Sharp Gårding inequality on compact Lie groups.

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15-19.10.2012, ESI, Wien, Phase space methods for pseudo-differential operators Ville Turunen, Aalto University, Finland (ville.turunen@aalto.fi) M. Ruzhansky, V. Turunen: Sharp Gårding inequality on compact Lie groups. Journal of Functional Analysis, 2011. (M. Ruzhansky, V. T.: Pseudo-Differential Operators and Symmetries. Birkhäuser 2010.) (M. Ruzhansky, V. T.:... IMRN 2012.) We establish the sharp Gårding inequality on compact Lie groups. The positivity condition is expressed in the noncommutative phase space in terms of the full symbol, which is defined using the representations of the group. Applications are given to the Sobolev boundedness of pseudo-differential operators. 1

Introduction to Sharp Gårding Gårding inequality [Gårding 1953]. Sharp Gårding inequality [Hörmander 1966]: If p S m 1,0 (Rn ) and p(x, ξ) 0 for every (x, ξ) R n R n then Re (p(x, D)u, u) L 2 (R n ) C u 2 H (m 1)/2 (R n ) for all u C 0 (Rn ). See also [Calderón 1968], [Nagase 1977], [Kumano-go 1981], [Taylor 1981]. Matrix case on R n [Lax & Nirenberg 1966, Friedrichs 1968, Vaillancourt 1970]. [Beals and Fefferman 1974], [Fefferman and Phong 1978]. The main difficulty on manifolds: the full symbol of an operator can not be invariantly defined via localisations. Hörmander Melin inequality on compact manifolds [Melin 1971, Hörmander 1977]. 2

m = 1 Sharp Gårding inequality on R n : if A Ψ m (R n ) = OpS m 1,0 (Rn ) such that σ A (x, ξ) 0 for all (x, ξ) R n R n = R n R n then A is almost positive : Re Au, u L 2 (R n ) constant u 2 L 2 (R n ) for every u C 0 (Rn ). m = 1 Sharp Gårding inequality on compact Lie group G: if A Ψ m (G) = OpS1,0 m (G) such that σ A (x, ξ) 0 for all (x, [ξ]) G Ĝ then A is almost positive : Re Au, u L 2 (G) constant u 2 L 2 (G) for every u C (G). 3

Applications of global Fourier analysis: 1. Ψ m (G)-ellipticity characterization, with applications to spectral problems. 2. Global hypoellipticity in Ψ m (G) (if Au C (G) then u C (G)); many examples on G = SU(2) = S 3 (Hörmander Sum of Squares Thm. Number theoretic conditions). 3. Sharp Gårding inequality etc. Notations: G compact Lie group with Haar measure & bi-invariant metric (G < U(N) C N N closed), n = dim(g) the real dimension, Ĝ the unitary dual. [ξ] Ĝ corresponds to a homomorphism ξ : G U(d ξ ) such that C d ξ = span{ξ(x)v : x G} if 0 v C d ξ. Non-commutative phase space G Ĝ. 4

Relevant Lie groups here: - R n - torus T n = R n /Z n - compact Lie groups G in general, e.g. SO(3): rotations of R 3 about the origin; SU(2): unitaries of C 2 General framework: A transitive action G M M of a Lie group G on a manifold M; calculus on M as a shadow from G. Example: - M = S n 1 = {x R n : x R n = 1}, especially S 2 or S 3 - G = SO(3) or G = SU(2) Basic tools for calculus on G: (a G ) Noncommutative Fourier transform on G (b G ) Taylor polynomials respecting G-operations 5

Ψ m (R n ) on the Euclidean space R n : f(ξ) = R n f(x) e i2πx ξ dx, Af(x) = ei2πx ξ σ A (x, ξ) f(ξ) dξ, R n ξ α β x σ A (x, ξ) constant αβ ξ m α [Kohn & Nirenberg, Hörmander -65]. Ψ m (T n ) on the torus T n = R n /Z n : f(ξ) = f(x) e i2πx ξ dx, T n Af(x) = e i2πx ξ σ A (x, ξ) f(ξ), ξ Z n α ξ β x σ A(x, ξ) constant αβ ξ m α [Agranovich -90], [McLean -91], [T. -00]. Ψ m (G) on a compact Lie group G: f(ξ) = f(x) G ξ(x) dx, Af(x) = d ξ Tr ( ξ(x) σ A (x, ξ) f(ξ) ) [ξ] Ĝ α ξ β x σ A (x, ξ) constant op αβ ξ m α. 6

