4-dimensional \(c\)-symplectic \(S^1\)-manifolds with non-empty fixed point set need not be \(c\)-Hamiltonian.

*(English)*Zbl 0994.57032
Oprea, John (ed.) et al., Homotopy and geometry. Proceedings of the workshop, Banach Center, Warsaw, Poland, June 9-13, 1997. Warsaw: Polish Academy of Sciences, Banach Cent. Publ. 45, 91-93 (1998).

It is known that if the circle group acts on a compact symplectic 4-manifold with non-empty fixed point set, then the action is necessarily Hamiltonian and this phenomenon does not occur in higher dimensions. Let \((M,x)\) be a cohomologically symplectic manifold, which means \(M\) is \(2n\)-dimensional and \(x\) is a cohomology class in \(H^2(M)\) such that \(x^n\neq 0\) in \(H^{2n}(M)\). Suppose that the group \(S^1\) acts on \(M\). The author addresses the following questions posed by J. Oprea: Whether \((M,x)\) is \(c\)-Hamiltonian when \(\dim M= Y\) and the fixed point set \(M^{S^1}\) is non-empty? Note that by definition \((M,x)\) is called \(c\)-Hamiltonian of \(\lambda(x)= 0\), where for \(x\in H^q(M)\), \(\lambda(x)\in H^{q-1}(M)\) defined by \(\varphi^*(x)= 1\otimes x+ u^*_1\otimes \lambda(x)\), where \(u^*_1\in H^1(S^1)\) is the standard generator. The author shows that there are examples in which \(\dim M= 4\), \(M^{S_1}\neq\emptyset\) but \((M,x)\) is not \(c\)-Hamiltonian.

For the entire collection see [Zbl 0906.00019].

For the entire collection see [Zbl 0906.00019].

Reviewer: Messoud Efendiev (Berlin)