Liouville Quantum Gravity on the Riemann sphere

Transcription

1 Liouville Quantum Gravity on the Riemann sphere Rémi Rhodes University Paris-Est Marne La Vallée Joint work with F.David, A.Kupiainen, V.Vargas

2 A.M. Polyakov: "Quantum geometry of bosonic strings", 1981 Liouville QFT Random Planar Maps Question: What is the geometry of space under the coupled effect of Quantum Gravity interacting with Conformal Field Theories (CFT)? Founding fathers: Polyakov, David, Distler, Kawai, Knizhnik, Zamolodchikov in the eighties

3 Plan of the talk Constructing the Liouville quantum field theory on the Riemann sphere Conjectures relating to random planar maps

4 Plan of the talk Constructing the Liouville quantum field theory on the Riemann sphere Conjectures relating to random planar maps

5 Uniformization and Liouville equation Problem: Given a 2d Riemann manifold (M, ĝ), find a metric g conformally equivalent to ĝ with constant Ricci scalar curvature µ Liouville equation: If X : M R solves the equation then the metric g = e X ĝ satisfies R g = µ ĝx Rĝ = µe X R g = µ Liouville action: Find such a X by minimizing the functional ( ) S L (X, ĝ) = ĝx 2 + 2RĝX + µe X dvĝ M Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

6 Liouville Quantum Field Theory Consider a 2d Riemann manifold (M, ĝ). Define a (probability) measure e β S L(X,ĝ) DX defined on the maps X : M R, where S L is the Liouville action S L (X, ĝ) = 1 ( ) ĝx 2 + 2RĝX + µe X dvĝ 4π and DX is the "uniform measure" on maps β fixed parameter and µ > 0. M X : M R, Get rid of the parameter β by making a change of variables X γx. Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

7 Liouville Quantum Field Theory Consider a 2d Riemann manifold (M, ĝ). Define a (probability) measure e β SL(X,ĝ) DX e SL(X,ĝ) DX defined on the maps X : M R, where S L is the Liouville action S L (X, ĝ) = 1 ( ) ĝx 2 + QRĝX + µe γx dvĝ 4π and DX is the "uniform measure" on maps M X : M R, and γ [0, 2] and Q = 2 γ + γ 2 and µ > 0. Get rid of the parameter β by making a change of variables X γx. Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

8 Liouville Quantum Field Theory Consider a 2d Riemann manifold (M, ĝ). Define a (probability) measure e β S L (X, ĝ)dx e SL(X,ĝ) DX defined on the maps X : M R, where S L is the Liouville action S L (X, ĝ) = 1 ( ) ĝx 2 + QRĝX + µe γx dvĝ 4π and DX is the "uniform measure" on maps M X : M R, and γ [0, 2] and Q = 2 γ + γ 2 and µ > 0. Under this (probability) measure, one studies the "physical" random metric e γx ĝ through its volume form, Brownian motion, Riemann distance,... Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

9 LQFT on the Riemann sphere Sphere = R 2 equipped with the round metric ĝ(x)dx 2 = 4 (1 + x 2 ) 2 dx 2. Define the path integral by viewing it as a perturbation of the Gaussian measure ( ) e SL(X,ĝ) DX = e 1 4π R 2 QRĝX+µe γx dvĝ e 1 4π R 2 ĝx 2 dvĝ }{{ DX } Gaussian measure Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

10 Gaussian part Focus on the "Gaussian measure" e 1 4π R 2 ĝx 2 dvĝ DX gives equal weights to constant functions not a probability measure! independent of the choice of the metric g conformally equivalent to the round metric ĝ g X 2 dv g = R 2 ĝx 2 dvĝ R 2 invariant under composition of X by Möbius transforms g (X ψ) 2 dv g = R 2 g X 2 dv g R 2 Notations: g = Laplacian, g = gradient, R g=ricci curvature, V g=volume form

11 GFF with vanishing g-mean Denote by m g (f ) the mean value of f in the metric g: 1 m g (f ) = V g (R 2 f dv g ) R 2 If we restrict the Gaussian measure e 1 4π R 2 ĝx 2 dvĝ DX to maps X : R 2 R with vanishing mean in the metric g then it corresponds to the Gaussian Free Field with vanishing g-mean X g. The GFF X g is a Gaussian random "function" (rigorously a distribution) X g : R 2 R s.t. E[X g (x)] = 0 E[X g (x)x g (y)] = G g (x, y) where G g = Green function of the Laplacian g with 0-mean g u = 2π f, m g (u) = 0.

