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Oct 6, 2023 at 15:41 history edited Nate River CC BY-SA 4.0
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Oct 6, 2023 at 15:22 answer added Uri Bader timeline score: 3
Oct 6, 2023 at 14:34 comment added Nate River @TerryTao Thank you very much, I will work out the equivalence.
Oct 6, 2023 at 14:25 comment added Terry Tao You should measure $A$ in $\mu^Y_y$ rather than $\mu^Y_{S^n y}$, and integrate in $y$ rather than ask for pointwise ae convergence, then it will be an equivalent form of relative weak mixing.
Oct 6, 2023 at 7:43 comment added Nate River @TerryTao Oh true... maybe the fibres need to be moving. According to Corollary 5.24 in Ward, we have $T_* \mu^Y_{y} = \mu^Y_{Sy}$ for $\nu$-a.e. $y$, so we can try to define relative weak mixing to be that the limit $$\lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^n |\mu^Y_{S^n y} (T^{-n} A \cap B) - \mu^Y_{S^n y} (A) \mu^Y_{S^n y} (B)|$$ exists and equals $0$ for $\nu$-almost every $y \in Y$ and $A, B \in \mathcal B_X$. Does this definition work?
Oct 6, 2023 at 6:07 comment added Terry Tao The measures $\mu^Y_y$ are almost never $T$-invariant (this basically only occurs when $S$ is trivial).
Oct 6, 2023 at 5:17 history edited Nate River CC BY-SA 4.0
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Oct 6, 2023 at 4:53 history edited Nate River CC BY-SA 4.0
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Oct 6, 2023 at 4:47 history asked Nate River CC BY-SA 4.0