Timeline for Does a random walk on a surface visit uniformly?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2021 at 21:59 | comment | added | Pierre PC | It should be good now. | |
| Jun 14, 2021 at 18:59 | history | edited | Pierre PC | CC BY-SA 4.0 |
Corrected the argument: one needs every point to be reacheable.
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| Jun 6, 2021 at 7:48 | comment | added | Pierre PC | @RW You are both right, I implicitly used the fact that one can go from a given point to any other one on the manifold using steps of size $\delta$, which is not obvious and actually false for some $\delta$. However, it will be true, and uniformly so in terms of how many steps, provided the walk can reach a neighbourhood of the initial point after two steps, which is true when $\delta$ is at most the injectivity radius. I'll try to edit later today. | |
| Jun 6, 2021 at 6:32 | comment | added | Leo Moos | It seems a bit weird that you say the size of $\delta$ 'plays no role'. If for example $S = \mathbf{S}^2$ and $\delta = 2\pi$ then $x_0 = x_1 = \cdots$ and there is no walk to speak off, no? | |
| Jun 6, 2021 at 5:06 | comment | added | R W | I am not convinced by your proof of ergodicity. It is a general property of Markov chains with a finite stationary measure that all invariant sets in the path space come from invariant sets in the state space. However, I do not see where you prove the absence of the latter sets. | |
| Jun 5, 2021 at 23:54 | history | edited | Pierre PC | CC BY-SA 4.0 |
added 7 characters in body
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| Jun 5, 2021 at 23:48 | history | answered | Pierre PC | CC BY-SA 4.0 |