Given a rational homogeneous manifold \(S=G/P\) of Picard number one and a Schubert variety \(S_0 \) of S, the pair \((S,S_0)\) is said to be homologically rigid if any subvariety of S having the same homology class as \(S_0\) must be a translate of \(S_0\) by the automorphism group of S. The pair \((S,S_0)\) is said to be Schur rigid if any subvariety of S with homology class equal to a multiple of the homology class of \( S_0 \) must be a sum of translates of \(S_0\). Earlier we completely determined homologically rigid pairs \((S,S_0)\) in case \(S_0 \) is homogeneous and answered the same question in smooth non-homogeneous cases. In this article we consider Schur rigidity, proving that \((S,S_0)\) exhibits Schur rigidity whenever \(S_0 \) is a non-linear smooth Schubert variety. Modulo a classification result of the first author’s, our proof proceeds by a reduction to homological rigidity by deforming a subvariety Z of S with homology class equal to a multiple of the homology class of \( S_0 \) into a sum of distinct translates of \( S_0\), and by observing that the arguments for the homological rigidity apply since any two translates of \(S_0\) intersect in codimension at least two. Such a degeneration is achieved by means of the \({\mathbb {C}}^*\)-action associated with the stabilizer of the Schubert variety \(T_0\) opposite to \(S_0\). By transversality of general translates, a general translate of Z intersects \(T_0\) transversely and the \({\mathbb {C}}^*\)-action associated with the stabilizer of \(T_0\) induces a degeneration of Z into a sum of translates of \(S_0\), not necessarily distinct. After investigating the Bialynicki-Birular decomposition associated with the \({\mathbb {C}}^*\)-action we prove a refined form of transversality to get a degeneration of Z into a sum of distinct translates of \(S_0\).

中文翻译：

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Picard一号有理齐次流形中Schubert变种的Schur刚度

给定Picard一号的有理齐次流形\（S = G / P \）和S的Schubert变量\（S_0 \），如果有任何子变量，则对\（（S，S_0）\）是同构刚性的的小号具有相同的同源性的等级\（S_0 \）必须是翻译的\（S_0 \）由构群的小号。如果对等（S，S_0）的同源性等于对（S_0 \）的同源性的倍数的S的任何子变体必须是\（S_0 \ ）。之前我们完全确定了同构刚性对\（（S，S_0）\）是\（S_0 \）是齐次的情况，并且在光滑非齐次情况下回答了相同的问题。在本文中，我们考虑了Schur刚度，证明\（（S，S_0）\）在\（S_0 \）是非线性光滑Schubert变体时都表现出Schur刚度。模的第一作者的分类结果，我们证明前进通过还原到同源刚性由变形的subvariety ž的小号具有同源性的类等于同源性类的倍数\（S_0 \）成不同的转换的总和\ （S_0 \），并观察到同质刚度的论点适用于\（S_0 \）的任意两个转换至少在两个维度上相交。通过与与（S_0 \）相反的舒伯特变种\（T_0 \）的稳定剂相关的\（{\ mathbb {C}} ^ * \）作用来实现这种退化。通过一般翻译的横向性，Z的一般翻译与\（T_0 \）横向相交，并且与\（T_0 \）的稳定子相关的\（{\ mathbb {C}} ^ * \）作用引起Z的退化到\（S_0 \）的翻译量之和，不一定是不同的。在调查与\（{\ mathbb {C}} ^ * \）相关的Bialynicki-Birular分解后动作我们证明了一种精细的横向形式，可以将Z退化为\（S_0 \）的不同平移之和。