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首页> 外文期刊>Compositio mathematica >Is the function field of a reductive Lie algebra purely transcendental over the field of invariants for the adjoint action?
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Is the function field of a reductive Lie algebra purely transcendental over the field of invariants for the adjoint action?

机译:归约李代数的函数域在伴随作用的不变量域上是纯粹超越的吗?

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摘要

Let k be a field of characteristic zero, let G be a connected reductive algebraic group over k and let g be its Lie algebra. Let k(G), respectively, k(g), be the field of k-rational functions on G, respectively, g. The conjugation action of G on itself induces the adjoint action of G on g. We investigate the question whether or not the field extensions k(G)/k(G)(G) and k(g)/k(g)(G) are purely transcendental. We show that the answer is the same for k(G)/k(G)(G) and k(g)/k(g)(G), and reduce the problem to the case where G is simple. For simple groups we show that the answer is positive if G is split of type A(n) or C(n), and negative for groups of other types, except possibly G(2). A key ingredient in the proof of the negative result is a recent formula for the unramified Brauer group of a homogeneous space with connected stabilizers. As a byproduct of our investigation we give an affirmative answer to a question of Grothendieck about the existence of a rational section of the categorical quotient morphism for the conjugating action of G on itself.
机译:令k为零特性场,令G为k上的连通还原代数群,令g为李氏代数。令k(G)分别为k(g),分别为G上的k有理函数的字段。 G对自身的共轭作用诱导G对g的伴随作用。我们研究以下问题:场扩展k(G)/ k(G)(G)和k(g)/ k(g)(G)是否纯粹是超越的。我们证明,对于k(G)/ k(G)(G)和k(g)/ k(g)(G),答案是相同的,并将问题简化为G简单的情况。对于简单的组,我们证明如果将G拆分为A(n)或C(n)类型,则答案为肯定,而对于其他类型的组(可能为G(2)),答案为否。证明阴性结果的关键因素是具有连接的稳定剂的均质空间的未分支Brauer组的最新公式。作为我们研究的副产品,我们对格洛腾迪克(Grothendieck)问题的肯定回答,该问题涉及对G自身共轭作用的分类商态的有理部分。

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