Let (V,Φ) be a root system

Contents

1. Day 7

• (a) Weyl group

• (b) Properties of root systems

• (c) Simple roots and Weyl chambers

1.1. Weyl group. Let (V,Φ) be a root system. The subgroupW ⊂GL(V) generated by the reflections s_α, for α ∈ Φ, permutes Φ, by definition. It follows thatW is finite (a subgroup of the symmetric group).

Lemma 1.1. Let Φbe a finite set spanning a Eucldean spaceV. SupposeΦ is invariant under the reflection s_α for all α ∈Φ. Let g ∈GL(V) stabilize Φ; assume g fixes some hyperplane P ⊂V and takes some α in Φ to −α.

Then g=sα and P =Pα is the hyperplane orthogonal to α.

Proof. Note that s²_α = 1. Letτ =gsα =gs⁻¹_α . Thus τ(Φ) = Φ, τ(α) = α, and τ acts as the identity on Rα as well as on the quotient V /Rα, since P ∩Rα = 0. Thus all the eigenvalues of τ equal 1, and so its minimal polynomial divides (X −1)^`. On the other hand, τ fixes the finite set Φ, so for β ∈Φ, some power τ^N(β) =β. We may assume N sufficiently large that τ^N(β) = β for all β ∈ Φ. Since Φ spans V, τ^N = 1; so its minimal

polynomial is in factX−1, i.e. τ = 1.

Lemma 1.2. Any element g ∈ GL(V) that leaves Φ invariant is an auto- morphism of the root system: in other words, for all β, α,

2hg(β), g(α)i

hg(α), g(α)i = 2hβ, αi hα, αi.

In particular, W is a group of automorphisms of the root system.

Proof. Letα, β ∈Φ. Then gs_αg⁻¹(g(β)) =gs_α(β) ∈Φ becauses_α(β) ∈Φ.

We rewrite the right-hand expression gsαg⁻¹(g(β)) =g(β)−2hβ, αi

hα, αiα=g(β)−2hβ, αi hα, αig(α).

Thus gsαg⁻¹ takes every element of Φ (sinceg(β) runs over all elements of Φ) to an element of Φ; i.e. gs_αg⁻¹ leaves Φ invariant. Moreover, it fixes the hyperplane orthogonal to g(α) pointwise and takes g(α) to −g(α); it

1

thus follows from the above lemma that it is the reflection in the hyperplane orthogonal to g(α) i.e.

gsαg⁻¹(g(β)) =g(β)−2hg(β), g(α)i hg(α), g(α)ig(α).

Comparing the two expressions, we find the equality 2hg(β), g(α)i

hg(α), g(α)i = 2hβ, αi hα, αi.

1.2. Properties of root systems. Let α, β ∈ Φ and suppose they are neither proportional nor orthogonal. By Cauchy-Schwartz, we know that

0< hα, βi²

hα, αihβ, βi <1;α^∨(β) := 2hα, βi hα, αi ∈Z and similarly β^∨(α)∈Z. Thus Cauchy-Schwartz gives

0≤α^∨(β)β^∨(α)≤4.

Up to replacing α by −α, we may assume hα, βi < 0, and without loss of generality we may assume hα, αi ≤ hβ, βi. Thus the term in the middle is a product of negative integers. So there are three possibilities.

(i) α^∨(β) =β^∨(α) =−1;hα, αi=hβ, βi.

(ii) α^∨(β) =−2, β^∨(α) =−1; 2hα, αi=hβ, βi.

(iii) α^∨(β) =−3, β^∨(α) =−1; ; 3hα, αi=hβ, βi.

In other words, in the plane spanned by α andβ, using the formula for the cosine, we see the angle between them is either ^2π₃ , ^3π₄ , or ^5π₆ .

In particular, in any irreducible root system of rank 2, there are roots α, β such that β^∨(α) = +1 (replaceβ by −β. Now

sα(β) =β−β^∨(α)α =β−α soα−β and β−α are roots. It follows that

Lemma 1.3. Let α and β be nonproportional roots in Φ. If hα, βi < 0 (resp. hα, βi>0) thenα+β ∈Φ (resp. α−β ∈Φ).

