When do skew-products exist?
Electronic Communications in Probability, 20, no. 54, 1-14, 2015 The classical skew-product decomposition of planar Brownian motion represents the process in polar coordinates as an autonomously Markovian radial part and an angular part that is an independent Brownian motion on the unit circle time-...
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Zusammenfassung: | Electronic Communications in Probability, 20, no. 54, 1-14, 2015 The classical skew-product decomposition of planar Brownian motion represents
the process in polar coordinates as an autonomously Markovian radial part and
an angular part that is an independent Brownian motion on the unit circle
time-changed according to the radial part. Theorem 4 of Liao (2009) gives a
broad generalization of this fact to a setting where there is a diffusion on a
manifold $X$ with a distribution that is equivariant under the smooth action of
a Lie group $K$. Under appropriate conditions, there is a decomposition into an
autonomously Markovian "radial" part that lives on the space of orbits of $K$
and an "angular" part that is an independent Brownian motion on the homogeneous
space $K/M$, where $M$ is the isotropy subgroup of a point of $x$, that is
time-changed with a time-change that is adapted to the filtration of the radial
part. We present two apparent counterexamples to Theorem 4 of Liao (2009). In
the first counterexample the angular part is not a time-change of any Brownian
motion on $K/M$, whereas in the second counterexample the angular part is the
time-change of a Brownian motion on $K/M$ but this Brownian motion is not
independent of the radial part. In both of these examples $K/M$ has dimension
$1$. The statement and proof of Theorem 4 from Liao (2009) remain valid when
$K/M$ has dimension greater than $1$. Our examples raise the question of what
conditions lead to the usual sort of skew-product decomposition when $K/M$ has
dimension $1$ and what conditions lead to there being no decomposition at all
or one in which the angular part is a time-changed Brownian motion but this
Brownian motion is not independent of the radial part. |
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DOI: | 10.48550/arxiv.1408.6507 |