Descent Theory and Its Applications in Lie Theory. Zhihua Chang. Bar-Ilan University. September 22, 2014

Transcription

1 GAP Seminar at USTC Descent Theory and Its Applications in Lie Theory Zhihua Chang Bar-Ilan University September 22, Department of Mathematics, Bar-Ilan University, Ramat-Gan, 52900, Israel 1 / 86

2 2 / 86 References A. Pianzola, (notes by Z. Chang) Descent Theory with an Application to Infinite Dimensional ebras, zhchang/research/ FieldsNotes_15X13.pdf. M. A. Knus, A. Merkurjev, M. Rost, and J. P. Tignol, The book of involution (Ch7) AMS, Providence, RI, M. A. Knus, M. Ojanguren Théorie de la descente et algèbres d Azumaya, Springer, New York, J. P. Serre Galois cohomology Springer-Verlag, Berlin, W. C. Waterhouse Introduction to affine group schemes (Ch17&18) Springer-Verlag, New York, 1979.

3 3 / 86 Assumptions All rings are assumed to be unital commutative associative rings. A ring homomorphism is assumed to maps the identity element to the identity element. A ring extension S/R means a ring homomorphism R S.

4 4 / 86 Basic Questions Let S/R be a ring extension and N an R module. We know that N R S is an S module. Questions Given an S module M, does there exist an R module N such that N R S = M? can we classify all R modules N such that N R S = M? Hopeless question in general. Example. R = Z and S = Q. For every finite abelian group N, N R S = 0. Nice answer when S/R is faithfully flat.

5 5 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

6 6 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

7 7 / 86 Faithfully flat modules Let E be an R module. We consider the functor: E : R-Mod R-Mod, N N E. Definition E is flat if the functor E is exact, i.e., the exactness of N N N in R Mod implies the exactness of N E N E N E.

8 8 / 86 Faithfully flat modules Definition E is faithfully flat if E is flat and the functor E is faithful, or equivalently, the functor E is exact and faithful. The functor E is faithful if it induces an injective homomorphism Hom R (N, N ) Hom R (N E, N E) for arbitrary R modules N and N,or equivalently, for a homomorphism α : N N of R modules, α 1 = 0 implies α = 0.

9 9 / 86 Faithfully flat modules Proposition For an R module E, the following are equivalent (FFa) E is faithfully flat. (FFb) N N N is exact iff N E N E N E is exact. (FFc) E is flat and for N R Mod, N E = 0 implies N = 0. (FFd) E is flat and for N α N R Mod, α 1 is injective implies α is injective. (FFe) E is flat and me E for every maximal ideal m of R.

10 10 / 86 Faithfully flat modules Examples E := p Spec(R) R p is a faithfully flat R module. If E is a faithfully flat R module, then for every ring extension R /R, E R R is a faithfully flat R module. A free module of finite rank is faithfully flat. Q is a flat Z-module but not faithfully flat.

11 11 / 86 Faithfully flat ring extension Definition A ring extension S/R is faithfully flat if S is a faithfully flat R module. Examples An arbitrary field extension is faithfully flat. Let k be a field and m N. The ring extension k[t ±1 ] k[t ± 1 m ] is faithfully flat.

12 12 / 86 Faithfully flat ring extensions Proposition Let S/R be a ring extension. The following are equivalent: S/R is faithfully flat. S/R is flat and for every R module N, the canonical map ι : N N S, n n 1 is injective. S/R is flat and the map Spec(S) Spec(R), q q R is surjective.

13 Faithfully flat ring extensions Proposition Let S/R be a faithfully flat ring extension and N an R module. Then 0 N ι N S p N 1 N S S p N 2 is exact, where p N 1 p N 2 :N S N S S, n s n s 1, :N S N S S, n s n 1 s. 13 / 86

14 14 / 86 Sketch of the Proof Since S/R is faithfully flat, it is equivalent to show 0 N S ι 1 N S S p N 1 1 p N 2 1 N S S S is exact. Im(ι 1) ker(p N 1 1 pn 2 1). Define β : N S S S N S S, n a b c n a bc.

15 15 / 86 Sketch of the Proof If n i a i b i ker(p N 1 1 pn 2 1), then ni a i 1 b i = n i 1 a i b i. Applying β, ni a i b i = n i 1 a i b i Im(ι 1).

