Chapter 13

Twisted affine Lie algebras

We give an explicit construction of twisted affine Lie algebras in this chapter. This will give us a description of the root and coroot systems and we will deduce a description of the . We end the chapter will some (very few) applications of the denominator identity as functional identities.

13.1 Construction

Let g be a simple finite dimensional and let σ be an automorphism of its . Remark that such an automorphism is the identity except if g is simply laced and that these auto- morphisms where already used to construct non simply laced simple finite dimensional Lie algebras. We start the construction with the L(g) = g ⊗ C[t, t −1] and the affine Lie algebra g = L(g) ⊕ Cc ⊕ Cd. We already saw that the automorphism of the Dynkin diagram σ defines an automorphism (still denoted σ) of g. We extend this automorphism to L(g) and g as follows: for bx ∈ g, a ∈ Z and λ, µ ∈ C we set 1 b σ(x ⊗ ta + λc + µd ) = σ(x) ⊗ ta + λc + µd ζa where ζ is a primitive r-th root of unity where r is as usual the order of σ.

Fact 13.1.1 This defines an automorphism (still denoted σ) of the Lie algebras L(g) and g.

L L Definition 13.1.2 Let us denote by (g, σ ) and g(σ) the σ-invariant subalgebra of (g)b and g re- spectively.

r−1 b b Proposition 13.1.3 Let us denote by g = ⊕i=0 g(j) the eigenspaces decomposition with respect to the action of σ (described in Theorem 12.2.6). We have the following decompositions:

L(g, σ ) = g(k¯) ⊗ tk and g(σ) = g(k¯) ⊗ tk ⊕ Cc ⊕ Cd ∈Z Ã ∈Z ! Mk Mk b where k¯ is the rest of k modulo r.

Proof : It is clear that these spaces are contained in the σ-invariant subalgebra. Conversely, if r−1 x ⊗ P + λc + µd is in g(σ). Then σ(x ⊗ P ) = x ⊗ P . Writing x = i=0 xi with xi ∈ g(i) and k i−k k k k P = pkt we get ζ xi ⊗ pkt = xi ⊗ pkt . In particular for i 6≡ k (mod r) we have xi ⊗ pkt = 0. k P The result follows. b ¤ P

107 108 CHAPTER 13. TWISTED AFFINE LIE ALGEBRAS

A2 13.1.1 All cases except 2n In this subsection we consider g a simple simply laced Lie algebra and σ a non trivial automorphism of the Dynkin diagram. These automorphisms were described in the last chapter. Let us denote by g′ the σ-invariant subalgebra of g. We also keep the notation of the previous chapter for root systems and basis elements of g and g′.

Theorem 13.1.4 Let g be a of type X and let σ be a rank r automorphism of the Dynkin diagram. The Lie algebra g(σ) is isomorphic to the affine Lie algebra of type Xr.

Proof : Let us fix the following additionalb notation: • We denote by n the rank of g and by m the rank of g′. ′ ′ • We keep the notation αi for the simple roots in ∆ . ′ ∨ ′ ′ • We denote by αi the coroot of αi a simple root in ∆ . ′ ′∨ • Denote by θ∨ the highest root of the dual root system ∆ , this is also the root rβ where β was ′ described in Lemma 11.1.5. We denote this root β by αθ. ′ ′ ′ ′ ′ • Set E = E ′ and F = −E− ′ for α a simple root. i αi i αi i