Moreover, theα-string through β, defined to be the set of all roots of the form β+iα∈Φ, with i∈Z, is unbroken – if β+iα∈Φ and β+jα ∈Φ then β+kα∈Φ for allk between i and j. The set of such i has at most 4 elements.

Proof. The first step follows as before: ifhα, βi>0 we may assumehα, βi= 1 and then α−β is a root; but then so is β−α. Replacing β by −β gives the result if hα, βi<0.

Now letr, q∈Nbe the smallest integers such that theα-string throughβ is contained in [β−rα, β+qα]. Suppose k∈[−r, q] and β+kα /∈Φ. Then we can find p < sin the interval such that

β+pα∈Φ, β+ (p+ 1)α /∈Φ, β+sα∈Φ, β+ (s−1)α /∈Φ.

Apply the lemma: since (β+pα) +α /∈Φ (resp. (β+sα)−α /∈Φ ) it follows that

hα, β+pαi ≥0;hα, β+sαi ≤0.

Thus 0≤ hα,(p−s)αi= (p−s)hα, αi<0 which is absurd. So the string is unbroken.

Nows_αeither adds or subtracts a multiple ofαto any root, and therefore fixes every α-string. Because it is a reflection that reverses the α-axis, it reverses the string. So

s_α(β+qα) =β−rα.

But the left hand side is

β−β^∨(α)α−qα

sor−q=β^∨(α)≤3. Thus the string has at most 4 elements.

We now classify the four possible root systems of rank 2. For this purpose, we chooseα, β above, with||α|| ≤ ||β||,hα, βi<0, and such that the angle betweenαandβis the largest possible between ^π₂ andπbutβ 6=−α. Orient the ambient space so that α is the coordinate vector on the x axis. In all cases, the smallest angle between any pair of roots is ^π₆. In particular, one cannot have both ^π₃ and ^π₄.

1.1. Root system A1×A1. Ifα and β are orthogonal, this means by max- imality that there is no root in the second quadrant (with angle strictly between ^π₂ and π). Then the only roots possible are on the axes, with β a multiple of the coordinate vector on the y axis. Up to scaling the inner product, we may assume ||β|| = ||α|| = 1; the four roots are ±α,±β, and this is clearly the product of two copies of the same rank 1 root system, which we call A1.

1.2. Root systemA2.

α^∨(β) =β^∨(α) =−1;hα, αi=hβ, βi

Now the angle is ^2π₃ , which implies by the above computation that||β||=

||α|| = 1. Again, there is no angle between ^2π₃ and π, which means the smallest angle between any two roots is ^π₃; otherwise, by applying simple reflections. On the other hand, we can add−αand −β, ands_α(β) =α+β, which adds −α−β. These are the six vertices of a regular hexagon. We obtain the root system ofsl(3). Indeed, dimsl(3) = 8, with diagonal Cartan subgroup of rank 2, which leaves 6 root spaces.

1.3. Root systemB2.

α^∨(β) =−2, β^∨(α) =−1; 2hα, αi=hβ, βi

The angle is now ^3π₄ , and ||β||² = 2||α||². The α-string through β has length 3, and one gets a total of eight roots. This is the Lie algebra ofso(5), which is isomorphic to the Lie algebra of sp(4).

1.4. Root systemG2.

α^∨(β) =−3, β^∨(α) =−1; ; 3hα, αi=hβ, βi.

Finally, the largest possible angle, namely ^5π₆ , and ||β||² = 3||α||². Then theα-string throughβhas length 4. All vertices of a dodecagon are occupied, but the roots are of two lengths. In other words, there are two hexagons, of different diameters, at an angle of ^π₆ relative to one another. This is the non-classical Lie algebra G2.