16 16 / 86 Properties of faithfully flat base change Proposition Let S/R be a faithfully flat ring extension and N an R module. Then N is of finite type (resp. finite presented, or projective of finite type) iff N S has the same property. (c.f. ChI, 3.7, Proposition 11 and 12 in [Bourbaki, Commutative algebras])

17 17 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

18 18 / 86 Assumption From now on, we assume that S/R is a faithfully flat ring extension.

19 19 / 86 Covering datum Let M be an S module. There are two S S module structures on M S: for a, b, s S and m M. (a b)(m s) = am bs, (a b).(m s) = bm as, Definition A covering datum on M is an S S module isomorphism ψ : M S M S between two S S module structures on M S.

20 20 / 86 Covering Data Definition A covering datum on M is an additive isomorphism such that for all a, b, s S and m M. ψ : M S M S ψ(am bs) = (b a)ψ(m s),

21 21 / 86 The standard covering datum Let N be an R module and M := N S. Then θ : N S S N S S, is a covering datum on M = N S. Facts: As an R module, N = {m M θ(m 1) = m 1} n s t n t s = { n i s i N S n i 1 s i = n i s i 1}.

22 22 / 86 Descent data Let M be an S module and ψ : M S M S a covering datum. There are three additive isomorphisms ψ i : M S S M S S, i = 0, 1, 2, defined as follows: if ψ(m a) = m i a i, for m M and a, u S. ψ 0 (m u a) = m i u a i, ψ 1 (m u a) = m i a i u, ψ 2 (m a u) = m i a i u, Definition A covering datum ψ : M S M S is called a descent datum if ψ 1 = ψ 0 ψ 2.

23 23 / 86 Descent data Example Let N be an R module and M := N S. Then the standard covering datum θ : N S S N S S, is a descent datum. n s t n t s

24 24 / 86 Descended modules Theorem Let M be an S module and ψ : M S M S a descent datum. Then N := {m M ψ(m 1) = m 1} is an R module and as S modules. N S = M

25 25 / 86 Sketch of the Proof N is an R module. For n N, r R, ψ(rn 1) = ψ((r 1)(n 1)) = (1 r)(n 1) = n r = rn 1. Definition of N gives an exact sequence 0 N M ι M S. ψι By flatness of S/R, we have an exact sequence 0 N S M S ι 1 M S S. ψι 1

26 Sketch of the Proof Since S/R is faithfully flat, we have an exact sequence 0 M M S pm 2 M S S. The descent condition ψ 1 = ψ 0 ψ 2 implies that the following diagram commutes: p M 1 ι 1 0 N S M S M S S ψι 1 ψ ψ 0 0 M M S pm 2 M S S. p M 1 From the diagram, we deduce N S = M. 26 / 86

27 27 / 86 Descended modules Given two S modules with covering data (M, ψ) and (M, ψ ), a morphism of covering data (M, ψ) (M, ψ ) is a homomorphism of S modules α : M M such that the diagram commutes. Theorem M S α 1 M S ψ ψ M S α 1 M S the category of R modules equivalence the category of S modules with descent datum

28 28 / 86 Descended algebras Let M be an S module and ψ : M S M S a descent datum. Assume in addition that M is an S algebra and ψ preserves the multiplication: ψ(m 1 m 2 a 1 a 2 ) = ψ(m 1 a 1 )ψ(m 2 a 2 ) for m 1, m 2 M and a 1, a 2 S. Then the descended module N = {m M ψ(m 1) = m 1} is actually an R algebra.

29 29 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

30 30 / 86 Faithfully flat twisted forms Our second question: Given an R module N, how can we classify R modules N such that N S = N S?

31 31 / 86 Faithfully flat twisted form Let N be an R module, M := N S, θ the standard covering datum θ : N S S N S S, n s t n t s. Fact ψ : N S S N S S is a covering datum iff ϕ := θ 1 ψ Aut S S-mod (N S S). Aim Formulate descent conditions on ϕ.

