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Sep 23, 2021 at 19:43 comment added Nicolast To precise Will Sawin's remark, Poincaré--Bendixson's theorem implies that the ergodic invariant measures of a flow on $\mathbb R^2$ or $\mathbb S^2$ are supported by cycles or fixed points, so many ``natural'' measures cannot be ergodic invariant measures.
Sep 13, 2021 at 20:12 comment added Will Sawin Slowly raising from the north pole to the south pole doesn't sound like it has an invariant measure supported on the whole space, not to mention being ergodic.
Sep 13, 2021 at 20:02 comment added Cupitor @WillSawin, I don't see why a slowly raising dynamical system from southern pole to the northern pole (northern pole is where the vector field vanishes and southern pole is the opposite) finally stopping at the northern pole doesn't do the job, especially with the comment I added under MartinMW's answer.
Sep 13, 2021 at 19:35 comment added Will Sawin For the uniform measure on $S^2$, you may run in trouble finding a continuous time dynamical system it is ergodic for, because of the hairy ball theorem.
Sep 13, 2021 at 18:55 comment added Cupitor @WillSawin, I guess I assume the measure has a connected support.
Sep 13, 2021 at 15:28 comment added Piyush Grover You will probably need a absolutely continuous measure w.r.t to Lebesgue.
Sep 13, 2021 at 0:53 answer added Martin M. W. timeline score: 3
Sep 10, 2021 at 22:26 comment added Martin M. W. Instead of looking at flows, maybe it would be simpler to ask about a discrete dynamical system generated by a continuous map?
Sep 10, 2021 at 22:07 comment added Will Sawin If the support has two disjoint components then the flow I don't think the flow can be a smooth function of time and also be ergodic since the flow would not be able to mix the two components. So there must be some criteria.
Sep 10, 2021 at 21:58 comment added Cupitor @WillSawin, thanks very right point. I am phrasing this question from pure intuition. I made corrections. I hope the new form makes sense.
Sep 10, 2021 at 21:57 history edited Cupitor CC BY-SA 4.0
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Sep 10, 2021 at 19:56 comment added Will Sawin What kind of measure spaces are you looking for distributions on? For many notions of "distribution" and "nice" there is a theorem of the form "there is a single distribution that every nice distribution is equivalent to" so the result you're looking for would be true, but basically trivial.
Sep 10, 2021 at 17:10 history edited YCor CC BY-SA 4.0
removed capitals from title
S Sep 10, 2021 at 16:20 review First questions
Sep 10, 2021 at 16:51
S Sep 10, 2021 at 16:20 history asked Cupitor CC BY-SA 4.0