Exposé XXI. Root data

by M. Demazure

¹ This Exposé collects, in the absence of a suitable reference,² some known results on root data (= "abstract" root systems), most of which will be used in what follows.

Notations. — We denote by $Q_{+}$ the set of positive (or zero) rational numbers; one has $\mathbb{Z}\cap Q_{+}=\mathbb{N}$. Let $V$ be a $\mathbb{Q}$-vector space; if $A$ (resp. $B$) is a subset of $\mathbb{Q}$ (resp. $V$), we denote by $A\cdot B$ the image of $A\otimes B$ under the morphism $\mathbb{Q}\otimes V\to V$, in other words the set of linear combinations of elements of $B$ with coefficients in $A$. We write $-B=-1\cdot B$. We denote by $E-F$ the set of elements of $E$ that do not belong to $F$.

1. Generalities

1.1. Definitions, first properties

Definition 1.1.1. Let $M$ and $M\ast$ be two free $\mathbb{Z}$-modules of finite type in duality. Write $V=M\otimes \mathbb{Q}$, $V\ast =M\ast \otimes \mathbb{Q}$; these are two $\mathbb{Q}$-vector spaces in duality. We identify $M$ (resp. $M\ast$) with a subset of $V$ (resp. $V\ast$). The canonical bilinear form on $M\ast \times M$ (resp. $V\ast \times V$) is denoted $(,)$.

Let $R$ be a finite subset of $M$. Suppose we are given a map $\alpha \mapsto \alpha \ast$ of $R$ into $M\ast$; the set of $\alpha \ast$, for $\alpha \in R$, is denoted $R\ast$. To each $\alpha \in R$ we associate the endomorphism $s_{\alpha }$ (resp. $s{\ast }_{\alpha }$) of $M$ and $V$ (resp. $M\ast$ and $V\ast$) given by the formulas:

(1)    s_α(x) = x − (α*, x)α,    i.e.   s_α = id − α* ⊗ α;
(1*)   s*_α(u) = u − (u, α)α*,   i.e.   s*_α = id − α ⊗ α*.

One says that the pair $(R,R\ast )$ (more precisely the pair $(R,R\to M\ast )$) is a root datum in $(M,M\ast )$, or that $(M,M\ast ,R,R\ast )$ is a root datum, if the following axioms are satisfied:

(DR I)   For each α ∈ R, one has    (α*, α) = 2.
(DR II)  For each α ∈ R, one has    s_α(R) ⊂ R,    s*_α(R*) ⊂ R*.

One says that $R$ is the root system of the root datum $\mathcal{R}=(M,M\ast ,R,R\ast )$. The elements of $R$ (resp. $R\ast$) are called the roots (resp. coroots*) of the root datum.*

Remark 1.1.2. Axiom (DR I) is equivalent to any of the following properties:

(2)   s_α s_α = id,        (2*)  s*_α s*_α = id,
(3)   s_α(α) = −α,         (3*)  s*_α(α*) = −α*.

Remark 1.1.3. Axioms (DR I) and (DR II) entail

R = −R,   R* = −R*,   0 ∉ R,   0 ∉ R*.

Lemma 1.1.4. The map $R\to R\ast$ is a bijection. More generally, if $\alpha ,\beta \in R$ and $(\alpha \ast ,x)=(\beta \ast ,x)$ for all $x\in R$, then $\alpha =\beta$.

Proof. Indeed, one then has $s_{\beta }(\alpha )=\alpha -2\beta$, $s_{\alpha }(\beta )=\beta -2\alpha$. One deduces at once

s_β s_α(α) = 2β − α = α + 2(β − α),    s_β s_α(β − α) = s_β(β − α) = β − α,

whence $(s_{\beta }{s}_{\alpha }{)}^n(\alpha )=\alpha +2n(\beta -\alpha )\in R$ by (DR II). Since $R$ is finite, one has $\beta -\alpha =0$.

Corollary 1.1.5. The inverse map $R\ast \to R$ defines a root datum

ℛ* = (M*, M, R*, R)

called the dual of $\mathcal{R}$.³

Definition 1.1.6. We denote by ${\Gamma }_0(R)$ the subgroup of $M$ generated by $R$. We denote by $V(R)$ the vector subspace of $V$ generated by $R$, that is to say ${\Gamma }_0(R)\otimes \mathbb{Q}$. Applying these definitions to $\mathcal{R}\ast$, one constructs similarly ${\Gamma }_0(R\ast )$ and $V(R\ast )$.

One calls reductive rank of $\mathcal{R}$ the number

rgred(ℛ) = rank(M) = dim(V) = dim(V*) = rank(M*) = rgred(ℛ*).

One calls semisimple rank of $\mathcal{R}$ the number

rgss(ℛ) = rank(R) = rank(Γ₀(R)) = dim(V(R)).

One thus has $rgss(\mathcal{R})⩽rgred(\mathcal{R})$.

We shall see below that $rgss(\mathcal{R})=rgss(\mathcal{R}\ast )$, that is to say that $V(R)$ and $V(R\ast )$ have the same dimension.

Definition 1.1.7. One says that $\mathcal{R}$ is semisimple (resp. trivial*) if $rgss(\mathcal{R})=rgred(\mathcal{R})$ (resp. $rgss(\mathcal{R})=0$). For $\mathcal{R}$ to be trivial it is therefore necessary and sufficient that $R$ be empty. The trivial root datum of reductive rank zero is denoted $0=(0,0,\varnothing ,\varnothing )$.*

Definition 1.1.8. We denote by $W(\mathcal{R})$ the group of transformations of $M$ generated by the $s_{\alpha }$, $\alpha \in R$. We call it the Weyl group of $\mathcal{R}$. We write

$$W\ast (\mathcal{R})=W(\mathcal{R}\ast ).$$

Then $W(\mathcal{R})$ operates on $R$, ${\Gamma }_0(R)$, $V(R)$, $M$ and $V$. If $w\in W(\mathcal{R})$ and $x\in M$ (resp. $x\in V$), one has $wx-x\in {\Gamma }_0(R)$ (resp. $wx-x\in V(R)$); this is immediate from formula (1). Likewise for $W\ast (\mathcal{R})$.

Lemma 1.1.9. For all $\alpha \in R$, $x\in V$, $u\in V\ast$, one has

(4)    (s*_α(u), s_α(x)) = (u, x).

Proof. Indeed, taking the scalar product of (1) and (1*), one finds that the left-hand side equals $(u,x)+(u,\alpha )(\alpha \ast ,x)((\alpha \ast ,\alpha )-2)=(u,x)$.

Remark 1.1.10. If one assumes $0\notin R$ and $0\notin R\ast$, then 1.1.9 is equivalent to (DR I).

Corollary 1.1.11. Denote by $h\mapsto h^{\vee }$ the isomorphism of $GL(M)$ onto $GL(M\ast )$ that associates to $h$ its contragredient. Then formula (4) is also written

$$(5)(s_{\alpha }{)}^{\vee }=s{\ast }_{\alpha }.$$

Corollary 1.1.12. The preceding isomorphism induces an isomorphism of $W(\mathcal{R})$ onto $W\ast (\mathcal{R})$.

Scholie 1.1.13. By virtue of the preceding result, we shall identify $W$ and $W\ast$, and we shall regard $W$ as a group of transformations of $R$, $R\ast$, $M$, $M\ast$, ${\Gamma }_0(R)$, ${\Gamma }_0(R\ast )$, $V$, $V\ast$, $V(R)$, $V(R\ast )$. We shall write $s_{\alpha }$ for $s{\ast }_{\alpha }$.

1.2. The map $p$

Lemma 1.2.1. Let $p:M\to M\ast$ (resp. $V\to V\ast$) be the linear map defined by

(6)    p(x) = Σ_{u ∈ R*} (u, x) u.

Write $\ell (x)=(p(x),x)$. One has the following properties:

(7)    ℓ(x) ⩾ 0,    ℓ(α) > 0    for α ∈ R.

(8)    ℓ(w x) = ℓ(x),    for w ∈ W.

(9)    (p(x), y) = (p(y), x),    for x, y ∈ V.

(10)   ℓ(α) α* = 2 p(α),    for α ∈ R.

