Lifting in Sobolev spaces of manifold-valued maps International Conference on PDEs and Applications In Honor of Prof. Haim Brezis Beijing Normal University Petru Mironescu
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(2) The lifting problem. Statement of the problem. · π : E → N a Riemannian covering · E , N connected Riemannian manifolds · N compact · Ω an N-dimensional simply connected smooth domain (or compact. manifold) · s > 0, 1 ≤ p < ∞ · Lifting problem: given any u ∈ W s ,p (Ω;N ), is it always possible to write u = π ◦ ϕ, with ϕ ∈ W s ,p (Ω;E )? · Subsequent questions: if yes, estimates? Uniqueness? If no, “optimal” smoothness of ϕ? How to detect the u’s that lift? · The compactness of E plays a crucial role · Compact case: E is compact/Non-compact case: E is non-compact. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 2 / 13.
(3) The lifting problem. Examples. · N = RP2 , N = S2 , π(x) = {x , −x }. Given a Sobolev non oriented vector field u : Ω → RP2 , find an orientation of ϕ : Ω → S2 of u, as smooth as u · An application. If h lifts g , then ½Z ¾ min |∇u |2 ;u : Ω → RP2 , u = g on ∂Ω = Ω ½Z ¾ min |∇ϕ|2 ; ϕ : Ω → S2 , ϕ = h on ∂Ω. (1). Ω. · N = S1 , E = S1 , π(z) = z d , d ≥ 2. Given u ∈ W s ,p (Ω; S1 ), is it possible to write u = ϕd with ϕ ∈ W s ,p (Ω; S1 ) (dth root problem,. Bethuel and Chiron)? · N = S1 , E = R, π(t) = e ı t . Given u ∈ W s ,p (Ω; S1 ), is it possible to write u = e ıϕ , with ϕ ∈ W s ,p (Ω; R)? + analogue of (1). Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 3 / 13.
(4) The lifting problem. An issue. · The definition of W s ,p (Ω;N ) and W s ,p (Ω;E )? · We embed N into some Rν and then. W s ,p (Ω;N ) = {u ∈ W s ,p (Ω; Rν ); u(x) ∈ N a.e.} · Definition independent of the embedding · E need not be compact: obvious definition of W s ,p (Ω;E ) only when s ≤ 1. E.g., when s < 1. W. s ,p. (Ω;E ) =. ½. p ϕ : Ω → E ; |ϕ|W s ,p. Ï. =. dE (ϕ(x), ϕ(y ))p dxdy < ∞ |x − y |m+sp. ¾. ·. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 4 / 13.
(5) Lifting of S1 -valued maps. The global picture. Theorem (Bourgain, Brezis, M 00) For the universal covering of S1 , the lifting property holds except when · 1 ≤ sp < 2 ≤ N = dim Ω · 0 < s < 1, 1 ≤ sp < N = dim Ω · Topological obstruction to lifting: u(z) = z/|z | (z ∈ C) s ,p. · Analytical obstruction to lifting: ϕ ∈ Wloc (Ω \ {a}) \ W s ,p (Ω) such that e ıϕ ∈ W s ,p (Ω) · It is possible to have the lifting property without estimates: in W 1/p ,p ((0, 1); S1 ), 1 < p < ∞, every map lifts, but no norm control if x ≤ 1/2 0,. · The kink: ϕε (x) =. 2π, affine,. satisfies, ∀ 1 < p < ∞, if x ≥ 1/2 + ε if 1/2 ≤ x ≤ 1/2 + ε ıϕε |ϕε |W 1/p,p → ∞, |e |W 1/p,p . 1. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 5 / 13.
(6) The lifting problem. The non-compact case. Theorem (Bethuel, Chiron 07) Let π : E → N be the universal covering of N , with π1 (N ) infinite. Then the lifting property holds except when · 1 ≤ sp < 2 ≤ N = dim Ω · 0 < s < 1, 1 ≤ sp < N = dim Ω · The topological obstruction and the kink exist without any assumption on π1 (N ) · The analytical obstruction is obtained from the existence of a ray in E. (isometrically embedded real line in E ), which requires E non-compact. · The same result holds (essentially with the same proof) for any. covering provided E is non-compact. · The space W s ,p (Ω;E ) is defined by embedding E · When E is compact, open case: the one connected to the analytical obstruction: 0 < s < 1, 2 ≤ sp < N = dim Ω. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 6 / 13.
(7) The lifting problem. The compact case. Theorem (M, Van Schaftingen 19) Assume E compact and π nontrivial. Then the lifting property holds except when 1 ≤ sp < 2 ≤ N = dim Ω · Thus the difference between the compact and the non-compact case occurs exactly in the range 0 < s < 1, 2 ≤ sp < N = dim Ω: existence in. the compact case, non-existence in the non-compact case · The result applies in particular to the dth root problem and to the orientation problem in RPk. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 7 / 13.