1.3. Simple roots and Weyl chambers. Let (Φ, V) be a root system.

Choose a linear formf ∈V^∗=Hom(V,Q) that is orthogonal to none of the roots. Then f partitions Φ into Φ⁺`

Φ⁻ where Φ⁺ ={α ∈Φ |f(α)>0}, Φ⁻ = −Φ⁺. Say α ∈ Φ⁺ is simple (relative to f) if it is not of the form β1+β2 withβi∈Φ⁺.

Proposition 1.4. Let∆be the set of simple roots relative tof. Then every α ∈ Φ⁺ is a sum of elements of ∆ with non-negative integer coefficients.

Moreover, ∆is a basis of V.

Proof. First, suppose there is α ∈ Φ⁺ that is not a positive linear combi- nation of elements of ∆. We may assume f(α) is minimal for α with this property. By hypothesis, α is not simple, henceα =β1+β2 withβi ∈Φ⁺. So f(α) =f(β₁) +f(β₂) and each of the summands is positive. This con- tradicts minimality.

Now the first property implies that ∆ contains a basis. We show linear independence. Note that, forα6=β ∈∆ we havehα, βi ≤0. If not, we have α−β ∈Φ. If α−β ∈Φ⁺ thenα =α−β+β is not simple; ifβ−α∈Φ⁺ thenβ is not simple.

SupposeP

α∈∆λαα= 0. Let ∆⁺={α |λα≥0, ∆⁻ = ∆\∆⁺. We may write the relation

X

α∈∆⁺

λ_αα= X

β∈∆⁻

λ_ββ =v, say. Then

||v||² =hv, vi= X

α,α⁰∈∆⁺

λαλ_α⁰hα, α⁰i ≤0

because each λα ≥0 and each hα, α⁰i ≤0. But thenv = 0. So 0 =f(v) = P

α∈∆⁺λ_αf(α) is a sum of non-negative terms; thus eachλ_α = 0. Similarly,

each λ_β = for β∈∆⁻.

Definition 1.5. A basis of Φ is a set of simple roots relative to some f ∈ V^∗\ ∪_αVα where Vα is the hyperplane vanishing on α.

InV_R, the complement of a finite set of hyperplanes through the origin is a finite union of open polyhedral cones. In this case, each set of simple roots determines a collection of dimV hyperplanesVα, each of which dividesV^∗\ V_α into theV_α⁺ ={f |f(α)>0} and V_α⁻. So the set of f corresponding to the basis ∆ is precisely the intersection∩_α∈∆V_α⁺. This is theWeyl chamber corresponding to ∆.

Given a basis ∆ of Φ, the set of simple reflexions (relative to ∆) is the set ofs_α, α∈∆.

Proposition 1.6. Every root is the image of a simple root by a product of simple reflexions. The Weyl group is generated by the set of simple reflex- ions.

The proof is based on the following basic lemma.

Lemma 1.7. Let α∈∆. Then s_α(Φ⁺\ {α}) = Φ⁺\ {α}.

Proof. Letγ ∈Φ⁺,γ 6=α. Writeγ =P

β∈∆n_ββ with each n_β ≥0, and at least onenβ >0 for β 6=α

σ_α(γ) =X

β∈∆

n_βσ_α(β) = X

β∈∆

n_ββ−uα

where u = P

β∈∆2n_β^hβ,αi_hα,αi = 2_hα,αi^hγ,αi. In particular, the coefficient of β doesn’t change for anyβ 6=α. Soσα(γ) has at least one positive coefficient, and is therefore not in Φ⁻; hence it must be in Φ⁺. Finally, σ_α(γ) 6= α

because γ6=−α.

Proof of Proposition. Writeγ >0 ifγ ∈Φ⁺,si =sαi,i= 1. . . , `for a basis α. Suppose we know the proposition for γ >0, say γ =Q

is_i(β) for some simpleβ. Then −γ =Q

isi(−β) =Q

isisβ(β).

Thus we may assume γ > 0, say γ = P

n_iα_i. Let h(γ) = P

n_i > 0 and induct onh(γ). Ifh(γ) = 1 thenγ ∈∆ and there is nothing to prove.