′ 1 r−1 −i ′ −1 r−1 i ′ i ′ i • Set E0 = E−α = i=0 ζ Eσ (−αθ) and F0 = −Eα = − i=0 ζ Eσ (αθ) where αθ is any root in ′θ ′ θ ∆ such that θ∨ = rα θ is r times the associated short root (we always have σ(αθ) 6= αθ). Remark ′ ′ P P that E0 and F0 and defined modulo a r-th root of 1 or equivalently modulo the choice of a root in the orbit of αθ but we only need that they are both defined by the same root. ′ ′ ′ ′ • Set ei = Ei ⊗ 1 ∈ g(σ) and fi = Fi ⊗ 1 ∈ g(σ) for i > 0. • Set e′ = E′ ⊗ t ∈ g(σ) and f ′ = F ′ ⊗ t−1 ∈ g(σ). 0 0 b 0 0 b • Set h(σ) = h(0) ⊕ Cc ⊕ Cd. We will denote by h(σ)∗ its dual. b b ′ ∗ ′ ′ ′ • Defineb δ ∈ h(σ) by δ |h(0) = Id h(0) , δ (c) = 0 andb δ (d) = 1. ′ ′ ′ • Define α0 =bδ − αθ. ′ ∨ r ′ ′ • Define α0 = a0 c − θ∨ = rc − θ∨. • We will view h(0) ∗ as being equal to h(0) thank to the invariant bilinear form and h(0) ∗ as a subspace of h(σ)∗ by setting λ(c) = λ(d) = 0 for λ ∈ h(0) ∗. We first need to define a realisation of the of type Xr in h(σ). For this let us define ∨ ′ ∨ b′ ∨ ′ ′ Π = {α0 , · · · , α m } and Π = {α0, · · · , α m}. b Lemma 13.1.5 The triple (h(σ), Π, Π∨) is a realisation of the affine Cartan matrix of type Xr.

Proof : The dimension of hb(σ) is m + 2 which is the right dimension. We need to check that the ′ ′ r ′ ∨ matrix ( hαi, α ji) is the Cartan matrix of type X . This is true for i > 0 and j > 0 because αi ′ ′ r and αj are the simple corootsb and roots of g which is of the type Y obtained from X by removing ′ ∨ ′ ′ ′ the zero vertex. We then compute hα0 , α ji = −h θ∨, α ji this gives the result for j > 0 because of our description of β (or more conceptually because we get the same added vertex as for Y ∨ but with ′ ∨ ′ ′ ′ ¤ reversed arrow. Now compute hα0 , α 0i = hθ∨, α θi = 2. c 13.1. CONSTRUCTION 109

′ ′ ′ ∨ ′ ′ ′ ′ ′ ′ We now verify the relations [ ei, f j] = δi,j αi , [ h, e i] = hαi, h iei and [ h, f i ] = −h αi, h ifi . The first ′ ′∨ relation is true for i > 0 and j > 0. Because θ∨ is the highest root of ∆ we get the result for ij = 0 ′ ′ ′ ′ ′ ′ i j and ( i, j ) 6= 0. Furthermore [ e0, f 0] = [ E0, F 0] + ( E0, F 0)c. Because σ (αθ) is positive when −σ (αθ) is i i j negative, we have [ Eσ (−αθ), E σ (αθ)] = δi,j σ (αθ) thus

r−1 ′ ′ i ′ ′ ′ ′ ′ ′ ′ ∨ [e0, f 0] = − σ (αθ) + ( E0, F 0)c = −rα θ + ( E0, F 0)c = rc − θ∨ = α0 i X=0 ′ ′ j−i i j because we have ( E0, F 0) = − i,j ζ (Eσ (−αθ), E σ (αθ)) = r (recall that ( Eα, E β) = −δα, −β). ′ ′ ′ ′ ′ ′ For the relations [ h, e i] = hαi, h iei, this is clear for i > 0 and h ∈ h(σ). For [ h, e 0] = hα0, h ie0, ′ P ′ ′ we start with h ∈ h . Now the relations in g, see Theorem 12.2.11, give [ h, e 0] = [ h, E 0] ⊗ t = 1 ′ ′ ′ ′ [h, E ′ ] ⊗ t = −(h, α )e = −(h, α )e and the result follows fromb the fact that ( δ , h ) = 0 for −θ∨ θ 0 θ 0 ′ ′ ′ h ∈ h . We are left with h = c or d. But ( c, α 0) = 0 and [ c, e 0] = 0, furthermore, ( d, α 0) = 1 and ′ ′ ′ ′ ′ ′ ′ [d, e 0] = [ d, E 0 ⊗ t] = E0 ⊗ t = e0. The same proof gives the relations [ h, f i ] = −h αi, h ifi These relations give a map ψ : g(Ar) → g(σ) where Ar is the Cartan matrix of type Xr. Now we need the eigenspace decomposition with respect to h: e b b b Proposition 13.1.6 We have the following decompositionb

