GEOMETRIC APPROACH TO GOURSAT FLAGS * Richard ...

More documents

Recommendations

Info

Proof of Proposition 6.1. The proof will consist of three steps.First step. We will show that for suciently large N the agF Nt : D sNt D s;1 D 1 D sNt =(D s ! Nt ) ! Nt = ! + t(! N ; !)is a Goursat ag for any t 2 [0 1]. To show this we have tocheck the following statements:N(a) ! Nt j DsNt (p) is a nonzero 1-form for any p 2 M, t 2 [0 1] and suciently large N(b) d! Nt j DsNt (p) is a nonzero 2-form for any p 2 M, t 2 [0 1] and suciently large(c) if is a 1-form annihilating the distribution D s;1 then dj DsNt (p) =0forany N,any p 2 M and t 2 [0 1].The statements (a) and (b) follow from the fact that they are valid for t =0,thecondition that ! N tends to ! in the C 1 -Whitney topology (here we usethat l 1intheformulation of Proposition 6.1), and the observation that the hyperplane D sNt (p) aswellas the restrictions of the forms ! Nt and d! Nt to this hyperplane depend on the 1-jet atp of the form ! Nt only.To prove(c)weconsider the space L(D s;1 )(p). By the sandwichlemma 2.1 it is a codimension2 subspace of D s;1 (p) and the 1-forms ! and ! N annihilate this space. Therefore! Nt annihilates L(D s;1 )(p) for all t, i.e. L(D s;1 )(p) isahyperplane in D sNt (p), independentofN. Because L(D s;1 )(p) isthekernel of the 2-form d restricted to D s;1 (p), where annihilates D s;1 but not D s;2 (see Lemma 2.2), any hyperplane in D s;1 (p) containingL(D s;1 )(p) isisotropic for d. Inparticular, D sNt (p) isisotropic for d.Second step. We have proved that F Nt is a Goursat ag for suciently large Nand all t 2 [0 1]. In what follows assume that N is suciently large. Now we start toconstruct a path Nt of global dieomorphisms such that ( Nt ) F Nt = F N0 = F and inparticular ( N1 ) F N = F . Weuse the homotopy method. The second step of the proofis to reduce the construction of Nt to the construction of a path X Nt of global vectorelds satisfying the linear equations(X Nt cd! Nt + ! N ; !)j DsNt =0 X Nt 2 L(D s;1 ): (6:1)Assume that X Nt satises (6.1). Consider the following ordinary dierential equation andthe initial condition with a parameter p 2 M:d Nt (p)dt= X Nt ( Nt (p)) N0 (p) =p p 2 M: (6:2)Since M is a compact manifold and t varies on the compact segment [0 1], the solutionof (6.2) is a path Nt of global dieomorphisms on M. Let us show that ( Nt ) F Nt =F N0 . The condition X Nt 2 L(D s;1 )impliesthat Nt preserves the distribution D s;1 .Therefore to show that ( Nt ) F t = F 0 it is suces to show that there exists a path H Ntof nonvanishing functions such that(H Nt Nt! Nt ; ! 0 )j Ds;1 0: (6:3)24
We willseek for H Nt in the form H Nt = e h Nt,where h N0 is a function identically equalto 1. Let A Nt = H Nt Nt ! Nt ; ! 0 .Then A N0 is the zero 1-form and therefore (6.3) canbe replaced by the equation ( dA Nt)jdt Ds;1 0. We havedA Ntdt= H Ntdh Ntdt Nt! Nt + H Nt Nt(L XNt ! Nt + d! Nt)dtwhere L XNt is the Lie derivative forthe vector eld X Nt . Let q Nt be apath of functionson M such that dh Nt= qdt Nt ( Nt ). Then the equation ( dA Nt)jdt Ds;1 0isequivalent tothe equation(q Nt ! Nt + L XNt ! Nt + ! N ; !)j Ds;1 =0 (6:4)with respect to the path of functions q Nt . By the sandwich lemmaL(D s;1 )isasubsetof D st for all t. Therefore ! Nt annihilates X Nt 2 L(D s;1 ). It follows that L XNt ! Nt =X Nt cd! Nt . Then (6.4) can be written in the form(q Nt ! Nt + X Nt cd! Nt + ! N ; !)j Ds;1 =0:This equation has a solution q Nt due to the relation (6.1), and the denition of D sNt .Third step.Note that the dieomorphisms Nt dened by the ordinary dierentialequation (6.2) tend to the identity dieomorphism as N !1in the same topology inwhich X Nt ! 0. Therefore to nish the proof of Proposition 6.1 it suces to prove that(6.1) has a solution X Nt tending to the zero vector eld as N !1in the C l -Whitneytopology. The third step of the proof is to construct such X Nt .Fix a Riemannian metrics on M. Let V Nt (p) D sNt (p) bethe orthogonal complementtoL(D sNt )(p) within D sNt (p) withrespect to this metric. By Lemmata 2.1and 2.2 dim V Nt (p) =2and rank(d! Nt )j VNt (p) =2. Therefore there is a unique vectorX pNt 2 V Nt (p) such that(X pNt cd! Nt + ! N ; !)j Vt (p) =0 p 2 Mt 2 [0 1]: (6:5)Set X Nt (p) =X pNt . Since ! N ; ! tends to 0 in the C l -Whitney topology, X Nt ! 0asN !1in the same topology. Wewillshow that the path X Nt satises (6.1). This willcomplete the proof of Proposition 6.1.Since L(D st ) V t = D sNt the rst condition in (6.1), which istosaythevalidity ofequation there, follows immediately from (6.5) once we have shown that all the forms inthat equation, namely d! Nt ! N and ! annihilate L(D st ). The fact that d! Nt annihilatesany vector in L(D st )iscontained in Lemma 2.2. To prove that ! and ! N annihilate, usethe sandwich lemma 2.1 twice to conclude that L(D s;1 )(p) iscontained in both D s (p) andin D sN (p). Therefore ! and ! N annihilate the space L(D s;1 )(p). But the sandwichlemmaalso gives L(D st )(p) L(D s;1 )(p), and therefore these forms annihilate L(D st )(p).It remains to prove the inclusion X Nt 2 L(D s;1 )ofequation (6.1). The validity ofthe rst equation in (6.1) and the fact that ! and ! N annihilate the space L(D s;1 )(p)imply(X pNt cd! Nt )j L(Ds;1)(p) =0: (6:6)25
Page 1 and 2: GEOMETRIC APPROACH TO GOURSAT FLAGS
Page 4 and 5: Theorem 1. Apply the Cartan prolong
Page 7 and 8: Lemma 2.2. Let (F) be the Goursat a
Page 9 and 10: If it were true that each Kumpera-R
Page 11: for any Goursat distribution D.On t
Page 14: transformation and a single reectio
Page 17 and 18: (B 1 D 1 E 1 ) D 5 (0) 6= L(D 3 )(0
Page 19 and 20: Returning to the general rank 2 pro
Page 21 and 22: Proof. Let E be the prolongation of
Page 23: Remarks.1. Note that in part 1 wema
Page 27 and 28: the point p 2 M. The denition of th
Page 29 and 30: d! Nt (Z Y Nt )=(1; t)d!(Z Y Nt )+t
Page 31 and 32: To prove (B.3) we use the relations
Page 33 and 34: and the local classication of arbit
Page 35 and 36: References1887 Frobenius F., Uber d

GEOMETRIC APPROACH TO GOURSAT FLAGS * Richard ...

Create successful ePaper yourself

Delete template?

Save as template?