Proof. The first three relations are immediate.⁴ Let us prove the last. By (1), for $u\in V\ast$:

(u, α)² α* = (u, α) u − (u, α) s_α(u)
           = (u, α) u + (u, s_α(α)) s_α(u)
           = (u, α) u + (s_α(u), α) s_α(u)    (since s_α² = id).

Since $s_{\alpha }$ is a permutation of $R\ast$ (by (DR II)), it remains only to sum over $u\in R\ast$ to conclude.

Scholie 1.2.2. Relation (10) says that $\alpha \mapsto \ell (\alpha )\alpha \ast$ is the restriction to $R$ of a linear map from $M$ into $M\ast$. In particular, one has

$$(-\alpha )\ast =-\alpha \ast .$$

Corollary 1.2.3. The map $p$ induces an isomorphism of $V(R)$ onto $V(R)\ast$.

Proof. Indeed, $p$ sends $V(R)$ into $V(R\ast )$. We thus have

$$\dim (V(R))⩾\dim (V(R\ast )).$$

Applying this inequality to the dual root datum, we deduce

$$\dim (V(R))=\dim (V(R\ast )),$$

so $p$, being surjective, is also bijective.

Corollary 1.2.4. One has $\dim (V(R))=\dim (V(R\ast ))$, hence

$$rgss(\mathcal{R})=rgss(\mathcal{R}\ast ).$$

Corollary 1.2.5. The bilinear form $(u,x)$ is nondegenerate on $V(R\ast )\times V(R)$, and so puts these $\mathbb{Q}$-vector spaces in duality.

Proof. Indeed, if $(u,x)=0$ for all $u\in R\ast$, then $p(x)=0$.

Corollary 1.2.6. The symmetric bilinear form (p(x), y) is positive nondegenerate on $V(R)$.

Corollary 1.2.7. $W$ operates faithfully on $R$ (and hence on the other sets of 1.1.13).

Proof. Indeed, let $u\in V\ast$. Let $w\in W$; suppose that $w(\alpha )=\alpha$ for all $\alpha \in R$, and let us prove that $w(u)=u$. One has

(w(u) − u, α) = (w(u), α) − (u, α) = (u, w⁻¹(α)) − (u, α) = 0.

But $w(u)-u\in V(R\ast )$. If it is orthogonal to all the roots, it is zero by 1.2.5.

Corollary 1.2.8. The group $W$ is finite.

Proposition 1.2.9. The operations of $W$ respect the correspondence between roots and coroots. In other words, for $\alpha \in R$ and $w\in W$, one has

$$w(\alpha \ast )=w(\alpha )\ast .$$

Proof. It suffices to verify this for $w=s_{\beta }$, $\beta \in R$, that is to say to verify the formula

$$s_{\beta }(\alpha \ast )=s_{\beta }(\alpha )\ast .$$

Now $\ell (s_{\beta }(\alpha ))\cdot s_{\beta }(\alpha )\ast /2$ equals:

p(s_β(α)) = Σ_{u ∈ R*} (u, s_β(α)) u = Σ_{u ∈ R*} (s_β(u), α) u = Σ_{u ∈ R*} (u, α) s_β(u) = s_β(p(α));

since $\ell (s_{\beta }(\alpha ))=\ell (\alpha )$, one obtains $s_{\beta }(\alpha )\ast =s_{\beta }(2p(\alpha )/\ell (\alpha ))=s_{\beta }(\alpha \ast )$.

Corollary 1.2.10. If $\alpha \in R$ and $w\in W$, one has

w s_α w⁻¹ = s_{w(α)}.

Proof. Indeed, $w{s}_{\alpha }{w}^{-1}(x)=x-(\alpha \ast ,w^{-1}(x))w(\alpha )=x-(w(\alpha \ast ),x)w(\alpha )$, and this last term equals, by 1.2.9:

x − (w(α)*, x) w(α) = s_{w(α)}(x).

Corollary 1.2.11. Let $R^{\prime }\subset R$. For $(M,M\ast ,R^{\prime },R^{\prime }\ast )$ to be a root datum, it is necessary and sufficient that $\alpha ,\beta \in R^{\prime }$ imply $s_{\alpha }(\beta )\in R^{\prime }$.

2. Relations between two roots

2.1. Proportional roots

Proposition 2.1.1. Let $\alpha$ and $\beta$ be two roots. The following conditions are equivalent:

(i) There exists $k\in \mathbb{Q}$ such that $\alpha =k\beta$.

(ii) $s_{\alpha }=s_{\beta }$.

Moreover, under these conditions one has $\alpha \ast =k^{-1}\beta \ast$, and $k$ is equal to one of the numbers 1, $-1$, 2, $-2$, 1/2, $-1/2$.

Proof. Suppose first (i). One has first

α* = ℓ(α)⁻¹ · 2 p(α) = k⁻² ℓ(β)⁻¹ · 2 k p(β) = k⁻¹ β*.

This entails at once $s_{\alpha }=s_{\beta }$. Conversely, if $s_{\alpha }=s_{\beta }$, then

α = s_α(−α) = s_β(−α) = (β*, α) β − α,

whence $2\alpha =(\beta \ast ,\alpha )\beta$, with $(\beta \ast ,\alpha )\in \mathbb{Z}$, so (ii) entails (i). Finally, if $\alpha =k\beta$, then $\alpha \ast =k^{-1}\beta \ast$, whence

(α*, β) = 2 k⁻¹,    (β*, α) = 2 k,

so 2k and $2k^{-1}$ are integers and we are done.

Application 2.1.2. The root data $\mathcal{R}$ with $rgss(\mathcal{R})=1$ are of one of the following two types:

(i) Type $A_1$: there exists $\alpha \in M$ such that the roots are $\alpha$ and $-\alpha$. The coroots are then $\alpha \ast$ and $-\alpha \ast$.

(ii) Type $A_1^{\prime }$: there exists $\alpha \in M$ such that the roots are $\alpha$, $-\alpha$, $2\alpha$, $-2\alpha$. The coroots are then $\alpha \ast$, $-\alpha \ast$, $\alpha \ast /2$, $-\alpha \ast /2$.

Definition 2.1.3. One says that $\alpha \in R$ is indivisible if $\alpha /2\notin R$. One says that $\mathcal{R}$ is reduced if every root is indivisible.

For $\mathcal{R}$ to be reduced it is necessary and sufficient that $\mathcal{R}\ast$ be so. If $\alpha$ is indivisible and if $w\in W$, then $w(\alpha )$ is indivisible.

Definition 2.1.4. Let $\alpha \in R$. If $\alpha$ is indivisible, set $ind(\alpha )=\alpha$. Otherwise, set $ind(\alpha )=\alpha /2$.

Corollary 2.1.5. If $\alpha \in R$ is indivisible and if $k\alpha \in R$, where $k\in \mathbb{Q}$, then $k\in \mathbb{Z}$.

Proposition 2.1.6. Let $\mathcal{R}$ be a root datum. Then

ind(ℛ) = (M, M*, ind(R), ind(R)*)

is a reduced root datum, and one has

$$W(ind(\mathcal{R}))=W(\mathcal{R}).$$

Proof. Indeed, $ind(\mathcal{R})$ is a root datum by 1.2.11, since the Weyl group permutes indivisible roots. The second assertion follows from 2.1.1.

Remark 2.1.7. If $\mathcal{R}$ is not reduced, one has $ind(R)\ast \ne ind(R\ast )$ and so $ind(\mathcal{R}\ast )\ne ind(\mathcal{R})\ast$.

2.2. Orthogonal roots

Lemma 2.2.1. Let $\alpha$ and $\beta$ be two roots. One has

(11)    ℓ(α) (α*, β) = ℓ(β) (β*, α).

Proof. This follows at once from 1.2.1, formulas (9) and (10).

Corollary 2.2.2. Let $\alpha ,\beta \in R$. The following conditions are equivalent:

(i) $(\alpha \ast ,\beta )=0$, (i bis) $(\beta \ast ,\alpha )=0$,

(ii) $(p(\alpha ),\beta )=0$,

(iii) $s_{\alpha }(\beta )=\beta$, (iii bis) $s_{\beta }(\alpha )=\alpha$,

(iv) $s_{\alpha }(\beta \ast )=\beta \ast$, (iv bis) $s_{\beta }(\alpha \ast )=\alpha \ast$,

(v) $s_{\alpha }\ne s_{\beta }$ and $s_{\alpha }$ and $s_{\beta }$ commute.