′ k k k rk rk g(σ) = h(σ) ⊕ CE ′ ⊕ h(k¯) ⊗ t ⊕ CE ⊗ t ⊕ CE ⊗ t α  α   α  Ãα′∈∆′ ! Ãk∈Z ! k∈Z, α ′∈∆′ k∈Z, α ′∈∆′ M M M s M l b     which is the eigenspace decomposition with respect to the action of h(σ). The respective weights are given by 0, α′ ∈ ∆′, kδ ′, α′ + kδ ′ and α′ + rkδ ′. b Proof : The decomposition of g(σ) and of the spaces g(j) give the decomposition. We only need to prove that these spaces have the right weights under the action of h(σ). This is clear for the first ′ ′ ′ k two terms (this comes from theb relations in g ). For h ∈ h(0) and h ∈ h(k¯) we have [ h, h ⊗ t ] = 0 and [ c, h ′ ⊗ tk] = 0 while [ d, h ′ ⊗ tk] = kh ′ ⊗ tk and the weight is bkδ ′. For h ∈ h(0), compute k k k k ′ k k k k k k k k [h, E α ⊗ t ] = ( h, α )Eα ⊗ t = ( h, α )Eα ⊗ t . Furthermore [ c, E α ⊗ t ] = 0 and [ d, E α ⊗ t ] = kE α ⊗ t . The weight is thus α′ + kδ ′. To conclude, we need to remark that if α′ is long, then for k such that k¯ 6= 0, we have: r−1 r−1 k −kj −kj E = ζ E j = ζ E = 0 . α σ (α)   (α) j j X=0 X=0   ¤

Let r be an ideal in g(σ) with trivial intersection with h(σ). Assume r is not trivial, then we get from the fact that r has a weight space decomposition according to the action of h(σ) that r∩h(k¯)⊗tk C k k ¯ −k or r ∩ Eα ⊗ t is non trivial. Take a non trivial x is thatb intersection, then [ x, h(−k) ⊗ t ] or −k −k [x, E −α ⊗ t ] is non trivial giving rise to a non trivial element in r ∩ h(σ). Thisb implies that the map ψ factors through g(Ar). Denote by ψ′ this new map. Because ψ′ is the identity on h(σ) this implies ′ ′ ′ b ′ that ψ is injective. Furthermore, the image of ψ contains the ei, the fi and h(σ) thus the image ′ ′ ′ ′ ′ ′ contains g . Now theb image contains e0 = E0 ⊗ t with E0 ∈ g (1). Because g (1)b is an irreducible g′-module we get all g(1) ⊗ t in the image. Now we prove that g(k¯) ⊗ tk is in theb image by induction ¯ k ′ ¯ k+1 on k ≥ 0. Take x ∈ g(k) ⊗ t and consider [ e0, x ]. This is an element in g(k + 1) ⊗ t and one more time because g(k¯ + 1) is a irreducible g′-module, the result follows. ¤ 110 CHAPTER 13. TWISTED AFFINE LIE ALGEBRAS

Corollary 13.1.7 The root system ∆( σ) of the twisted affine Lie algebra g(σ) is given by

′ Z ′ ′ Z ′ ′ ′ ′ Z ′ ′ ∆( σ) = {kδ / k ∈ , k 6= 0 } ∪ { bα + kδ / k ∈ , α ∈ ∆s} ∪ { α + rkδb / k ∈ , α ∈ ∆l}.

The positiveb roots are given by ′ ′ Z ′ ′ ′ ′ ′ ′ ′ ′ ∆( σ)+ = ∆ + ∪ { kδ / k > 0 ∈ } ∪ { α + kδ / k > 0, α ∈ ∆s} ∪ { α + rkδ / k > 0, α ∈ ∆l}.

′ ′ ′ Theb multiplicity of the roots α + kδ is 1 (this root is real. The multiplicity kδ is m for k¯ = 0 and n − m otherwise. r − 1

Proof : The only subtlety is for the multiplicity of imaginary roots. For k¯ = 0 the eigenspace is h′ = h(0) of dimension m the rank of g′. For k¯ 6= 0 the eigenspace is h(j) whose basis is given by the (j) ′ ′ elements αi for αi ∈ ∆s a simple root. The dimension is s the number of short simple roots.If l is the number of long roots, we have n = l + ( r − 1) s and m = l + s. The result follows. ¤

