Singularities of bihamiltonian systems and the multidimensional rigid body, страница 7

An integrable linear pencil Πg,A is non-degenerate if and onlyif the set of operators{adξ |g/Ker A }ξ∈Ker Ais a Cartan subalgebra in sp(g/Ker A, A). The type of Sing(Πg,A ) coincideswith the type of this subalgebra.Remark 5.1.1. Since A is a skew-symmetric 2-form, the space g/Ker A issymplectic. The condition of compatibility of A with the Lie-Poisson bracketimplies that all operators adξ , where ξ ∈ Ker A, are skew-symmetric withrespect to A. Therefore, they generate an Abelian (see Proposition 2.1.3)subalgebra in sp(g/Ker A, A). Now we see that non-degeneracy is equivalentto the fact that this subalgebra is a Cartan subalgebra.Corollary 5.1.1.

If Πg,A is non-degenerate, then Ker A consists of adsemisimple elements.Since Ker A is a commutative subalgebra (see Proposition 2.1.3) whichconsists of semisimple elements, all operators adξ , ξ ∈ Ker A may be simultaneously diagonalized (over C). Now we can consider “root” decompositionof g:nXg = Ker A +(Vλi + V−λi ),i=1where each V±λi is spanned by one common eigenvector corresponding to theeigenvalue ±λ(ξ). Eigenvalues enter in pairs because the operators adξ aresymplectic.Remark 5.1.2. Note that all V±λi are one-dimensional by definition.Proposition 5.1.1. If Ker A is diagonalizable, then the pencil is nondegenerate if and only if λ1 , . . .

λn are linearly independent.50Type ofSing(Πg,A ) is (ke , kh , kf ) where ke is the number of pure imaginary λi , khis the number of real λi , and kf is the number of pairs of complex conjugateλi .Proof. Consider the mapad : Ker A → sp(g/Ker A, A),which sends ξ to the operator adξ . We want the image of this map to be aCartan subalgebra (see Lemma 5.1.1). Since it is abelian and diagonalizable,it is Cartan if and only if the dimension of it equalsn=1dim(g/Ker A).2Obviously,dim ad(Ker A) = dimC hλi i,where hλi i is the subspace in (Ker A)∗C spanned by λ1 , . .

. , λn . Therefore,ad(Ker A) is a Cartan subalgebra if and only if the roots are linearly independent.The second statement directly follows from the definition of type.Corollary 5.1.2. If Πg,A is non-degenerate, then Ker A is a Cartan subalgebra.5.2Classification of non-degenerate linearpencils: the complex caseΠg,A is non-degenerate ⇒ g is non-degenerate, and Ker A is a Cartansubalgebra.51Taking into account Corollary 5.1.2 and Proposition 5.1.1, it suffices toshow that if a complex Lie algebra g admits a root decompositiong=h+nX(Vλi + V−λi ),(5.1)i=1with linearly independent λi , then g is non-degenerate.By definition we have[h, h] = 0,[h, x] = λ(h)x for h ∈ h, x ∈ Vλ .The following is standardProposition 5.2.1. If eα ∈ Vα , eβ ∈ Vβ , then [eα , eβ ] ∈ Vα+β .Proof.

Let h ∈ h. By definition we have[h, eα ] = α(h)eα ,[h, eβ ] = α(h)eα .Therefore[h, [eα , eβ ]] = −[eα , [eβ , h]] − [eβ , [h, eα ]] = β[eα , eβ ] + α[eα , eβ ],q.e.d.Since the roots are independent, α + β is a root if and only if β = −α.Consequently, we have the following relations[Vλi , V−λi ] ∈ h,[Vλi , V±λj ] = 0.Let eλ be a basis vector in Vλ . Denoting hλi = [eλi , e−λi ], we will have[hλi , eλj ] = [[eλi , e−λi ], eλj ] = 052if i 6= j due to the Jacobi identity.