Proof. All the equivalences are immediate, except those involving (v). Let us show that (i) (and (i bis)) entail (v). One has

s_α s_β(x) = x − (β*, x) β − (α*, x) α + (β*, x) (α*, β) α.

If $(\alpha \ast ,\beta )=0$, then $(\beta \ast ,\alpha )=0$ and $s_{\alpha }{s}_{\beta }(x)=s_{\beta }{s}_{\alpha }(x)$.

Suppose conversely (v). One has

s_α = s_β s_α s_β = s_{s_β(α)}    (by 1.2.10).

By 2.1, there exists $k\in \mathbb{Q}$ such that $\alpha =k{s}_{\beta }(\alpha )=k\alpha -k(\beta \ast ,\alpha )\beta$. Since $s_{\alpha }\ne s_{\beta }$, $\beta$ and $\alpha$ are not proportional by 2.1.1, hence $(\beta \ast ,\alpha )=0$.

Definition 2.2.3. Two roots satisfying the equivalent conditions of 2.2.2 are called orthogonal.

Remark 2.2.4. The roots $\alpha$ and $\beta$ are orthogonal if and only if the coroots $\alpha \ast$ and $\beta \ast$ are orthogonal.

Lemma 2.2.5. If $\alpha$ and $\beta$ are two orthogonal roots, then $\alpha +\beta \in R$ if and only if $\alpha -\beta \in R$.

Proof. Indeed, $s_{\beta }(\alpha +\beta )=s_{\beta }(\alpha )+s_{\beta }(\beta )=\alpha -\beta$.

Lemma 2.2.6. Let $\alpha$ and $\beta$ be two non-orthogonal roots. If one defines $\ell (\gamma \ast )$ for a coroot $\gamma \ast$ as $\ell$ of the root $\gamma \ast$ of $\mathcal{R}\ast$, one has the relation

(12)    ℓ(α) ℓ(α*) = ℓ(β) ℓ(β*).

Proof. Indeed, multiplying formula (11) by the corresponding formula for $\mathcal{R}\ast$, and using the equality $(\gamma \ast )\ast =\gamma$ for all $\gamma \in R$, one finds:

(β*, α) (α*, β) ℓ(α) ℓ(α*) = (β*, α) (α*, β) ℓ(β) ℓ(β*).

2.3. General case

Proposition 2.3.1. If $\alpha$ and $\beta$ are any two roots, one has

$$(13)0⩽(\alpha \ast ,\beta )(\beta \ast ,\alpha )⩽4.$$

If $\alpha$ and $\beta$ are neither proportional nor orthogonal, one has

$$1⩽(\alpha \ast ,\beta )(\beta \ast ,\alpha )⩽3.$$

Proof. Indeed, one has $4(p(\alpha ),\beta {)}^2=\ell (\alpha )\ell (\beta )(\alpha \ast ,\beta )(\beta \ast ,\alpha )$. On the other hand, by 1.2.6 the symmetric bilinear form (p(x), y) is positive nondegenerate on $V(R)$, whence $(p(x),y{)}^2⩽\ell (x)\ell (y)$.⁵

Corollary 2.3.2. Let $\alpha$ and $\beta$ be two non-orthogonal roots. If $\ell (\alpha )=\ell (\beta )$, there exists $w\in W$ such that $w(\alpha )=\beta$.

Proof. Indeed, if $\alpha$ and $\beta$ are proportional, then since $\ell (\alpha )=\ell (\beta )$, one has $\alpha =\beta$ or $\alpha =-\beta$; in this case one takes $w=1$ or $w=s_{\alpha }$. If $\alpha$ and $\beta$ are neither proportional nor orthogonal, then by formula (11) and 2.3.1:

(β*, α) = (α*, β) = ±1.

If $(\beta \ast ,\alpha )=(\alpha \ast ,\beta )=1$, one takes $w=s_{\beta }{s}_{\alpha }{s}_{\beta }$. If $(\beta \ast ,\alpha )=(\alpha \ast ,\beta )=-1$, one takes $w=s_{\alpha }{s}_{\beta }$.

Corollary 2.3.3. If $\alpha$ and $\beta$ are two roots, if $\alpha \ne \beta$ and $(\beta \ast ,\alpha )>0$ (resp. if $\alpha \ne -\beta$ and $(\beta \ast ,\alpha )<0$), then $\alpha -\beta$ (resp. $\alpha +\beta$) is a root.

Proof. The second case is deduced from the first by changing $\beta$ into $-\beta$. If $\alpha$ and $\beta$ are proportional and $(\beta \ast ,\alpha )>0$, then $\beta =\alpha$, $2\beta =\alpha$, or $\beta =2\alpha$. The first case is excluded. In the others, one has respectively $\alpha -\beta =\beta \in R$ or $\alpha -\beta =-\alpha \in R$. If $\alpha$ and $\beta$ are not proportional, $(\alpha \ast ,\beta )$ and $(\beta \ast ,\alpha )$ are two strictly positive integers whose product is at most 3. One of them is therefore equal to 1. If $(\beta \ast ,\alpha )=1$, one has $\alpha -\beta =s_{\beta }(\alpha )\in R$; if $(\alpha \ast ,\beta )=1$, one has $\alpha -\beta =-s_{\alpha }(\beta )\in R$.

Lemma 2.3.4. Let $\alpha$ and $\beta$ be two non-proportional roots. If $\beta -\alpha$ is not a root, then $\beta +k\alpha \in R$ for $k=-(\alpha \ast ,\beta )$, but not for $k=-(\alpha \ast ,\beta )+1$.

Proof. Indeed, $\beta -(\alpha \ast ,\beta )\alpha =s_{\alpha }(\beta )\in R$, but $\beta +(-(\alpha \ast ,\beta )+1)\alpha =s_{\alpha }(\beta -\alpha )\notin R$.

Proposition 2.3.5. Let $\alpha$ and $\beta$ be two non-proportional roots. The set of integers $k$ such that $\beta +k\alpha \in R$ is an interval [p, q] ($p⩽0$, $q⩾0$), and one has $p+q=-(\alpha \ast ,\beta )$.

Proof. For the first assertion it suffices, for example, to prove that if $\beta +k\alpha \in R$, $k$ integer > 0, then $\beta +(k-1)\alpha \in R$. If $k=1$, this is trivial. If $k⩾2$, one has

(α*, β + k α) = (α*, β) + 2k ⩾ −3 + 4 > 0,

and one concludes by 2.3.3. Let then [p, q] be the interval in question. Applying 2.3.4 to $\beta +p\alpha$, one finds

q − p = −(α*, β + p α) = −(α*, β) − 2p.

⁶

Remark 2.3.6. The previous formula contains the qualitative statements 2.2.5 and 2.3.3.

Complements 2.3.7.⁷ By 1.2.1 (9) and 1.2.6, the bilinear form on $V(R)$ defined by

⟨x, y⟩ = (p(x), y)

is symmetric and positive definite. By 1.2.1 (10), for any $\alpha \in R$ and $y\in V(R)$,

⟨α, y⟩ = (ℓ(α)/2) (α*, y),

where $\ell (\alpha )=⟨\alpha ,\alpha ⟩$. Consequently, one deduces from 2.3.3 the following corollary.

Corollary. Let $\alpha \ne \beta$ in $R$. If $\alpha -\beta \notin R$, then $⟨\alpha ,\beta ⟩⩽0$.

3. Simple roots, positive roots

3.1. Systems of simple roots

Lemma 3.1.1. Let $\alpha$ and ${\alpha }_i$, $i=1,\cdots ,n$, be roots. Assume $\alpha$ is distinct from the ${\alpha }_i$. If one has a relation

α = Σᵢ qᵢ αᵢ,    qᵢ ∈ Q₊,

there exists an index $i$ such that $q_i\ne 0$, $(\alpha \ast ,{\alpha }_i)>0$, and $\alpha -{\alpha }_i\in R$.

