On the algebra generated by μ ¯ , ∂ ¯ , ∂ , μ

Theorem 2.1

The associative algebra A generated by μ ¯ , ∂ ¯ , ∂ , and μ is the universal enveloping algebra of the graded Lie algebra g they generate.

Proposition 2.2

Suppose R ⊂ L S . Then, T S / I R is the universal enveloping algebra of L S / J R , where J R is the Lie ideal of L S generated by R .

Proof

Consider the graded Lie algebra homomorphism L S ↪ T S → T S / I R (since I R is a two-sided ideal, it is also a Lie ideal). Since R ⊂ L S is mapped to zero in T S / I R , we have the following commutative diagram of graded Lie algebras:

This, in turn, yields, by the universal property of universal enveloping algebra, the following commutative diagram of associative algebras:

On the other hand, consider the algebra map T S = U L S → U ( L S / J R ) , which takes R ⊂ T S to zero. Therefore, we have a map ψ : T S / I R → U ( L S / J R ) , and the following diagram commutes:

Note that both vertical maps are surjective: the right vertical map is surjective by construction and the left vertical map is surjective due to the Poincaré-Birkhoff-Witt theorem. To see the latter, one can choose an ordered basis of L S / J R , lift it to L S , and extend it to an ordered basis of L S . Now the commutativity of the diagram and the surjectivity of the vertical maps force ϕ and ψ to be inverses.□

Lemma 2.3

The Lie subalgebra h ⊂ g generated by ∂ , ∂ ¯ is a Lie ideal.

Proof

Since g is generated by μ ¯ , ∂ ¯ , ∂ , and μ and [ ∂ ¯ , − ] and [ ∂ , − ] preserve h , it suffices to show [ μ ¯ , − ] and [ μ , − ] preserve h . Furthermore since h is generated by ∂ ¯ and ∂ , it suffices to check that [ μ ¯ , ∂ ¯ ] , [ μ ¯ , ∂ ] , [ μ , ∂ ¯ ] , and [ μ , ∂ ] are contained in h , which obviously follows from (2).□

Corollary 2.4

g = h in degrees ≥ 2 , and g / h is isomorphic to the free graded abelian Lie algebra on two (equivalent classes of) generators μ ¯ and μ .

Proof

From

we see g 2 = h 2 = span { [ ∂ , ∂ ] , [ ∂ ¯ , ∂ ¯ ] , [ ∂ , ∂ ¯ ] } . Then, inductively using the above lemma and that

we have g = h in degrees ≥ 2 . The second assertion follows easily.□

Theorem 2.5

The Lie subalgebra h of g generated by ∂ ¯ and ∂ is a free graded Lie algebra.

Proof

Let h ′ be the free graded Lie algebra generated on symbols ∂ ¯ and ∂ . We define a derivation D μ ¯ on h ′ by first setting

and then extending it to a derivation. It is easy to see that [ D μ ¯ , D μ ¯ ] = 2 D μ ¯ 2 = 0 on ∂ ¯ and ∂ . Since [ D μ ¯ , D μ ¯ ] is a derivation on h ′ that vanishes on the generators, [ D μ ¯ , D μ ¯ ] must vanish identically on h ′ .

Define g ″ to be the semi-direct product h ′ ⊕ C D μ ¯ with Lie bracket inherited from that of h ′ , the action of D μ ¯ on h ′ , and [ D μ ¯ , D μ ¯ ] = 0 . More precisely, for homogeneous x , y ∈ h ′ , and s , t ∈ C , we define

This Lie bracket indeed satisfies the Jacobi identity since D μ ¯ is a derivation on h ′ . We can similarly define a derivation D μ on g ″ by

Since D μ 2 vanishes on ∂ ¯ and ∂ , and

We see [ D μ , D μ ] vanished identically on g ″ . Define g ′ to be the semi-direct product g ″ ⊕ C D μ .

