Constructing the 2d Liouville Model

Constructing the 2d Liouville Model Antti Kupiainen joint work with F. David, R. Rhodes, V. Vargas Porquerolles September 22 2015

γ = 2, (c = 2) Quantum Sphere

γ = 2, (c = 2) Quantum Sphere

Planar maps Let T N = triangulations of S 2 with N faces with 3 marked ones T T N is a graph with topology of S 2 and all faces triangles For µ 0 > 0 consider the probability P µ0 on T = N T N : It is known (Tutte): E µ0 F := 1 e µ 0N Z µ0 N T T N F(T ) 1 = N 1 2 e µn (1 + o(1)), T T N = P µ0 is defined for µ 0 > µ and lim µ 0 µ E µ 0 T = N N 1 2 =.

Planar maps + matter For each γ [ 2, 2] and each T T one may define a critical lattice model on the graph T. γ = 2 spanning trees, γ = 8/3 percolation, γ = 3, Ising model, γ = 2 GFF Let Z γ (T ) be the partition function of the model on graph T : e.g. Z 2 (T ) = det T, Z 8/3 (T ) = 1, Z 2(T ) = det 1 2 T

Planar maps + matter Define E µ0,γf := 1 Z µ0,γ e µ 0N N T T N Z γ (T )F(T ) It is conjectured Z γ (T ) = N 1 4 γ 2 e µn (1 + o(1)). T T N so again lim µ 0 µ E µ 0,γ T N N 2 4 γ 2 =

Random measure on S 2 Conformal structure on T : triangles equilateral area 1. Map T conformally to S 2 s.t. marked faces map to z 1, z 2, z 3. Image of volume on T is a measure ν T (dz) on S 2. Under P µ0,γ, ν T becomes a random measure ν µ0,γ on S 2.

Scaling limit Recall: as µ 0 µ typical size of triangulation diverges. Let µ > 0 and define ρ (ɛ) µ,γ := ɛν µ+ɛµ,γ Conjecture. ρ (ɛ) µ,γ converges in law as ɛ 0 to a random multifractal measure ρ µ,γ on S 2. Since ɛν T (S 2 ) = ɛn the law of total volume ρ (ɛ) µ,γ(s 2 ) is: E[F(ρ (ɛ) µ,γ(s 2 ))] = 1 e µɛn N 1 4 γ 2 F (ɛn). Z ɛ It converges to Γ(2 4 γ 2, µ) as ɛ 0. We will construct a measure Liouville quantum gravity on S 2 with this law for its total mass. N

History of LQG Polyakov 1981: Liouville theory as model for 2d gravity (random geometry) Kniznik, Polyakov, Zamolochikov 1988: KPZ relations for critical exponents David, Distler, Kawai, 1988, lots of physics work thereafter Duplantier, Sheffield 2008: Reinterpretation in terms of Gaussian multiplicative chaos

2d Gravity a la Polyakov Polyakov, KPZ, David: our random measure is given by ρ µ,γ (dz) = e γx(z) dz where the law of X is formally given by E γ,µ f (X) = Z 1 f (X) e S(X,g) DX (1) where S is the Liouville action functional: ( S(X, g) := g X 2 + QR g X + µe γx ) dλ g S 2 g = g(z)dz 2 is any smooth conformal metric on S 2 R g = log g scalar curvature, dλ g (z) = g(z)dz volume form. Q = 2/γ + γ/2 Aim: give precise meaning to the law (1) and study its properties: conformal invariance, KPZ scaling,...

Classical Liouville theory If we take Q = 2/γ extrema of S(X, g) are solutions of Liouville equation R e γx g = 1 2 µγ2. Solutions define metrics e γx g with constant negative curvature and lead to the uniformisation theorem of Riemann surfaces. For the quantum (random) case we need to take Q = 2/γ + γ/2. The resulting quantum geometry has interesting parallels with the classical one.

Quantum Liouville theory We want to give meaning to the integral F(X)e S 2 ( g X 2 +QRX+µe γx )dλ g DX by viewing it as a perturbation of which looks like a Gaussian measure. e 1 4π S 2 ( g X 2 dλ g DX (2) Indeed, one could interpret (2) as the Gaussian Free Field on S 2 i.e. the Gaussian measure with covariance ( g ) 1.

Constant mode However, GFF is defined only up to an additive constant since g annihilates constants. We want to include constant fields to DX: the "gaussian" measure will not be a probability measure Quadratic part of action is independent on metric: g X 2 dλ g = X 2 dλ. S 2 S 2 so look for a measure having this property.