Example. For A, B Ψ (R n ), σ AB (x, ξ) α 0 1 α! ( α ξ σ A)(x, ξ) ( α x σ B )(x, ξ). For A, B Ψ (G), σ AB (x, ξ) 1 α 0 α! ( α ξ σ A)(x, ξ) ( x (α) σ B )(x, ξ). Especially, for G = T 1 we have Ĝ = Z 1, ( ξ σ)(ξ) = σ(ξ + 1) σ(ξ), α ξ f(ξ) = x (α) f(ξ). x (α) = ( e i2πx 1 ) α, (α) x = x ( x 1) ( x α + 1). 7

Theorem [Ruzhansky-T.-Wirth]. Let A : C (G) D (G) be linear and continuous. Then the following statements are equivalent: (A) A Ψ m (G). (B) admissible 1,..., n diff 1 (Ĝ), α, β : α ξ β x σ A (x, ξ) op C αβ ξ m α, and sing suppr A (x, ) {e}. (C) For strongly admissible 1,..., n diff 1 (Ĝ), α, β : α ξ β x σ A (x, ξ) op C αβ ξ m α. 8

On proving previous theorem Main tool: commutator characterization (see e.g. [T. 2000], [Ruzhansky-T. 2010]). It is sufficient to show that 1. any symbol σ A satisfying condition (B) gives rise to a bounded linear operator A : H m (G) L 2 (G); 2. the symbol σ [X,A] of the commutator [X, A] = XA AX also satisfies (B), for any left-invariant vector field X. σ [X,A] = σ XA σ AX σ X σ A σ A σ X + = σ X σ A σ A σ X. α >1 1 α! ( α ξ σ X)( (α) x σ A ) 9

Main result Sharp Gårding Thm [Ruzhansky-T. 2011]. Let A Ψ m (G) be such that σ A (x, ξ) 0 for all (x, [ξ]) G Ĝ. Then constant < u C (G): Re(Au, u) L 2 (G) constant u 2 H (m 1)/2 (G). Cor. 1. Let A Ψ 1 (G) be such that (x, [ξ]) G Ĝ : σ A(x, ξ) op C. Then A is bounded on L 2 (G). Cor. 2. Let A Ψ m (G) and let ( M = ξ m ) σ A (x, ξ) op. sup (x,[ξ]) G Ĝ Then s R C > 0 u C (G): Au 2 H s (G) M 2 u 2 H s+m (G) +C u 2 H s+m 1/2 (G). 10

Tools for obtaining sharp Gårding Recall: characterization of Ψ m (G) [Ruzhansky-T.-Wirth]. Defn. A matrix-valued symbol σ A belongs to Sρ,δ m (G) if it is smooth in x and if for a strongly admissible collection 1,..., n diff 1 (Ĝ) α ξ β x σ A(x, ξ) op C αβ ξ m ρ α +δ β for all α, β N n 0, uniformly in (x, [ξ]) G Ĝ. Thm [Ruzhansky-T. 2011]. Let A : C (G) C (G) be continuous and linear. Let µ, C α R such that α x σ A (x, ξ) op C α ξ µ for all (x, [ξ]) G Ĝ and all multi-indices α. Then A extends to a bounded operator H s (G) H s µ (G) for all s R. 11

Amplitudes on G Let 0 δ, ρ 1. An amplitude a A m ρ,δ (G) is a mapping defined on G G Ĝ, smooth in x and y, such that a(x, y, ξ) : H ξ linear H ξ, and for a strongly admissible collection of difference operators α ξ we have the amplitude inequalities α ξ β x y γ a(x, y, ξ) C op αβγ ξ m ρ α +δ β+γ, for all multi-indices α, β, γ, and all (x, [ξ]) G Ĝ. For an amplitude a, the amplitude operator Op(a) : C (G) D (G) is defined by Op(a)u(x) := ( dim(η) Tr η(x) a(x, y, η) u(y) G η(y) dy [η] Ĝ Notice that if a(x, y, η) = σ A (x, η) then Op(a) = A. 12 ).