Proof. Indeed, one writes $2=(\alpha \ast ,\alpha )={\Sigma }_i{q}_i(\alpha \ast ,{\alpha }_i)$, which proves the first two assertions. The third then follows from 2.3.3.

Proposition 3.1.2. Let $\alpha$ and ${\alpha }_i$, $i=1,\cdots ,n$, be roots. If

α = Σᵢ mᵢ αᵢ,    mᵢ ∈ ℕ₊,

there exists a sequence ${\beta }_1,\cdots ,{\beta }_m$ of roots taken from the ${\alpha }_i$ (not necessarily pairwise distinct) such that if one denotes

γ_p = Σ_{i=1}^{p} βᵢ,    p = 1, …, m,

one has ${\gamma }_p\in R$ and ${\gamma }_m=\alpha$.

Proof. We argue by induction on the integer $\Sigma m_i=m^{\prime }$. If $\alpha$ equals one of the ${\alpha }_i$, say ${\alpha }_{i_0}$ (which is automatic if $m^{\prime }=1$), one takes $m=1$, ${\beta }_1={\alpha }_{i_0}$. Otherwise, one applies the previous lemma and there exists an index $i$ such that $m_i\ne 0$ and $\alpha -{\alpha }_i$ is a root. One then has $(m_i-1)\in \mathbb{N}$ and

α − αᵢ = (mᵢ − 1) αᵢ + Σ_{j ≠ i} mⱼ αⱼ.

It remains only to apply the induction hypothesis to $\alpha -{\alpha }_i$.

Corollary 3.1.3. Let $R^{\prime }\subset R$. The following conditions are equivalent:

(i) If $\alpha ,\beta \in R^{\prime }$ and $\alpha +\beta \in R$, then $\alpha +\beta \in R^{\prime }$.

(ii) $(\mathbb{N}\cdot R^{\prime })\cap R\subset R^{\prime }$.

Proof. Indeed, one has clearly (ii) ⇒ (i). The converse follows at once from the proposition.

Definition 3.1.4. A set of roots satisfying the conditions of 3.1.3 is called closed.

Proposition 3.1.5. Let $\Delta \subset R$ be a set of roots. The following conditions are equivalent:

(i) The elements of $\Delta$ are indivisible, linearly independent, and

R ⊂ (Q₊ · Δ) ∪ (−Q₊ · Δ).

(ii) The elements of $\Delta$ are linearly independent and

R ⊂ (ℕ · Δ) ∪ (−ℕ · Δ).

(iii) Every root is written in a unique way as a linear combination of the elements of $\Delta$, with integer coefficients all of the same sign.

⁸ One has obviously (ii) ⇒ (iii). Denote by ${\alpha }_1,\cdots ,{\alpha }_n$ the (distinct) elements of $\Delta$.

Proof of (i) ⇒ (ii). Let $\alpha \in R$. One has thus a unique expression

ε α = Σᵢ qᵢ αᵢ    qᵢ ∈ Q₊, ε = ±1.

Let us show that the qᵢ are integers. This is certainly true if all but one of them are zero (cf. 2.1.5). Otherwise, $\alpha$ is distinct from the ${\alpha }_i$ and, applying 3.1.1, one finds $i_0$ such that ${\alpha }^{\prime }=ϵ\alpha -{\alpha }_{i_0}\in R$ and $q_{i_0}\ne 0$. This gives

α' = (q_{i₀} − 1) α_{i₀} + Σ_{i ≠ i₀} qᵢ αᵢ.

Since at least one of the qᵢ, $i\ne i_0$, is nonzero, (i) entails $q_{i_0}-1⩾0$. One repeats the operation for ${\alpha }^{\prime }$ and after finitely many steps one has shown that the qᵢ are integers.

Proof of (iii) ⇒ (i). Let us show that ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$ are linearly independent over $\mathbb{Q}$. In the contrary case, one would have an equality

x = Σ_{i ∈ I} aᵢ αᵢ = Σ_{j ∈ J} bⱼ αⱼ,

where $I$, $J$ are two disjoint subsets of $1,\cdots ,n$, at least one of them, say $I$, being nonempty, and $a_i,b_j\in \mathbb{N}\ast$. By Corollary 2.3.7, one has $⟨{\alpha }_i,{\alpha }_j⟩⩽0$ if $i\in I$ and $j\in J$, whence $⟨x,x⟩⩽0$ and so $x=0$. Let $i_0\in I$; then the equalities

α_{i₀} = α_{i₀} + Σ_{i ∈ I} aᵢ αᵢ = α_{i₀} + Σ_{j ∈ J} bⱼ αⱼ

entail (by (iii)) $a_i=0=b_j$ for all $i$, $j$, a contradiction. This shows that the elements of $\Delta$ are linearly independent over $\mathbb{Q}$; let us show they are also indivisible.

So let $\beta \in \Delta$ be divisible. One has $\beta /2\in R$, whence

β/2 = ε Σ_{α ∈ Δ} m_α α,    m_α ∈ ℕ, ε = ±1,

so also $\beta =2ϵ{\Sigma }_{\alpha }{m}_{\alpha }\alpha$, whence by uniqueness $m_{\alpha }=0$ if $\alpha \ne \beta$ and $2ϵm_{\beta }=1$, a contradiction.

Definition 3.1.6. A set $\Delta$ of roots satisfying the conditions of 3.1.5 is called a system of simple roots, or a base of $\mathcal{R}$.

If $w\in W$ and if $\Delta$ is a system of simple roots, then $w(\Delta )$ is a system of simple roots.

Remark 3.1.7. This definition involves only $R$ and not $\mathcal{R}$, and in fact involves only $ind(R)$.

Remark 3.1.8. If $\Delta$ is a system of simple roots, $\Delta$ is a basis of the free abelian group ${\Gamma }_0(R)$. One has thus $Card(\Delta )=rgss(\mathcal{R})$.

Remark 3.1.9. The conditions of 3.1.5 are then evidently equivalent to:

(i′) The elements of $\Delta$ are indivisible, of number $⩽rgss(\mathcal{R})$, and $R\subset (Q_{+}\cdot \Delta )\cup (-Q_{+}\cdot \Delta )$.

(ii′) $Card(\Delta )⩽rgss(\mathcal{R})$ and $R\subset (\mathbb{N}\cdot \Delta )\cup (-\mathbb{N}\cdot \Delta )$.

Corollary 3.1.10. If $\Delta$ is a system of simple roots, then $ind(\Delta \ast )$ is a system of simple coroots (i.e. a system of simple roots of $\mathcal{R}\ast$).

Proof. Indeed, if $\beta ={\Sigma }_{\alpha \in \Delta }{a}_{\alpha }\alpha$ ($a_{\alpha }\in \mathbb{N}$), then $p(\beta )={\Sigma }_{\alpha \in \Delta }{a}_{\alpha }p(\alpha )$, whence, by 1.2.1 (10):

β* = Σ_{α ∈ Δ} a_α (ℓ(α)/ℓ(β)) α*,

which shows that $ind(\Delta \ast )$ satisfies (i′).

⁹ By 2.1.1, if $\alpha \ast$ is not indivisible, then $\alpha \ast =2u\ast$, where $u\ast \in ind(\Delta \ast )$ (and $u=2\alpha$). One deduces:

Corollary 3.1.11. If $\Delta$ is a system of simple roots and if $\beta ={\Sigma }_{\alpha \in \Delta }{a}_{\alpha }\alpha$ ($a_{\alpha }\in \mathbb{Z}$) is the expression of $\beta$ along $\Delta$, then $2a_{\alpha }\ell (\alpha )$ is divisible by $\ell (\beta )$ (and even by $2\ell (\beta )$ if $\alpha \ast$ is indivisible).

Before continuing to state the properties of systems of simple roots, let us show that such systems exist.

3.2. Systems of positive roots

Definition 3.2.1. A subset $R_{+}\subset R$ is called a system of positive roots of $R$ (or of $\mathcal{R}$, cf. Remark 3.2.2) if it satisfies the following conditions:

(P 1) $R_{+}\cap -R_{+}=\varnothing$.

(P 2) $R_{+}\cup -R_{+}=R$.

(P 3) $(Q_{+}\cdot R_{+})\cap R\subset R_{+}$.

