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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Lefschetz manifold ： ウィキペディア英語版 Lefschetz manifold In mathematics, a Lefschetz manifold is a particular kind of symplectic manifold $(M^,\; \backslash omega)$, sharing a certain cohomological property with Kähler manifolds, that of satisfying the conclusion of the Hard Lefschetz theorem. More precisely, the strong Lefschetz property asks that for $k\; =\; 1\; \backslash ldots\; n$, the cup product :$\backslash cup\; ():\; H^(M,\; \backslash mathbb\; R)\backslash to\; H^(M,\; \backslash mathbb\; R)$ be an isomorphism. The topology of these symplectic manifolds is severely constrained, for example their odd Betti numbers are even. This remark leads to numerous examples of symplectic manifolds which are not Kähler, the first historical example is due to William Thurston. ==Lefschetz maps== Let $M$ be a ($2n$)-dimensional smooth manifold. Each element :$()\; \backslash in\; H\_^2\; (M)$ of the second de Rham cohomology space of $M$ induces a map :$L\_:\; H\_\; (M)\; \backslash to\; H\_\; (M),\; ()\; \backslash mapsto\; (\backslash wedge\; \backslash alpha)$ called the Lefschetz map of $()$. Letting $L\_^i$ be the $i$th iteration of $L\_$, we have for each $0\; \backslash leq\; i\; \backslash leq\; n$ a map :$L\_^i\; :\; H\_^(M)\; \backslash to\; H\_^(M).$ If $M$ is compact and oriented, then Poincaré duality tells us that $H\_^(M)$ and $H\_^(M)$ are vector spaces of the same dimension, so in these cases it is natural to ask whether or not the various iterations of Lefschetz maps are isomorphisms. The Hard Lefschetz theorem states that this is the case for the symplectic form of a compact Kähler manifold. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Lefschetz manifold」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.