Suppose h(γ) > 1. Note that 0 < ||γ||² = P

inihγ, α_ii, so hγ, α_ii > 0 for somei. Thens_i(γ)>0 by the previous lemma, but

si(γ) =X

j6=i

njαj+ (ni−2hγ, α_ii hα_i, αiiαi

and h(si(γ))< h(γ). Thus we conclude by induction.

By definition, W is generated by the s_γ for γ ∈ Φ. By the first part of the proposition, we may supposeγ =w(α) withα∈∆ andw a product of simple reflections. Thens_γ =ws_αw⁻¹is also a product of simple reflections.

We draw a corollary from the Lemma. Letδ = ¹₂P

α∈Φ⁺α. This depends on the choice of basis.

Corollary 1.8. For any simple α∈∆, sα(δ) =δ−α.

Theorem 1.9. The Weyl group acts simply transitively on the Weyl cham- bers, or equivalently on the bases of Φ.

Proof. Clearly the Weyl group permutes bases of Φ. We show the action is transitive and then simply transitive.

Fix a basis ∆. For transitivity, it suffices to show that, for any element f ∈V^∗\ ∪_αVα, there is aσ∈W such that hσ(f), αi>0 for allα∈∆. This implies thatW acts transitively on the Weyl chambers. Chooseσ ∈W such thathσ(f), δi>0is as big as possible. Ifα∈∆, then by hypothesis

hσ(f), δi ≥ hσ_ασ(f), δi=hσ(f), σα(δ)i=hσ(f), δi − hσ(f), αi.

Thus hσ(f), αi ≥0 for all α ∈ ∆. Since f is not orthogonal to anyα, and sinceW permutes theα-hyperplanes, we must in fact havehσ(f), αi>0 for all α. Thus σ(f) is in the Weyl chamber corresponding to ∆.

Now to prove that the action is simply transitive, we supposew∈W,w6=

1, withw=s1· · · · ·sN, a product of simple reflections withN minimal. We show thatw(α_N)∈Φ⁻, which implies thatwdoes not fix ∆. Suppose not, so w(α_N)>0. Letjbe the smallest integerisuch thatβ_i =s_i·· · ··s_N(α_N)>0.

SincesN(αN)<0, we havej < N; thusβj+1 is well-defined and an element of Φ⁻. Moreover, s_j(β_j) = β_j+1 < 0, and since β_j > 0 the lemma implies thatβj =αj. Hence

s_j =s_βj = (s_j· · · · ·s_N)s_N(s_j· · · · ·s_N)⁻¹; in other words

s_j(s_j · · · · ·s_N) = (s_j· · · · ·s_N)s_N;s_j+1· · · · ·s_N =s_j· · · · ·sN−1

after cancellation. This means

w=s₁· · · · ·s_js_j· · · · ·sN−1=s₁· · · · ·sj−1s_j+1· · · · ·sN−1

has a factorization of length N −2, which contradicts the minimality of

N.

2. Day 8

• (a) Cartan matrices

• (b) Dynkin diagrams

• (c) The Lie algebra of a root system

• (d) Proof of Serre’s theorem (sketch)

2.1. Cartan matrices. The classification of semisimple Lie algebras is based on the following proposition:

Proposition 2.1. Letg⊃hand g⁰ ⊃h⁰ be pairs consisting of a semisimple Lie algebra and a Cartan subalgebra. LetΦandΦ⁰ be the corresponding root systems, ∆ ⊂ Φ, ∆⁰ ⊂ Φ⁰ sets of simple roots, and for each α ∈ ∆ (resp.

α⁰ ∈ ∆⁰) choose a generator Eα ∈ g^α (resp. E_α⁰ ∈ g^0,α0. Suppose there is an isomorphism of root systems f : Φ → Φ⁰ such that f(∆) = ∆⁰. Then there is a unique isomorphism of Lie algebrasr:g → g⁰ takinghtoh⁰, with r(E_α) =E_α⁰, r(H_α) =H_α⁰.

