Let X_1 and X_2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann(CR) manifolds of dimensions 2m-1 and 2n-1 in C^(m+1)and C^(n+1), respectively. We introduce the ThomSebastiani sum X = X_1 ⊕X_2...Let X_1 and X_2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann(CR) manifolds of dimensions 2m-1 and 2n-1 in C^(m+1)and C^(n+1), respectively. We introduce the ThomSebastiani sum X = X_1 ⊕X_2which is a new compact connected strongly pseudoconvex embeddable CR manifold of dimension 2m+2n+1 in C^(m+n+2). Thus the set of all codimension 3 strongly pseudoconvex compact connected CR manifolds in Cn+1for 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 X_1 ⊕ X_2. 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 = X_1 ⊕ X_2, then the Kohn-Rossi cohomology of the X is the product of those Kohn-Rossi cohomology coming from X_1 and X_2 provided that X_2 admits a transversal holomorphic S^1-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 X_1 and X_2 be two compact connected strongly pseudoconvex embeddable Cauchy-Riemann(CR) manifolds of dimensions 2m-1 and 2n-1 in C^(m+1)and C^(n+1), respectively. We introduce the ThomSebastiani sum X = X_1 ⊕X_2which is a new compact connected strongly pseudoconvex embeddable CR manifold of dimension 2m+2n+1 in C^(m+n+2). Thus the set of all codimension 3 strongly pseudoconvex compact connected CR manifolds in Cn+1for 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 X_1 ⊕ X_2. 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 = X_1 ⊕ X_2, then the Kohn-Rossi cohomology of the X is the product of those Kohn-Rossi cohomology coming from X_1 and X_2 provided that X_2 admits a transversal holomorphic S^1-action.
