Non - Realizable CR Structures

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Thus both Re/ and Im/ must have zeroes in T. Our next resuU comes from applying this to/=g/i„.

Lemma L7. Let g be a non-negative Junction with support in 7 (but g not

identically zero). If iLH-g^^J h = 0 then Re/i^^ and Imh^ both have zeroes in T.

We are now ready to show that any operator which is strongly geneously solvable may be perturbed to obtain one which is not.

^ г

Theorem L Let L = ^ a, -— have the property that L, L, and [L, L] are linearly

1=1 ^ ^i independent at some point p. Assume the equation Lh = 0 has two solutions h^ and h2 with dh^ a dh2^0 at p. Tlien there exists a C^ function f vanishing to infinite order at p and a real vector field Q such that if v^ and V2 are both C^ solutions to the equation {L-\-fQ)v = 0 then at p, dv^ Adv2=0.

Titus L IS the induced CR operator Jor some M^aC^ but L+fQ is not such an induced CR operator.

Proof . Starting with L we find M and introduce the coordinates (z, z, u). The closed curves Г(/) fill out some neighborhood Q of the origin. Take a sequence of curves C, in the y = 0 plane in M^ converging to the origin and such that for each curve we get a topological torus S^ foliated by curves F^. In particular each Cj is chosen so as not to include any points on F^. Let T^ be the open sohd torus bounded by S,. Let / be a non-negative C^ function with supp/ = closure of IJ Tj. Note that/and all its derivatives are zero at the origin.

Now assume that Lh-\-fh^ = 0 in some neighborhood of the origin. By shrinking Q and omitting some of the first C, we may assume Lh-\-fh^ = ^ in Q where Q contains \JT^. Now

j h dz

is holomorphic for all Я for which r(/l)c=int(f2 —IJlj and since any such F{X) may be deformed in mi[Q — [jT) to a point of F^ (using the method of Lemma 1.4) we have that

j hdz = 0.

Г ( Я )

Thus

\\\fhjxdydz = 0. 7,

Therefore in each 7^ there must be points p, and q^ for which Re/7„(pj) = 0 and Im/î^(gJ = 0. Since p,^0 and ^,-^0 we see that h^ = 0 at the origin. But then from the equation also h^ = 0. That is dh Adz = 0 at the origin. This implies that

for any two solutions v^ and V2 of (l+/—-) i; = 0 we have dv^ Adv^=0 at the origin. \ dul

In [6] Nirenberg showed how to obtain a stronger conclusion in rem 1 (for the Lewy operator) by using a more complicated perturbation. His ideas may be applied also to our more general situation.