An example of a factor map without a saturated compensation function (Q2773062)
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scientific article; zbMATH DE number 1709206
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | An example of a factor map without a saturated compensation function |
scientific article; zbMATH DE number 1709206 |
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5 August 2002
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symbolic dynamics
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saturated compensation function
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factor map
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subshifts of finite type
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non-Markovian factor map
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0.80377203
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0.76127815
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0.75392187
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0.7501993
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An example of a factor map without a saturated compensation function (English)
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Let \((X,S)\) and \((Y,T)\) be two topological dynamical systems. Let \(\pi:X\to Y\) be a factor map. A continued function \(F\) of \(X\) is a compensation function if for all continued function \(\varphi\) of \(Y\), we have \(P_X [F+\varphi\circ \pi]= P_Y(\varphi)\) where \(P\) is the pressure function. It is known that there always exists a compensation function when \(X\) and \(Y\) are irreducible subshifts of finite type. Moreover a compensation function is a saturated compensation function if there exists a continued function \(G\) of \(Y\) such that \(F= G\circ \pi\). NEWLINENEWLINENEWLINEIn this paper the author discusses the existence of saturated compensation function. In the first part of the paper, the author presents an example of a factor map between two subshifts of finite type that does not have a saturated compensation function. He also presents an example of a non-Markovian factor map with a saturated compensation function. NEWLINENEWLINENEWLINEIn the second part of the paper, the author gives a necessary and sufficient condition for a certain type of factor map to have a saturated compensation function.
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