12 Free GFF Rule for conformal change of metrics X g law = Xg m g (X g ) Integrate with respect to Lebesgue to get rid of the metric dependency X = c + X g where c is "distributed" as the Lebesgue measure. Free GFF Infinite measure on maps R 2 R, conformally invariant F(X) e 1 4π gx 2 dv g DX = E[F (c + X g )] dc The definition does not depend on g. Remark: coincides with Lebesgue measure when restricted to constant functions (called zero mode in physics) R

13 Gaussian multiplicative chaos Recall the strategy Recall that ( ) e SL(X,ĝ) DX = e 1 4π R 2 QRĝX+µe γx dvĝ X = c + Xĝ e 1 4π R 2 ĝx 2 dvĝ }{{ DX } Gaussian measure with c distributed as Lebesgue measure and Xĝ is the GFF with vanishing ĝ-mean. construct the random measure e γx ĝ(x) dvĝ Problem: the GFF Xĝ is a distribution, not a fairly defined function E[Xĝ(x)Xĝ(y)] = ln 1 + smooth function x y

14 Gaussian multiplicative chaos Construct the random measure e γx ĝ(x) dvĝ Regularization: X ɛ ĝ = X ĝ ρ ɛ where ρ ɛ = ɛ 2 ρ ( ɛ) is a mollifying sequence. Kahane 85: for γ [0, 2[, ɛ γ2 lim ɛ 0 eγx ĝ (x) 2 E[(X ɛ ĝ (x))2] Vĝ(dx) = M γ ĝ (dx) in probability. The random measure M γ ĝ has finite mass, full support, is diffuse and has carrier with Hausdorff dimension 2 γ2 2.

15 Path integral Now we are in position to define the formal integral ) F(X)e 1 4π (QRĝX+4πµe R 2 γx dvĝ e 1 4π R 2 ĝx 2 ĝ dv ĝ }{{ DX } Gaussian part where X = c + Xĝ. In the round metric, Rĝ = 2, and the curvature term becomes 1 QRĝ(c + Xĝ) dvĝ = 2Qc 4π R 2 so that the integration measure is µ L (dx) =e 2Qc µeγc M γ ĝ (R2) dc dp Xĝ Problem: µ L is not a probability measure because µ L (dx) e 2Qc µeγc dc = + R Law of X under µ L invariant under the Möbius group SL 2 (C) non compact!

16 Back to classical uniformization theory Liouville action S L (X, ĝ) = 1 ( ) ĝx 2 + 2RĝX + µe X dvĝ 4π M Saddle point If g = e ϕ ĝ then ĝϕ Rĝ = 2πµe ϕ. R g = 2πµ < 0 No such a metric on the sphere Liouville action is not bounded from below on the sphere

17 Back to classical uniformization theory Remedy: Insert strong "bumps of positive curvature" at some fixed places to compensate for the lack of positive curvature n ĝϕ Rĝ = 2πµe ϕ 2π α i δ zi i=1 z i = location of the bump α i = "strength of the bump" The metric g = e ϕ ĝ has curvature 2πµ everywhere except at the places (z i ) i where g(x) x z i α i when x z i Troyanov 91: solvable if i, α i < 2 and At least 3 insertions! Quantum analog? α i > 2 i

18 Correlation functions To get a probability measure, define "quantum insertions" Define the vertex operators α2 αx(z) V α (z) = e 2 E[X ĝ(z) 2 ]+ αq 2 ln ĝ Correlation functions Π (z i,α i ) i µ,γ (F ) = ( n F(X) i=1 ) V αi (z i ) µ L (dx) Liouville partition function with n vertex operators on the sphere Π (z i,α i ) µ,γ (1) =C ( {z i } ) [ ] e ( i α i 2Q)c E e µeγc Z (zi,α i )(R 2 ) dc R where Z (zi,α i )(dx) = e γ i α i Gĝ(x,z i ) M γ ĝ (dx)

