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The Gaussian free field is a random distribution, so its purported level curve cannot be defined by evaluating its height at every point. A coupling must instead be stated using conditional means and covariances. The SLE4 coupling with a Gaussian free field does exactly this.

Choose the Green-function normalizationA Zero-boundary Gaussian free field $H_0$ is characterized on real compactly supported smooth test functions byEquivalently its energy normalization is the Dirichlet inner product $(2\pi {)}^{-1}\int \nabla f\cdot \nabla gdA$. This fixes the otherwise convention-dependent height constant.

Set $\lambda =\pi /2$ andThe field $H$ has prescribed Dirichlet values $-\lambda$ on the negative half-line and $+\lambda$ on the positive half-line. Subtracting $m_0$ recovers the zero-Dirichlet field $H_0$. It is the shifted field $H$, rather than an unshifted zero-boundary field, whose zero-height interface has ordinary chordal ${\mathrm{SLE}}_4$ law.

Let $\eta$ be chordal ${\mathrm{SLE}}_4$ from $0$ to $\infty$, with Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-W_t)$ and $W_t=2{\beta }_t$. Write $D_t=\mathbb{H}\setminus \eta [0,t]$ and $f_t=g_t-W_t$. The coupling statement is:where, conditionally on the curve, $H_0^{D_t}$ is an independent zero-boundary field in the slit domain. The equality is for restrictions to test functions in that domain, and extends to suitable stopping times. Its covariance is $G_{D_t}(z,w)=G_{\mathbb{H}}(f_t(z),f_t(w))$ by conformal invariance. The two sides of the revealed slit have heights $-\lambda$ and $+\lambda$, in the order specified by their real images under $f_t$. This is the continuum meaning of a Gaussian free field level line.

The fundamental calculations explain why the parameter is four. For $Z_t=f_t(z)$, the Itô formula givesAt $\kappa =4$ the drift vanishes. ThusEach harmonic mean is a bounded martingale before approaching its point. Differentiating the explicit conformally transformed Green function gives the Loewner variation of the Dirichlet Green functionThereforeIn words, the variance learned from the evolving mean exactly equals the covariance lost when the slit is removed. For a Green kernel $cG$ the same calculation gives $\lambda =(\pi /2)\sqrt{c}$; using a different Green normalization changes the height constant, not the SLE parameter.

To construct the coupling, first sample the SLE curve. In the two components to its left and right, sample independent zero-boundary fields and add the corresponding constant heights. Extend these as distributions to obtain the candidate full field. The following martingale identity proves its marginal law rather than merely matching its first two moments.

For a real test function $\phi$, put $M_t=(m_t,\phi )$ andLocalization permits integration of the pointwise identities. They yield $d⟨M{⟩}_t=-d{V}_t$. Henceis a complex local martingale: the $-\frac{1}{2}d{V}_t$ drift cancels the $-\frac{1}{2}d⟨M{⟩}_t$ Itô correction. Since $V_t\ge 0$, $|{\mathcal{Z}}_t|\le 1$, making it a true martingale. At the complete-curve limit, $m_t$ becomes the constant height in each component and the remaining Green kernel becomes that of those components. Thus ${\mathcal{Z}}_{\infty }$ is the conditional characteristic function of the sampled candidate field. Taking expectations givesApplying this to every linear combination of test functions proves the entire Gaussian law, not only its covariance. The conditional version $\mathbb{E}({\mathcal{Z}}_{\infty }\mid {\mathcal{F}}_t)={\mathcal{Z}}_t$ proves the displayed conditional-field statement. Exhaustion by compact test supports justifies passage across the slit and the limiting distributional extensions.

This revealed curve is a local set of a Gaussian free field: conditionally on it, the remaining field is a zero-boundary field plus a specified harmonic function. The spatial Markov property is thereby preserved at random domains, a property not available for arbitrary field-dependent sets. One can strengthen the coupling to a curve measurable from the field. The proof explores compatible interfaces in small subdomains and uses the conditional boundary heights and monotonicity to show that two such interfaces for the same field cannot separate; a countable exhaustion gives uniqueness. This supplies a rigorous replacement for the informal phrase “draw the zero contour.” It does not assert that the distribution has pointwise values.

The coupling is useful in both directions: SLE martingales give exact conditional Gaussian free field data, while the Gaussian Markov structure explains the SLE domain Markov property and its distinguished parameter. With the above normalization, the height jump is $2\lambda =\pi$, and the corresponding interface is chordal SLE4.