We return to the proof if we have time. For the moment, we see that the classification of semisimple Lie algebras is reduced to two steps: (a) classification of irreducible root systems; and (b) determination of those irreducible root systems that correspond to Lie algebras. The answer to (b) is simple (though not simple to prove): all irreducible root systems come from Lie algebras.

Step (a) is carried out by rewriting the axioms of root systems in combi- natorial form.

Definition 2.2. Let ∆ be a finite set. A Cartan matrix for ∆ is a map c: ∆×∆→ Z such that

(C1) For all α, β∈∆,c(α, α) = 2;c(α, β)∈ {0,−1,−2,−3} ifα6=β;

(C2) c(α, β) = 0⇔c(β, α) = 0.

(C3) The quadratic form Q(x) = X

α∈∆

x²_α−X

α6=β

pc(α, β)·c(β, α)xαxβ

on the spaceR^∆ is positive-definite.

The Cartan matrices c and c⁰ for ∆ and ∆⁰ are equivalent if there is a bijection ∆ → ∆⁰ compatible with the constants cand c⁰.

Lemma 2.3. Let Φbe a root system.

(a) For any choice ∆ of simple roots of Φ, the function c(α, β) =α^∨(β) is a Cartan matrix C(Φ,∆).

(b) If ∆⁰ is another choice, then C(Φ,∆) and C(Φ,∆⁰) are equivalent.

(c) The matrix C(Φ,∆) determinesΦ up to isomorphism.

(d) Finally, Φ is reducible if and only if C(Φ,∆) is a sum of (at least) two diagonal blocks.

Proof. (a) Property (C1) was shown in the previous lecture (Day 7). Since α^∨(β) = 2_hα,αi^hα,βi, property (C2) is clear. (C3) is equivalent to the positive- definiteness of h•,•i. More precisely, if we writey=P

α∈∆xα

α α∈R^∆, then one computes that

pc(α, β)p

c(β, α) =−2 hα, βi phα, αihβ, βi which easily implies that

Q(x) =hy, yi.

(b) There is a unique w ∈ W such that w(∆) = ∆⁰, and because the Killing form is invariant, it associates the Cartan integers:

c(w(α), w(β)) = 2hw(α), w(β)i

hw(α), w(α)i = 2hα, βi

hα, αi =c(α, β).

(c) We know that the set of Weyl group translates of ∆ equals Φ, and for each α ∈ ∆, sα(β) = β−c(α, β)α is determined by C(Φ,∆). So Φ is recovered from ∆ and c.

(d) One direction is clear. SupposeC is the sum of two diagonal blocks.

Thus ∆ = ∆₁`

∆₂ withc(α, β) = 0 ifα∈∆₁,β∈∆₂. Note that ifα∈∆₁ then sα acts trivially on the span of ∆2. Thus W =W1×W2 where Wi is generated by thes_αwithα∈∆_i,i= 1,2. It follows that Φ = Φ₁`

Φ₂ with Φ_i=W_i(∆_i), and from there the result is easy.

Example 2.4. If g = sl(n), we can take ∆ = {α₁, . . . , αn−1 with α_i = ei−ei+1 and the Euclidean norm. Then hα_i, αji = 2 if i = j and = −1 otherwise.

2.2. Dynkin diagrams.

Lemma 2.5. Let C= (c(α, β))be a Cartan matrix on the finite set ∆. Let

∆⁰ ⊂ ∆ and let C⁰ be the corresponding submatrix. Then C⁰ is a Cartan matrix.

If c(α, β)6= 0, then either c(α, β) =−1 or c(β, α) =−1.

If C⁰ is any matrix on ∆ such that c⁰(α, β) ≥c(α, β), and satisfies (C1) and (C2), thenC⁰ is again a Cartan matrix.

Proof. The first condition is clear (a positive definite quadratic form re- stricted to a subspace remains positive-definite.). We apply this to the subset ∆⁰ = {α, β}: we see that Q⁰(x, y) = x²−p

c(α, β)·c(β, α)xy+y² has discriminant c(α, β)c(β, α)−4. In order for Q⁰ to be positive definite, we need c(α, β)c(β, α)−4 < 0, which in view of (C2), implies the second condition.