32 32 / 86 The functor GL(N) GL(N) : R-alg grp, R Aut R -mod(n R ). Given a homomorphism of R-algebras f : R R, we have a group homomorphism GL(N)(f ) : Aut R -mod(n R ) Aut R -mod(n R ), φ φ R, where ϕ R : N R N R is defined by if ϕ(n 1) = n i r i. ϕ R (n r ) = n i f (r i)r

33 33 / 86 Descent conditions on ϕ Consider three homomorphisms of R-algebras: S S S S S p 12 : s t s t 1, p 13 : s t s 1 t, p 23 : s t 1 s t. By the functoriality of GL(N), we get three group homomorphisms d jk := GL(N)(p jk ) : GL(N)(S S) GL(N)(S S S) for 1 i < j 3.

34 34 / 86 Descent conditions on ϕ Applying d 12, d 23 and d 13 to ϕ, we get d jk ϕ GL(N)(S S S), 1 j < k 3. Explicitly, for n N and a, b S, if ϕ(n a b) = n i a i b i, then (d 12 ϕ)(n a b u) = n i a i b i u, (d 13 ϕ)(n a u b) = n i a i u b i, (d 23 ϕ)(n u a b) = n i u a i b i. Proposition ψ := θ ϕ : N S S N S S is a descent datum iff d 13 ϕ = (d 23 ϕ)(d 12 ϕ).

35 35 / 86 1-cocycle and its descended module Definition An element ϕ GL(N)(S S) is called a 1-cocycle if d 13 ϕ = (d 23 ϕ)(d 12 ϕ). A 1-cocycle ϕ GL(N)(S S) determines a descent datum ψ := θϕ, and determines a descended R module N : = { n i s i N S ψ( n i s i 1) = n i s i 1} = { n i s i N S ϕ( n i s i 1) = n i 1 s i }.

36 36 / 86 Equivalent 1-cocycles Question Two 1-cocycles ϕ and ϕ determines two descended modules N and N respectively. When are N and N isomorphic? If N is isomorphic to N, we have a commutative diagram N S = N S. = N S λ = = N S Hence, we obtain λ GL(N)(S).

37 37 / 86 Equivalent 1-cocycles Consider the following two homomorphisms of R algebras: S S S, p 1 : s s 1, p 2 : s 1 s. By the functoriality of GL(N), we get d i = GL(N)(p i ), i = 1, 2. Applying to λ, we obtain d 1 λ, d 2 λ GL(N)(S S). Explicitly, if λ(n a) = n i a i, then (d 1 λ)(n a u) = n i a i u, (d 2 λ)(n u a) = n i u a i.

38 38 / 86 Equivalent 1-cocycles Moreover, N is isomorphic to N implies that which yields that (λ 1)θϕ = θϕ (λ 1), ϕ = (d 2 λ)ϕ(d 1 λ) 1.

39 The first non-abelian cohomology set Let G : R-alg grp be a functor. Define the set of 1-cocycles Z 1 (S/R, G) = {ϕ G(S S) d 13 ϕ = (d 23 ϕ)(d 12 ϕ)}. For ϕ, ϕ Z 1 (S/R, G), ϕ ϕ if there exists λ G(S) such that (d 2 λ)ϕ(d 1 λ) 1 = ϕ. is an equivalence relation on Z 1 (S/R, G). The set of equivalent classes H 1 (S/R, G) = Z 1 (S/R, G)/ is called the first non-abelian cohomology set. 39 / 86

40 40 / 86 Twisted forms and their classification Definition Let N be an R module and S/R a faithfully flat ring extension. An R module N is called an S/R twisted form of N if as S modules. N S = N S Theorem The set of isomorphism classes S/R twisted forms of N 1-1 H 1 (S/R, GL(N))

41 41 / 86 Twisted forms of algebras If N has additional structures, the descent formalism can be rewritten with respect to certain structures on N. For example, if N is a Lie algebra over R, we define Then Aut(N) : R-alg grp, R Aut R -Lie(N R ). The set of isomorphism classes of R Lie algebras N such that N S = N S as S-Lie algebras 1-1 H 1 (S/R, Aut(N))

42 42 / 86 Twisted forms of algebras Example Let R and C be the field of real and complex numbers, and so 3 (R): the Lie algebra R 3 under the cross product, sl 2 (R): the Lie algebra of trace zero 2 2-matrices under the commutator. Then so 3 (R) C = sl 2 (R) C as Lie algebras over C. Hence, so 3 (R) is a C/R form of sl 2 (R). Note that so 3 (R) = sl2 (R).