In particular, such a set is closed. We shall see later (3.3.8) that in fact a closed subset satisfying (P 1) and (P 2) also satisfies (P 3), and hence is a system of positive roots. If $w\in W$ and if $R_{+}$ is a system of positive roots, then $w(R_{+})$ is a system of positive roots.

Remark 3.2.2. This definition involves only $R$. One will also say that $R_{+}$ is a system of positive roots of $\mathcal{R}$.

Remark 3.2.3. From (P 1) and (P 2), one obtains at once

$$Card(R_{+})=Card(R)/2.$$

It follows that if $R_{+}$ and $R_{+}^{\prime }$ are two systems of positive roots with $R_{+}\subset R_{+}^{\prime }$, then $R_{+}=R_{+}^{\prime }$.

Remark 3.2.4. If $R_{+}$ is a system of positive roots, $R{\ast }_{+}$ is a system of positive coroots (i.e. a system of positive roots of $\mathcal{R}\ast$).

This follows at once from 1.1.4 and 1.2.2.

Definition 3.2.5. Let $\Delta$ be a system of simple roots. Set

P(Δ) = (Q₊ · Δ) ∩ R = (ℕ₊ · Δ) ∩ R.

Proposition 3.2.6. If $\Delta$ is a system of simple roots, $P(\Delta )$ is a system of positive roots. If $\Delta$ is a system of simple roots and $R_{+}$ is a system of positive roots, one has the equivalence:

Δ ⊂ R₊  ⟺  R₊ = P(Δ).

Proof. The first assertion is immediate. If $\Delta \subset R_{+}$, then $P(\Delta )\subset R_{+}$ by (P 3), so $P(\Delta )=R_{+}$ by 3.2.3. The rest is trivial.

Remark 3.2.7. Systems of positive roots exist: let $⩾$ be a structure of totally ordered vector space on $V(R)$. The set of roots $⩾0$ for this order relation is a system of positive roots.

Theorem 3.2.8. Let $R_{+}$ be a system of positive roots. There exists a unique system of simple roots $\mathcal{S}(R_{+})$ such that $\mathcal{S}(R_{+})\subset R_{+}$, i.e. such that $R_{+}=P(\mathcal{S}(R_{+}))$.

Proof. Uniqueness follows at once from:

Lemma 3.2.9. Let $\Delta$ be a system of simple roots. Then for $\alpha \in P(\Delta )$ to belong to $\Delta$, it is necessary and sufficient that $\alpha$ not be a sum of two elements of $P(\Delta )$.

Proof. This lemma follows at once from the definitions and from 3.1.2.

Let us now prove the existence of $\mathcal{S}(R_{+})$. Consider the set of subsets $T$ of $R_{+}$ such that $T=ind(T)$ and $(Q_{+}\cdot T)\supset R_{+}$. This set is nonempty, since it contains $ind(R_{+})$. Let $\Delta$ be a minimal element of this set for the inclusion relation. We show that $\Delta$ is a system of simple roots, that is to say, by 3.1.5 (i), that $\Delta$ is a free subset of $M\otimes \mathbb{Q}$.

Lemma 3.2.10. If $\alpha ,\beta \in \Delta$ and $q\alpha +q^{\prime }\beta \in R$, where $q,q^{\prime }\in \mathbb{Q}$, then $q{q}^{\prime }⩾0$.

Proof. Indeed, if $q{q}^{\prime }<0$, one can write (possibly exchanging $\alpha$ and $\beta$)

a α − b β ∈ R₊,    a, b ∈ Q*₊,

so by hypothesis there exists a relation

a α − b β = Σ_{γ ∈ Δ} c(γ) γ,    c(γ) ∈ Q₊.

If $a⩽c(\alpha )$, it is written

−β = ((c(α) − a)/b) α + Σ_{γ ≠ α} (c(γ)/b) γ.

Then this element belongs to $(Q_{+}\cdot \Delta )\cap R$, which is contained in $R_{+}$ by (P 3). But then $\beta \in R_{+}\cap -R_{+}$, which contradicts (P 1).

If, on the contrary, $a>c(\alpha )$, one writes

(a − c(α)) α = b β + Σ_{γ ≠ α} c(γ) γ,

which proves $\Delta \subset Q_{+}\cdot (\Delta -\alpha )$, contrary to the minimal character of $\Delta$.

Recall (cf. 2.3.7) that the bilinear form on $V(R)$ defined by $⟨x,y⟩=(p(x),y)$ is a Euclidean inner product; moreover, for $\alpha \in R$ one has $⟨\alpha ,y⟩=\ell (\alpha )(\alpha \ast ,y)/2$, so that $⟨\alpha ,y⟩$ and $(\alpha \ast ,y)$ have the same sign.

Lemma 3.2.11. If $\alpha ,\beta \in \Delta$, then $(\beta \ast ,\alpha )⩽0$ and so $⟨\beta ,\alpha ⟩⩽0$.

Proof. Indeed, $s_{\beta }(\alpha )=\alpha -(\beta \ast ,\alpha )\beta \in R$, whence $(\beta \ast ,\alpha )⩽0$ by 3.2.10.

Let us now prove that $\Delta$ is free. In the contrary case, 3.2.11 entails, as in the proof of 3.1.5, the existence of a non-trivial relation

Σ_{α ∈ Δ} m(α) α = 0,    m(α) ∈ ℕ,

whence $-{\alpha }_0=(m({\alpha }_0)-1){\alpha }_0+{\Sigma }_{\alpha \ne {\alpha }_0}m(\alpha )\alpha$, if $m({\alpha }_0)⩾1$. Then, by (P 3), ${\alpha }_0$ belongs to $R_{+}\cap -R_{+}$, contradicting (P 1).

This shows that $\Delta$ is a base of $\mathcal{R}$ and completes the proof of Theorem 3.2.8.

Corollary 3.2.12. Let $R_{+}$ be a system of positive roots, $\Delta$ a base of $\mathcal{R}$, and $w\in W$. One has:

Δ ⊂ R₊  ⟺  R₊ = P(Δ)  ⟺  Δ = 𝒮(R₊)

P(ind(Δ*)) = P(Δ)*,    𝒮(R*₊) = ind(𝒮(R₊)*);

𝒮(w(R₊)) = w(𝒮(R₊)),    P(w(Δ)) = w(P(Δ)).

Definition 3.2.13. If one has chosen a system of simple roots $\Delta$, the elements of $P(\Delta )$ will be called positive. If one has chosen a system of positive roots $R_{+}$, the elements of $\mathcal{S}(R_{+})$ will be called simple.

Corollary 3.2.14. Let $R_{+}$ be a system of positive roots. Let $\alpha \in R_{+}$. The following conditions are equivalent:

(i) $\alpha$ is simple (i.e. $\alpha \in \mathcal{S}(R_{+})$).

(ii) $\alpha$ is not a sum of two elements of $R_{+}$.

(iii) $R_{+}-\alpha$ is closed (cf. 3.1.4).

Proof. The equivalence of (i) and (ii) follows at once from 3.2.9. The equivalence of (ii) and (iii) is immediate.

Definition 3.2.15. Let $\Delta$ be a system of simple roots. The sum of the coefficients of the decomposition of a root $\alpha$ along $\Delta$ is called the order of $\alpha$ relative to $\Delta$ and is denoted $or{d}_{\Delta }(\alpha )$.

One has the equivalences:

α ∈ P(Δ)  ⟺  ord_Δ(α) > 0,    α ∈ Δ  ⟺  ord_Δ(α) = 1.

Lemma 3.2.16. Let $\Delta$ be a system of simple roots and $\alpha \in P(\Delta )$. There exists a sequence ${\alpha }_i\in \Delta$, $i=1,\cdots ,m$, such that

α₁ + α₂ + ⋯ + α_p ∈ P(Δ)    for p = 1, …, m,
α₁ + α₂ + ⋯ + α_m = α.

Moreover, for every such sequence ${\alpha }_i$, one has $m=or{d}_{\Delta }(\alpha )$.

Proof. Trivial by 3.1.2.

3.3. Characterization and conjugacy of systems of positive roots

Lemma 3.3.1. If $\alpha \in P(\Delta )$, $\beta \in \Delta$, and if $\alpha$ and $\beta$ are not proportional, then $s_{\beta }(\alpha )\in P(\Delta )$.