Therefore,λj (hλi ) = 0if i 6= j.Now suppose that λi (hλi ) 6= 0 for some value of i. Then the tripleeλi , e−λi , hλi generate a subalgebra isomorphic to so(3, C). Let us show thatit admits a complementary subalgebra in g.Leteh = {h ∈ h : λi (h) = 0}.Denoteeg=eh+X(Vλj + V−λj ).j6=iProposition 5.2.2. g = eg ⊕ so(3, C)Proof.

It is obvious that g can be decomposed into the sum of eg and so(3)as a vector space. Therefore, it suffices to show that eg is a subalgebra and[eg, so(3)] = 0.1. eg is a subalgebra.Indeed,eg=eh+X(Vλj + V−λj )j6=iand[eh,eh] = 0,[V±λj , V±λk ] = 0 j 6= k.Therefore, it is sufficient to show that [Vλj , V−λj ] ∈ eh if j 6= i. But[Vλj , V−λj ] is generated by hλj and λi (hλj ) = 0, q.e.d.532. [eg, so(3)] = 0.Indeed,eg=eh+X(Vλj + V−λj ),j6=iso(3) = hhλi i + (Vλi + V−λi )and[eh, hλi ] = 0, because h is abelian,[eh, Vλi + V−λi ] = 0, because λi (eh) = 0,[hλi , Vλj + V−λj ] = 0, because λj (hλi ) = 0,[Vλj +V−λj , Vλi + V−λi ] = 0.Separating so(3) summands for all i such that λi (hλi ) 6= 0, we obtain thefollowingLemma 5.2.1.

There exists a decomposition(g, h) = (g1 , h1 ) ⊕ (g2 , h2 ),where1. g1 =Lso(3, C) and h1 is a Cartan subalgebra.2. g2 is solvable and admits decomposition (5.1) with respect to h2 .Proof. After the separation of all so(3) summands we will have λi (hλi ) = 0for all i. It is easy to see that the third derived subalgebra of such an algebrais zero.54Now it suffices to study the solvable case.As the first step we shall separate an Abelian summand.

Decompose thecenter of g into a direct sumZ(g) = (Z(g) ∩ [g, g]) ⊕ V.It is obvious that V can be separated from g as a direct summand.After separating an Abelian summand we may assume thatZ(g) ⊂ [g, g].This means that the center is generated by hi = [ei , e−i ].Now decompose subalgebra h as followsh = hh1 , . . . hk i ⊕ T.Since λi (hλj ) = 0 for all i and j, all λi are linearly independent in T ∗ .Moreover, for each t ∈ T there is i such that λi (t) = 0 (otherwise t belongsto the center, which is not possible, because the center is generated by hi ).Therefore, the set of λi is a basis in T ∗ and we can choose a basis t1 , . . .

tk inT such thatλi (tj ) = δij .Consequently, g is generated by ei , e−i , hi , ti with the following relations[ei , e−i ] = hi ,[ei , ej ] = 0 if j 6= −i,[ti , ei ] = ei ,[ti , e−i ] = −ei ,[ti , ej ] = 0 if j 6= ±i,[hi , g] = 0,55[ti , tj ] = 0,[ti , hj ] = 0If all hi were linearly independent, our algebra could be decomposed into adirect sum of gC♦ -subalgebras. Since this is not necessarily the case, g is aquotient of such a direct sum by some central ideal.Therefore, g is indeed non-degenerate.g is non-degenerate and Ker A is a Cartan subalgebra ⇒ Πg,A isregular, integrable and non-degenerate.First note that if g is non-degenerate and Ker A is a Cartan subalgebra,then ind g = corank A. Moreover, the central extension gA is again nondegenerate and the kernel of the lift of A onto gA is again a Cartan subalgebra.Therefore, ind gA = corank A.

