By Xiaoman Chen, Kunyu Guo

The seminal 1989 paintings of Douglas and Paulsen at the idea of Hilbert modules over functionality algebras brought about a couple of significant learn efforts. This in flip resulted in a few interesting and worthy effects, really within the parts of operator conception and useful research. With the sphere now commencing to blossom, the time has come to gather these ends up in one quantity. Written through of the main energetic and often-cited researchers within the box, Analytic Hilbert Modules deals a transparent, logical survey of modern advancements, together with advances made by way of authors and others. It offers much-needed perception into functionality concept of numerous variables and comprises major effects released the following for the 1st time in components similar to attribute area thought, pressure phenomena, the equivalence challenge, Arveson modules, extension thought, and reproducing Hilbert areas on n-dimensional complicated house.

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7 Let M be a submodule of an analytic Hilbert module X with finite rank and let I be a finite codimensional ideal of C. Then dim M/[IM ] ≤ codim(I) rank(M ). Proof. We assume rank(M ) = m with a generating set {f1 , f2 , · · · , fm }. Now let P : M → M [IM ] be the orthogonal projection. We write f˜i = P fi for i = 1, 2, · · · , m. Since I annihilates the quotient module M/[IM ], one can consider M/[IM ] as a C/I-module with a generating set {f˜1 , f˜2 , · · · , f˜m }. Since C/I is finite dimensional, this ensures dim M/[IM ] < ∞, and actually dim M/[IM ] ≤ codim(I) rank(M ).

Moreover, the same argument as the above proof enables us to obtain the following. 4 Let both M1 and M2 be submodules of X. If M1 ⊇ M2 and dim M1 /M2 = k < ∞, then we have (1) Z(M2 )\Z(M1 ) ⊆ σp (Mz1 , Mz2 , · · · , Mzn ) ⊂ Ω, (2) dim M1 /M2 ≥ dim M2λ /M1λ = card(Z(M2 )\Z(M1 )), λ∈Z(M2 )\Z(M1 ) and “equal” if and only if M2 is an AF-cosubmodule. We end with two examples. 5 Let f be analytic on a neighborhood of the origin in Cn , and let f (z) = m fm (z) be the homogeneous expansion of f at the origin.

This implies that there is some gs such that gs (λ) = 0. From the equality pps = f gs one has [pps ]λ = [f ]λ . 1, [f ]eλ = [pps ]eλ = q1 ps C. However, for each i, [f ]λ ⊂ [ppi ]λ and hence for every i, [f ]eλ ⊃ [ppi ]eλ = q1 pi C. Thus, for every i, q1 ps C ⊃ q1 pi C. Thus each pi is divisible by ps . Thus, every pi is divisible by ps . This is impossible. Therefore, p1 , p2 , · · · , pk have no common zero in Ω, that is, Z(L) ∩ Ω = ∅. The proof is complete. 5 Let I = pL be the Beurling form of the ideal I.