Proof. Indeed, $s_{\beta }(\alpha )=\alpha -(\beta \ast ,\alpha )\beta$. Since at least one simple root other than $\beta$ appears in the decomposition of $\alpha$ with a nonzero (hence strictly positive) coefficient, it also appears in the decomposition of $s_{\beta }(\alpha )$ with the same coefficient, so $s_{\beta }(\alpha )$ is also positive.

Corollary 3.3.2. If $\beta \in \Delta$, the symmetry $s_{\beta }$ interchanges the elements of $P(\Delta )$ not proportional to $\beta$.

Lemma 3.3.3. If $\alpha \in P(\Delta )-\Delta$, $\alpha$ indivisible, there exists $\beta \in \Delta$ such that $s_{\beta }(\alpha )\in P(\Delta )$ and $or{d}_{\Delta }(s_{\beta }(\alpha ))<or{d}_{\Delta }(\alpha )$.

Proof. Indeed, by 3.1.1 there exists $\beta \in \Delta$ such that $(\beta \ast ,\alpha )>0$. Since $\alpha \notin \Delta$ and is indivisible, $\alpha$ cannot be proportional to $\beta$. Hence $s_{\beta }(\alpha )\in P(\Delta )$, by 3.3.1, and one has ord_Δ(s_β(α)) = ord_Δ(α) − (β*, α) < ord_Δ(α).

Corollary 3.3.4. If $\alpha \in P(\Delta )$ is indivisible, there exists a sequence ${\beta }_p\in \Delta$, $p=1,\cdots ,q$, such that

α_p = s_{β_p} ⋯ s_{β_1}(α) ∈ P(Δ)    for p = 1, …, q,

and ${\alpha }_q\in \Delta$.

Proof. This follows from 3.3.3 by induction on $or{d}_{\Delta }(\alpha )$.

Proposition 3.3.5. The Weyl group is generated by the $s_{\alpha }$, for $\alpha \in \Delta$. Every indivisible root is conjugate to a simple root by an element of the Weyl group.

Proof. The second assertion follows at once from 3.3.4. The first follows from it by 1.2.10 and 2.1.1.

Proposition 3.3.6. Let $R_{+}$ be a system of positive roots. Let $P^{\prime }\subset R$ satisfy (P 2) and be closed. Then there exists $w\in W$ such that $w(R_{+})\subset P^{\prime }$.

Let us state the corollaries at once.

Corollary 3.3.7. The Weyl group operates transitively on the set of systems of positive roots (resp. of systems of simple roots).

Corollary 3.3.8. For a subset of $R$ to be a system of positive roots, it is necessary and sufficient that it satisfy (P 1) and (P 2) and be closed.

Corollary 3.3.9. If one endows ${\Gamma }_0(R)$ with a structure of ordered group such that every root is > 0 or < 0, the set of positive roots for this order structure is a system of positive roots.

Proof of 3.3.6. One can find $w\in W$ such that $Card(w(R_{+})\cap P^{\prime })$ is maximal, hence by replacing $R_{+}$ by $w(R_{+})$ one may assume that

(∗)    Card(R₊ ∩ P') ⩾ Card(R₊ ∩ s_α(P'))

for all $\alpha \in \mathcal{S}(R_{+})=\Delta$. Let us show that $R_{+}\subset P^{\prime }$. Otherwise, $P^{\prime }$ being closed, there exists $\alpha \in \Delta$, $\alpha \notin P^{\prime }$. But $P^{\prime }$ satisfying (P 2), one has then $-\alpha \in P^{\prime }$ (so $-2\alpha \in P^{\prime }$ if $2\alpha$ is a root). For every $\beta \in R_{+}\cap P^{\prime }$, one has $\beta \ne \alpha$; if $2\alpha$ is not a root, then $s_{\alpha }(\beta )\in R_{+}$ (by 3.3.1), so

s_α(R₊ ∩ P') ⊂ R₊ ∩ s_α(P');

but $R_{+}\cap s_{\alpha }(P^{\prime })$ also contains $\alpha =s_{\alpha }(-\alpha )$, contradicting inequality (∗). If $2\alpha$ is a root, one argues similarly.

To study sets of roots satisfying (P 2) and closed, one is therefore reduced to the case where they contain a set of positive roots.

Proposition 3.3.10. Let $R_{+}$ be a system of positive roots and $P^{\prime }$ a closed subset of $R$ containing $R_{+}$. If one denotes $\Delta =\mathcal{S}(R_{+})$ and ${\Delta }^{\prime }=\Delta \cap -P^{\prime }$, then $P^{\prime }$ is the union of $R_{+}$ and the set of roots that are linear combinations with negative coefficients of the elements of ${\Delta }^{\prime }$.

Proof. We prove the assertion by induction on the order of a root $\gamma \in P^{\prime }-R_{+}$. If $or{d}_{\Delta }(\gamma )=-1$, then $-\gamma \in \Delta$ and $\gamma \in -{\Delta }^{\prime }$. If $or{d}_{\Delta }(\gamma )<-1$, there exists, by 3.1.2, $\beta \in \Delta$ such that $-\gamma -\beta \in R$; then

0 < ord_Δ(−γ − β) = ord_Δ(−γ) − 1 < ord_Δ(−γ)

and the first inequality shows that $-\gamma -\beta \in R_{+}$. Hence, since $R_{+}\cap -R_{+}=\varnothing$ and $\gamma +\beta$ is a sum of two roots of $P^{\prime }$, it is an element of $P^{\prime }-R_{+}$ with $or{d}_{\Delta }(\gamma +\beta )>or{d}_{\Delta }(\gamma )$. So, by the induction hypothesis, $\gamma +\beta$ is a linear combination with negative coefficients of the elements of ${\Delta }^{\prime }$. Since $\gamma =(\gamma +\beta )-\beta$, it suffices to verify that $\beta \in -P^{\prime }$. Now $\gamma \in P^{\prime }$ and $-(\gamma +\beta )\in R_{+}\subset P^{\prime }$, so $-\beta =\gamma -(\gamma +\beta )$ belongs to $P^{\prime }$.

Definition 3.3.10.1.¹⁰ One says that a subset $R^{\prime }$ of $R$ is symmetric if $-R^{\prime }=R^{\prime }$.

Scholie 3.3.11. Let $P^{\prime }$ be a set of roots satisfying (P 2) and closed. There exists a system of simple roots $\Delta$ and a subset ${\Delta }^{\prime }$ of $\Delta$ such that

P' = R ∩ (ℕ · Δ ∪ −ℕ · Δ').

If one denotes $R_{{\Delta }^{\prime }}=(\mathbb{Z}\cdot {\Delta }^{\prime })\cap R$, which is a closed and symmetric subset of $R$, then $P^{\prime }$ is the disjoint union of $R_{{\Delta }^{\prime }}$ and of the closed part of $P(\Delta )$ formed by the $\alpha \in P(\Delta )-\mathbb{N}\cdot {\Delta }^{\prime }$, i.e. the positive roots which in the decomposition along $\Delta$ "contain at least one root of $\Delta -{\Delta }^{\prime }$".

3.4. Closed and symmetric sets of roots

Proposition 3.4.1. Let $\mathcal{R}=(M,M\ast ,R,R\ast )$ be a root datum and $R^{\prime }$ a closed and symmetric subset of $R$. Then:

(i) ${\mathcal{R}}^{\prime }=(M,M\ast ,R^{\prime },R^{\prime }\ast )$ is a root datum;

(ii) for every system of positive roots $R_{+}$ of $R$, $R_{+}^{\prime }=R_{+}\cap R^{\prime }$ is a system of positive roots of $R^{\prime }$;

(iii) the Weyl group $W({\mathcal{R}}^{\prime })$ of ${\mathcal{R}}^{\prime }$ is the subgroup of $W(\mathcal{R})$ generated by the $s_{\alpha }$, for $\alpha \in R^{\prime }$.

Proof. The first assertion is trivial by 1.2.11, the second follows from 3.3.8, the third is evident.