(8) The lifting problem. The compact case. Sketch of proof · We consider only the new the case 0 < s < 1, 2 ≤ sp < N = dim Ω · Strategy: establish an a priori estimate for u in a dense subset of W s ,p (Ω;N ), having the lifting property: linear estimate · The maps continuous outside some (N − bsp c − 1)-skeleton are dense. (Brezis, M 15) · Such maps do lift · The a priori estimate relies on a one-dimensional reverse oscillation inequality for f ∈ C 0 (R, R), 0 < s < 1, sp > 1: Z Z [osc p [x , y ] f ] R R. |y − x |1+sp. Z Z. dxdy ≤ Cs ,p. Petru Mironescu (Université Claude Bernard Lyon 1). R R. |f (y ) − f (x)|p |y − x |1+sp. dxdy. July 3rd, 2019. 8 / 13.
(9) The lifting problem. Topological vs analytical obstructions. · When N = dim Ω ≥ 2 and π is non-trivial, topological obstructions exist when 1 ≤ sp < 2 · Maps including such obstructions do not belong to the closure of. smooth maps and, in general, they do not lift · What happens in absence of topological obstructions? Other types of. obstructions? Yes, in the non-compact case. In the compact case: Theorem (M, Van Schaftingen 19) Let 0 < s < 1, 1 < sp < 2, N = dim Ω ≥ 2. In the case of the universal covering with compact E , the following are equivalent for u ∈ W s ,p (Ω;N ): (a) u is in the strong closure of C ∞ (Ω;N ) (b) u is in the weak closure of C ∞ (Ω;N ) (c) u lifts · Assuming only E compact, we have (b) =⇒ (c) Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 9 / 13.
(10) The lifting problem. Topological vs analytical obstructions. Theorem (M, Van Schaftingen 19) Let 1 < p < ∞, N = dim Ω ≥ 2. Assume π non-trivial. Then there exists some u ∈ W 1/p ,p (Ω;N ) such that: · u is in the strong closure of C ∞ (Ω;N ) · u does not lift · Thus an analytical obstruction persists in the limiting case sp = 1 · In this case, the compactness of E plays no role · u can be chosen to be smooth in Ω except at some point a ∈ Ω. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 10 / 13.
(11) Loss of control. The kink. · The kink satisfies, ∀ 1 < p < ∞, |ϕε |W 1/p,p → ∞, |e ıϕε |W 1/p,p . 1 z −a · An analogue on S1 : Ma (z) = , |a| < 1: |Ma |W 1/p,p . 1 and the az − 1 phase of Ma blows up in W 1/p ,p near a as |a| → 1. Theorem (M 15) Let 1 < p < ∞. Given M > 0, each u ∈ W 1/p ,p (S1 ; S1 ) with |u |pW 1/p,p ≤ M can be written as u=. K Y. (Mak )±1 e ıψ ,. k =1. with K ≤ F (M), |ψ|W 1/p,p ≤ G (M) · Proof “by induction” on M, via a bubbling analysis · Proof by contradiction, giving a (linear) control on F (M), but not on. G (M) Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 11 / 13.
(12) Loss of control. The kink. Theorem (Bourgain, Brezis 03) Let 1 < p ≤ 2. W 1/p ,p ((0, 1); S1 ) can be written as. Each u ∈. u = e ı (ϕ+ψ) , |ϕ|W 1/p,p . |u |W 1/p,p , |ψ|W 1,1. (1) . |u |pW 1/p,p. (2). Theorem (Nguyen 08) The above holds in the full range 1 < p < ∞ · Bourgain, Brezis: proof by explicit decomposition (works in any. dimension) · Uses only u ∈ L∞ ∩ W 1/p ,p , not the full strength of |u | = 1 · Nguyen: Proof by duality (works only in dimension one) · Uses |u | = 1. Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 12 / 13.
(13) Loss of control. Factorization. Theorem (M 08–...) Let 1 ≤ p < ∞, s > 0. Each u ∈ W s ,p ((0, 1); S1 ) can be written as u = v e ıϕ ,. (3). |ϕ|W s ,p . |u |W s ,p , |v |B sp. 1, 1. . |u |pW s ,p. (4). sp. · B1,1 is a Besov space. It coincides with the Sobolev space W sp ,1 for. non integer sp. Is strictly smaller than W sp ,1 for integer sp. · Proof by explicit decomposition. Works in any dimension, with the estimate |v |B sp . |u |pW s ,p + |u |pW s ,p 1, 1. · Uses |u | = 1 · “Converse” is true: if 0 < s < 1,. [v ∈ B1sp,1 (Ω; S1 ), ϕ ∈ W s ,p (Ω; R)] =⇒ v e ıϕ ∈ W s ,p (Ω; S1 ) · (But v ∈ W 1,1 (Ω; S1 ) does not imply v ∈ W 1/p ,p (Ω; S1 ), 1 < p < ∞) Petru Mironescu (Université Claude Bernard Lyon 1). July 3rd, 2019. 13 / 13.
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