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Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

Write $D=\mathbb{H}\setminus K$. By definition of a compact H-hull, $D$ is a proper simply connected domain, and it agrees with the complex upper half-plane outside a sufficiently large disc. The Riemann mapping theorem supplies a conformal isomorphism $f:D\to \mathbb{H}$. Choose its boundary normalization so that the prime end at infinity maps to infinity. This prime end is unambiguous because the domain has an ordinary straight real boundary near infinity.

The Schwarz reflection principle, applied there in the coordinate $1/z$, extends $f$ across the real boundary near infinity. It has a simple pole there, with Laurent seriesThe pole is simple because the reflected map is locally conformal at infinity; its leading coefficient is positive because it maps the upper side to the upper side. Postcomposing with the half-plane automorphism $w\mapsto (w-b)/a$ givesThis is hydrodynamic normalization at infinity.

If $g_1,g_2$ both satisfy it, $g_2\circ g_1^{-1}$ is a conformal automorphism of the upper half-plane. The asymptotics force it to fix infinity, so it has the form $w\mapsto \alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. The same asymptotics force $\alpha =1$, $\beta =0$. Hence the normalized mapping-out function exists and is unique. The argument requires reflection only outside a large disc, not regularity of the hull's entire boundary.

The two curves have different destinations, so the equality of their initial segments is not an equality of their complete laws. At the first contact with $[1,\infty )$, the initial hull separates the target $1$ from infinity. Except for the null event of hitting the fixed point $1$ itself, this happens by a contact at a real point strictly larger than $1$. The fact that a fixed nonzero real point is almost surely not visited is the SLE does not hit a fixed nonzero boundary point property; it is consistent with the strictly positive probability of swallowing an entire interval at once in Question 3.

The original curve, whose target is infinity, continues in the closure of the unbounded surviving component. The curve targeted at $1$ continues in the surviving component whose boundary contains its target, which is now separated from infinity. Its domain normalization must follow that component. Ultimately one curve tends to infinity and the other to $1$. This is target-dependent continuation of SLE6.

The precise failure of the argument is the loss of the conformal conjugacy between the same surviving components. In the transformed construction, at time $S$ the pole $-1$ is swallowed; its image is infinity. The composition $g_t^{\ast }\circ \Phi \circ g_t^{-1}$ was a conformal automorphism of the upper half-plane only while both maps described corresponding surviving domains. After the boundary contact, the image of the original target-containing domain is the component containing $1$, whereas $g_t^{\ast }$ in the ordinary half-plane-capacity convention maps out the hull relative to the component containing infinity. These are no longer the same component. Equivalently, the swallowed pole collides with the growing hull, and the analytic-neighbourhood assumption at the Loewner driving function cannot be continued by the preceding argument.

Hence neither the given analytic identities nor the Brownian motion description obtained from their capacity clock licenses a continuation of the comparison beyond the stopping time. The zero drift before contact does not remove this geometric obstruction. The restriction to pre-contact segments is essential, even though the curves can continue in their respective target components.

The conformal automorphism of the upper half-plane with the indicated three boundary values isIts Möbius transformation matrix has determinant one, so it preserves $\mathbb{H}$. The inverse is ${\Phi }^{-1}(w)=w/(1-w)$. In particularon the extended real boundary. Therefore the image trace's first contact with the latter ray corresponds to the original time $S$. The map sends the original target at infinity to the new target $1$; it does not preserve the target of the chordal curve.

We use the image half-plane capacity as its time parameter:Since ${\Phi }_t$ is a conformal automorphism of the upper half-plane analytic at ${\xi }_t$, its real derivative there is strictly positive. Hence $u$ is continuous and strictly increasing before $S$. The original time $t$ and image time $u$ must be kept distinct throughout the comparison.

Write $D=\mathbb{H}\setminus K$. The mapping-out function is a conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinityEquivalently, $g_K(z)-z\to 0$. The Riemann mapping theorem gives a map onto the half-plane. Since $K$ is bounded, infinity has an analytic real-boundary neighborhood. After sending its image to infinity, the Schwarz reflection principle gives a Laurent series $az+b+O(z^{-1})$ there, with $a>0$ and $b$ real. A real affine automorphism removes $a$ and $b$, giving the stated normalization. Reflection also makes $a_K$ real.

If $g_1,g_2$ both have this normalization, $g_2{g}_1^{-1}$ is a conformal automorphism of the upper half-plane. Such maps are real Möbius transformations. The expansion forces it to fix infinity, have leading coefficient $1$, and have zero translation. It is therefore the identity. Thus the normalized mapping-out function exists and is unique.

A conformal automorphism of the upper half-plane is a real Möbius transformationwith coefficients determined up to a common nonzero factor. The boundary interval $(\infty ,1)$ is the oriented real boundary arc from infinity to $1$, namely the finite real interval $(-\infty ,1)$. The map must take the left endpoint $-1$ to infinity and the right endpoint infinity to $1$, while fixing zero. These three boundary values determine the Möbius transformation uniquely:Its determinant is positive, $\mathrm{Im}\varphi (z)=\mathrm{Im}z/|1+z{|}^2>0$, and on $(-1,\infty )$ its derivative is $(1+x{)}^{-2}>0$. Its endpoint limits are $-\infty$ at $-1+$ and $1$ at $+\infty$, so it has exactly the required boundary extension.

Put ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$ and $p_t={\varphi }_t^{\prime }({\xi }_t)$. The given conformal change of half-plane capacity makes the image mapping-out equationIndeed its capacity speed is $2p_t^2$, and the stated Loewner local growth property identifies the indicated single driver. For $t<T$, differentiate the conjugacy identityat a fixed original point. The Chordal Loewner equation then gives the conformal conjugacy derivative for the chordal Loewner equationThe change of variable $z=g_t(w)$ covers the whole upper half-plane. The conformal automorphism of the upper half-plane ${\varphi }_t$ is a real Möbius transformation. Its pole is the image of the still-unswallowed boundary point $-1$, so it is holomorphic in a neighbourhood of ${\xi }_t$ before $T$.

To evaluate the apparent singularity, write $h=z-{\xi }_t$, $p={\varphi }_t^{\prime }({\xi }_t)>0$ and $q={\varphi }_t^{\prime \prime }({\xi }_t)$. The Taylor series isThus the first term in the derivative is $2p/h-q+O(h)$, while the second is $2p/h+2q+O(h)$. Their poles cancel and their constant terms differ by $-3q$. Therefore

Changing the target boundary point by a conformal automorphism of the upper half-plane preserves the local chordal SLE law at parameter six, up to a capacity time change and until the relevant boundary point is swallowed. The transformed SLE driving function has zero drift at six; the Dambis-Dubins-Schwarz theorem makes it Brownian in image capacity time.

At parameter six, changing a boundary target preserves the initial unparameterized SLE law until the growing hull separates the two possible targets. After separation, each curve continues in the complementary component whose boundary contains its own target. A curve targeted at infinity follows the unbounded component, whereas a finite separated target lies on a different component. The target-change locality of SLE6 proof no longer applies: its conformal automorphism of the upper half-plane conjugates corresponding surviving domains only before this separation. A zero drift before contact does not give equality of complete curve laws afterwards.