Finally, for x =P

xαα ∈ R^∆, let |x| =P

|x_α|α. Then Q⁰(x) ≥ Q⁰(|x|) because this is true for each term in the sum−P

α6=β

pc(α, β)·c(β, α)xαxβ. Similarly, Q⁰(|x|)≥Q(|x|) becausec⁰(α, β)≥c(α, β) implies

−p

c⁰(α, β)·c⁰(β, α)≤ −p

c(α, β)·c(β, α)

(pay attention to signs!) SoC⁰ again satisfies (C3).

To each Cartan matrixC we associate the Dynkin diagram: it is a (par- tially) oriented graph whose vertices are ∆ and such that the numbern(α, β) of edges betweenαand β is given byc(α, β)c(β, α), with an arrow pointing from the shorter to the longer root if the number of edges is >1. In other words, α→ β ifn(α, β)>1 and c(α, β) =−1.

Conversely, suppose Γ is a finite graph, withn(α, β) edges betweenα and β, and an orientation if n(α, β) > 1. Say it is a Dynkin diagram if the quadratic form

Q(x) =X

x²_α−X

α6=β

pn(α, β)xαx_β

is positive-definite. We associate to Γ its set ∆ of vertices and a matrix C=c(α, β) by writingc(α, β) =−n(α, β) ifn(α, β)<2 orβ → α; otherwise c(α, β) = −1. Then C is a Cartan matrix (and indeed n(α, β) < 4 for all α, β, by definiteness).

It is easy to see that we have a bijection between Cartan matrices and Dynkin diagrams. Under this bijection, irreducible Cartan matrices cor- respond to connected Dynkin diagrams. The properties of the preceding lemma will be used to eliminate candidate diagrams.

Theorem 2.6. Every irreducible Dynkin diagram belongs to the following list:

Here n≥1 for An,n≥2 for Bn and Cn, and n≥3 for Dn.

Proof. The idea of the proof is that the positivity of the Cartan matrix excludes certain configurations. Let ∆ = {α_i, i = 1, . . . , `} and let e_i =

αi/||α_i|| be the corresponding unit vector. Recall that the ratio of lengths (||α_i||/||α_j||)² is always 1, 2, or 3 (or the reciprocal). Moreover, ifi6=jthen he_i, eji ≤0 and

4he_i, e_ji²= 4 hα_i, αji

hα_i, α_iihα_j, α_ji =α^∨_i(α_j)α^∨_j(α_i) = 0,1,2,3 is the number of edges connectingei toej.

We start with the following observation.

(1) The number of pairs of vertices in Γ connected by at least one edge is strictly less than `. Indeed, let a=a(Γ) be this number, and lete=P

e_i. Since they are linearly independent, e6= 0, so

0<he, ei=`+ 2X

i<j

he_i, e_ji

Ifhe_i, e_ji 6= 0 thenhe_i, e_ji² = ¹₄,¹₂,³₄ which together with negativity implies 2he_i, eji ≤ −1. Thus

0< `+ 2X

i<j

he_i, e_ji ≤`−a(Γ)

which implies` > a(Γ).

In particular, Γ contains no cycles; otherwise, we could replace Γ by the subgraph Γ⁰ of `<sup>0</sup> vertices of the cycle, for which a(Γ<sup>0</sup>)≥`⁰.

(2) No vertex lies on more than three edges. Fix e= ei and assumevi, i = 1, . . . k, are the vertices attached to e; in particular he, v_ji < 0. Since there are no cycles, thev_i are not linked, sohv_i, v_ji= 0 ifi6=j. LetV be the space spanned by e and the vi, and letv0 ∈V be a unit vector orthogonal to all thev_i; thushe, vi 6= 0, and{v₀, v₁, . . . , v_k} is an orthonormal basis for V. It follows that

1 =he, ei=he,

k

X

i=0

he, v_iiv_ii=

k

X

0

(he, v_ii)². Thus Pk

i(he, v_ii)² < 1 (strict inequality), or again Pk

i 4(he, v_ii)² <4. But 4(he, v_ii)² is the number of edges connecting etovi.