Now we show that g ′ is isomorphic to g . First, consider the map from the free graded Lie algebra generated by the symbols μ ¯ , ∂ ¯ , ∂ , and μ onto g ′ by taking μ ¯ , ∂ ¯ , ∂ , μ to D μ ¯ , ∂ ¯ , ∂ , D μ , respectively. This map descends to an epimorphism g → g ′ since, by the construction of g ′ , all the relations in (2) are mapped to zero in g ′ . Moreover, this map g → g ′ by construction takes h onto h ′ . We claim the map h → h ′ is an isomorphism. Indeed, since h ′ is free, we have a unique graded Lie algebra homomorphism h ′ → h defined by sending the symbols ∂ ¯ and ∂ in h ′ to ∂ ¯ and ∂ in h , respectively. This map is inverse to the previous homomorphism h → h ′ since the two compositions of these two morphisms are identities on the corresponding generators. Consequently, the map g → g ′ yields h ≅ h ′ . Combining this with the above corollary, we conclude g ≅ g ′ .□

Remark 2.6

The Lie subalgebra of g generated by μ ¯ and μ is isomorphic to the Heisenberg (graded) Lie algebra. That is, it is the three-dimensional graded Lie algebra spanned by μ ¯ , μ , and [ μ ¯ , μ ] , whose center is spanned by [ μ ¯ , μ ] , and in which [ μ ¯ , μ ¯ ] = [ μ , μ ] = 0 .

Corollary 2.7

U g is a free (left and right) module over U h with basis 1 , μ ¯ , μ , μ ¯ μ .

Proof

U g is a free module over U h follows from the Poincaré-Birkhoff-Witt theorem (see [3, pp. 288]). To prove the second assertion, note that h being a g -ideal implies that U h is a (two-sided) U g -ideal. The quotient U g / U h is the free associative algebra generated by μ ¯ , ∂ ¯ , ∂ , μ modulo the relations ∂ ¯ = 0 , ∂ = 0 , [ μ ¯ , μ ] = 0 . This quotient algebra has a basis 1 , μ ¯ , μ , μ ¯ μ , thus proving 1 , μ ¯ , μ , μ ¯ μ is a U h -basis for U g .□

Theorem 3.1

The natural quotient morphism f : ( g , ad d ) → ( g hol , ad ∂ ¯ + ∂ ≡ 0 ) is a quasi-isomorphism of differential graded Lie algebras.

Proposition 3.2

H k ( h , ad μ ¯ ) = 0 for k > 2 .

Proof

First, we note ad μ ¯ ∂ ¯ = 0 and

We leave it to the reader to check when k = 1 , 2 , H k ( h , ad μ ¯ ) is one-dimensional and spanned by (the equivalence classes of) ∂ ¯ and [ ∂ , ∂ ¯ ] , respectively. Next, we observe that H ( h , ad μ ¯ ) is a Lie algebra and H 1 ( h ) ⊕ H 2 ( h ) forms an abelian Lie subalgebra since [ ∂ ¯ , ∂ ¯ ] = − 2 [ μ ¯ , ∂ ] is ad μ ¯ -exact, and [ ∂ ¯ , [ ∂ , ∂ ¯ ] ] = 0 , [ [ ∂ , ∂ ¯ ] , [ ∂ , ∂ ¯ ] ] = 0 by Jacobi identities.

Now consider the universal enveloping algebra U H ( h , ad μ ¯ ) of H ( h , ad μ ¯ ) . It contains the universal enveloping algebra of the abelian Lie subalgebra H 1 ( h ) ⊕ H 2 ( h ) , which is the free graded commutative algebra Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) generated by ∂ ¯ and [ ∂ , ∂ ¯ ] . Meanwhile, since the universal enveloping algebra functor commutes with cohomology (see, e.g., [6, Appx B. Prop. 2.1]), we have U H ( h , ad μ ¯ ) = H ( U h , ad μ ¯ ) , where ad μ ¯ on U h is the extended adjoint action. So we obtain

By Poincaré-Birkhoff-Witt theorem, our proposition is equivalent to Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) = H ( U h , ad μ ¯ ) . This equality clearly holds in degrees ≤ 2 . We will prove this by induction on degree, but we need to make some preparations.

For simplicity of notation, denote B = U h , which is the free tensor algebra on ∂ ¯ and ∂ by Theorem 2.5. Under isomorphism

the differential ad μ ¯ ∣ B k can be written as the matrix

by using the relations (2). To see this, we compute for x , y ∈ B k − 1

In particular, by setting x = 0 , we see ad μ ¯ skew commutes with ∂ ¯ . This means both ± ∂ ¯ are morphisms of cochain complexes ± ∂ ¯ : B • → B • [ 1 ] , where B • = ( B , ad μ ¯ ) . Moreover, (3) shows that the mapping cone of − ∂ ¯ is isomorphic to B • [ 2 ] by ϕ . Then, the inclusion of B • [ 1 ] into the mapping cone of − ∂ ¯ is identified with ∂ ¯ : B • [ 1 ] → B • [ 2 ] , and the projection from the mapping cone of − ∂ ¯ onto B • [ 1 ] is identified with δ : B • [ 2 ] → B • [ 1 ] , which takes ∂ x + ∂ ¯ y to x . It follows that we have an exact triangle

