h. Choose F ∈ g−α so that B(Eα, F) = 1, Hα0 = [Eα, F]. We have Hα0 ∈ g0 = h.

Moreover, for allH ∈h,

B(H, Hα0) =B(H,[Eα, F]) =B([H, Eα], F) =α(H)B(Eα, F) =α(H).

We show thatα(Hα0)6= 0. In any case, there exists some β∈Φ such that β(Hα0) 6= 0. Consider W = ⊕n∈Ngβ+nα. This subspace is invariant under adEα andadF, as well as under adH0α; but

T rW(ad(Hα0)) =T rW([ad(Eα, ad(F)]) = 0.



0 =X


(β+nα)(Hα0) dimgβ+nα. Ifα(Hα0) = 0 then this isβ(Hα0)P

ndimgβ+nα6= 0, which is a contradiction.

Thusα(Hα0)6= 0.

Next, we show that [Eα,g−α] ⊂ KHα0. Indeed, if Y ∈ g−α then the character

H 7→B(H,[Eα, Y]) =B([H, Eα], Y) =α(H)B(Eα, Y)

is a multiple ofα(H) =B(H, Hα0). SinceB is nondegenerate onh, [Eα,g−α] is a multiple of Hα0.


V =KEα⊕KHα0 ⊕M



This space is stable under ad(F) and ad(Hα0) and we have seen it is stable under Eα. (So we can use the representation theory of sl(2).) As above, T rV(ad(Hα0) = 0. Thus

α(Hα0) +X


nα(Hα0) dimg= 0

It follows that g= 0 for n <−1 and dimg−α = 1.

Corollary 1.11. The restriction to h of the Killing form is given by the following formula:

∀H, H0∈h, B(H, H0) =X



Proof. We can choose a basis of g with respect to which the action ofh is diagonal: If dimh=`, let {ei, i= 1, . . . `} be a basis ofh,eα a basis of gα for all α ∈ Φ. Then for H, H0 ∈ h, adH ◦adH0(ei) = 0 for all i, whereas adH ◦adH0(eα) =α(H)α(H0)eα. The formula follows.

Corollary 1.12. Let α∈Φ. There exists a uniqueHα∈[gα,g−α]such that α(Hα) = 2. Given any Eα 6= 0 in gα, there exists a unique Fα ∈g−α such that[Eα, Fα] =Hα. The set (Hα, Eα, Fα) is ansl(2)-triple, and generates a Lie subalgebra ofg isomorphic tosl(2).

Proof. ChooseEα ∈gα as in the proof of the Proposition, and let F denote the F of that proof; let Hα0 = [Eα, F]. We have seen this is non-zero and since dimgα = dimg−α = 1, Hα0 spans [gα,g−α]. Moreover, we showed that α(Hα0) 6= 0; so let Hα = 2Hα0/α(Hα0). Then α(Hα) = 2 and it is the unique vector in the one-dimensional space [gα,g−α] with this property.

This implies that

[Hα, E] = 2E,[Hα, F] =−2F,∀E ∈gα, F ∈g−α.

There is then a unique Fα ∈ g−α such that [Eα, Fα] =Hα. By definition

(Hα, Eα, Fα) is ansl(2)-triple.


1.3. Abstract root systems. Notation is as above. The next step is to use the representation theory of sl(2) to characterize the set Φ. First, we have

Proposition 1.13. Let β ∈Φ, β 6=α,−α. Then (i) β(Hα)∈Zand the linear form β−β(Hα)α ∈Φ.

(ii) The set ofn∈Z such that β+nα∈Φ is an interval in Z. (iii) Ifα+β ∈Φ then[gα,gβ] =gα+β.

Proof. Let V = ⊕n∈Zgβ+nα ⊂ g. This is invariant under the action of each member of the sl(2)-triple (Hα, Eα, Fα), hence is a sum of irreducible representations ofsl(2). The weights ofKHα onV are of the formβ(Hα) + 2n. The classification of representations ofsl(2) shows that each of these is an integer, hence β(Hα) ∈Z. Moreover, each such weight has multiplicity one and are of the same parity; hence V is an irreduciblerepresentation of sl(2). Let N(V) be the set of nsuch that gβ+nα6= 0. There is thus m∈N such that

{β(Hα) + 2n, n∈N(V)}={−m,−m+ 2, . . . , m−2, m}.

