Geometric aspects of the p -Laplacian on complete manifolds Stefano - PowerPoint PPT Presentation

Geometric aspects of the p -Laplacian on complete manifolds Stefano Pigola Università dell’Insubria, Como Grenoble, 5-9 September 2011

I. Introductory examples We are given an m -dimensional Riemannian manifold ( X m ; h ; i ) . A natural way to detect the geometry and the topology of X is to view X either as the domain or as the target space of some interesting class of maps. Clearly, the Riemannian structure adds information on X and therefore the interesting maps should take them into account. Let us consider a couple of (classical) examples to give some ‡avour of ideas and techniques and to introduce (some of) the main ingredients.

Let M; N be compact, with Sec N � 0 . Let f : M ! N be a smooth map. Then we have Th. 1 (Eells-Sampson, Hartman) �Z � Z M j du j 2 = min M j dh j 2 : h homotopic to f 9 u : M ! N : : The minimizer u satis…es the (system of) equations � u := div ( du ) = 0 i.e. u is a harmonic map. Note: u is smooth by elliptic regularity. In particular, the validity of a Liouville type result � u = 0 = ) u = const gives that f is topologically trivial. For instance, we have the following

Th. 2 (Eells-Sampson) M cmpt, Ric M � 0 and N cmpt, Sec N � 0 . (a) If Ric M ( p 0 ) > 0 for some p 0 2 M = ) Liouville for harmonic maps = ) every smooth f : M ! N is homotopically trivial. (b) If Sec N < 0 then either the harmonic map u : M ! N is constant or u ( M ) = � a closed geodesic of N . Proof. Let u : M ! N be harmonic. The Bochner-Weitzenböck formula states X 1 2� j du j 2 = j Ddu j 2 + h du ( Ric M ( E i )) ; du ( E i ) i i � � X 2 : � � � Sec N ( du ( E i ) ^ du ( E j )) � du ( E i ) ^ du ( E j ) � i;j

Since Ric M � 0 and Sec N � 0 , � j du j 2 � 0 ; equality holding i¤ Ddu = 0 . Use Stokes theorem with X = j du j 2 r j du j 2 : Z Z � � r j du j 2 � 2 � 0 ) j du j � const : � � 0 = M div ( X ) � � M and du is parallel. If Ric M ( p 0 ) > 0 then d p 0 u = 0 and this implies du = 0 . � � Similarly if Sec N < 0 and du 6 = 0 , since du ( E i ) ^ du E j = 0 we obtain that u ( M ) is 1 -dimensional. Since Ddu = 0 ) u maps geodesics into geodesics ) u ( M ) � � geodesic. Assume � simple, otherwise more tricky. If � is not closed then u is homotopically trivial. But ( M cmpt) it can be shown that u minimizes energy in its homotopy class ) u � const . Contradiction. It is now easy to obtain u ( M ) = � . Now, some classical applications.

Application I. We …rst illustrate a use of X as a target space. Th. 3 (Preissman) X cmpt, Sec < 0 . Then Z 2 6� � 1 ( X ) . By contradiction, Z 2 � � 1 ( X ) . Proof. Fix any injective homomorphism � : � 1 ( T 2 ) ' Z 2 ! � 1 ( X ) with T 2 the ‡at torus. Since Sec X � 0 , by the general theory of aspherical spaces, we can assume that 9 smooth nonconst map u : T 2 ! X which induces � up to some � 2 Aut ( � 1 ( X )) , say � � � = u # . By Eells-Sampson-Hartman, we can take u harmonic. Liouville Theorem ) u ( T 2 ) = closed geodesic of X: Therefore, u # maps the generators of � 1 ( T 2 ) onto a single loop ) u # is not injective. Contradiction. The ‡at-torus theorem by Lawson-Yau and Gromoll-Wolf can be obtained along the same line.

Application II. Now we illustrate a use of X as a source space. Th. 4 Let X be cmpt with Ric X � 0 and Ric X > 0 somewhere. Then, every homomorphism � : � 1 ( X ) ! � 1 ( N ) where N cmpt and Sec N � 0 , must be trivial: � � 1 : Proof. As above, we can assume that 9 smooth harmonic map u : X ! N such that � � � = u # , for some � 2 Aut ( � 1 ( N )) . Since Ric X > 0 at some x 0 2 X , by the Liouville thm u is constant ) u # � 1 ) � � 1 . A consequence. There is no metric g on R m s.t.: (a) g = g Eu on R m n B 1 (0) ; (b) Ric g � 0 on R m ; (c) Ric g > 0 at some x 0 2 B 1 (0) . Cut a cube around B 1 (0) , periodise it to get an m -torus X with Ric X � 0 and Ric ( x 0 ) > 0 . Let N be the ‡at torus. By the thm, the homomorphism id : � 1 ( X ) ! � 1 ( N ) is trivial. Contradiction. ( Remark. By Lohkamp, there exist Ric < 0 balls!!!)