In particular, G₂ is the only connected graph with a triple edge, and no graph has two double edges coming from a single vertex.

(3) Let Γ⊃Γ^∗where Γ^∗is the graph on a chain of verticesu₁, . . . , u_k, with each ui connected to ui+1 by a single edge. Let Γ⁰ be the graph obtained from Γ by shrinking Γ^∗ to a point and replacing{u₁, . . . , u_k} by u=P

ui. Then Γ⁰ is again a Dynkin diagram.

Let the vertices of Γ be e1, . . . , e_b;u1, . . . , u_k, ` = b+k; so the vertices of Γ<sup>0</sup> are e<sub>1</sub>, . . . , e<sub>b</sub>, u, which are vectors in the subspace they span in R<sup>`</sup>. Obviously they are linearly independent. Moreover, 2hu_i, ui+ii = −1 for i= 1, . . . , k−1; so

hu, ui=k+ 2X

i<j

hu_i, uji=k−(k−1) = 1.

Thusuis again a unit vector. Since there are no cycles, any ej is connected to at most oneui, and for thisiwe have 4(hu, e_ji)²= 4(hu_i, ej)²i= 0,1,2,3.

Combining (2) and (3), we can eliminate the following diagrams:

In particular, no connected diagram has more than one double edge.

It remains to show that the only diagrams containing a double edge are of the typeB_n=C_norF₄, and that the only diagrams with a branch point are of the type Dn or En with n = 6,7,8 (E5 = D5). In other words, we need to eliminate the four following diagrams:

For each of these we can exhibit a nonzero vector v₀ of length zero, which shows that the quadratic form is not positive-definite. For ˜F5 v0 = (1,2,3,2√

2,√

2) (in the order of the vertices from left to right). For the ˜En

we can use the following vectors (with the graphs labelled):

but it is more traditional to argue as follows. Let the branch node be labelled x and let the three chains leaving x be labelled u₁, . . . , up−1, v1, . . . , vq−1, w1, . . . , wr−1. We need to bound the lengths p, q, r relative to each other. Let u =P

iu_i, v =P

jv_j,w =P

kw_k. They are mutually orthogonal and linearly independent andxis not in their span. We compute

(hu, u/i) =

p−1

X

i=1

i²−

p−2

X

i=1

i(i+ 1) =p(p−1)/2

because all the edges are single. Similarly (hv, v/i) =q(q−1)/2, (hw, w/i) = r(r−1)/2.

Let θ1 be the angle between x and u, θ2 between x and v,θ3 between x and w. Since hx, xi= 1 it follows, as in the verification of (2), that

1>(hx, u/||u||i)²+ (hx, v/||v||i)²+ (hx, w/||w||i)² =

3

X

1

cos(θi)².

We compute

cos(θ₁)²= hx, u/i²

hu, u/ihx, x/i = (p−1)²hx, u_p−1i² hu, u/i

because all the other hx, u_ii = 0. Since x is connected to up−1 by a single edge, we have hx, u_p−1i=−¹₂ and so

cos(θ₁)²= (p−1)² 1/4

p(p−1)/2 = p−1 2p = 1

2(1−1 p) Similarly

cos(θ2)² = 1 2(1−1

q); cos(θ3)² = 1 2(1−1

r).

Thus we have 1 2(3−1

p −1 q −1

r)<1⇒ 1 p+1

q +1 r >1.

We may assumep≥q ≥r≥2; if any of them equals 1, then the diagram is an An. Then ³₂ ≥ ³_r ≥ ¹_p + ¹_q +¹_r >1, hencer = 2. Thus ¹_p +¹_q > ¹₂, hence

2

q > ¹₂, so 2≤q <4. If q = 3 then ¹_p > ¹₆ and sop= 3,4,5. So the possible triples (p, q, r) are of the form

(p,2,2) =D_p+2; (3,3,2) =E₆; (4,3,2) =E₇; (5,3,2) =E₈.