## 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)].

### MSC:

 53D05 Symplectic manifolds (general theory)

### Citations:

Zbl 0810.53056; Zbl 1442.53026; Zbl 0883.53032
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