But this implies regularity of Πg,A taking intoaccount Proposition 2.1.2.Now note that we do not need to prove integrability, because it automatically follows from non-degeneracy at the origin (which follows from thelinear independence of the roots). Indeed, we can always find a regular pointin the neighbourhood of a non-degenerate point. Due to analyticity regularpoints are everywhere dense, q.e.d.5.3Classification of non-degenerate linearpencils: the real caseIt is possible to classify real non-degenerate linear pencils by classifying thereal forms of complex non-degenerate algebras.

However it seems to be betterfor the logic of the text to do it explicitly.56Πg,A is non-degenerate ⇒ g is non-degenerate and Ker A is aCartan subalgebra.By Corollary 5.1.2 and Proposition 5.1.1 the subalgebra h = Ker A is aCartan subalgebra and the roots are linearly independent. These roots havethe form ±λ1 , . . . , ±λk , ±ν1 i, . .

. , ±νl i, ±µ1 ± ξ1 i, . . . , ±µm ± ξm i. We canwriteg = h + he±1 , . . . , e±k , f±1 , . . . , f±l , g±1 , h±1 , . . . , g±m , h±m i,where[x, ei ] = λi (x)ei for x ∈ h,[x, fi ] = νi (x)f−i for x ∈ h,[x, gi ] = µi (x)gi − ξi (x)hi for x ∈ h,[x, hi ] = ξi (x)gi + µi (x)hi for x ∈ h,where, by definition,λ−i = −λi , ν−i = −νi ,µ−i = −µi , ξ−i = −ξi .It is easy to check that[ei ,e−i ] ∈ h,[fi ,f−i ] ∈ h,[gi , g−i ] = −[hi , h−i ] ∈ h,[gi , h−i ] = [hi , g−i ] ∈ h,and all other commutators vanish.57Suppose that λi ([ei , e−i ]) 6= 0 for some i.In this case the tripleei , e−i , [ei , e−i ] generate a subalgebra isomorphic to sl(2, R). It can be shownthat this subalgebra can be separated as a direct summand, analogous to thecomplex case.Similarly, if νi ([fi , f−i ]) 6= 0, we obtain a summand isomorphic to so(3, R)if νi ([fi , f−i ]) > 0 and sl(2, R) if νi ([fi , f−i ]) < 0.Further, we haveξi ([gi , g−i ])gi + µi ([gi , g−i ])hi = [[gi , g−i ], hi ] == −[[g−i , hi ], gi ] = µi ([g−i , hi ])gi − ξi ([g−i , hi ])hi ,thereforeξi ([gi , g−i ]) = µi ([g−i , hi ]),µi ([gi , g−i ]) = −ξi ([g−i , hi ]),which means that ξi and νi are either linearly independent on the subspace h[gi , g−i ], [g−i , hi ]i or both vanish.In the first case the elementsgi , g−i , hi , h−i , [gi , g−i ], [g−i , hi ] generate a subalgebra isomorphic to so(3, C).After separating all described summands, we see that[ei ,e−i ] ∈ Z(g),[fi ,f−i ] ∈ Z(g),[gi , g−i ] = −[hi , h−i ] ∈ Z(g),[gi , h−i ] = [hi , g−i ] ∈ Z(g).In a way absolutely similar to the complex case it can be shown that g isdecomposed into a sum of an abelian algebra and a quotient of a sum ofseveral copies of gh♦ , g♦ , gC♦ by some central ideal.58g is non-degenerate and Ker A is a Cartan subalgebra ⇒ Πg,A isregular, integrable and non-degenerate.The proof is similar to the complex case.5.4Proof of the second part of Theorem 6In this section we will keep the notations of Section 5.3.By Proposition 5.1.1 the type of Sing(Πg,A ) is (ke , kh , kf ), where ke is thenumber of pairs of roots of type ±νj i, kh is the number of pairs of type ±λj ,kf is the number of quadruples of type ±µj ± ξj i.Now note that in the proof of Theorem 6 only complex roots gave riseto summands of type so(3, C), gC♦ .