Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 203 / 2

By the Phase classification of the SLE trace, chordal SLE is a simple curve forIf the convention restricts the notation to $\kappa >0$, the range is $0<\kappa \le 4$. The degenerate case $\kappa =0$ is a deterministic vertical slit and is simple as well.

By the Phase classification of the SLE trace, the trace has self-intersections but is not a space-filling curve forThe endpoint $\kappa =4$ remains simple, while $\kappa =8$ belongs to the space-filling regime.

The Phase classification of the SLE trace gives a space-filling curve precisely forIn particular, the critical value $\kappa =8$ is included.

Start the Bessel process at $x>0$ and let ${\tau }_0$ be its first hit of zero. Away from zero its stochastic differential equation isApply the Itô formula to $f(x)=x^{2-\delta }$, localizing first to a compact interval in $(0,\infty )$. Its drift cancels:Thus the Bessel power local martingale satisfiesFor $\delta =2$, this power is simply the constant one. For $\delta >2$, it is a continuous local martingale for all time, as zero is inaccessible.

Here is a proof of that inaccessibility and of the stronger minimum assertion. For $0<r<x<R$, stop the power at ${\tau }_r\wedge {\tau }_R$. The stopped process is bounded, so the optional stopping theorem givesExit from the bounded interval occurs almost surely. Taking $r\downarrow 0$ with $R$ fixed shows that zero cannot be hit before $R$, hence cannot be hit in finite time. Taking $R\to \infty$ instead givesIf the all-time infimum were zero, every level $r>0$ below $x$ would be hit, by continuity. Letting $r\downarrow 0$ proves the all-time minimum of a transient Bessel process assertion:The first claim in the printed question needs a qualification when $\delta <2$: it holds before the first hit of zero, or for the process stopped there, rather than for the usual reflecting continuation. For example, at $\delta =1$ the proposed power is $X$ itself, a Reflected Brownian motion; the Tanaka formula contains a nonzero local time of a semimartingale term, so it is not a local martingale after reflection. A positive starting point is also needed for the negative power when $\delta >2$; an entrance process started at zero would give an infinite value at time zero.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

The Scaling invariance of SLE says that $r^{-1}\gamma (r^{2}t)$ has the same law as $\gamma (t)$. Therefore its first time reaching height one is $r^{-2}{\sigma }_r$, andfor every $r>0$.

To identify $p$, it suffices to look at small heights. The trace must have a point of positive imaginary part: otherwise it lies entirely on the real axis and its hull in $\mathbb{H}$ is empty, contradicting $\mathrm{hcap}(A_t)=2t>0$. By continuity from $\gamma (0)=0$, any positive height attained by the trace forces it to pass through every smaller positive height.

Consequently the increasing union of events $\{{\sigma }_{1/n}<\infty \}$ has probability one. Each event has probability $p$, so continuity of probability gives $p=1$. This proves that SLE reaches every positive height:Equivalently, a positive random height supremum whose law is invariant under every dilation must be infinite almost surely. This argument uses neither Transience of chordal SLE nor any assertion about where the trace goes as time tends to infinity.