43 43 / 86 Hilbert s Theorem 90 Theorem Let R be a local ring, and GL m,r := GL(R m ) for n 1. Then, for every faithfully flat ring extension S/R, Sketch of the Proof H 1 (S/R, GL m,r ) = 1. Let N := R n. Then GL m,r = GL(N). It suffices to show that if an R module N such that N S = N S = S n, then N = R n.

44 44 / 86 Sketch of the Proof Since S/R is faithfully flat and N S = S n is free of finite rank, N is projective of finite type. Since R is local, N = N.

45 45 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

46 46 / 86 Galois extension Recall that a field extension L/K is Galois if it is normal and separable. normal: L is the splitting field of a family of polynomials in K[X]. separable: algebraic and the minimal polynomial of any α L has no multiple root. L/K is Galois iff L/K is algebraic and K = {α L γ(α) = α, γ Gal(L/K)}.

47 47 / 86 Galois ring extension Definition Let S/R be a ring extension and Γ a finite subgroup of Aut R (S). We say that S/R is Galois with Galois group Γ if both of the following conditions are satisfied: (Gal1) S/R is faithfully flat. (Gal2) The map ϱ : S S Γ S, a b (γ(a)b) γ Γ, is an isomorphism of S algebras.

48 48 / 86 Galois ring extension Facts Let S/R be a Galois ring extension with Galois group Γ. Then ϱ : S S Γ S, a b (γ 1 (a)b) is an isomorphism of S algebras. R = {s S γ(s) = s, γ Γ}. Indeed, since S/R is faithfully flat, there is an exact sequence 0 R S p 1 p 2 S S. The assertion follows by applying the isomorphism ϱ to the fourth item.

49 Then S/R is a Galois extension with Galois group Γ. 49 / 86 Galois ring extension Example: a usual Galois field extension Let L/K be a Galois field extension with Galois group Γ. Then L = K[x]/(f (x)), where f (x) = σ Γ (x σ(a)) for some a L. Hence, L K L = L K (K[x]/(f (x))) = L[x]/(f (x)) = L. Example S = R R, Γ = {1, σ} = Z/2Z, and σ(a, b) = (b, a).

50 50 / 86 Galois ring extension Example: Laurent polynomial ring R = C[t ±1 1,..., t±1 N ], S = C [t ± 1 m 1 1,..., t ± 1 N m N ], Γ = Z/m 1 Z Z/m N Z, for (ī 1,..., ī N ) Γ, (ī 1,..., ī N ).t 1 m j j = ζ i j 1 m m j t j j, j = 1,..., N. Then S/R is a Galois extension with Galois group Γ.

51 51 / 86 Sketch of the Proof in Case N = 1 S = C[t ± 1 m ] is a free R = C[t ±1 ] module, so S/R is faithfully flat. We can show that ϱ : S S γ Γ S, a b (γ(a)b) γ Γ has the inverse map S S S γ Γ (s γ ) γ Γ γ Γ m 1 γ 1 (t i i m ) t m sγ, i=0

52 52 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

53 53 / 86 Assumption & Question Assumption In this section, we assume S/R is a Galois extension with Galois group Γ. Question How can we reformulate the descent conditions in the case of Galois extension?

54 54 / 86 Galois Covering Data Let M be an S module. Recall that a covering datum is an additive isomorphism ψ : M S M S such that for m M and a, b, s S. Since S/R is Galois, ψ(am bs) = (b a)ψ(m s). ϱ M :M S Γ ϱ M :M S Γ M, m s (γ(s)m) γ Γ, M, m s (γ 1 (s)m) γ Γ, are both additive isomorphisms.

55 Galois Covering data A covering datum ψ uniquely determines a map h : Γ M Γ M such that the following diagram commutes: M S ψ M S ϱ M ϱ M Γ M h Γ M. Moreover, h is componentwise. It determines a family of additive isomorphisms h γ : M M, γ Γ such that each h γ is γ-semi-linear, i.e., h γ (sm) = γ(s)h γ (m). 55 / 86

56 56 / 86 Galois Covering Data Proposition covering data ψ : M S M S 1-1 Γ -tuples of additive isomorphisms (h γ : M M) γ Γ such that h γ is γ-semi-linear.