Amplitude operator Op(a) : C (G) D (G) is defined by Op(a)u(x) := ( dim(η) Tr η(x) a(x, y, η) u(y) G η(y) dy [η] Ĝ ). Prop. Let 0 δ < 1 and 0 ρ 1, and let a A m ρ,δ (G). Then Op(a) is a continuous linear operator from C (G) to C (G). Prop. Let 0 δ < ρ 1 and let a A m ρ,δ (G). Then A = Op(a) is a pseudodifferential operator on G with a matrix symbol σ A Sρ,δ m (G). Moreover, σ A has the asymptotic expansion σ A (x, ξ) α 0 1 α! (α) y α ξ a(x, y, ξ) y=x. 13

Proof of the sharp Gårding inequality If Q : H (m 1)/2 (G) H (m 1)/2 (G) is a bounded linear operator (e.g. Q Ψ m 1 (G)), then Re(Qu, u) L 2 (G) (Qu, u) L 2 (G) Qu H (m 1)/2 (G) u H (m 1)/2 (G) Q H (m 1)/2 H (m 1)/2 u 2 H (m 1)/2. Hence the sharp Gårding inequality on G would follow if we can show that A = P + Q, where P is positive (on C (G) L 2 (G)) and Q : H (m 1)/2 (G) H (m 1)/2 (G) is bounded. 14

G < GL(N, R) R N N, Lie algebra g R N N, g = R n. V g, V = B(0, r) = {z R n : z < r}, diffeomorphism exp : V exp(v ). φ : [0, r) [0, ) smooth, supp(z φ( z )) V, φ(s) = 1 for small s > 0. [ φ( X ξ 1/2 ) 2 ξ n/2 w ξ (x) := R n φ( Z )2 dz det D exp(x) f(x) where x = exp(x), D exp is the Jacobi matrix of exp, and f(x) is density w.r.t Lebesgue measure of Haar measure pulled back by exp. ]1/2, Remark: E.g. w ξ C (G), w ξ L 2 (G) = 1, ( (x, ξ) wξ (x)i dim(ξ) ) S n/4 1,1/2 (G). 15

Let σ A S1,0 m (G) and p(x, y, ξ) := w ξ(xz 1 ) w ξ (yz 1 ) σ A (z, ξ) dz, G where w ξ C (G) is as before. Prop. Let P u(x) = Op(p)u(x) := G [ξ] Ĝ dim(ξ)tr ( ξ(y 1 x)p(x, y, ξ) ) u(y) dy. Then p A m 1,1/2 (G) and P = Op(p) 0. Lemma. p(x, x, ξ) σ A (x, ξ) is a symbol of a bounded operator H s (G) H s (m 1) (G) for any s R. Lemma. σ P (x, ξ) p(x, x, ξ) is a symbol of a bounded operator H s (G) H s (m 1) (G) for any s R. 16

And the main result was: Sharp Gårding Thm [Ruzhansky-T. 2011]. Let A Ψ m (G) be such that σ A (x, ξ) 0 for all (x, [ξ]) G Ĝ. Then constant < u C (G): Re(Au, u) L 2 (G) constant u 2 H (m 1)/2 (G). Cor. 1. Let A Ψ 1 (G) be such that (x, [ξ]) G Ĝ : σ A(x, ξ) op C. Then A is bounded on L 2 (G). Cor. 2. Let A Ψ m (G) and let ( M = ξ m ) σ A (x, ξ) op. sup (x,[ξ]) G Ĝ Then s R C > 0 u C (G): Au 2 H s (G) M 2 u 2 H s+m (G) +C u 2 H s+m 1/2 (G). 17