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Jun 20, 2016 at 7:33 comment added Andreas Thom Right, as I said, it is equivalent.
Jun 20, 2016 at 6:30 comment added Li Jingyang Hi, Andreas. sorry, i can not understand your reply of my comment. I only want to say that Furstenberg's conjecture is about 2-3-invariant $\text{ergodic}$ probability measure but not 2-3-invariant probability measure.
Jun 20, 2016 at 6:15 comment added Andreas Thom @LiJingyang: The only other ergodic 2-3-invariant measures are conjectured to be concentrated on periodic orbits. Since there are only countable many periodic orbits, this and absense of atoms already implies that the measure must be Lebesgue measure (and hence be ergodic).
Jun 20, 2016 at 3:54 comment added Li Jingyang I think you missed ergodicity in Furstenberg's conjecture: Furstenberg's conjecture is about 2 -3-invariant $\text{ergodic}$ probability measure.
Oct 31, 2010 at 10:07 vote accept Andreas Thom
Oct 31, 2010 at 1:44 answer added John Griesmer timeline score: 21
Oct 30, 2010 at 12:00 history edited Andreas Thom CC BY-SA 2.5
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Oct 30, 2010 at 11:53 history edited Andreas Thom CC BY-SA 2.5
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Oct 30, 2010 at 10:54 history edited Andreas Thom CC BY-SA 2.5
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Oct 30, 2010 at 10:32 history asked Andreas Thom CC BY-SA 2.5