Analysis of the Hodge Laplacian on the Heisenberg group - download pdf or read online

By Detlef Muller, Marco M. Peloso, Fulvio Ricci

ISBN-10: 1470409399

ISBN-13: 9781470409395

The authors ponder the Hodge Laplacian ? at the Heisenberg workforce H n , endowed with a left-invariant and U(n) -invariant Riemannian metric. For 0=k=2n 1 , permit ? ok denote the Hodge Laplacian limited to ok -forms. during this paper they deal with 3 major, similar questions: (1) no matter if the L 2 and L p -Hodge decompositions, 1 (2) even if the Riesz transforms d? -12 ok are L p -bounded, for 1<<8 ; (3) tips to turn out a pointy Mihilin-Hormander multiplier theorem for ? ok , 0=k=2n 1

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10) = T −4 (−4iT Γ)2 (−T 2 ) ΔH − T 2 + (2m − ) = 16(ΔH − T 2 + m2 ) ΔH − T 2 + (2m − ) , ¨ D. MULLER, M. M. PELOSO, and F. 11) R22 = 16(ΔH − T 2 + m2 ) ΔH − T 2 + (2m − ) −1 R11 . Moreover, by the injectivity of R11 , ker R22 = ker R11 R22 = ker ⊕ ker . In order to repeat the same argument used above for R11 , we start from the operator R22 = R22 + δp,0 C + δq,0 C¯ (with δ denoting the Kronecker symbol) acting on scalar-valued functions. 11), R22 in invertible on S0 and, after tensoring and restricting, it is also invertible on W0p,q .

3. For every λ > 0 and σ ∈ Σ± , dπ±λ,σ (Δk ) is invertible and for every pair of elements u, v ∈ Eσ± , π±λ,σ (Δk )u, v is a polynomial in λ. For maps S0 Λk into itself. every α > 0, Δ−α k 1 Proof. 1), Δk2 ω 2 ≥ Tω 2 for every ω ∈ S0 Λk . This implies that 1 πλ,σ (Δk ) 2 ξ ≥ |λ| ξ , ξ ∈ Eσsgn λ , for every λ, σ with λ = 0. The rest is obvious. D. We call Riesz transforms the operators − 12 Rk = dΔk : S0 Λk −→ S0 Λk+1 , and their adjoints −1 Rk∗ = Δk 2 d∗ : S0 Λk+1 −→ S0 Λk . 4. 4) ∗ =I . Rk∗ Rk + Rk−1 Rk−1 In particular, Rk Rk−1 = 0.

UNITARY INTERTWINING OPERATORS AND PROJECTIONS 57 where P22 = P21 . 19), with s = p + q in place of k, we have ¯ P21 = iT −1 ( + 1)(∂ − ∂) (ΔH − T 2 ) + T −1 ∂( + i( + 1)T ) − ∂¯ ¯ = iT −1 (∂ − ∂) − ∂( − iT −1 ∂ = −∂ c (ΔH − T 2 ) + ( ΔH − T 2 − T c + ( + 1)∂ + iT −1 ∂¯ + i( + 1)T ) − ∂¯ = −c∂ (ΔH − T 2 ) − ∂( − i( + 1) c + c + c Δ ( + 1) (ΔH − T 2 ) Δ + i( + 1)T ) − ∂¯ + i( + 1)T )(c + c + c Δ c Δ + ∂¯ (c + c This proves the lemma. Δ . D. From the previous results we immediately get an explicit formula for U2, , at least when p = 0 and q = 0.

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Analysis of the Hodge Laplacian on the Heisenberg group by Detlef Muller, Marco M. Peloso, Fulvio Ricci


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