19 Seiberg s bounds Liouville partition function with n vertex operators on the sphere Π (z i,α i ) µ,γ (1) =C ( ) [ ] (z i ) i e ( i α i 2Q)c E e µeγc Z (zi,α i )(R 2 ) dc R where Z (zi,α i )(dx) = e γ i α i Gĝ(x,z i ) M γ ĝ (dx) Theorem (DKRV 14) The Liouville partition function is well defined, i.e. 0 < Π (z i,α i ) µ,γ (1) <, iff α i > 2Q and i, α i < Q. Remark: at least 3 vertex operators needed i

20 Seiberg s bounds Liouville partition function with n vertex operators on the sphere Π (z i,α i ) µ,γ (1) =C ( ) [ ] (z i ) i e ( i α i 2Q)c E e µeγc Z (zi,α i )(R 2 ) dc R where Z (zi,α i )(dx) = e γ i α i Gĝ(x,z i ) M γ ĝ (dx) Theorem (DKRV 14) The Liouville partition function is well defined, i.e. 0 < Π (z i,α i ) µ,γ (1) <, iff α i > 2Q and i, α i < Q. Remark: at least 3 vertex operators needed i Zero mode contribution R e ( i α i 2Q)c e µeγc dc < + i α i > 2Q

21 Seiberg s bounds Liouville partition function with n vertex operators on the sphere Π (z i,α i ) µ,γ (1) =C ( ) [ ] (z i ) i e ( i α i 2Q)c E e µeγc Z (zi,α i )(R 2 ) dc R where Z (zi,α i )(dx) = e γ i α i Gĝ(x,z i ) M γ ĝ (dx) Theorem (DKRV 14) The Liouville partition function is well defined, i.e. 0 < Π (z i,α i ) µ,γ (1) <, iff α i > 2Q and i, α i < Q. Remark: at least 3 vertex operators needed i the random measure Z (zi,α i )(dx) is of finite total mass if and only if i, α i < Q.

22 KPZ Theorem (DKRV 14)) (a) KPZ scaling laws: scaling w.r.t the cosmological constant µ Π (z i,α i ) µ,γ (ĝ, 1) = µ 2Q i α i γ Π (z i,α i ) 1,γ (ĝ, 1). (b) Conformal covariance: let ψ be Möbius. Then ( ) Π (ψ(z i ),α i ) i µ,γ (ĝ, 1) = ψ (z i ) 2 α i Π (z i,α i ) i µ,γ (ĝ, 1) where the conformal weights are defined by α = α 2 (Q α 2 ). (c) Weyl invariance: Let g = e h ĝ i ln Π(z i,α i ) i µ,γ (e h ĝ, 1) = c ( ) L ĝh 2 Π (z dvĝ + 2Rĝh dvĝ i,α i ) i µ,γ (ĝ, 1) 96π R 2 R 2 where the central charge of the Liouville theory is c L = 1 + 6Q 2

23 Liouville measure It is the "volume form" Z associated to the (physical) metric e γx dvĝ, where the law of X is ruled by Π (z i,α i ) i µ,γ (ĝ, ). Explicit expression [ E γ,µ (z i,α i ) [F(Z (A))] = CE F ( ξ Z (z i,α i )(A) ) 1 ] Z (zi,α i )(R 2 ) Z (zi,α i )(R 2 i ) α i 2Q γ where C is a renormalization constant and ξ has Gamma law Γ( (reminder) i α i 2Q γ, µ) independent of Z (zi,α i )(dx) Z (zi,α i )(dx) = e γ i=1 α i Gĝ(x,z i ) M γ ĝ (dx) Conformally invariant and metric independent.

24 Plan of the talk Constructing the Liouville quantum field theory on the Riemann sphere Conjectures relating to random planar maps

25 Triangulations of the sphere Triangulations of the sphere: glue N triangles together along their edges so as to get a shape with the topology of a sphere. Combinatorics: finite number of such objects up to angle preserving homeomorphisms.

26 Discrete LQG Conformal structure of triangulations see each face of the triangulation as an equilateral triangle and glue them along their edges so as to get a conformal structure with the topology of a sphere.

27 Discrete LQG Conformal structure of triangulations I see each face of the triangulation as an equilateral triangle and glue them along their edges so as to get a conformal structure with the topology of a sphere. Conformal map I N scaling limit of random triangulations conformally embedded onto the sphere S2? I In which sense? Which notion of randomness?

