Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 27 / 3

The complex form of the Chordal Loewner equation, driven by a continuous real Loewner driving function, isFor $z\in \mathbb{C}\setminus \{{\xi }_0\}$ this is an ordinary differential equation up to its maximal lifetime, when the solution reaches the driving singularity. The coefficients respect complex conjugation, so the lower-half-plane flow is the conjugate of the upper-half-plane flow. On surviving real points it is the real boundary flow. The hydrodynamic normalization at infinity and factor two correspond to half-plane-capacity parameterization $\mathrm{hcap}(K_t)=2t$.

Assume $\kappa >0$ and $x\ne 0$, since the initial driving point does not admit the displayed ordinary boundary flow. The sign of the driver in this part is negative, soAfter division by $\sqrt{\kappa }$, the Boundary-point Bessel flow for SLE isFor $x>0$ this is a Bessel process of dimension $\delta =1+2a$. For $x<0$, $-X_t$ obeys the same equation driven by $-W_t$ until hitting zero. It suffices to treat a positive initial value.

The infinitesimal generator is $\mathcal{L}f=\frac{1}{2}{f}^{\prime \prime }+(a/u)f^{\prime }$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <x<b$, optional stopping theorem gives the boundary hitting probability from a diffusion scale functionWhen $a<1/2$, put $\beta =1-2a>0$. The inner boundary is reached in finite time: the nonnegative functionvanishes at $0,b$ and satisfies $\mathcal{L}v=-1$. Applying the Itô formula before exiting $(\epsilon ,b)$ gives $\mathbb{E}({\tau }_{\epsilon }\wedge {\tau }_b)\le v(x)$. As $\epsilon \downarrow 0$, these times increase to a finite limiting exit time almost surely. Thus the scale limit is an actual hitting event, not just asymptotic approach to zero, andLet $b\uparrow \infty$ to obtain ${\mathbb{P}}_x({\tau }_0<\infty )=1$.

If $a>1/2$, then $s(\epsilon )\to -\infty$, and the same formula makes the probability of hitting zero before any fixed $b$ equal to zero. A finite-time hit would occur before reaching some integer upper level, because the stopped path is continuous and bounded on a finite interval. Taking the countable union over those levels proves that no hit occurs. At $a=1/2$, $s(u)=\log u$ gives the same conclusion. ThereforeThe source's $x=0$ case must be excluded: zero is then already the initial value, and the displayed singular flow is undefined.

The Phase classification of the SLE trace has its simple rangewith $\kappa =0$ the deterministic vertical slit. For positive $\kappa$, the dividing parameter is precisely $\delta =1+4/\kappa =2$ in the Boundary-point Bessel flow for SLE.

Here is the reason this diffusion threshold controls simplicity. For $\kappa \le 4$, no nonzero real boundary point is swallowed. It suffices to check rational boundary points: a first meeting with either nonzero real half-axis would close a boundary crosscut and swallow a nonempty real interval, including a rational point. Thus the Loewner trace stays in the complex upper half-plane apart from its starting point. By the domain Markov property of a chordal Loewner chain, after any fixed rational time $s$ the future mapped and centred Loewner trace has the same boundary-avoidance property.

Suppose two Loewner trace times $r<r^{\prime }$ had the same image. Positive half-plane capacity growth rules out constancy on a nonempty time interval, so continuity supplies a rational $s\in (r,r^{\prime })$ with ${\gamma }_s\ne {\gamma }_r$. In the mapped future, the point at $r^{\prime }$ is either in the open upper half-plane or at the starting boundary point $0$; it cannot lie at another real point. The first case puts ${\gamma }_{r^{\prime }}$ inside the surviving domain at $s$, whereas ${\gamma }_r$ lies on the past Loewner trace. The second gives ${\gamma }_{r^{\prime }}={\gamma }_s$, also contradicting the choice of $s$. Boundary continuity of the inverse mapping-out function makes these identifications valid. This proves that the Loewner trace has no repeated points.

For $\kappa >4$, part (ii) makes a fixed positive real point have finite swallowing time, so the SLE boundary swallowing criterion ensures that the Loewner trace hits the positive real axis. In fact positive boundary-interval hitting probability for SLE above parameter four holds: any interval $J\subset (0,\infty )$ has positive hitting probability. To prove this, cover the positive axis by countably many dilates of $J$. If $J$ had zero hitting probability, Scaling invariance of SLE would give zero probability for every dilate, contradicting the almost sure hit of the positive axis. Reflection gives the same conclusion for negative intervals.

At a fixed positive time, if the past Loewner trace already repeats a point there is nothing to prove. Otherwise, boundary continuity of the inverse mapping-out function supplies a nonempty real interval, away from the current driving point, mapped back into the earlier Loewner trace in the open upper half-plane. Such an interval exists because positive capacity growth creates a genuine Loewner trace boundary in the interior; choose an accessible boundary point away from the tip and then a small interval around its preimage. Conditional on the past, the domain Markov property of a chordal Loewner chain gives a future centred SLE. With positive conditional probability its Loewner trace hits this interval, by the preceding boundary-interval argument. Mapping back then gives a visit to the earlier Loewner trace. Thus a repeated point occurs by some finite time with positive probability.

Finally let $E_t$ be the event of a repeated point by time $t$. Scaling invariance of SLE makes $\mathbb{P}(E_t)$ the same for every $t>0$. The positive finite-time probability just proved makes this common value positive. Hence ${\bigcap }_n{E}_{1/n}$ also has positive probability. This event belongs to the Brownian germ sigma-field; the Blumenthal zero-one law forces its probability to be one. Consequently the Loewner trace is not simple almost surely for every $\kappa >4$. At $\kappa =4$, the logarithmic scale function gives non-hitting of zero, so equality belongs to the simple range.