The extension problem for Lie algebroids on schemes

The extension problem for Lie algebroids on schemes

Ugo Bruzzo

SISSA (International School for Advanced Studies), Trieste Universidade Federal da Para´ıba,

Jo˜ao Pessoa, Brazil

S˜ao Paulo, November 14th, 2019

2nd Workshop of the S˜ao Paulo Journal of Mathematical Sciences:

J.-L. Koszul in S˜ao Paulo, His Work and Legacy

Lie algebroids

X: a differentiable manifold, or complex manifold, or a noetherian separated schemeover an algebraically closed field kof

characteristic zero.

Lie algebroid: a vector bundle/coherent sheaf C with a morphism ofOX-modules a:C →Θ_X and ak-linear Lie bracket on the sections ofC satisfying

[s,ft] =f[s,t] +a(s)(f)t for all sections s,t of C andf of OX.

a is a morphism of sheaves of Liek-algebras kerais a bundle of Lie OX-algebras

Examples

A sheaf of Lie algebras, with a= 0 ΘX, with a= id

More generally, foliations, i.e., a is injective Poisson structures Ω¹_X −→^π Θ_X,

Poisson-Nijenhuis bracket

{ω, τ}= Lie_π(ω)τ −Lie_π(τ)ω−dπ(ω, τ) Jacobi identity⇔ [[π, π]] = 0

Lie algebroid morphisms

f:C →C⁰ a morphism ofOX-modules & sheaves of Liek-algebras C ^{f //}

a C⁰

a⁰

Θ_X

⇒kerf is a bundle of Lie algebras

Derived functors

Aan abelian category,A∈Ob(A)

Hom(−,A) :→Ab is a (contravariant) left exact functor, i.e., if

0→B⁰ →B →B⁰⁰→0 (∗) is exact, then

0→Hom(B⁰⁰,A)→Hom(B,A)→Hom(B⁰,A) is exact

Definition

I ∈Ob(A) isinjective ifHom(−,I) is exact, i.e., for every exact sequence (*),

0→Hom(B⁰⁰,I)→Hom(B,I)→Hom(B⁰,I)→0 is exact

Definition

The categoryAhas enough injectives if every object in Ahas an injective resolution

0→A→I⁰ →I¹ →I²→. . . Aabelian category with enough injectives

F:A→B left exact functor Derived functorsRⁱF:A→B

RⁱF(A) =Hⁱ(F(I^•))

Example: Sheaf cohomology. X topological space,A=Sh_X, B=Ab,F = Γ (global sections functor)

RⁱΓ(F) =Hⁱ(X,F)

The category Rep( C )

From now on,X will be a scheme (with the previous hypotheses) Given a Lie algebroidC there is a notion ofenveloping algebra U(C)

It is a sheaf of associativeOX-algebras with ak-linear augmentationU(C)→OX

Rep(C)'U(C)-mod

⇒ Rep(C) has enough injectives

Universal enveloping algebraU(L) of a (k,A)-Lie-Rinehart algebraL

Ak-algebra U(L) with an algebra monomorphism ı:A→U(L) and ak-module morphism:L→U(L), such that

[(s), (t)]−([s,t])= 0, s,t ∈L,

[(s), ı(f)]−ı(a(s)(f))= 0, s ∈L,f ∈A (∗) Construction: standard enveloping algebraU(AoL) of the semi-direct productk-Lie algebraAoL

U(L) =U(AoL)/V, V =hf(g,s)−(fg,fs)i U(L) is anA-module via the morphism ı

due to (*) the left and rightA-module structures are different morphism ε:U(L)→U(L)/I =A(the augmentation

morphism) where I is the ideal generated by(L). Note thatε is a morphism of U(L)-modules but not of A-modules, as ε(fs) =a(s)(f) when f ∈A,s ∈L.