This already shows (ii). Moreover, ifiis in the right-hand interval, so is−i.

In particular, taking n= 0, we have i=β(Hα) and so −i =−β(Hα) is of the formβ(Hα)+2nfor a uniquen∈N(V). This implies thatn=−β(Hα), in other words thatβ−β(Hα)α∈Φ. Thus we have (i).

Finally, if α+β ∈ Φ, so n = 1∈ N(V), and gβ+α is the eigenspace for β+ 2 forHα. We know thatEα takes theβ-weight space in the (irreducible) representationV to theβ+ 2-weight space; hence

[gα,gβ] =adEα(gβ) =gα+β.

Let hQ (resp. h

Q denote the Q-subspace of h(resp. of h) generated by theHα, α∈Φ (resp. generated by α∈Φ).

Proposition 1.14. Let `= dimKh. Then `= dimQhQ = dimQh

Q. More- over, the pairing h⊗h → K restricts to a perfectQ-bilinear pairing

hQ⊗hQ → Q.

The Killing form restricts to a positive-definite rational bilinear pairing h,i onhQ.

Proof. We have already seen that Φ contains a basis ofh, say α1, . . . , α`. Then Hα0i,i= 1, . . . , `, forms the dual basis ofhwith respect to h,i:

hHα0, Hi=α(H).

Moreover, recall that, for any α ∈ Φ, Hα = hH02

α,Hα0iHα0, so the Hαi form a basis ofh. We need to show thatHα belongs to theQ-span of theHi=Hαi for all α. Write Hα = P

λiHi with λi ∈ K. Consider the matrix M =


(hHi, Hji). This is invertible because the bilinear form is nondegenerate and theHi form a basis. Moreover, we have

hHα, Hii= X



and we have seen that eachβ(Hα)∈Z. SoM ∈GL(`,Q) and we have hHα, Hji=X

λihHi, Hji, j = 1, . . . , `

is a system of`linear equations overQwith a unique solution (λ1, . . . , λ`) It follows that theλi∈Q. By duality, we see that the α are all in theQ-span of the αi.

It remains to show that hiis positive definite. But for anyH∈h, hH, Hi=X



which is positive for H in the real span of theHα. We can now define an abstract root system.

Definition 1.15. LetV be a finite-dimensional euclidean vector space over R, with inner producth,i. A (reduced) root system in V is a subset Φ∈V in bijection via a map α7→α, such that

• ”(i)” Φ is finite, spansV, and 0∈/ Φ;

• ”(ii)” For allα, β ∈Φ, β(α) := 2hα,βihβ,βi ∈Z;

• ”(iii)” For allα∈Φ, and sα(Φ) = Φ, where sα(v) =v−2hv, αi

hα, αiα

is the Euclidean reflection through the hyperplane orthogonal toα.

• ”(iv)” Ifα∈Φ andλα∈Φ for some λ∈R, thenλ=±1.

The rank of the root system is the dimension ofV. The α (resp. α :=

hα,αi) are called roots (resp. coroots).

An isomorphism (Φ, V) −→(Φ 0, V0) of root systems is an isomorphism g:V → V0 of vector spaces that induces a bijection Φ→ Φ0 and such that for all α, β∈Φ

2hg(β), g(α)i

hg(α), g(α)i = 2hβ, αi hα, αi.

(it is not assumed to be an isometry). If (Φ, V),(Φ0, V0) are two root systems, the direct sum is the pair (Φ`

Φ0, V ⊕V0). The root system (Φ, V) is irreducibleif it is not the direct sum of two non-trivial root systems.

Exercise 1.16. Show that there is a unique root system of rank 1, up to isomorphism.


The main example of a root system is that associated to a pairh⊂gof a Cartan subalgebra contained in a semisimple Lie algebra. We do not prove the following theorem:

Theorem 1.17. Let g be a semisimple Lie algebra, Aut(g) the subgroup of GL(g) preserving the Lie algebra structure. Then any two Cartan subalge- bras of g are conjugate under Aut(g).

In particular, the root system attached to the pair (g,h), where h is any Cartan subalgebra, depends only ong.

In the next section we classify irreducible root systems of rank 2. In the last section we will sketch the proof of the theorem that every irreducible root system comes from a simple Lie algebra.


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