AI 中文总结
本文证明了紧致离散赋值环上的中心傅里叶代数不可amenable,通过研究交换轨道超群的傅里叶代数,证实了Alaghmandan-Spronk猜想,并建立了与经典情形的对偶同构。
AI 中文摘要
设 $R$ 为紧致离散赋值环,$G = R^d \rtimes \operatorname{GL}_d(R)$。我们证明了中心傅里叶代数 $\operatorname{ZA}(G)$ 不是可amenable的,从而再次证实了Alaghmandan和Spronk的一个猜想。我们的方法归结为研究交换轨道超群 $H = R^d/\operatorname{GL}_d(R)$ 的傅里叶代数。在此过程中,我们还证明了任何交换轨道超群 $H$ 的对偶是对应对偶作用的超群,这进而表明 $\operatorname{A}(H) \cong \operatorname{L}^1(\widehat{H})$,这与经典情形一致。
英文摘要
Let $R$ be a compact discrete valuation ring and $G = R^d \rtimes \operatorname{GL}_d(R)$. We show that the central Fourier algebra $\operatorname{ZA}(G)$ is not amenable, reaffirming a conjecture of Alaghmandan and Spronk. Our methods reduce to studying the Fourier algebra of the commutative orbit hypergroup $H = R^d/\operatorname{GL}_d(R)$. Along the way, we also show that dual of any commutative orbit hypergroup $H$ is the hypergroup of the corresponding dual action, and this in turn shows that $\operatorname{A}(H) \cong \operatorname{L}^1(\widehat{H})$, which aligns with the classical setting.
Comments11 pages