Gaussian Free Field (GFF) Let X g be Gaussian Free Field with zero λ g -average: m g (X g ) := 1 λ g (S 2 X g dλ g = 0 ) S 2 This means: covariance of X g is Green function of the problem g u = 2πf, udλ g = 0 As g varies the fields differ by random additive constants: Our GFF is then defined as where c is uniform on R. "Law" of X is independent of g X g m g (X g ) law = X g. X = X g + c

Liouville as perturbation of GFF Thus we should view the integral F(X)e S 2 ( g X 2 +QR gx+µe γx )dλ g DX as E(F(X)e S 2 (QR g(x)+µe γx )dλ g )dc where X = X g + c. However, EX g (z)x g (w) = log z w + smooth i.e. X g (z) is a.s. not defined pointwise. We need to give a meaning to e γxg.

Multiplicative Chaos Introduce small scale cutoff ɛ, e.g. X g,ɛ (x) = 1 2π X g (x + ɛe iθ ) dθ 2π 0 and define random measure on S 2 Then in probability as ɛ 0. γ2 γxg,ɛ M g,ɛ := e 2 E[X g,ɛ 2 ] λ g M g,ɛ M g M g is an example of Gaussian multiplicative chaos, a random multifractal measure on S 2. We have EM g = λ g and hence M g (S 2 ) < a.s.

Liouville field We define the formal integral F(X)e S 2 ( g X 2 +QRX+µe γx )dλ g DX as R E g (F (X)e Q S 2 R gx dλ g µe γc M g(s 2 ) ) dc where X = c + X g. In the round metric g = ĝ = 1 (1 + zz) 2 π Rĝ = 2, vol(s 2 ) = 1 and the curvature term becomes S 2 QRĝ(Xĝ + c)dλĝ = 2Qc so that the integration measure is dµ L (X) := e 2Qc e µeγc Mĝ(S 2) dc dp(xĝ)

Zero mode Problem: µ L is not a probability law. The normalization ( ) Eĝ e 2Qc e µeγc Mĝ(S 2 ) dc = R since Mĝ(S 2 ) is a.s. finite and c-integral diverges at. Divergence is geometric: by Gauss-Bonnet 2Q = Q R g λ g = 2(1 genus Σ )Q (Σ = S 2 ) Σ Extrema of S L are negative curvature metrics: none exist on S 2 and Liouville action S L is not bounded from below.

Möbius invariance Another way to see this: classically the Möbius group SL(2, C) acts on the set of conformal metrics e γx(z) g(z) dz 2 e γx(ψ(z)) g(ψ(z)) ψ (z) 2 dz 2 i.e. the field φ = X + 1 γ log g transforms as The quantum version of this is φ φ ψ + 2 γ log ψ. Proposition. The Liouville field φ = X + Q 2 log g is Möbius invariant under µ L : F(φ)dµ L = F(φ ψ + Q log ψ )dµ L

Punctures We are interested in S 2 with 3 marked points z 1, z 2, z 3. To have a constant negative curvature metric on S 2 one needs to include conical singularities. Let us do this at the points z 1, z 2, z 3. Such metrics are extrema of the functional S(X, g, α i, z i ) = S(X, g) 3 α i X(z i ) i=1 suitably renormalized. The angle of the cone at z i is Ω i = 2π(1 α i /Q class ), Q class = γ/2 and one needs (2π Ω i ) > 4π i.e. i αi > 2Q class. hence since Ω i > 0 need at least three conical singularities.

Vertex operators Do the same in the random case. The density e S(X,g,α i,z i ) = 3 e α i X(z i ) e S(X,g) i=1 is defined in terms of the vertex operators V α (z) := lim e γxg,ɛ γ 2 2 E[X g,ɛ] 2 ɛ 0 Define the probability for punctured sphere with normalization Z αi,z i = dp αi,z i (X) := 1 Z αi,z i R ( Eĝ 3 V αi (z i )dµ L (X) i=1 n ) V αi (z i )e 2Qc e µeγc Mĝ(S 2 ) dc i=1

Seiberg bounds Now the c-integral becomes e ( α i 2Q)c µe γc Mĝ(S 2) dc This is finite if and only if α i > 2Q. R Theorem Let α i > 2Q. Then Z < and Z > 0 if and only if α i < Q for all i. Corollary. Indeed, need at least three vertex operators as in the classical case.

