GEOMETRIC APPROACH TO GOURSAT FLAGS * Richard ...

Returning to the general rank 2 prolongation PD,let! 1 :::! s be one-forms whosevanishing denes D. Complete these forms to a local co-framing of all of T M by addingtwo other one-forms, say dx and dy. Restricted to D m , the forms dx and dy form a linearcoordinate system. Then any line ` D m can be expressed in the form adx + bdy =0,with (a b) 6= 0. Thus [a b] form homogeneous coordinates on the projective linePD m .One obtains a ber ane coordinate by writing [a b] =[z1]. This z is dened away fromthe \vertical line" dx =0and is the negative oftheslope: z = ;dy=dx. Therefore z formsan ane ber coordinate for the bundle PD ! M. The Pfaan system describing theprolonged distribution on PD is ! i , i =1:::s together with! s+1 = dx + zdy:The coordinate z breaks down in a neighborhood of the vertical lines. There we mustswitch tothe other ane coordinate ~z which isrelated to z by ~z = ;dx=dy =1=z in theircommon domain. In such a\vertical" neighborhood we must use the form ~zdx+dy insteadof dx + zdy.Proposition 5.1. The prolongation E of a Goursat distribution D of rank k andcorank s on a manifold M is a Goursat distribution of rank k and corank s +1 on themanifold PD. Itsatises E 2 = D. If rank(D) =2then L(E 2 )=ker(d), the verticalspace for the bration PD ! M.Proof. We only give the proof in the case rank(D) =2. E is rank 2, so E 2 has rankat most 3. Now E D,where D is the rank 3 distribution on PD dened by thevanishing of the ! i as above. Indeed, in terms of our coordinatesE = fv 2 D : ! s+1 (v) =0gwith ! s+1 = dx + zdy as above. E 2 = D because d! s+1 = dz ^ dy 6= 0mod! s+1 .(Seethe proof of lemma 2.2.) Now E j = D j;1 j =3:::, and they have theright rank, sothe rest of the Goursat conditions follow. E is Goursat.By denition, the vertical space ker(d) belongs to D,andis involutive. Thusker (d) L(E 2 ). The equality ker(d) =L(E 2 )now follows from the sandwich lemmaand a dimension count. Alternatively, toget equality, use the fact that E = D is denedby the vanishing of the ! i ,and these forms are independent ofthe vertical direction.Consequently L(E 2 )=ker(d). Q.E.D.5.2. DeprolongationThe reverse of prolongation is deprolongation. Suppose that E is a distribution onamanifold Q, and that L(E 2 )isaconstant rank foliation. Let us suppose that the leafspaceM = Q=L(E 2 )is a manifold, and that the projection : Q ! M19