Fix $u\in \mathbb{R}$ and let $s\to t$. For any $\delta >0$, use $|e^{ia}-e^{ib}|\le \min (2,|a-b|)$ to obtainThe probability tends to zero by stochastic continuity. Taking the limit superior and then letting $\delta \downarrow 0$ provesThis proves continuity of Lévy characteristic functions from the elementary estimate, without requiring moments or replacing convergence in probability by an unjustified almost sure limit. At $t=0$, time approaches from the right. If stochastic continuity is formulated only at zero, stationary increments give the same argument at every $t$: the absolute value of $X_s-X_t$ has the law of $|X_{|s-t|}|$. For $u=0$ the characteristic function is identically one.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

For the construction above, independence of the normal random variables makes $S_N(h)$ normal with mean zero and variance $v_N={\sum }_{j\le N}⟨h,e_j{⟩}^2$. Its characteristic function is $\exp (-{\theta }^2{v}_N/2)$. The $L^2$ convergence gives $L^1$ convergence, andSince $v_N\to \Vert h{\Vert }^2$ by the Parseval identity for a Hilbertian basis, the limiting characteristic function identifiesThis includes $h=0$, where the normal distribution is degenerate at zero.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

At the terminal time, $M_T=e^{i\theta X_T}$, so the martingale from part (c) has terminal value $e^{-{\theta }^2⟨X{⟩}_T/2}$. Its initial value is $M_0$, since $X_0=⟨X{⟩}_0=0$. Taking expectations givesHere $M_0$ may be a nonconstant ${\mathcal{F}}_0$-measurable variable; the tower property of conditional expectation still gives $\mathbb{E}{M}_0=\mathbb{E}{e}^{i\theta X_T}$. This characteristic function under conditionally symmetric martingale increments identity relates the characteristic function of the terminal martingale to the Laplace transform of a nonnegative random variable given by its quadratic variation.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,

For a finite list of real coefficients $(a_j{)}_{j\le m}$,A deterministic Itô integral is centered Gaussian: first verify it for step functions as a linear combination of independent Gaussian Brownian increments, then pass to an L2 approximation using the Itô isometry and characteristic functions. Its variance is the squared L2 norm of the integrand, which is ${\sum }_j{a}_j^2$ by the supplied orthonormality. ThereforeThis factors as the joint characteristic function of independent standard normals. Since every finite subfamily has this law, the entire sequence is independent and each ${\xi }_n$ has law N(0,1). This is the Gaussian coordinates of deterministic orthonormal Wiener integrands principle.

Use the intrinsic definition of a Lévy process: $X_0=0$ almost surely, its increments over disjoint time intervals are independent, their probability distributions depend only on interval length, and $X$ has stochastic continuity. In symbols,with times restricted to the half-line. Such a stochastic process has a càdlàg modification, and is usually represented by that version. Requiring càdlàg paths in the definition is a common equivalent convention at the level of modifications; it is important to distinguish this from a claim about the paths of an arbitrary supplied version.

For the characteristic function, writeHere $\psi (0)=0$ and $\psi$ is the positive-time-sign characteristic exponent of a Lévy process. The independent increments and stationary increments give ${\varphi }_{s+t}(\theta )={\varphi }_s(\theta ){\varphi }_t(\theta )$; stochastic continuity gives continuity in time and ${\varphi }_0=1$. This continuous multiplicative semigroup has the stated exponential form. Its exponent has the Lévy–Khintchine formulawhere $b\in \mathbb{R}$, ${\sigma }^2\ge 0$, and the Lévy measure $\nu$ satisfies $\int (1\wedge y^2)\nu (dy)<\infty$. The truncation convention fixes the drift coefficient $b$; the displayed sign convention agrees with $\Psi =-\psi$.

Fix $s<t$ and an event $A\in {\mathcal{F}}_s$. Multiplying the conditional characteristic function identity by $1_A$ and taking expectation givesThe left side is the Fourier transform of the finite measureBy the uniqueness theorem for characteristic functions, this measure equals $\mathbb{P}(A)$ times the $N(0,t-s)$ distribution. Thus, for every Borel set $D$,Taking $A$ to be the whole sample space identifies the increment's law, and the full identity proves independence from ${\mathcal{F}}_s$. Together with the assumed adaptation, continuity, and initial value, these are exactly the defining Brownian properties. This proves the conditional characteristic-function criterion for Brownian increments.

Choose independent random variables ${\xi }_i\sim N(0,1)$ on a countable product probability space. For each fixed $t$, the diagonal assumption givesThe partial sums ${\sum }_{i\le n}{k}_i(t){\xi }_i$ are therefore Cauchy in $L^2$. DefineChoose a representative of this limit for each $t$. No path continuity or simultaneous series convergence over all uncountably many times is being asserted.

For any finite list $t_1,\dots ,t_m$ and real coefficients $a_j$, the linear combination ${\sum }_j{a}_j{X}_{t_j}$ is the $L^2$ limit of centered Gaussian variablesTheir variances converge, so their characteristic functions converge to that of a centered normal distribution. This proves that every finite-dimensional vector is Gaussian and hence that $X$ is a Gaussian process. Taking $L^2$ limits also givesThis is the Gaussian process construction from square-summable features.