Corollary 3.4.2. Let $w\in W({\mathcal{R}}^{\prime })$. The order of $w$ is the smallest integer $n>0$ such that $w^n({\alpha }^{\prime })={\alpha }^{\prime }$ for every ${\alpha }^{\prime }\in R^{\prime }$.

Proof. It suffices to apply 1.2.7 to the root datum ${\mathcal{R}}^{\prime }$.

Corollary 3.4.3. Let $\alpha$ and $\beta$ be two non-proportional roots. Let $n$ be the smallest integer > 0 such that $(s_{\alpha }{s}_{\beta }{)}^n(\alpha )=\alpha$ and $(s_{\alpha }{s}_{\beta }{)}^n(\beta )=\beta$. Then the subgroup $W_{\alpha ,\beta }$ of $W$ generated by $s_{\alpha }$ and $s_{\beta }$ is defined by the relations:

s_α² = 1,    s_β² = 1,    (s_α s_β)ⁿ = 1.

Proof. Taking 3.4.2 into account, it suffices to verify:

Lemma 3.4.4. Let $G$ be the group generated by two elements $x$ and $y$ subject to the relations $x^2=y^2=1$. Every normal subgroup of $G$ containing neither $x$ nor $y$ is generated (as a normal subgroup) by an element of the form $(xy{)}^n$, where $n>0$.

¹¹ Proof. Indeed, every element of $G$ is written $(xy{)}^n$, or $(yx{)}^n=(xy{)}^{-n}$, or:

(a)    x (yx)^{2n}    or    y (xy)^{2n+1}

(b)    y (xy)^{2n}    or    x (yx)^{2n+1},

where $n\in \mathbb{N}$. Now the elements of type (a) (resp. (b)) are conjugate to $x$ (resp. to $y$).

Remark 3.4.5. One computes the integer $n$ immediately: if one sets

(α*, β) = p,    (β*, α) = q,

one has¹²

(s_α s_β)(α) = (pq − 1) α − q β,    (s_α s_β)(β) = p α − β.

The integer $n$ is therefore the order of the matrix

⎛pq − 1   p⎞
⎝  −q   −1⎠.

If $pq=0$, then $p=q=0$, by 2.2.2, whence $n=2$. Otherwise, by 2.3.1, pq equals 1, 2, or 3, and one finds, respectively, $n=3$, 4, or 6.

N.B. By writing that the order of the previous matrix is finite, one recovers inequality (13) of 2.3.1.

Definition 3.4.6. Let $\Delta$ be a system of simple roots and ${\Delta }^{\prime }\subset \Delta$. We write

R_{Δ'} = R ∩ (ℚ · Δ') = R ∩ (ℤ · Δ').

Lemma 3.4.7. $R_{{\Delta }^{\prime }}$ is closed and symmetric, ${\Delta }^{\prime }$ is a system of simple roots of the root datum $(M,M\ast ,R_{{\Delta }^{\prime }},R{\ast }_{{\Delta }^{\prime }})$, whose Weyl group is the subgroup $W_{{\Delta }^{\prime }}$ of $W$ generated by the $s_{\alpha }$, for $\alpha \in {\Delta }^{\prime }$. One has $\Delta \cap R_{{\Delta }^{\prime }}={\Delta }^{\prime }$.

Proof. Trivial.

Proposition 3.4.8. Let $R^{\prime }\subset R$. The following conditions are equivalent:

(i) There exists a vector subspace $V^{\prime }$ of $V$ (or of $V(R)$) such that $R^{\prime }=R\cap V^{\prime }$.

(ii) There exists a system of simple roots $\Delta$ of $R$ and a subset ${\Delta }^{\prime }$ of $\Delta$ such that $R^{\prime }=R_{{\Delta }^{\prime }}$.

More precisely, under these conditions, every system of simple roots ${\Delta }^{\prime }$ of $(M,M\ast ,R^{\prime },R^{\prime }\ast )$ is contained in a system of simple roots $\Delta$ of $R$, and one has $R^{\prime }=R_{{\Delta }^{\prime }}$.

Proof. One has obviously (ii) ⇒ (i). Suppose (i) verified: then $R^{\prime }$ is closed and symmetric, so $(M,M\ast ,R^{\prime },R^{\prime }\ast )$ is a root datum. Let ${\Delta }^{\prime }$ be a system of simple roots of this datum and $R_{+}^{\prime }=P({\Delta }^{\prime })$. If $V^{\prime }=V$, then ${\Delta }^{\prime }$ is a system of simple roots of $\mathcal{R}$ and we are done. Otherwise, there exists $x\in V\ast$ such that

(x, V') = {0},    (x, α) ≠ 0    for all α ∈ R − R'.

Put $R_{+}(x)=\alpha \in R|(x,\alpha )>0$ and $R_{+}=R_{+}(x)\cup R_{+}^{\prime }$. For every $\alpha \in R$, one has the equivalences

(x, α) > 0  ⟺  α ∈ R₊(x),
(x, α) < 0  ⟺  α ∈ −R₊(x),
(x, α) = 0  ⟺  α ∈ R'.

It follows at once from 3.3.8 that $R_{+}$ is a system of positive roots of $R$. Put $\Delta =\mathcal{S}(R_{+})$. It evidently suffices to prove ${\Delta }^{\prime }\subset \Delta$. Otherwise let $\alpha \in {\Delta }^{\prime }-\Delta$. Then, by 3.2.14, there exist $\beta ,\gamma \in R_{+}$ such that $\alpha =\beta +\gamma$. If $\beta ,\gamma \in R_{+}(x)$, one has $\alpha \in R_{+}(x)$, which is absurd since $(x,{\Delta }^{\prime })=0$. If $\beta$ or $\gamma$, say $\beta$, belongs to $R_{+}^{\prime }$, then, since $R^{\prime }$ is symmetric and closed, $\gamma =\alpha -\beta$ belongs to $R_{+}\cap R^{\prime }=R_{+}^{\prime }$; but then $\alpha$ is not simple in $R_{+}^{\prime }$.

Lemma 3.4.9. Under the preceding conditions, let $\alpha \in P(\Delta )-R^{\prime }$. For every $w\in W_{{\Delta }^{\prime }}$, one has $w(\alpha )\in P(\Delta )-R^{\prime }$.

Proof. It indeed suffices to verify this for $w=s_{\beta }$, $\beta \in {\Delta }^{\prime }$, in which case it follows from 3.3.1 and from the fact that $s_{\beta }(R^{\prime })=R^{\prime }$.

Lemma 3.4.10. Let $w\in W$. Under the conditions of 3.4.8, the following conditions are equivalent:

(i) $w\in W_{{\Delta }^{\prime }}$.

(ii) For every $m\in M$, $w(m)-m\in V^{\prime }$.

(iii) For every $\alpha \in R$, $w(\alpha )-\alpha \in V^{\prime }$.

Proof. One has obviously (i) ⇒ (ii) ⇒ (iii). Let us prove (iii) ⇒ (i). Let then $w\in W$ be such that $w(\alpha )-\alpha \in V^{\prime }$ for all $\alpha \in R$. Write $w=s_{{\alpha }_n}\cdots s_{{\alpha }_1}$, with ${\alpha }_i\in \Delta$, and let us prove by induction on $n$ that each ${\alpha }_i\in {\Delta }^{\prime }$. One may assume that $w^{\prime }=s_{{\alpha }_{n-1}}\cdots s_{{\alpha }_1}\in W_{{\Delta }^{\prime }}$. One has then, for all $\alpha \in R$,

w(α) − α = w'(α) − α − (α*_n, w'(α)) α_n.

Taking $\alpha =w^{\prime -1}({\alpha }_n)$, one finds $2{\alpha }_n\in V^{\prime }$, whence ${\alpha }_n\in {\Delta }^{\prime }$, so $w=s_{{\alpha }_n}{w}^{\prime }$ belongs to $W_{{\Delta }^{\prime }}$.

3.5. Miscellaneous remarks

Proposition 3.5.1. Let $R_{+}$ be a system of positive roots. Denote

ρ_{R₊} = (1/2) Σ_{α ∈ ind(R₊)} α.

Then $(\beta \ast ,{\rho }_{R_{+}})=1$ for every $\beta \in \mathcal{S}(R_{+})$.