Seiberg bounds Proof. Recall V αi Z αi,z i = R = const. e α i X(z i ) and ( Eĝ n ) V αi (z i )e 2Qc e µeγc Mĝ(S 2 ) dc i=1 Girsanov: get rid of V αi by Xĝ Xĝ + H with H(x) = i α i Gĝ(z i, x) = i α i log z i x 1 + regular. Hence Z αi,z i = e 1 2 i j α i α j Gĝ(zi,z j ) R dc e ( α i 2Q)c Eĝ e µeγc S 2 e γh dmĝ

Seiberg bounds Modulus of continuity of the Chaos measure: for any δ > 0 Mĝ(B r ) C(ω)r γq δ Singularity of the integrand at z i : e γhĝ(z) e γα i log z i x 1 = z z i γα i Integrable if γα i < γq i.e. α i < Q Also z z i γα i dmĝ = B r (z i ) a.s. if α i Q, Remark. Classically z z i γα i integrable iff α i < 2/γ.

Liouville measure Consider the law of the chaos measure Mĝ under P. Let A = Mĝ(S 2 ) be the "volume of the universe". Let s = ( i α i 2Q)/γ Theorem Under P αi,z i the law of A is Γ(s, µ): EF(A) = µs Γ(s) 0 F(y)y s e µy dy. Remark. Taking n = 3, α i = γ this agrees with Planar Maps.

CFT The Liouville measure µ L gives rise to a Conformal Field Theory: The correlation functions Z ({z i }, {α i }, ĝ, µ) := k V αi (z i )dµ L (X) transform covariantly under conformal maps and satisfy scaling relations Using Reflection Positivity one can construct a Hilbert space carrying a representation of the Virasoro Algebra with central charge c = 1 + 6Q 2. There is a beautiful axiomatic-algebraic approach to the CFT and its correlation functions. Our aim is to derive this from the probabilistic setup. i=1

KPZ Theorem (a) Conformal covariance. Let ψ be Möbius. Then Z ({ψ(z i )}, {α i }, ĝ, µ) = i ψ (z i ) 2 α i Z ({z i }, {α i }, ĝ, µ) where α = α 2 (Q α 2 ). (b) Weyl invariance. Let g = e φ ĝ. Then Z ({z i }, {α i }, g, µ) = e c L 1 96π ( φ 2 dλ+ 2Rĝφ dλĝ) Z ({z i }, {α i }, ĝ, µ) where c L = 1 + 6Q 2 is the central charge of the Liouville theory.

Liouville field Given α i, z i we can define the probability measure dp αi,z i := Z ({z i }, g, γ) 1 n Then the Liouville field satisfies i=1 V αi (z i )e 1 QRgX 4π e µeγc M g(s 2) dcdp g φ := X + Q 2 log g φ Pαi,z i law = (φ ψ + Q log ψ ) Pαi,ψ(z) i

3-point function 3-point function believed to determine whole Liouville theory. Z ({z i }, g, µ) = z 1 z 2 2 12 z 2 z 3 2 23 z 1 z 3 2 13 C γ (α 1, α 2, α 3 ) where 12 = α3 α1 α2 etc. and Z = C γ (α 1, α 2, α 3 ) = µ s Γ(s) E Z s z α 1γ z 1 α 2γĝ(z) γ 4 3 i=1 α i Mĝ(dz). For s = k N, E Z k can be computed explicitly in terms of Selberg-Dotsenko-Fateev integrals. Formal "analytical continuation" in k yields explicit expression for C γ (α 1, α 2, α 3 ) the DOZZ Conjecture. Prove it!

γ = 2 barrier Multiplicative chaos has a freezing transition at γ = 2 (Remi s talk). For γ = 2 need extra renormalization log ɛ 1 : e γx : γ > 2 the Liouville theory freezes to the γ = 2. Is there a duality γ 4/γ in the Liouville model?

Limit law for metric Metric spaces: Consider Riemannian metric ɛ 2α (ψ 1 T ) 2 dz 2 (lengths of the edges of the image triangles are ɛ α ). Let d T,ɛ be the corresponding distance function on S 2 and d (ɛ) µ,γ the corresponding random metric on S 2 Question: Find a α s.t. the metric space (S 2, d (ɛ) µ,γ) converges in law as ɛ 0 towards a metric space (S 2, d µ,γ ). For γ = 8/3 one expects α = 1/4 and the limit related to the Brownian map. Is there a direct construction of the limiting metric in LQG?