Kolmogorov automorphism

In mathematics, a Kolmogorov automorphism, K-automorphism, K-shift or K-system is an invertible, measure-preserving automorphism defined on a standard probability space that obeys Kolmogorov's zero–one law.[1] All Bernoulli automorphisms are K-automorphisms (one says they have the K-property), but not vice versa. Many ergodic dynamical systems have been shown to have the K-property, although more recent research has shown that many of these are in fact Bernoulli automorphisms.

Although the definition of the K-property seems reasonably general, it stands in sharp distinction to the Bernoulli automorphism. In particular, the Ornstein isomorphism theorem does not apply to K-systems, and so the entropy is not sufficient to classify such systems – there exist uncountably many non-isomorphic K-systems with the same entropy. In essence, the collection of K-systems is large, messy and uncategorized; whereas the B-automorphisms are 'completely' described by Ornstein theory.

Formal definition

Let ${\displaystyle (X,{\mathcal {B}},\mu )}$ be a standard probability space, and let ${\displaystyle T}$ be an invertible, measure-preserving transformation. Then ${\displaystyle T}$ is called a K-automorphism, K-transform or K-shift, if there exists a sub-sigma algebra ${\displaystyle {\mathcal {K}}\subset {\mathcal {B}}}$ such that the following three properties hold:

${\displaystyle {\mbox{(1) }}{\mathcal {K}}\subset T{\mathcal {K}}}$

${\displaystyle {\mbox{(2) }}\bigvee _{n=0}^{\infty }T^{n}{\mathcal {K}}={\mathcal {B}}}$

${\displaystyle {\mbox{(3) }}\bigcap _{n=0}^{\infty }T^{-n}{\mathcal {K}}=\{X,\varnothing \}}$

Here, the symbol ${\displaystyle \vee }$ is the join of sigma algebras, while ${\displaystyle \cap }$ is set intersection. The equality should be understood as holding almost everywhere, that is, differing at most on a set of measure zero.

Properties

Assuming that the sigma algebra is not trivial, that is, if ${\displaystyle {\mathcal {B}}\neq \{X,\varnothing \}}$, then ${\displaystyle {\mathcal {K}}\neq T{\mathcal {K}}.}$ It follows that K-automorphisms are strong mixing.

All Bernoulli automorphisms are K-automorphisms, but not vice versa.

References

1. Peter Walters, An Introduction to Ergodic Theory, (1982) Springer-Verlag ISBN 0-387-90599-5

Further reading

• Christopher Hoffman, "A K counterexample machine", Trans. Amer. Math. Soc. 351 (1999), pp 4263–4280.