Timeline for Is there something like a "self-avoiding Markov chain" on a continuous space?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2022 at 17:11 | comment | added | Christophe Leuridan | I do not know. Compare your simulations with T Kennedy's ones and try larger values of $\kappa$. I have no other idea on this question. | |
| Dec 18, 2022 at 16:38 | comment | added | 0xbadf00d | @ChristopheLeuridan Yes, it should be space-filling. But as you can see the map to $[0,1]^2$ gives a poor result (hopefully I did not make any mistake in the implementation, but I don't see where it should be) | |
| Dec 18, 2022 at 16:29 | comment | added | Christophe Leuridan | @0xbadf00d I am not a specialist on this topic. According to Kostya_I, $SLE_\kappa$ is space filling as soon as $\kappa \ge 8$. This may be more visible for larger values of $\kappa$. You can find other simulations there math.arizona.edu/~tgk/sle | |
| Dec 17, 2022 at 15:51 | comment | added | 0xbadf00d | @ChristopheLeuridan So, as long as I didn't made any mistake in my conformal map to $[0,1]^2$, this seems to be bad for my purposes. What I actually wanted to obtain is a space-filling curve on $[0,1]^2$. What do you think? | |
| Dec 17, 2022 at 15:50 | comment | added | 0xbadf00d | @ChristopheLeuridan Here is a plot of my implementation of the SLE for $\kappa=8$, $10.000$ points and up to time $t=1$: i.sstatic.net/orBbG.png (the time steps are uniform; I know that it is better to adapteively refine depending on the sampled path of the Brownian motion, but that's not the point in the sequel). Now, I've applied a conformal map from $\mathbb H$ to $[0,1]^2$ and obtained i.sstatic.net/m4gTL.png. As you can see, everything seems to get concentraited on the right (and this gets even worse for larger $t$). | |
| Dec 14, 2022 at 21:02 | comment | added | 0xbadf00d | @ChristopheLeuridan I was able to implement the evolution on $\mathbb H$, but I really got some trouble to even understand the definition of the conformal map to $(0,1)^2$ given in the Wikipedia article. Please take a look at the question I've asked for that: mathoverflow.net/q/436590/91890. | |
| Dec 11, 2022 at 15:53 | comment | added | Christophe Leuridan | To get a conformal bijection between the half upper plane and a square, look at en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping | |
| Dec 10, 2022 at 22:52 | comment | added | 0xbadf00d | @Kostya_I If $\gamma_t$ is the curve in $\mathbb H$, how do we obtain the curve in $(0,1)^2$ from that? How is the conformal map defined you are talking about? If you got a reference for me, that would already been really helpful. | |
| Dec 10, 2022 at 15:18 | comment | added | Kostya_I | @0xbadf00d the SLE process in $(0,1)^2$ is the image of the process in $\mathbb{H}$ under a conformal map. (Alternatively, you can conformally transplant the Lowener ODE into $(0,1)^{2}$ by a conformal map and solve the resulting equation) As for the second question, look up numerical simulations of SLE, people have done it and it's not trivial. | |
| Dec 10, 2022 at 12:11 | comment | added | 0xbadf00d | @Kostya_I Did I understand this correctly: Let $(\xi_t)_{t\ge0}$ denote a "driving function" (for example, $\xi_t=\sqrt\kappa B_t$ for some Brownian motion $B$) and $(g_t)_{t\ge0}$ denote the solution of $$g_t(z)=z+\int_0^t\frac2{g_s(z)-\xi_s}\:{\rm d}s.$$ Then the curve we are actually considering is $\gamma_t:=g_t^{-1}(\xi_t)$. Now, there are two issues: (a) $\gamma_t$ moves in $\mathbb H:=\{z\in\mathbb C:\Re(z)>0\}$, but I need a cruve in $[0,1)^2$. How do I fix that? (b) Everything can be easily obtained numerically, but how should I numerically compute the inverse $γ_t=g_t^{-1}(\xi_t)$? | |
| Dec 8, 2022 at 10:26 | comment | added | Kostya_I | @0xbadf00d, Loewner evolution is a (rather convoluted) method to encode a planar curve (say in the upper half-plane) by a real-valued "driving function" $\zeta(t)$, $t\geq 0$. The example is for the case $\zeta(t)\equiv 0$, which encodes the vertical straight line ray. If you take $\zeta$ to be a random function, then the curve will also be random. | |
| Dec 8, 2022 at 9:32 | comment | added | 0xbadf00d | I already looked at the wikipedia page and I thinnk it is rather confusing. I don't even understand the example. "Loewner's equation" is a differential equation relating functions $f,g,\zeta$. Why is $\zeta$ suddenly a Brownian motion? Where does stochastic come into the game at all? | |
| Dec 8, 2022 at 7:47 | comment | added | Christophe Leuridan | Look at the wikipedia page added in the references. | |
| Dec 8, 2022 at 7:46 | history | edited | Christophe Leuridan | CC BY-SA 4.0 |
I added a reference
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| Dec 7, 2022 at 21:52 | comment | added | 0xbadf00d | Thank you for your answer. Unfortunately, in your references there is a lot of theory involved which I'm not familiar with and I would need to dive into. Could you shortly explain how these evolutions could be used to reach my goal? Can we generate such an evolution (on $[0,1)^2$) numerically? | |
| Dec 7, 2022 at 18:06 | history | answered | Christophe Leuridan | CC BY-SA 4.0 |