57 57 / 86 The standard covering datum Let N be an R module and M := N S. Then the standard covering datum θ : N S S N S S, n s t n t s. corresponds to for γ Γ. h γ : M M, n s = n γ(s),

58 58 / 86 Galois Descent Data Proposition Let ψ : M S M S be a covering datum, and h γ : M M, γ Γ the maps determined by ψ. Then ψ is a descent datum iff for all σ 1, σ 2 Γ. h σ1 σ 2 = h σ1 h σ2

59 Sketch of the Proof Recall that ψ is a descent datum iff ψ 1 = ψ 0 ψ 2, where, for m M and a, u S, ψ 0 (m u a) = m i u a i, ψ 1 (m u a) = m i a i u, ψ 2 (m a u) = m i a i u, if ψ(m a) = m i a i. Since S/R is Galois, ϱ M : M S S Γ Γ M, m a b (γ 1 (a)γ 2 (b)m) (γ1,γ 2 ) Γ 1 Γ 2 is an additive isomorphism. 59 / 86

60 Sketch of the Proof Each ψ i determines an additive isomorphism h i satisfying the commutative diagram: M S S ψ i M S S ϱ M ϱ M M Γ Γ h i M Γ Γ h 0, h 1, h 2 satisfy h 0 ( (δ σ1,γ 1 δ σ2,γ 2 m) (γ1,γ 2 ) Γ Γ) = (δσ 1 2 h 1 ( (δ σ1,γ 1 δ σ2,γ 2 m) (γ1,γ 2 ) Γ Γ) = (δσ 1 2 δ σ 1,γ 1 σ 1 h,γ 2 2 σ 1 (m)) (γ1,γ 2 ) Γ Γ 2 δ,γ 1 σ 1 h σ 2 1,γ 2 σ 1 (m)) (γ1,γ 2 ) Γ Γ 2 h 2 ( (δ σ1,γ 1 δ σ2,γ 2 m) (γ1,γ 2 ) Γ Γ) = (δσ 1 1 for σ 1, σ 2 Γ and m M. δ,γ 1 σ 1 h σ 1 2,γ 2 σ 1 (m)) (γ1,γ 2 ) Γ Γ 1 60 / 86

61 61 / 86 Galois Descended Modules Let M be an S module and ψ : M S M S a descent datum. Let (h γ ) γ Γ be the maps corresponding to ψ. Then the descended module determined by ψ is N : = {m M ψ(m 1) = m 1} = {m M h γ (m) = m, γ Γ}.

62 62 / 86 Galois Descended Data Let (h γ ) γ Γ be a Galois descent datum. Then γ Γ M M, (γ, m) m is an action of Γ on M. It is semi-linear, i.e., Proposition γ (sm) := h γ (m), γ = γ(s) m, s S, m M. Galois descent data (h γ : M M) γ Γ 1-1 semi-linear Γ-actions on M.

63 63 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

64 Assumption & Aim Assumption S/R is a Galois extension with Galois group Γ. N is an R module. Aim Reformulate 1-cocycle conditions in the case of Galois extensions. Possible Directions Galois Descent Data Faithfully Flat Twisted Forms Galois Twisted Forms 64 / 86

65 65 / 86 Galois 1-cocycles Let G : R-alg grp be a functor such that G(R R ) = G(R ) G(R ). Recall that a 1-cocyle ϕ Z 1 (S/R, G) is an element ϕ G(S S) such that G(p 13 )(ϕ) = G(p 23 )(ϕ)g(p 12 )(ϕ), where S S S S S p 12 : s t s t 1, p 13 : s t s 1 t, p 23 : s t 1 s t.

66 66 / 86 Galois 1-cocycles Since S/R is Galois, we have additive isomorphisms ϱ :S S Γ S, a b (γ(s)t) γ Γ, ϱ :S S S Γ Γ S, a b c (γ 1 (a)γ 2 (b)c) (γ1,γ 2 ) Γ Γ. By the functoriality of G, we obtain group homomorphisms G(ϱ) :G(S S) Γ G(S), G(ϱ ) :G(S S S) Γ Γ G(S).