Lie algebroid cohomology

Given a representation (ρ,M) of M define

M^C(U) ={m∈M(U)|ρ(C)(m) = 0}

and a left exact functor

I^C: Rep(C) → k-mod M 7→ Γ(X,M^C) Definition (B 2016¹)

H^•(C;M)'R^•I^C(M)

(¹) J. of Algebra483(2017) 245–261

Grothendieck’s thm about composition of derived functors

A−→^F B−^G→C A,B,C, abelian categories

A,B with enough injectives

F and G left exact,F sends injectives to G-acyclics(i.e., RⁱG(F(I)) = 0 fori >0when I is injective)

Theorem

For every object A inAthere is a spectral sequence abutting to R^•(G◦F)(A) whose second page is

E₂^pq =R^pF(R^qG(A))

Local to global

Rep(C) ⁽⁻⁾

C//

I^C &&

kX-mod

Γ

k-mod

Grothendieck’s theorem on the derived functors of a composition of functors implies:

Theorem (Local to global spectral sequence)

There is a spectral sequence, converging toH^•(C;M), whose second term is

E₂^pq =H^p(X,H^q(C;M))

Hochschild-Serre

Extension of Lie algebroids

0→K →E →Q →0 K is a sheaf of Lie OX-algebras

Rep(E) ⁽⁻⁾

K//

I^E %%

Rep(Q)

I^Q

k-mod

Moreover, the sheaves H^q(K;M) are repre- sentations of Q

Theorem (Hochschild-Serre type spectral sequence)

For every representationM of E there is a spectral sequence E converging toH^•(E;M), whose second page is

E₂^pq =H^p(Q;H^q(K;M)).

The extension problem

An extension

0→K →E −→^π Q→0 (1)

defines a morphims

α:Q → Out(Z(K)) (2)

α(x)(y) = {y,x⁰} where π(x⁰) =x The extension problem is the following:

Given a Lie algebroidQ, a coherent sheaf of LieOX-algebras K, and a morphismαas in (2), does there exist an extension as in (1) which induces the givenα?

We assumeQ is locally free

Abelian extensions

IfK is abelian, (K, α) is a representation of Q onK , and one can form the semidirect product

E =K oαQ,

E =K ⊕Q asOX-modules,

{(`,x),(`⁰,x⁰)}= (α(x)(`<sup>0</sup>)−α(x<sup>0</sup>)(`),{x,x⁰})

Theorem (²)

IfK is abelian, the extension problem is unobstructed; extensions are classified up to equivalence by the

hypercohomology groupH²(Q;K)⁽¹⁾α

E1

0 ^//K

BB

Q ^//0

E2

CC

(²) U.B., I. Mencattini, V. Rubtsov, and P. Tortella, Nonabelian holomorphic Lie algebroid extensions, Internat. J. Math. 26(2015) 1550040

M a representation of a Lie algebroidC. Sharptruncation of the Chevalley-Eilenberg complexσ^≥1Λ^•C^∗⊗M defined by

0 ^//0 ^//C^∗⊗M ^//Λ²C^∗⊗M ^//· · ·

degree 1

OO

We denoteHⁱ(C;M)⁽¹⁾:=Hⁱ(X, σ^≥1Λ^•C^∗⊗M) Derivation ofC in M: morphism d:C →M such that

d({x,y}) =x(d(y))−y(d(x))

Proposition

The functorsHⁱ(C;−)⁽¹⁾ are, up to a shift, the derived functors of Der(C;−) : Rep(C) → k-mod

M 7→ Der(C,M) i.e.,

RⁱDer(C;−)'Hⁱ⁺¹(C;−)⁽¹⁾

Realize the hypercohomology using ˇCech cochains: ifUis an affine coverofX, andF^• a complex of sheaves on X, thenH^•(X,F^•) is isomorphic to the cohomology of the total complexT of

K^p,q= ˇC^p(U,F^q)

0 ^//K|U_i //E|U_i π //Q|U_i //

si

kk 0 (3)

IfU_i ∈U, Hom(Q|Ui,E|Ui)→Hom(Q|Ui,Q|Ui) is surjective, so that one has splittingss_i, and one can define

{φ_ij =s_i−s_j} ∈Cˇ¹(U,K ⊗Q^∗)

This is a 1-cocycle, which describes the extensiononly as an extension ofOX-modules

0→K(U_i)→E(U_i)→Q(U_i)→0

is an exact sequence of Lie-Rinehart algebras (over (k,OX(U_i))) which is described by a2-cocycleψ_i in the Chevalley-Eilenberg (-Rinehart) cohomology ofQ(U_i) with coefficients in K(U_i)

(φ, ψ)∈Cˇ¹(U,K ⊗Q^∗)⊕Cˇ⁰(U,K ⊗Λ²Q^∗) =T²

δφ= 0, dφ+δψ= 0, dψ= 0

⇒cohomology class in H²(Q;K)⁽¹⁾α

The nonabelian case

Theorem (^2,3)

IfK is nonabelian, the extension problem is obstructed by a class ob(α)in H³(Q;Z(K))⁽¹⁾α .

