Symplectic domination. (English) Zbl 1472.53087

In this short note the authors prove that every compact oriented symplectic manifold \(M\) of even dimension is dominated by a symplectic manifold \(S\) of the same dimension, that is to say that there exists a map of positive degree \(f\,:\,S \rightarrow M\).
The sketch of the proof deserves to be briefly recalled. The first remark is that the twistor bundle \(Z\rightarrow M\) of an oriented even-dimensional Riemannian manifold has model fiber \(F:=\operatorname{SO}(2n)/\operatorname{U}(n)\) which has the structure of an homogeneous integral symplectic manifold. Thus, for each model fiber \(F\) we have the circle bundle of the quantization \(P_F\rightarrow F\) which induces a circle bundle \(P\rightarrow Z\); by using the Levi-Civita connection for defining the horizontal distribution transverse to the fibers of \(Z\rightarrow N\) we have a connection for the circle bundle \(P\rightarrow Z\) whose curvature determines a closed integral two form \(\omega\), which is not symplectic in general. By a result of A. G. Reznikov [Ann. Global Anal. Geom. 11, No. 2, 109–118 (1993, Zbl 0810.53056)] \(\omega\) is symplectic when the metric on \(M\) is a sufficiently pinched negatively curved metric; this is not restrictive since, by a result of P. Ontaneda [Publ. Math., Inst. Hautes Étud. Sci. 131, 1–72 (2020, Zbl 1442.53026)], there exists a manifold \(N\) of the same dimension of \(M\) and degree one map \(f\,:\,N\rightarrow M\) such that the sectional curvature of \(N\) is in \([-1-\epsilon,\,-1]\). So, up to replacing \(M\) by \(N\) we have a map \(f\,:\,Z \rightarrow N\), the last part of the proof consists in considering a Donaldson hypersurface in \(Z\) and studying the degree of the resulting map [S. K. Donaldson, J. Differ. Geom. 44, No. 4, 666–705 (1996, Zbl 0883.53032)].


53D05 Symplectic manifolds (general theory)
Full Text: DOI arXiv