Timeline for In the plane, does complement of Brownian path have infinitely many connected components?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2015 at 19:16 | comment | added | Serguei Popov | "I don't see where you used the information about dimension in your argument." - here: "I think it's evident that the Brownian trajectory on the time interval $[0,1]$ intersects itself with positive probability." | |
| Oct 25, 2015 at 19:05 | comment | added | Serguei Popov | Douglas Zare, but it's known that a BM's trajectory doesn't hit any fixed point a.s., so the complement is clearly dense in $\mathbb{R}^2$. But I agree that instead of just "intersects itself" better write something like "goes around a small ball". | |
| Oct 25, 2015 at 19:02 | comment | added | Liviu Nicolaescu | A proof of the existence of self-intersections in dimensions $\leq 3$ can be found in Sec. 9.3 of Morters and Peress' book research.microsoft.com/en-us/um/people/peres/brbook.pdf | |
| Oct 25, 2015 at 19:01 | comment | added | Douglas Zare | Step 3 should have an argument, too, to rule out something like a space-filling curve that intersects itself a lot but which has empty complement. | |
| Oct 25, 2015 at 18:46 | comment | added | Liviu Nicolaescu | In dimensions $\geq 4$ a.s. a Brownian path will not have self-intersections according to Dvoretsky, Erdos and Kakutani. The fact that the dimension is $\leq 2$ must enter the argument. I don't see where you used the information about dimension in your argument. | |
| Oct 25, 2015 at 17:25 | comment | added | Serguei Popov | By conformal invariance, this probability must be the same for all intervals (consider the time interval $[0,c]$; then the mapping $z\mapsto z/\sqrt{c}$ sends the trajectory on $[0,c]$ to trajectory on $[0,1]$). I think it's evident that the Brownian trajectory on the time interval $[0,1]$ intersects itself with positive probability. | |
| Oct 25, 2015 at 12:27 | comment | added | Liviu Nicolaescu | Why is (1) "easy to see"? | |
| Oct 25, 2015 at 11:58 | history | answered | Serguei Popov | CC BY-SA 3.0 |