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Schramm's formula on left passage probabilities of $SLE_k$ is stated for $k \in [0,8)$ in theorem 2 here. However, after the statement he remarks that the formula simplifies to $1/2$ for $k = 8$. It seems that at the time of that paper, it was not known that the $SLE_k$ trace existed for $k = 8$ (second paragraph of third page of Schramm's article). However, it was later discovered that there is a continuous trace for $k = 8$, which arises as the scaling limit of the Peano curve of a uniform spanning tree.

Question: Is it true that the left-passage probability for the $SLE_8$ trace is $1/2$? Is there a reference that proves this?

Edit: Dmitry Krachun points out that it's not clear that there is a good meaning for left-passage probabilities, since $SLE_8$ is space filling. Schramm's definition for left-passage probabilities makes sense for any $\kappa < 8$, because for $\kappa < 8$ and any fixed $z_0$, $\mathbb{P}( z_0 \in \gamma( [0, \infty)) | \gamma \in SLE_{\kappa} ) = 0$. Nonetheless it seems possible that there is some interpretation of the left-passage probability being $1/2$, perhaps arising from the Peano curve approximation.

In the wikipedia article for the left-passage probabilities, no restrictions on $\kappa$ are given. Is this correct? No reference is provided.

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  • $\begingroup$ Note that $SLE_8$ is space-filling, so I'm not sure what is meant by the left passage probability. $\endgroup$ Jul 9, 2019 at 18:25
  • $\begingroup$ @DmitryKrachun That's a good point. Perhaps one interpretation could be via the Peano curves converging to the trace - maybe as one goes out in the sequence of curves, a fixed point is on the left half of the time? Or if one takes a UST from a sufficiently fine lattice, the probability that a given point is on the left of the Peano curve occurs roughly half of the time? $\endgroup$
    – Elle Najt
    Jul 9, 2019 at 18:42
  • $\begingroup$ well, if you look into how the UST Peano curve is drawn, you can see that actually each point in the plane is either deterministically on the right, or deterministically on the left (and there are half of points of each type), which perhaps is what explains this 1/2 answer. $\endgroup$
    – Kostya_I
    Jul 20, 2019 at 21:27
  • $\begingroup$ I suppose what you can try to do for the actual SLE${}_8$ curve is this: look at the function in the Wikipeda article, drop the 1/2 term and the gamma prefactors (that are equal to 0 for $\kappa=8$) and plug $x_t$ and $y_t$ instead of $x_0$ and $y_0$ (where $x_t+iy_t=g_t(z)$). What you get is a local martingale. It is unbounded, so, it should either have a limit at the swallowing time, or oscillate infinitely many times between arbitrary large positive and negative values. I have no idea which is the case. $\endgroup$
    – Kostya_I
    Jul 20, 2019 at 21:40

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