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Nov 24, 2021 at 11:30 comment added Gabe K @AliTaghavi A distribution $f d \mu$ is uniform with respect to a measure $\mu$ if and only if it satisfies the identity $\int_E f d \, \mu = C \mu(E) $ for all measurable sets $E$. In other words, you require the Radon-Nikodym derivative $\frac{ f d \mu}{d \mu}$ to be a constant value $C$.
Nov 24, 2021 at 11:09 comment added Ali Taghavi But I agree with the first comment by @MattF. What is a relation to the initial definition of uniform distribution?
Nov 24, 2021 at 11:06 comment added Ali Taghavi Thank you for the edit. but I am not convinced yet that this is the definition of uniformicity.
Nov 23, 2021 at 18:49 comment added Gabe K It's easy to miss the forest for the trees in the calculation, but if you think about what the scalar curvature is actually measuring, it should give some intuition for why this must be the case.
Nov 23, 2021 at 18:45 comment added Gabe K @MattF. The $C$ in the above formula comes from the definition of a distribution being uniform, so this identity must hold for all $\varepsilon$. If you want a more explicit estimate on the $O(\varepsilon^{n+1})$ term, from the first integral identity in the answer, you can replace the $O(\varepsilon)$ term by $\pm \|\nabla S \|_\infty \varepsilon$. Going through the rest of the calculation, this shows that the $O(\varepsilon^{n+1})$ term at the end can be bounded by $ \pm \|\nabla S \|_\infty \frac{\pi^{n/2}}{\Gamma\bigl(\tfrac n2 + 1\bigr)} \varepsilon^{n+1}$
Nov 23, 2021 at 17:37 comment added user44143 @GabeK, I think an estimate for the needed $\epsilon$ in terms of $C$ would convince me.
Nov 23, 2021 at 15:38 comment added Gabe K @MattF. In order for the distribution to be uniform, we must have that $\int_{B_\epsilon(p)} S d \mu = C Vol({B_\epsilon(p)})$ for some constant $C$. This calculation shows that if the scalar curvature is not constant, this is not the case because the integral is $S(p) Vol({B_\epsilon(p)}) + O(\varepsilon^{n+1})$
Nov 23, 2021 at 15:24 comment added user44143 @GabeK, how does this use the uniformity of the distribution?
Nov 23, 2021 at 14:51 comment added Gabe K @AliTaghavi I've edited the answer to provide more detail why it is necessary for the scalar curvature to be constant.
Nov 23, 2021 at 14:49 history edited Gabe K CC BY-SA 4.0
I gave more details for why it is necessary for the scalar curvature to be constant for the measure to be uniform.
Nov 23, 2021 at 12:52 comment added Ali Taghavi Thank you very much and my previous +1 for your attention to my question and your answer.I would appreciate if you explain the details. I thought about your answer but i do not get the idea how to conclude the imposibility of a uniform distribution.
Nov 23, 2021 at 12:26 comment added Ali Taghavi @MattF. Let the image of the curvature is $[a,b]$ and the curvature has uniform distribution. According to the formula $log (V_{g}(\epsilon)(p)/V_{E}(\epsilon)p) \to k(p)$ how can we get a contradiction?
Nov 23, 2021 at 11:34 comment added Ali Taghavi @MattF. Yes that s is my question. but the answer by Gabe K says that it is impossible. The random variable is constant.
Oct 16, 2021 at 12:45 comment added Gabe K That's correct. The observation about the geodesic balls shows that in order for the measure to be uniform, the scalar curvature must be constant.
Oct 16, 2021 at 12:04 comment added user44143 I think the question is asking about a distribution that is uniform between some $a$ and $b$ — i.e. for a manifold where, whenever $x<b$, the area with curvature between $a$ and $x$ is proportional to $x-a$.
Oct 16, 2021 at 11:52 history edited Gabe K CC BY-SA 4.0
Made the epsilons consistent
Oct 16, 2021 at 11:44 history edited YCor CC BY-SA 4.0
formatting
Oct 16, 2021 at 11:34 history answered Gabe K CC BY-SA 4.0