II. p -harmonic functions and maps The previous examples involve (2 - ) harmonic maps. The concept was introduced by Eells-Sampson in the mid ’60s and extends the notion of harmonic function. Let u : ( M m ; h ; i M ) ! ( N n ; h ; i N ) be a smooth map. The Hilbert-Schmidt � � T � M � u � 1 TN norm of its di¤erential du 2 � is denoted by j du j . Let p > 1 . Def. 1 The map u is said to be p -harmonic if � � j du j p � 2 du � p u := div = 0 ; where � div is the formal adjoint of d with respect to the standard L 2 -inner product on vector valued 1 -forms. The operator � p u is called the p -Laplacian (or p -tension …eld) of u .

In case u 2 C 1 the above condition has to be interpreted in the sense of distributions, i.e., Z D E j du j p � 2 du; d� (� p u; � ) = � = 0 ; M 8 � 2 � c ( u � 1 TN ) . In local coordinates the above writes Z � � p � 2 n� @ � D � Eo � � + � ; � ! � @ � � @ � ! ! u ; @ � ! ! u ; @ � ! ! � u � � u � = 0 ; where � ! � is an R n -valued quadratic form (involving N � A BC ). Note also the relation between � p and � : � r j du j p � 2 � � p u = j du j p � 2 � u + du : In the special case N = R one can also speak of p -subharmonic function whenever � p u � 0 and of p -superharmonic function if � p u � 0 .

II.a. p -harmonic maps as “canonical” representatives We are interested in complete non-compact domains. It is then natural to prescribe asymptotic (decay) properties to maps, more precisely on the energy of f j p 2 L 1 ( M ) . According maps. Say that f : M ! N has …nite p -energy if j d to results by R. Schoen and S.T. Yau, F. Burstall, B. White, S.W. Wei, p - harmonic maps can be considered as canonical representatives of homotopy class of maps with …nite p -energy into nonpositively curved targets. Th. 5 Let ( M; h ; i M ) be complete and ( N; h ; i N ) be compact with Sec N � f j p 2 L 1 ( M ) , 0 . Fix a smooth map f : M ! N with …nite p -energy j d p � 2 . Then, in the homotopy class of f , there exists a p -harmonic map u 2 C 1 ;� ( M; N ) with j du j p 2 L 1 ( M ) . If p = 2 then u 2 C 1 ( M; N ) :

Some consequences and questions that arise naturally from the existence thm: (a) Trivial homotopy type . Liouville type thms under geometric assumptions on M ) a map f : M ! N with …nite p -energy must be topologically trivial. (b) Comparison of homotopic p -harmonic maps . How many p -harmonic maps with …nite p -energy are there in a given homotopy class ? In case p = 2 (harmonic case) both questions in the non-compact setting are answered in deep seminal works by Schoen-Yau (the compact case is due to P. Hartman). They proved: (A) vanishing results for harmonic maps assuming that either Ric M � 0 or M is a stable minimal hypersurface in R m +1 ; (B) comparison of homotopic harmonic maps and uniqueness of the harmonic representative, assuming vol ( M ) < + 1 .

II.b. Vanishing for p -harmonic maps Schoen-Yau vanishing results alluded to in (A) are uni…ed and extended by al- lowing a controlled amount of negative Ricci curvature (and di¤erent energies). The negative part of the curvature is measured via a spectral assumption. Suppose Ric M � � a ( x ) ; a ( x ) � 0 . Let L H = � � � Ha ( x ) ; H 2 R is a parameter. By de…nition �R � jr ' j 2 � Ha ( x ) ' 2 : ' 2 C 1 R � 1 ( L H ) := inf c ( M ) n f 0 g : ' 2 Intuitively, � 1 ( L H ) � 0 relies on the fact that a ( x ) is small in some inte- gral sense. In the terminology of P. Li and J. Wang, � 1 ( L H ) � 0 ( ) a weighted Poincaré inequality holds.

Th. 6 Let M be complete, noncmpt, Ric � � a ( x ) with � 1 ( L H ) � 0 for some H > ( m � 1) =m . Let N be complete, Sec N � 0 . Then every harmonic map u : M ! N with …nite energy j du j 2 L 2 must be constant. In particular, f j 2 L 2 is homotopically trivial. every map f : M ! N with j d Rmk 1 a ( x ) � 0 ) � 1 ( L H ) � 0 is weaker than � 1 ( L 1 ) � 0 previously considered e.g. by [P.-Rigoli-Setti, JFA ’05]. Proof. Starting point: Bochner formula+re…ned Kato (RHS) j du j � j du j +a ( x ) j du j 2 = j Ddu j 2 � jr j du jj 2 � 1 m jr j du jj 2 . By applying the next vanishing result with = j du j , A = � 1 =m and p = 2 we deduce that j du j � const and either j du j � 0 or a � 0 , i.e., Ric � 0 . Suppose 0 6 = j du j 2 L 2 . Then vol M < + 1 and Ric � 0 . Contradiction.

Th. 7 (Bérard, P.-Veronelli) Let M be complete, let 0 � be a Lip loc solution of � + a ( x ) 2 + A jr j 2 � 0 ; with 0 � a ( x ) 2 C 0 ( M ) , A 2 R . Assume (i) � 1 ( L H ) � 0 for some H > A + 1 > 0 , L H = � � � Ha ( x ) . R B R 2 p = o ( R 2 ) , for some p 0 < p < p 1 , where p 0 � H � p 1 roots of (ii) q 2 � 2 Hq + H ( A + 1) = 0 : Then � const. and either � 0 or a � 0 .