This exact triangle induces a long exact sequence in cohomology

Now we can inductively prove H ( B • ) = Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) . We note that Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) is one-dimensional in each degree and spanned by powers of [ ∂ , ∂ ¯ ] and ∂ ¯ times powers of [ ∂ , ∂ ¯ ] . Assume the desired equality is proved in degrees < k . Observe that ∂ ¯ vanishes on ( Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) ) odd , so if k is even, then from the above long exact sequence, we can infer that δ : H k ( B • ) → H k − 1 ( B • ) is an isomorphism. On the other hand, if k is odd, then ∂ ¯ : H k − 1 ( B • ) → H k ( B • ) is monic since ∂ ¯ takes ( Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) ) even injectively into ( Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) ) odd ⊂ H odd ( B • ) , and thus the above long exact sequence implies ∂ ¯ : H k − 1 ( B • ) → H k ( B • ) is an isomorphism. So in either case, we have H k ( B • ) ≅ H k − 1 ( B • ) , and in particular, H k ( B • ) is one-dimensional. But H k ( B • ) ⊃ ( Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) ) k , therefore H k ( B • ) must be equal to ( Λ ( ∂ ¯ , [ ∂ , ∂ ¯ ] ) ) k . This finishes the inductive step and thus completes the proof.□

Proof of Theorem 3.1

One can explicitly find

and H 2 ( g , ad d ) = H 2 ( g hol , ad d ) = 0 . So f induces isomorphisms on H 1 and H 2 . It remains to show f also induces an isomorphism on H k for k > 2 . But since g hol is concentrated in degrees ≤ 2 , it suffices to prove H k ( g , ad d ) = 0 for k > 2 . For this, we use the previous proposition.

Note that by Corollary 2.4, g = h in degrees ≥ 2 , so we have H k ( g , ad μ ¯ ) = H k ( h , ad μ ¯ ) = 0 for k > 2 . We claim the generalized Frölicher spectral sequence of Cirici and Wilson [2] implies H k ( g , ad d ) = 0 for k > 2 . Indeed, their spectral sequence, even though designed for studying almost complex manifolds, applies more generally to (non-negatively) bigraded g -modules and in particular to the adjoint action of g on itself. Thus, we have a spectral sequence

Therefore, the claim follows, and the proof is complete.□

Proposition 3.3

The map P 1 → P MC ( g ) , [ s : t ] ↦ d s , t = s 3 μ ¯ + s 2 t ∂ ¯ + s t 2 ∂ + t 3 μ is an isomorphism. In particular, P MC ( g ) is a twisted cubic.

Proof

We express a in the basis a = x μ ¯ + y ∂ ¯ + z ∂ + w μ . Then, by expanding the (quadratic) equation [ a , a ] = 0 , we have that a is a solution to this Maurer-Cartan equation if and only if the following three quadratic equations are satisfied:

These are exactly the equations in P 3 defining the standard twisted cubic. The map [ s : t ] ↦ d s , t = s 3 μ ¯ + s 2 t ∂ ¯ + s t 2 ∂ + t 3 μ is a parametrization of the twisted cubic.□

Remark 3.4

Compare the above to [9, Lemma 3.1], where a different parametrization of MC ( g ) is obtained. In addition, under the parametrization given in Proposition 3.3, we have d 1 , 0 = μ ¯ , d 0 , 1 = μ , and d 1 , 1 = d .

Lemma 3.5

For s t ≠ 0 , d s , t and

span H 1 ( g , d s , t ) . Moreover,

1. d s , t is a real operator (i.e., it is conjugate to itself) if and only if d s , t J is a real operator;

2. [ d s , t J , d s , t J ] = 0 if and only if [ ∂ , ∂ ] = [ ∂ ¯ , ∂ ¯ ] = [ ∂ , ∂ ¯ ] = 0 .

Proof

Note that

The rest is straightforward.□

Remark 3.6

In particular, H 1 ( g , d ) is spanned by d and d J = − 1 ( 3 μ ¯ + ∂ ¯ − ∂ − 3 μ ) . The authors learned from Scott Wilson why this is geometrically significant: the action of the operator d J on differential forms of an almost complex manifold ( M , J ) coincides with the (graded) commutator ℒ J = [ d , J ] , where J acts on differential forms by extending its action on 1-form to a derivation on the de Rham algebra of all forms. See [1] for more on operator ℒ J .