A2 13.1.2 The case 2n ′ ′ ′ A very similar proof but with some changes in the definition of the elements θ∨, E0 and F0 would lead to the following:

Theorem 13.1.8 Let g be s simple Lie algebra of type A2n and let σ be the non trivial involution of 2 the Dynkin diagram. The Lie algebra g(σ) is isomorphic to the affine Lie algebra of type A2n.

re im The root system is described as follows:b ∆( σ) = ∆( σ) ∪ ∆( σ) . Recall that we proved that ∆( σ)im = {kδ ′ / k ∈ Z, k 6= 0 }. We have b b b 1 ∆(b σ)re = { (α′+(2 k−1) δ′ / k ∈ Z, α ′ ∈ ∆′}∪{ α′+kδ ′ / k ∈ Z, α ′ ∈ ∆′ }∪{ α′+rkδ ′ / k ∈ Z, α ′ ∈ ∆′}. 2 l s l Theb positive roots are given as in the other cases. The multiplicity of reals is 1 and the multiplicity n − m kδ ′ is m for k¯ = 0 and otherwise. r − 1

13.1.3 Further constructions The Dynkin diagrams of affine Lie algebras have more symmetries than the classical ones. We may reproduce the construction by invariants for these Lie algebras. We get in this way another construc- tion, starting from simply laced affine Lie algebras of all affine Lie algebras. We could also construct double affine Lie algebras and double twisted affine Lie algebras by reproducing the construction with the loop algebra.

13.2 Back to the Weyl group

A2 13.2.1 All cases except 2n Let W ′ be the Weyl group of g′. It is a finite group. Consider Q′ the root lattice i.e. the Z submodule ′ ′ ′∨ of h generated by the simple roots αi for i ∈ [1 , m ]. Remark that the coroot lattice Q is contained ∗ in Q′ (when identifying h′ and h′ using the bilinear form ( , )). Any root α′ ∈ ∆′ lies in Q′. In ′ ′ ′ ′ ′ particular for any w ∈ W , we have w(αi) ∈ Q thus W acts on Q . 13.2. BACK TO THE WEYL GROUP 111

Definition 13.2.1 The twisted affine Weyl group W ′ is the semidirect product of the Weyl group W ′ by the root lattice Q′. In symbols: W ′ = W ′ ⋉cQ′. ′ ′ ′ ′ For h ∈ Q the corresponding element in W will be denoted th. It acts on h (or Q ) by translation. ′ ′ c ′ ′ For α a root of g , we will denote by sα′ the reflection in W corresponding to the reflection sα′ ∈ W . c To describe the action of the Weyl group on h∗(σ) we use the element Λ ′ ∈ h∗(σ) defined by ′ ′ ∨ ′ b ′ c′ ∗ hΛ , α i i = δ0,i and hΛ , d i = 0. The elements (( αi)i∈[0 ,n ], Λ ) form a basis of h (σ). As a consequence, ′ ′ ′ ∗ b C ′ C ′ b ′ the elements (( αi)i∈[1 ,n ], δ , Λ ) form a basis of h (σ) and δ ⊕ Λ is a supplementary of h in h(σ). b ′ ′ Remark 13.2.2 Consider the Weyl group Wbaff of the Kac-Moody Lie algebra g(A ) = g(σb) and ′ ′ ′ consider the simple reflections s0, · · · , s m. Let us denote by W the subgroup of Waff generated by ′ ′ ′ ′ ′ C ′ C ′ s1, · · · , s m. The action of si on δ and Λ is trivial thus W acts trivially on δ ⊕ Λb andb stabilises ′ ′ ∗ ′ h . This implies (because Waff acts faithfully on h (σ)) that W acts faithfully on h . We can thus identify W ′ with W acting on h(s) by its action on h′.