Proof. Indeed, one can write

2 ρ_{R₊} = β + Σ_{α ∈ ind(R₊), α ≠ β} α,

so $s_{\beta }(2{\rho }_{R_{+}})=2{\rho }_{R_{+}}-2\beta$, by 3.3.2.

Corollary 3.5.2. Put $\rho {\ast }_{R_{+}}=(1/2){\Sigma }_{\alpha \ast \in ind(R{\ast }_{+})}\alpha \ast$. Then:

(i) $(\rho {\ast }_{R_{+}},\alpha )>0$ for every $\alpha \in R_{+}$ (i.e. $\rho {\ast }_{R_{+}}\in \mathcal{C}(R_{+})$, cf. 3.6.8).

(ii) For every $\alpha \in R$, one has $or{d}_{\mathcal{S}(R_{+})}(\alpha )=(\rho {\ast }_{R_{+}},\alpha )$.¹³

Remark 3.5.3. If $w\in W$, one has ${\rho }_{w(R_{+})}=w({\rho }_{R_{+}})$ and $\rho {\ast }_{w(R_{+})}=w(\rho {\ast }_{R_{+}})$.

Proposition 3.5.4. Let $\alpha$ and $\gamma$ be two non-proportional roots, with $\alpha$ indivisible. There exists a system of simple roots containing $\alpha$ and a root $\beta$ such that $\gamma =a\alpha +b\beta$, with $a,b\in \mathbb{N}$.

Proof. Indeed, let us construct a basis of the vector space $V(R)$ containing ${\alpha }_1=\alpha$, ${\alpha }_2=\gamma$. Consider the lexicographic order with respect to this basis. Denoting by $R_{+}$ the set of roots > 0, it is clear that $R_{+}$ is a system of positive roots and¹⁴ that the smallest element of $R_{+}$ not proportional to $\alpha$ is simple. This element is of the form

β = p α + q γ,    0 < q ⩽ 1.

One thus has $\gamma =q^{-1}\beta -q^{-1}p\alpha$, and since $q>0$, one has $q^{-1}\in \mathbb{N}\ast$ and $-q^{-1}p\in \mathbb{N}$.

Let us finally make two remarks about the group ${\Gamma }_0(R)$.

Proposition 3.5.5. Let $G$ be an abelian group and $f:R\to G$ a map satisfying the following two conditions:

(i) If $\alpha \in R$, $f(-\alpha )=-f(\alpha )$.

(ii) If $\alpha ,\beta ,\alpha +\beta \in R$, $f(\alpha +\beta )=f(\alpha )+f(\beta )$.

Then there exists a unique group homomorphism $\bar{f}:{\Gamma }_0(R)\to G$ such that $\bar{f}(\alpha )=f(\alpha )$ for $\alpha \in R$.

Proof. Indeed, if $\Delta$ is a system of simple roots of $R$, and if $\beta \in R$ is written ${\Sigma }_{\alpha \in \Delta }a(\alpha )\alpha$, it follows at once from 3.2.16 that $f(\beta )={\Sigma }_{\alpha \in \Delta }a(\alpha )f(\alpha )$. Now $\Delta$ is a basis of ${\Gamma }_0(R)$.

Proposition 3.5.6. Let $\Delta$ be a system of simple roots. There exists on ${\Gamma }_0(R)$ a structure of totally ordered group such that the roots > 0 are the elements of $P(\Delta )$ and $\alpha \mapsto or{d}_{\Delta }(\alpha )$ is an increasing function.

Proof. Indeed, let ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$ ($n=rgss(\mathcal{R})$) be the elements of $\Delta$. For $x\in {\Gamma }_0(R)$, one has a decomposition

x = Σᵢ mᵢ(x) αᵢ.

It suffices to take the lexicographic order with respect to the functions $\Sigma m_i$, $m_n$, $m_{n-1}$, …, $m_2$.

Remark 3.5.7. The first roots are in order:

$${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n;$$

one then has (if these are roots) $2{\alpha }_1,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,\cdots$.

3.6. Weyl chambers

Lemma 3.6.1. Let $V$ be a finite-dimensional $\mathbb{R}$-vector space.¹⁵ Let fᵢ be independent linear forms. Set

C = {x ∈ V | fᵢ(x) > 0}.

Then $C$ is a maximal convex subset of $X=V-{\bigcup }_i{f}_i^{-1}(0)$.

Proof. Trivial.

Definition 3.6.2. A subset $C$ of $V$ describable by the procedure of 3.6.1 will be called (here) a chamber of $V$.

Definition 3.6.3. One says that the hyperplane $H$ of $V$ is a wall of $C$ if $H\cap (\bar{C}-C)$ contains a nonempty open subset of $H$.

Remark 3.6.4. For a convex subset, the closure is described without appeal to the topology of $V$: it is the set of endpoints of all open segments contained in the given subset.

Lemma 3.6.5. Under the conditions of 3.6.1, one has

C̄ = {x ∈ V | fᵢ(x) ⩾ 0}.

The walls of $C$ are the hyperplanes $f_i^{-1}(0)$.

Proof. The first assertion is clear. The second then follows from the fact that $\bar{C}-C\subset {\bigcup }_i{f}_i^{-1}(0)$ and that the $f_i^{-1}(0)$ are obviously walls of $C$.

Proposition 3.6.6. Let $C$ be a chamber of $V$. If Hᵢ, $i=1,2,\cdots ,n$, are the distinct walls of $C$, then for every system of linear forms $u_i$ such that $H_i=u_i^{-1}(0)$, there exist ${ϵ}_i\in -1,+1$ such that $C$ is defined by

C = {x ∈ V | εᵢ uᵢ(x) > 0}.

For every wall $H$ of $C$, one has $H\cap C=\varnothing$ and $H\cap \bar{C}=C_H$, where C_H is a chamber in $H$. The walls of $C_{H_i}$ are the $H_j\cap H_i$, for $j\ne i$.

Proof. This follows trivially from the lemma.

Definition 3.6.7. The $C_{H_i}$ are the faces of $C$.

Let now $\mathcal{R}=(M,M\ast ,R,R\ast )$ be a root datum. We put $V{\ast }_{\mathbb{R}}=M\ast \otimes \mathbb{R}$.

Definition 3.6.8.¹⁶ For every $\alpha \in R$, set

H_α = {x ∈ V*_ℝ | (x, α) = 0}.

We denote $X=V{\ast }_{\mathbb{R}}-{\bigcup }_{\alpha \in R}{H}_{\alpha }$. For every $x\in X$, set

R₊(x) = {α ∈ R | (x, α) > 0}.

For every system of positive roots $R_{+}$, denote

𝒞(R₊) = {x ∈ V*_ℝ | (x, α) > 0 for all α ∈ R₊}.

Proposition 3.6.9. (i) For every $x\in X$, $R_{+}(x)$ is a system of positive roots. For every system of positive roots $R_{+}$, $\mathcal{C}(R_{+})$ is a chamber in $V{\ast }_{\mathbb{R}}$. The $\mathcal{C}(R_{+})$ are the maximal convex subsets of $X$.

(ii) Let $\Delta$ be a system of simple roots. One has

𝒞(P(Δ)) = {x ∈ V*_ℝ | (x, α) > 0    for all α ∈ Δ}.

The walls of $\mathcal{C}(P(\Delta ))$ are the hyperplanes $H_{\alpha }={\alpha }^{-1}(0)$, for $\alpha \in \Delta$; its faces are the

C_α = {x ∈ V*_ℝ | (x, α) = 0, (x, β) > 0 for β ∈ Δ, β ≠ α}.

(iii) One has the equivalence

R₊(x) = R₊  ⟺  x ∈ 𝒞(R₊).

Proof. It is first clear that $R_{+}(x)$ is a system of positive roots. Since $x\in \mathcal{C}(R_{+}(x))$, the union of the $\mathcal{C}(R_{+}(x))$ is $X$. Property (iii) is immediate; it follows that the $\mathcal{C}(R_{+})$ form a partition of $X$. The first assertion of (ii) is evident. It follows at once that $\mathcal{C}(R_{+})$ is a chamber in $V$, which proves the rest of (i). Putting $C=\mathcal{C}(P(\Delta ))$, it remains only to remark that