Heisenberg calculus and spectral theory of hypoelliptic by Raphael S. Ponge

By Raphael S. Ponge

This memoir bargains with the hypoelliptic calculus on Heisenberg manifolds, together with CR and phone manifolds. during this context the most differential operators at stake contain the Hormander's sum of squares, the Kohn Laplacian, the horizontal sublaplacian, the CR conformal operators of Gover-Graham and the touch Laplacian. those operators can't be elliptic and the correct pseudodifferential calculus to check them is equipped through the Heisenberg calculus of Beals-Greiner and Taylor.

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Then the following are equivalent: (i) P is a ΨH DO of order m, m ∈ C. 33) kP (x, y) = |εx |KP (x, −εx (y)) + R(x, y), ˆ with KP ∈ Km (U × Rd+1 ), m ˆ = −(m + d + 2), and R ∈ C ∞ (U × U ). Furthermore, if (i) and (ii) hold and KP ∼ j≥0 KP,m+j , Kl ∈ Kl (U × Rd+1 ), ˆ d+1 then P has symbol p ∼ j≥0 pm−j , pl ∈ Sl (U ×R ), where pm−j is the restriction ∧ to U × (Rd+1 \0) of [KP,m+j (x, φ−1 ˆ x (y))]y→ξ . 17. Let a ∈ U . 34) −1 kP˜ (0, y) = |εa |−1 kP (ε−1 a (0), εa (y)) = KP (a, −y). Moreover, as we are in Heisenberg coordinates already, we have ψ0 = ε0 = φ0 = id.

The subspace of vectors ξ ∈ Hπ so that x → π(x)ξ is smooth from G to Hπ . 6 ([CGGP]). 1) The domain of πP a always contains Cπ∞ (Ea ). 2) If m ≤ 0 then the operator πP a is bounded. 3) We have (πP a )∗ = (πP a )∗ . 4) If P1 and P2 are ΨDO’s on M then π(P1 P2 )a = πP1a πP2a . 7. If Ea = C and P a is a differentiable operator then, as it is leftinvariant, P a belongs to the enveloping algebra U(g) of the Lie algebra g = ga M of G. In this case πP a coincides on Cπ∞ with the operator dπ(P a ), where dπ is the representation of U(g) induced by π.

10) (∂ α K)|S |K|k = |c(K)| + L2 (S) , K ∈ Ak . |k ’s give rise to a system of semi-norms on K−(d+2) (Rd+1 ) whose corresponding topology is the weakest topology making the maps K → c(K) and K → K|S be continuous from K−(d+2) (Rd+1 ) to C and C ∞ (S) respectively. a denote the convolution operator In the sequel for a ∈ U and K ∈ Ak we let PK a d+1 a PK f = K ∗ f , f ∈ S0 (R ). 13 ([KS, Thm. 1], [FS2, Thm. 19], [Ch2]). Let k be an integer greater than or equal to d + 3. Then: a (i) For any a ∈ U and any K ∈ Ak the operator PK extends to a bounded 2 d+1 operator on L (R ).

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