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arXiv 0903.4918math.SGmath.AT
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关于Petrie猜想的一个辛推广

On a symplectic generalization of Petrie's conjecture

Susan Tolman

摘要:

受Petrie猜想的启发,我们考虑以下问题:设一个圆以Hamilton方式作用在一个紧辛流形$(M,ω)$上,该流形对所有$i$都满足$H^{2i}(M;\R) = H^{2i}(\CP^n,\R)$. $H^j(M;\Z) = H^j(\CP^n;\Z)$是否适用于所有$j$?$M$的全Chern类是否由上同调环$H^*(M;\Z)$决定?我们回答这些问题,在六维的情况下,$H^j(M;\Z)$是等于$H^j(\CP^3;\Z)$的所有$j$,通过证明,只有四个上同调环可以出现,并通过计算总陈类在每种情况下。我们还证明了不存在奇异动作。更准确地说,如果$H^*(M;\Z)$同构于$H^*(\CP^3;\Z)$或$H^*(\Tilde{G}_2(\R^5);\Z)$,那么固定分量的表示与标准作用之一相容;在剩下的两种情况下,表示严格由上同调环确定。最后,我们的结果提出了一个自然的问题:剩下的两个上同调环实际上出现了吗?这个问题与辛拓扑学中的一些有趣的问题密切相关,如椭球的嵌入。

英文摘要:

Motivated by the Petrie conjecture, we consider the following questions: Let a circle act in a Hamiltonian fashion on a compact symplectic manifold $(M,ω)$ which satisfies $H^{2i}(M;\R) = H^{2i}(\CP^n,\R)$ for all $i$. Is $H^j(M;\Z) = H^j(\CP^n;\Z)$ for all $j$? Is the total Chern class of $M$ determined by the cohomology ring $H^*(M;\Z)$? We answer these questions in the six dimensional case by showing that $H^j(M;\Z)$ is equal to $H^j(\CP^3;\Z)$ for all $j$, by proving that only four cohomology rings can arise, and by computing the total Chern class in each case. We also prove that there are no exotic actions. More precisely, if $H^*(M;\Z)$ is isomorphic to $H^*(\CP^3;\Z)$ or $H^*(\Tilde{G}_2(\R^5);\Z)$, then the representations at the fixed components are compatible with one of the standard actions; in the remaining two case, the representation is strictly determined by the cohomology ring. Finally, our results suggest a natural question: do the remaining two cohomology rings actually arise? This question is closely related to some interesting problems in symplectic topology, such as embeddings of ellipsoids.

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