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Multiplicities, invariant subspaces and an additive formula
Proceedings of the Edinburgh Mathematical Society ( IF 0.7 ) Pub Date : 2021-04-30 , DOI: 10.1017/s0013091521000146
Arup Chattopadhyay , Jaydeb Sarkar , Srijan Sarkar

Let $T = (T_1, \ldots , T_n)$ be a commuting tuple of bounded linear operators on a Hilbert space $\mathcal{H}$. The multiplicity of $T$ is the cardinality of a minimal generating set with respect to $T$. In this paper, we establish an additive formula for multiplicities of a class of commuting tuples of operators. A special case of the main result states the following: Let $n \geq 2$, and let $\mathcal{Q}_i$, $i = 1, \ldots , n$, be a proper closed shift co-invariant subspaces of the Dirichlet space or the Hardy space over the unit disc in $\mathbb {C}$. If $\mathcal{Q}_i^{\bot }$, $i = 1, \ldots , n$, is a zero-based shift invariant subspace, then the multiplicity of the joint $M_{\textbf {z}} = (M_{z_1}, \ldots , M_{z_n})$-invariant subspace $(\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{\perp }$ of the Dirichlet space or the Hardy space over the unit polydisc in $\mathbb {C}^{n}$ is given by \[ \mbox{mult}_{M_{\textbf{{z}}}|_{ (\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{{\perp}}}} (\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{{\perp}} = \sum_{i=1}^{n} (\mbox{mult}_{M_z|_{\mathcal{Q}_i^{{\perp}}}} (\mathcal{Q}_i^{\bot})) = n. \]A similar result holds for the Bergman space over the unit polydisc.

中文翻译:

多重性、不变子空间和加法公式

$T = (T_1, \ldots , T_n)$是希尔伯特空间上的有界线性算子的交换元组$\数学{H}$. 的多样性$T$是关于最小生成集的基数$T$. 在本文中,我们建立了一类可交换运算符元组的多重性的加法公式。主要结果的一个特例如下:让$n \geq 2$, 然后让$\mathcal{Q}_i$,$i = 1, \ldots , n$, 是 Dirichlet 空间或 Hardy 空间在单位圆盘上的一个闭移位协不变子空间$\mathbb {C}$. 如果$\mathcal{Q}_i^{\bot }$,$i = 1, \ldots , n$, 是一个从零开始的移位不变子空间,那么关节的多重性$M_{\textbf {z}} = (M_{z_1}, \ldots , M_{z_n})$- 不变子空间$(\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{\perp }$Dirichlet 空间或 Hardy 空间在单位 polydisc 上的$\mathbb {C}^{n}$是(谁)给的\[ \mbox{mult}_{M_{\textbf{{z}}}|_{ (\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{{\perp}}} } (\mathcal{Q}_1 \otimes \cdots \otimes \mathcal{Q}_n)^{{\perp}} = \sum_{i=1}^{n} (\mbox{mult}_{M_z| _{\mathcal{Q}_i^{{\perp}}}} (\mathcal{Q}_i^{\bot})) = n。\]类似的结果也适用于单位多圆盘上的伯格曼空间。
更新日期:2021-04-30
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