Fredholmness of fringe operators over the bidisk

Fredholmness of fringe operators over the bidisk Let M be an invariant subspace of $$H^2$$ H 2 over the bidisk. Associated with M, we have the fringe operator $$F^M_z$$ F z M on $$M\ominus w M$$ M ⊖ w M . For $$A\subset H^2$$ A ⊂ H 2 , let [A] denote the smallest invariant subspace containing A. Assume that $$F^M_z$$ F z M is Fredholm. If h is a bounded analytic function on $$\mathbb {D}^2$$ D 2 satisfying $$h(0,0)\not =0$$ h ( 0 , 0 ) ≠ 0 , then $$F^{[h M]}_z$$ F z [ h M ] is Fredholm and $$\mathrm{ind}\,F^{[h M]}_z=\mathrm{ind}\,F^M_z$$ ind F z [ h M ] = ind F z M . http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Archiv der Mathematik Springer Journals

Fredholmness of fringe operators over the bidisk

, Volume 109 (3) – Aug 4, 2017
8 pages

/lp/springer_journal/fredholmness-of-fringe-operators-over-the-bidisk-LwH0BpjRS0
Publisher
Springer International Publishing
Subject
Mathematics; Mathematics, general
ISSN
0003-889X
eISSN
1420-8938
D.O.I.
10.1007/s00013-017-1075-7
Publisher site
See Article on Publisher Site

Abstract

Let M be an invariant subspace of $$H^2$$ H 2 over the bidisk. Associated with M, we have the fringe operator $$F^M_z$$ F z M on $$M\ominus w M$$ M ⊖ w M . For $$A\subset H^2$$ A ⊂ H 2 , let [A] denote the smallest invariant subspace containing A. Assume that $$F^M_z$$ F z M is Fredholm. If h is a bounded analytic function on $$\mathbb {D}^2$$ D 2 satisfying $$h(0,0)\not =0$$ h ( 0 , 0 ) ≠ 0 , then $$F^{[h M]}_z$$ F z [ h M ] is Fredholm and $$\mathrm{ind}\,F^{[h M]}_z=\mathrm{ind}\,F^M_z$$ ind F z [ h M ] = ind F z M .

Journal

Archiv der MathematikSpringer Journals

Published: Aug 4, 2017

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