About solvable algebras and solvable structures

See also about Lie algebras and involutive sets of vector fields.

  1. Lie algebras are defined like vector spaces with a binary operation [,], bla, bla, bla. The result of the bracket has to be expressed as linear combinations with coefficients in R, of course. Some Lie algebras are finite dimensional and other infinite dimensional.
  2. Some Lie algebras are solvable, in the sense that the derived series arrives at 0 in a finite number of steps. As far as I know, they can be finite or infinite dimensional.
  3. An example of an infinite dimensional Lie algebra: vector fields on a manifold X(M). If we fix a distribution ZX(M) in M, there is a (possibly infinite dimensional) Lie subalgebra consisting of all the symmetries of the distribution Sym(Z). Remember: they are vector fields X such that
[X,Z]Z$$sofor$ZZ$,$[X,Z]$isalinearcombinationwithcoefficientsin$C(M)$.Thedistribution$Z$willbe,inthecontextofanODE,theonegeneratedby$A$,and$Sym(S({A}))$willconsistofthegeneralizedsymmetries(afterprolongation).4.In[Lychagin1991]itissaid,looselyspeaking,thatifwefocusonafinitedimensionalLiesubalgebra$L$of$Sym(Z)$whichis"pointwiseindependentof$Z$",which"completes$Z$to$TM$"andwhichissolvablethenwecanfindtheintegralmanifoldsof$Z$byquadratures.ThisisLietheory,butwithdistributions.For$XL,YL$wehavethat$[X,Y]$hasconstantcoefficients.Butfor$XL,ZZ$wehavethat$[X,Z]$has"functionalcoefficients."5.Ifaboveweconsider$Z=S(A)$andfocusonthesubalgebraof$Sym(S(A))$thatcorrespondstoprolongedLiepointsymmetrieswegettheLie[[CONCEPTS/symmetryalgebrasymmetryalgebra]].AsfarasIknow,itisnotnecessarilyfinitedimensional[Olver1986]corollary2.40.6.Ontheotherhand,solvablestructuresaremadeofvectorfieldsinsuchawaythat:Thefirstoneisasymmetryofdistribution$Z$.TheLiebracketconditionis,therefore,"withfunctionalcoefficients".Thefollowingvectorfieldsaresymmetriesofthepreviousdistributions,sotheLiebracketconditionisagain"withfunctionalcoefficients".7.Inconclusion,thefreedomobtainedwithsolvablestructuresvssolvablealgebrascomesfromtwofacts:(1)wecanlookforvectorfieldswhosebracketdependsnotonlyontheothervectorfieldsofthesetbutalsoonthevectorfieldsofthedistribution(thevectorfieldoftheequationifweareinthecontextofanODE)and(2)thecoefficientscanbefunctions.Seealso[[MYRESEARCH/REFLECTIONS/ComparisonofstructuresComparisonofstructures]].