arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 1702.01263math.FAmath.CVmath.OA
收藏

余双交换子模的秩为2

Rank of a co-doubly commuting submodule is 2

Arup Chattopadhyay, B. Krishna Das, Jaydeb Sarkar

摘要:

证明了非平凡余双交换子模的秩是$2$ .更准确地说,假设$φ, ψ\in H^\infty(\mathbb{D})$是两个内部函数。如果$\mathcal{Q}_φ = H^2(\mathbb{D})/ φH^2(\mathbb{D})$和$\mathcal{Q}_ψ = H^2(\mathbb{D})/ ψH^2(\mathbb{D})$,则\[ \mbox{rank~}(\mathcal{Q}_φ \otimes \mathcal{Q}_ψ)^\perp = 2. \]一个直接的结果如下:设$\mathcal{S}$是$H^2(\mathbb{D}^2)$的余双交换子模。则$\mbox{rank~} \mathcal{S} = 1$当且仅当$\mathcal{S} = ΦH^2(\mathbb{D}^2)$对于某个单变量内部函数$Φ\in H^\infty(\mathbb{D}^2)$。这回答了R. G. Douglas和R.杨

英文摘要:

We prove that the rank of a non-trivial co-doubly commuting submodule is $2$. More precisely, let $φ, ψ\in H^\infty(\mathbb{D})$ be two inner functions. If $\mathcal{Q}_φ = H^2(\mathbb{D})/ φH^2(\mathbb{D})$ and $\mathcal{Q}_ψ = H^2(\mathbb{D})/ ψH^2(\mathbb{D})$, then \[ \mbox{rank~}(\mathcal{Q}_φ \otimes \mathcal{Q}_ψ)^\perp = 2. \] An immediate consequence is the following: Let $\mathcal{S}$ be a co-doubly commuting submodule of $H^2(\mathbb{D}^2)$. Then $\mbox{rank~} \mathcal{S} = 1$ if and only if $\mathcal{S} = ΦH^2(\mathbb{D}^2)$ for some one variable inner function $Φ\in H^\infty(\mathbb{D}^2)$. This answers a question posed by R. G. Douglas and R. Yang.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