′ b Theorem 13.2.3 Let Waff theb Weyl group associated to the Kac-Moody Lie algebra g(σ), then there ′ is a unique isomorphism of groups φ : Waff → W such that φ(s0) = tθ∨ sθ and φ(si) = si for i ∈ [1 , n ]. ∗ b Proof : Let γ ∈ h′ and define the following element T ∈ End( h∗(σ)): c γ b b 1 T (λ) = λ + hλ, rc iγ − (( λ, γ ) + (γ, γ )bhλ, rc i)δ′. γ 2 ′ ′ ′ Remark that for λ such that hλ, c i = 0 we get Tγ(λ) = λ−(λ, γ )δ . In particular Tγ(δ ) = δ . Compute Tγ ◦ Tγ′ , we get: ′ ′ ′ ′ ′ ′ 1 Tγ ◦ Tγ (λ) = Tγ(λ + hλ, rc iγ − (( λ, γ ) + 2 (γ , γ )hλ, rc i)δ ) 1 ′ = λ + hλ, rc iγ − (( λ, γ ) + 2 (γ, γ )hλ, rc i)δ +hλ, rc i(γ′ − (γ′, γ )δ′) ′ 1 ′ ′ ′ −(( λ, γ ) + 2 (γ , γ )hλ, rc i)δ ′ ′ 1 ′ ′ ′ = λ + hλ, rc i(γ + γ ) − (( λ, γ + γ ) + 2 (γ + γ , γ + γ )hλ, rc i)δ = Tγ+γ′ (λ).

∗ ′∗ ∗ In particular Tγ ∈ Aut( h (σ)) (its inverse is T−γ). This gives an embedding of h in Aut( h (σ)) (it is injective because as a group morphism we need to look at the kernel. It is given by the elements ′∗ ′∗ γ ∈ h such that Tγ = Id.b This gives for all λ ∈ h that ( λ, γ ) = 0 thus γ = 0.) b ′ ∗ ′∗ ∗ Recall that Waff is a subgroup of Aut( h (σ)). We assert that Q ⊂ h ⊂ Aut( h (σ)) is contained in ′ ′ ′∨ ′ ′ Waff . Remark that θ∨ is the highest root for ∆ and that its coroot is αθ. We thus have hαθ, θ ∨i = 2 ′ ′ ′ ′ ′ and hθ∨, λ i = 2( αθ, λ )/(αθ, α θ) = ( λ, α θ).b Let us compute the following b ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ 1 ′ Tα ◦ sα (λ) = sα (λ) + hsα (λ), rc iαθ − (( sα (λ), α θ) + 2 (αθ, α θ)hsα (λ), rc i)δ θ θ θ ′ ′θ ′ θ ′ ′ θ ′ = λ − h λ, θ ∨iαθ + hλ, rc iαθ − (( λ, α θ) − 2hλ, θ ∨i + hλ, rc i)δ ′ ′ ′ = λ − h λ, rc − θ∨i(δ − αθ) ′ = sα0 (λ). ′ We get in particular that T ′ ∈ W . Furthermore, for w ∈ W and λ such that hλ, c i = 0, we have αθ aff −1 −1 −1 ′ wT γw (λ) = w(w (λ) − h w (λ), γ iδ ) = λ − h λ, w (γ)iδ) = Tw(γ)(λ). 112 CHAPTER 13. TWISTED AFFINE LIE ALGEBRAS

−1 1 ′ 1 ′ −1 But wT γw (Λ) = w(Λ+ γ− 2 (γ, γ )δ ) = Λ+ w(γ)− 2 (γ, γ )δ . We deduce that wT γw (Λ) = Tw(γ)(Λ) −1 ′ ′ and wT γw = T . We deduce that W · T ′ ∈ W . To prove that Q ⊂ W , it suffices to show w(γ) αθ aff aff ′ ′ ′ ′∨ that W · αθ generates Q. But αθ is the coroot of θ∨ the highest root of the root system ∆ . Applying Lemma 11.2.20 we get the result. ′ ′ ′ Now Q is contained in Waff but this subgroup is normalised by W in Waff . Moreover, because ′ ′ ′ all elements in Q are of infinite order and because W is finite we have W ∩ Q = {1} in Waff . Now ′ ′ ′ ′ in the subgroup generated by W and Q, which is W , we have all the reflections si for i 6= 0 and s0 because of a previous computation. ¤ c A2 13.2.2 Case 2n In this case we have the same result:

′ 2 Theorem 13.2.4 Let Waff the Weyl group associated to the Kac-Moody Lie algebra g(σ) of type A2n, ′ ′ then there is a unique isomorphism of groups φ : Waff → W ⋉ Q . b Bibliography

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Nicolas Perrin , HCM, Universit¨atBonn. email : [email protected]

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