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Nov 23, 2022 at 23:39 history became hot network question
Nov 23, 2022 at 21:43 answer added Misha Verbitsky timeline score: 7
Nov 23, 2022 at 21:12 comment added Ben McKay Robert's example is often called a Hopf manifold.
Nov 23, 2022 at 15:55 comment added Robert Bryant Well, it doesn't have to be Kähler. For example, if you consider the compact manifold that you get by removing the origin from $\mathbb{C}^n$ ($n>1$) and dividing by the $\mathbb{Z}$-action of $n\cdot z = 2^n\,z$, that manifold, diffeomorphic to $S^1\times S^{2n-1}$, has $TX\simeq T^*X$ as complex bundles, but it is not holomorphic symplectic (or even Kähler).
Nov 23, 2022 at 15:37 history asked Jost Schultze CC BY-SA 4.0