Ifob(α) = 0, the space of equivalence classes of extensions is a torsor onH²(Q;Z(K))⁽¹⁾α .

Proof

Q can be written as a quotient of afree Lie algebroid F

(³) E. Aldrovandi, U.B., V. Rubtsov, Lie algebroid cohomology and Lie algebroid extensions, J. of Algebra 505 (2018) 456–481

0

J

0 ^//N ^//U(F) ^//

U(Q) ^//0 OX

0

Nfⁱ =N ⁱ/N ⁱ⁺¹, Jfⁱ =NⁱJ/Nⁱ⁺¹J, for i = 0, . . . Locally free resolution

· · · →Nf² →Jf¹→Nf¹→Jf⁰ →J →0

As Hom_U(Q)(J,Z(K))'Der(Q,Z(K)), applying the functor Hom_U(Q)(−,Z(K)) we obtain

0→Der(Q,Z(K))→Hom_U(Q)(Jf⁰,Z(K))−→^d1

Hom_U(Q)(Kf¹,Z(K))−→^d2 Hom_U(Q)(Jf¹,Z(K))−→^d3 Hom_U(Q)(Kf²,Z(K ))→. . . The cohomology of this complex is isomorphic toH^•+1(Q;Z(K)).

Pick a lift ˜α:F →Der(K) of α and get commutative diagram

0 ^//T ^//

β

F ^//

˜

α

Q ^//

α

0

0 ^//Z(K ) ^//K ^{ad ////}Der(K) ^//Out(K) ^//0 whereβ is the induced morphism.

Define a morphism

o:Jf¹→Z(K) (4) It is enough to defineo on an element of the typeyx, wherex is a generator ofF, and y is a generator of T

o(yx) =β({x,y})−α(x)(β(y)).˜ Note thato ∈Hom_U(Q)(Jf¹,Z(K )).

Lemma

d3(o) = 0. Moreover, the cohomology class of [o]∈H³(Q;Z(K))⁽¹⁾ only depends onα.

Part I of the proof: if an extension exists consider the diagram

0 ^//T ^//

β

F ^//

γ

Q ^//0

0 ^//K ^//E ^//Q ^//0

Define

˜

α:F →Der(K,K), α˜=−ad◦γ

Then ˜α is a lift of α, and for all sectionst ofT andx ofF

β({x,t})−α(x)(β(t)) = 0˜ (5)

so that the obstruction class ob(α) vanishes.

Conversely, assume thatob(α) = 0, and take a lift

˜

α:F →Der(K,K). The corresponding cocycle lies in the image of the morphismd2, so it defines a morphism β:T →K, which satisfies the equation (5). Again, we consider the extension

0→T →F →Q→0.

Note thatK is anF-module via F →Q. The semidirect productK oF contains the sheaf of Lie algebras

H ={(`,x)|x ∈T, `=β(x)}.

The quotientE =K oF/H provides the desired extension

Part II of the proof: reduction to the abelian case Proposition

Once a reference extensionE0 has been fixed, the equivalence classes of extensions ofQ byK inducing α are in a one-to-one correspondence with equivalence classes of extensions ofQ by Z(K) inducingα, and are therefore in a one-to-one

correspondence with the elements of the groupH²(Q;Z(K))⁽¹⁾

C1,C2 Lie algebroids with surjective morphismsfi:Ci →Q.

AssumingZ(kerf1)'Z(kerf2) =Z define C1?C2 =C1×_QC2/Z, whereZ →C1×_QC2 byz 7→(z,−z) Fix a reference extensionE0 of Q byK Lemma

(1) Any extensionE ofQ byK is equivalent to a product E0?D whereD is an extension of Q by Z(K)

(2) Given two extensionsD1,D2 of Q by Z(K), the extensions E1=E0?D1 andE2=E0?D2 are equivalent if and only ifD1 and D2 are equivalent

Muito obrigado pela aten¸c˜ ao!!