67 67 / 86 Galois 1-cocycles Denote the canonical projections: p σ,τ : Γ Γ S S, Now, ϕ Z 1 (S/R, G) iff p γ : Γ S S. G(p σ,τ ϱ p 13 )(ϕ) = G(p σ,τ ϱ p 23 )(ϕ)g(p σ,τ ϱ p 12 )(ϕ), for all σ, τ Γ.

68 68 / 86 Galois 1-cocycles Let ϕ γ := G(p γ ϱ)(ϕ) G(S). Define a Γ-action on G(S) by γ λ := G(γ)(λ) = (1 γ)λ(1 γ 1 ), γ Γ. ϕ Z 1 (S/R, G) iff τ ϕ τσ = ϕ τ ϕ σ, σ, τ Γ.

69 69 / 86 Galois 1-cocycles Two 1-cocycles ϕ, ϕ Z 1 (S/R, G) are equivalent if there exists λ G(S) such that which is equivalent to (d 2 λ)ϕ(d 1 λ) 1 = ϕ, γ λ ϕ γ λ 1 = ϕ γ, γ Γ.

70 70 / 86 The first Galois cohomology Definition A Galois 1-cocycle is a map Γ G(S), γ ϕ γ such that τ ϕ τσ = ϕ τ ϕ σ, σ, τ Γ. The set of Galois 1-cocycles is denoted by Z 1 (Γ, G(S)). Two Galois 1-cocycles (ϕ γ ) and (ϕ γ) are equivalent if if there exists λ G(S) such that γ λ ϕ γ λ 1 = ϕ γ, γ Γ.

71 Galois descended modules The set H 1 (Γ, G(S)) = Z 1 (Γ, G(S))/ is called the first Galois cohomology set. Let N be an R module and (ϕ γ ) γ Γ be the Galois 1 cocycle associated to the 1 cocycle ϕ GL(S S). The descended module determined by (ϕ γ ) γ Γ is N : = { n i s i N S ϕ( n i s i 1) = n i 1 s i } = { n i s i ϕ γ ( n i γ(s i )) = n i s i }. The Galois descent datum associated to (ϕ γ ) γ Γ is (h γ ) γ Γ, where h γ : M M, m ϕ γ ( γ m), for γ Γ. 71 / 86

72 72 / 86 Hilbert s Theorem 90: Classical Statement Hilbert s Theorem 90 Let L/K be a cyclic Galois extension of fields with Galois group Γ := Gal(L/K) generated by σ. If a L such that N L/K (a) = 1, then there existed b L such that a = σ(b) b. Sketch of the Proof H 1 (Γ, L ) = H 1 (L/K, GL 1 ) = 1. Since Γ is cyclic, every 1-cocycle α : Γ L is determined by c := α(σ). Indeed, α(σ i ) = c σ(c) σ i 1 (c).

73 73 / 86 Sketch of the Proof If σ has order m, then 1 = α(σ m ) = c σ(c) σ m 1 (c) = N L/K (c). Conversely, an element a L such that N L/K (a) = 1 determines a 1-cocycle α : σ a. H 1 (Γ, L ) = 1 implies that the 1-cocycle α is equivalent to the trivial 1-cocyle, i.e., there is b L such that a = b 1 σ(b).

74 74 / 86 Contents Faithfully flat modules and rings Faithfully flat descent Faithfully flat twisted forms Galois Ring Extension Galois Descent Galois Twisted Forms Applications to Lie theory

75 75 / 86 Untwisted loop Lie algebras Let g be a finite dimensional Lie algebra over C. Then g C C[t ±1 ] is a Lie algebra over C under the multiplication for x, y g and m, n Z. [x t m, y t n ] = [x, y] t m+n The Lie algebra g C [t ±1 ] is called a untwisted loop Lie algebra.

76 76 / 86 Twisted loop Lie algebras Let g a finite dimensional simple Lie algebra over C, σ an automorphism of g of order m. The twisted loop algebra L(g, σ) is the subalgebra L(g, σ) = g i C Ct i m g C C[t ± 1 m ], i Z where g i = {x g σ(x) = ζ i mx}, ζ m is a m-th primitive root of unit. In particular, L(g, id) = g C C[t ±1 ].

