11 One-Forms and Deformed de Rham Complex 2050
α 1 α 2 α 2 +α 3α 1 +α 2 α 1 +α 2 +α 3α 3
Fig. 11.2.The finite subsetΩg⊂H^1 (g)Ωg={αi 1 +···+αis| 1 ≤i 1 <···<is≤n, s=1,...,n}. (11.30)
It follows that if
−ω/∈Ωg
then the total cohomologyHω∗(g) is trivial:Hω∗(g)≡0.
One can easily remark that the subsetΩgis well defined and does not
depend on the ordering of weightsαi.
LetG/Γbe a compact solvmanifold, whereGis a completely solvable Lie
group. Then the left-invariant closed 1-forms fromΩgdefine a finite subset
inH^1 (G/Γ,R). We denote this subset byΩG/Γ.Letωbe a closed 1-form on
G/Γ. If the cohomology class
−[ω]∈/ΩG/Γthen the total cohomologyHω∗(G/Γ,R) is trivial:H∗ω(G/Γ,R)≡0. The subset
ΩG/Γ is well defined in terms of the corresponding Lie algebrag. The corre-
sponding Lie algebragmust to be unimodular, i.e. the left-invariantn-form
e^1 ∧···∧endetermines non-exact volume form onG/Γand hence
α 1 +α 2 +···+αn=0.
IfG/Γis a compact nilmanifold then all the weightsαi,i=1,...,nare
trivial and thereforeΩG/Γ ={ 0 }. Hence the cohomologyH∗ω(G/Γ,R)ofa
nilmanifoldG/Γis trivial if and only if the formωis non-exact.
Let us consider a 3-dimensional solvmanifoldG 1 /Γ 1 defined in Sect. 11.6.
We recall that the corresponding Lie algebrag 1 is defined by its basise 1 ,e 2 ,e 3
and the following non-trivial brackets:
[e 1 ,e 2 ]=ke 2 , [e 1 ,e 3 ]=−ke 3.For the dual basis of left-invariant 1-formse^1 =dz,e^2 =e−kzdx, e^3 =ekzdy
we had
de^1 =0,de^2 =−ke^1 ∧e^2 ,de^3 =ke^1 ∧e^3.
Henceα 1 =0,α 2 =−ke^1 ,α 3 =ke^1 andα 2 +α 3 = 0 (Fig. 11.3).
So it is easy to see that
ΩG 1 /Γ 1 ={±k[e^1 ]}