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The derived BV bracket is an odd Poisson bracket

Statement

Let (A,Δ) be as in BV algebra with a degree-one generator. For homogeneous a,b,c∈A, the derived bracket has degree ∣{a,b}∣=∣a∣+∣b∣+1 and satisfies {a,b}=−(−1)(∣a∣+1)(∣b∣+1){b,a}, {a,bc}={a,b}c+(−1)(∣a∣+1)∣b∣b{a,c}, and {a,{b,c}}={{a,b},c}+(−1)(∣a∣+1)(∣b∣+1){b,{a,c}}. Thus it is an odd Poisson bracket, meaning an odd-skew bracket that is a derivation in each argument and satisfies the displayed odd Jacobi identity.

Facts & Assumptions

Given: A degree-+1 BV generator Δ on a graded-commutative algebra A over a field of characteristic 0, and homogeneous a,b,c∈A.

[F1]

BV algebra with a degree-one generator: Δ(1)=0, Δ2=0, [[[Δ,ma],mb],mc]=0, and {a,b}=(−1)∣a∣(Δ(ab)−(Δa)b−(−1)∣a∣aΔb).

Proof

1.1F1algebra

Put m=∣a∣, n=∣b∣, δa=[Δ,ma], and Ta,b=[δa,mb]. By [F1], [Ta,b,mc]=0 for every homogeneous c, so Ta,b(c)=(−1)∣Ta,b∣∣c∣cTa,b(1)=Ta,b(1)c; the two signs cancel by graded commutativity, hence Ta,b=mTa,b(1). Direct evaluation gives Ta,b(1)=Δ(ab)−(−1)maΔb−(−1)(m+1)nbΔa=Δ(ab)−(−1)maΔb−(Δa)b=(−1)m{a,b}. Thus [δa,mb]=m(−1)m{a,b}.

1.2F1algebra

Graded commutativity expands the defining formula to {b,a}=(−1)n+mnΔ(ab)−(−1)n+mn+maΔb−(−1)mn+n(Δa)b. Multiplication by −(−1)(m+1)(n+1) reduces these coefficients, modulo 2, to (−1)m, −1, and −(−1)m, respectively, giving exactly {a,b}. Each term of the bracket formula has degree ∣a∣+∣b∣+1. This proves odd skew-symmetry and the degree assertion.

2.1F1step 1.1algebra

Define Ha=δa−mΔa. Then Ha(1)=0 and [Ha,mb]=m(−1)m{a,b} by step 1.1 because multiplication operators gradedly commute. Evaluating at 1 gives Ha(b)=(−1)m{a,b}; evaluating at homogeneous c gives Ha(bc)−(−1)(m+1)nbHa(c)=Ha(b)c. Multiplying by (−1)m proves {a,bc}={a,b}c+(−1)(m+1)nb{a,c}.

2.2F1step 1.1step 1.2algebra

Expansion of the graded commutators gives [Δ,[Δ,mb]]=[Δ2,mb]=0 and [δa,δb]=[Δ,[ma,δb]], because the other graded-Jacobi term is a signed multiple of [ma,[Δ,δb]]=0. By step 1.1 and odd skew-symmetry from step 1.2, [ma,δb]=−(−1)m(n+1)[δb,ma]=−m{a,b}, so [δa,δb]=−[Δ,m{a,b}].

2.3F1step 1.1algebra

Rearranging the defining bracket gives Δ(uv)=(Δu)v+(−1)∣u∣uΔv+(−1)∣u∣{u,v}. Apply this to (Δa)b and a(Δb) when expanding Δ{a,b}; the two (Δa)(Δb) terms cancel and Δ2=0 leaves Δ{a,b}={Δa,b}−(−1)m{a,Δb}.

3.1step 1.2step 2.1algebra

By steps 1.2 and 2.1, express {ab,c}=−(−1)(∣a∣+∣b∣+1)(∣c∣+1){c,ab}, expand {c,ab} by the derivation rule, and use odd skew-symmetry on the two terms; reducing exponents modulo 2 gives {ab,c}=a{b,c}+(−1)∣b∣(∣c∣+1){a,c}b. Hence the bracket is also a derivation in its first argument and is a biderivation.

4.1F1step 1.1step 1.2step 2.2step 2.3algebra∎

Expanding [Ha,Hb] using steps 1.1 and 2.2 gives [Ha,Hb]=−[Δ,m{a,b}]−(−1)mm{a,Δb}+(−1)(m+1)(n+1)+nm{b,Δa}. By step 1.2, {Δa,b}=−(−1)m(n+1){b,Δa}, and the last coefficient equals −(−1)m(n+1); step 2.3 therefore makes the last two terms mΔ{a,b}, so [Ha,Hb]=−H{a,b}. Thus Da={a,−}=(−1)mHa satisfies [Da,Db]=(−1)m+n+1H{a,b}=D{a,b}; applying this to c and rearranging is the stated odd Jacobi identity.

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