Let X1 and X2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann （CR） manifolds of dimensions 2m - 1 and 2n - 1 in Cm＋l and Cn＋l, respectively. We introduce the Thom- Sebastiani sum X = X1 X2...Let X1 and X2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann （CR） manifolds of dimensions 2m - 1 and 2n - 1 in Cm＋l and Cn＋l, respectively. We introduce the Thom- Sebastiani sum X = X1 X2 which is a new compact connected strongly pseudoconvex embeddable CR manifold of dimension 2m＋2n＋ 1 in Cm＋n＋2. Thus the set of all codimension 3 strongly pseudoconvex compact connected CR manifolds in C^n＋1 for all n ≥ 2 forms a semigroup. X is said to be an irreducible element in this semigroup if X cannot be written in the form X1 X2. It is a natural question to determine when X is an irreducible CR manifold. We use Kohn-Rossi cohomology groups to give a necessary condition of the above question. Explicitly, we show that if X = X1 X2, then the Kohn-Rossi cohomology of the X is the product of those Kohn-Rossi cohomology coming from X1 and X2 provided that X2 admits a transversal holomorphic Sl-action.展开更多
基金supported by National Natural Science Foundation of China(Grant Nos.11531007 and 11401335)Start-Up Fund from Tsinghua University and Tsinghua University Initiative Scientific Research Program
文摘Let X1 and X2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann （CR） manifolds of dimensions 2m - 1 and 2n - 1 in Cm＋l and Cn＋l, respectively. We introduce the Thom- Sebastiani sum X = X1 X2 which is a new compact connected strongly pseudoconvex embeddable CR manifold of dimension 2m＋2n＋ 1 in Cm＋n＋2. Thus the set of all codimension 3 strongly pseudoconvex compact connected CR manifolds in C^n＋1 for all n ≥ 2 forms a semigroup. X is said to be an irreducible element in this semigroup if X cannot be written in the form X1 X2. It is a natural question to determine when X is an irreducible CR manifold. We use Kohn-Rossi cohomology groups to give a necessary condition of the above question. Explicitly, we show that if X = X1 X2, then the Kohn-Rossi cohomology of the X is the product of those Kohn-Rossi cohomology coming from X1 and X2 provided that X2 admits a transversal holomorphic Sl-action.