77 77 / 86 Affine Kac-Moody algebras Theorem (Kac, Moody, 1960s) Every affine Kac-Moody algebra over C (derived modulo center) is isomorphic to a twisted loop Lie algebras.

78 78 / 86 Key observations on the twisted loop algebras Theorem (Pianzola, 2005) Every twisted loop Lie algebra L(g, σ) is a Lie algebra over C[t ±1 ]. A twisted loop Lie algebra L(g, σ) is a twisted form of g C C[t ±1 ] with respect to the Galois ring extension C[t ±1 ] C[t ± 1 m ], i.e., L(g, σ) C[t ±1 ] C[t ± 1 m ] = (g C C[t ±1 ]) C[t ±1 ] C[t ± 1 m ] = g C C[t ±1 ]

79 79 / 86 Twisted loop Lie algebras as twisted forms A 1-cocycle associated to L(g, σ) is ϕ : g C C[t ± 1 m ] C[t ±1 ] C[t ± 1 m ] g C C[t ± 1 m ] C[t ±1 ] C[t ± 1 m ] if a i g i. a i r s a i t i m r t i m s A Galois 1-cocycle associated to L(g, σ) is the Galois 1-cocycle z : Z/mZ Aut(g C C[t ±1 ]) determined by z( 1) = σ 1 1.

80 Review of progresses in Lie theory with the use of descent theory central extensions: [Pianzola-Prelat-Sun 2007], [Sun 2009]. derivations: [Pianzola 2010]. conjugacy of Cartan subalgebras: [Pianzola 2004], [Chernousov-Gille-Pianzola 2011], [Chernousov-Gille-Pianzola-Yahorau 2014]. finite dimensional irreducible representations: [Lau 2010], [Lau-Pianzola 2013]. invariant bilinear forms: [Neher-Pianzola-Prelat-Sepp 2013]. classification of twisted loop Lie conformal superalgebra: [Kac-Lau-Pianzola 2009], [Chang-Pianzola 2011], [Chang-Pianzola-2013] 80 / 86

81 81 / 86 Assumptions Let R = C[t ±1 ] and S = C[t ± 1 m ]. Let g be a finite dimensional simple Lie algebra over C. L = L(g, σ) a twisted loop Lie algebra.

82 Application I: Perfectness Proposition L is perfect. i.e. [L, L] = L. Proof Since L R S = g C S, [L, L] R S = [L R S, L R S] = [g C S, g C S] = g C S. Hence, L [L, L] R S = Since S/R is faithfully flat, L R S [L, L] R S = g C S g C S = 0. L = [L, L]. 82 / 86

83 83 / 86 Application II: Centroid Definition Let k be a ring and A a Lie algebra over k. Then the centroid of A is Ctd k (A) = {χ End k (A) χ([a, b]) = [χ(a), b] = [a, χ(b)], a, b A}. Facts Ctd k (A) is commutative if A is perfect. There is a canonical homomorphism k Ctd k (A) since for every r k, χ r : a ra is an element of Ctd k (A). If k /k is a ring extension and A is a k algebra, then Ctd k (A) Ctd k (A). The canonical map R = CtdC (g C R) is an isomorphism.

84 84 / 86 Application II: Centroid Proposition The canonical map R Ctd C (L) is an isomorphism. Proof Since Ctd C (g C R) = R and L R S = g C S, Ctd C (L) R S = Ctd C (L R S) = Ctd C (g C S) = S. Hence, R R S Ctd C (L) R S is an isomorphism. Since S/R is faithfully flat, R Ctd C (L) is an isomorphism.

85 85 / 86 Application III: Derivations Proposition Der R (L) = IDer(L). Proof. Using Whitehead s first lemma, we know that Since L R S = g C S, we have Der R (g C S) = IDer(g C S). IDer(L) R S = IDer(L R S) = IDer(g C S). Hence, Der R (L) IDer(L) R S = Der R(L) R S IDer(L) R S = Der S(g C S) IDer(g C S) = 0. Since S/R is faithfully flat, Der R (L) = IDer(L).

86 86 / 86 Thank You!