28 Random Planar Maps (Pure discrete gravity) Let T 3 N be the set of N-triangulations with 3 marked faces Partition function: Z µ = N e µn Card(T 3 N ) Asymptotically (Tutte) Card(T 3 N ) N 1 2 e µ cn for µ > µ c, define a probability measure on N T 3 N E µ [F(T )] = 1 Z µ N e µn T T 3 N F(T ) Remark: If µ µ c, triangulations with large number of faces are more likely to be sampled.

29 Random Planar Maps+Matter Consider a lattice model (percolation, Ising,...) on triangulations at its critical point Let Z (T ) be the partition function of the model on the triangulation T. Conjecturally, for some γ > 0 depending on the model, Z N = T T 3 N Z (T ) N 1 4 γ 2 e µcn for µ > µ c, define a probability measure on N T 3 N E µ [F (T )] = 1 Z µ N e µn T T 3 N Z (T )F(T ) If γ [ 2, 2], then triangulations with large number of faces are more likely to be sampled when µ µ c. γ = 8/3 Percolation, γ = 3 Ising, γ = 2 GFF,...

30 Scaling limit of random measures Fix the conformal map so that the three marked faces are sent to three fixed points z 1, z 2, z 3 on the sphere S 2. Give a mass a to each face of the triangulation and push to a measure ν T,a on the sphere Sample T according to P µ, hence Law of ν T,a : E µ [F (ν µ,a (dx))] = 1 Z µ N e µn Z (T )F(ν T,a (dx)) T T 3 N

31 Scaling limit of random measures Fix the conformal map so that the three marked faces are sent to three fixed points z 1, z 2, z 3 on the sphere S 2. Give a mass a to each face of the triangulation and push to a measure ν T,a on the sphere Sample T according to P µ, hence Law of ν T,a : E µ [F (ν µ,a (dx))] = 1 Z µ N e µn T T 3 N Z (T )F(ν T,a (dx)) Conjecture (DKRV) Fix µ > 0. In the regime a 0 and µ µ c + aµ, ν µ,a converges in law towards the Liouville measure Z of Liouville QFT with parameters (γ, µ) and three vertex insertions (z i, γ) i=1,2,3.

32 Scaling limit of random measures Conjecture (DKRV) Fix µ > 0. In the regime a 0 and µ µ c + aµ, ν µ,a converges in law towards the Liouville measure Z of Liouville QFT with parameters (γ, µ) and three vertex insertions (z i, γ) i=1,2,3. The Liouville measure (for (z i, γ) i=1,2,3 ): [ E γ,µ (z i,γ) [F(Z (A))] = CE F ( ξ Z (z i,γ)(a) ) Z (zi,γ)(r 2 Z(zi,γ)(R 2 ) (3 2Q )] γ ) where ξ has Gamma law Γ(3 2Q γ, µ) and Z (zi,γ)(dx) = e γ 3 i=1 α i Gĝ(x,z i ) M γ ĝ (dx)

33 Open question: DOZZ formula Fix 3 points z 1, z 2, z 3 and consider the 3-point function Π (z i,α i ) µ,γ (1) = µ 2Q i α i with 12 = α3 α1 α2,... γ z 1 z z 2 z z 1 z C γ (α 1, α 2, α 3 ) Believed to determine the whole Liouville theory. Conjectural explicit analytic expression for C γ (α 1, α 2, α 3 ). Prove it! Similars results on the Random Planar Maps (without matter): - for 2-point distances (Ambjørn-Watabiki, Di Francesco-Guitter ). - for 3-point distances (Bouttier-Guitter 08)

34 Connection with 2-point quantum sphere Can one give sense to the 2-point correlation functions (Seiberg 92)? Makes sense only for equal weights (z 1, γ) (z 2, γ) The resulting Liouville measure must be invariant under dilation Dilemma: not a true probability measure or not conformally invariant Recent rigorous construction by Duplantier-Miller-Sheffield 2014 Possible to recover the 2 point function from the three point functions: take three vertex operators (0, γ) (, γ) (z, ɛ) and send ɛ 0. Convergence modulo multiplicative constants of the unit volume Liouville measure

35 Thanks!