Surfaces in spheres: biharmonic and CMC E. Loubeau (joint with C. Oniciuc) Université de Bretagne Occidentale, France May 2014
Biharmonic maps Definition Bienergy functional at φ : (M, g) (N, h) E 2 (φ) = 1 τ(φ) 2 v g. 2 (τ(φ) = trace g dφ) Critical points of E = bi-harmonic maps. M
Biharmonic maps First variation Euler-Lagrange equation associated to E 2 : τ 2 (φ) = φ τ(φ) trace g R N (dφ, τ(φ))dφ, ) where φ = trace g ( φ φ φ is the Laplacian on sections of φ 1 TN.
Biharmonic maps Bochner formulas If M compact and Riem N 0 then biharmonic = harmonic (Jiang). A biharmonic isometric immersion is minimal if τ(φ) 2 is constant and Riem N 0 (Oniciuc). study biharmonic maps into spheres. Not satisfactory.
Biharmonic maps Main examples in spheres Harmonic maps are bi-harmonic, but S n ( 1 2 ) S q ( 1 2 ) S n+q+1 (n q) is bi-harmonic but not harmonic (Jiang). 45 th -parallel : S n ( 1 2 ) S n+1 Let M be a compact manifold and ψ : M S n ( 1 2 ) a nonconstant map. The map φ = i ψ : M S n+1 is nonharmonic biharmonic if and only if ψ is harmonic and e(ψ) is constant.
Other tool : stress-energy tensor Biharmonic stress-energy tensor Let φ : (M, g) (N, h). S 2 (X, Y ) = [ ] τ(φ) 2 /2 + dφ, τ(φ) g(x, Y ) dφ(x), Y τ(φ) dφ(y ), X τ(φ)
Biharmonic stress-energy tensor Biharmonic stress-energy tensor F (g) = 1 2 τ(φ) 2 v g. Derives from a metric variational problem : δf(g t ) = 1 2 S 2, ω v g M M
Biharmonic stress-energy tensor Hilbert s criterion div S 2 = dφ, τ 2 (φ)
Harmonic stress-energy tensor Definition Let φ : (M, g) (N, h), its (harmonic) stress-energy tensor is where e(φ) = 1 2 dφ 2. Hilbert s principle S = e(φ)g φ h, div S = dφ, τ(φ)
Harmonic stress-energy tensor Harmonic maps Let φ : (M 2, g) (N, h) be a harmonic map then : ( R S 1 )(X, Y ) = 2K M S 1 Take φ : (M 2, g) (N, h) harmonic map, M compact, K M 0, strictly at some point (e.g. the round sphere). Then S 1 = 0, that is φ is weakly conformal.
Harmonic stress-energy tensor 2-Torus If φ : (T 2, g 0 ) (N, h) is a harmonic map then either φ is weakly conformal or φ 1 TN admits a nowhere zero section. If φ : (T 2, g 0 ) (N 2, h) is a harmonic map then either φ is weakly conformal or φ 1 TN is trivial and (deg φ)c 1 (TN) = 0. There is no harmonic map φ : T 2 S 2 of degree ±1 (whatever the metrics).
Biharmonic stress-energy tensor On surfaces Take φ : (M 2, g) (N, h) biharmonic map : { R S 2 = 2K M S 2 + d( τ(φ) 2 )+ K M τ(φ) 2 + τ(φ) 2} g,
Biharmonic stress-energy tensor On compact surfaces M 2 is compact. Then S 2 2 v g +2 M M K M( S 2 2 τ(φ) 4 ) v g = d( τ(φ) 2 ) 2 v g, 2 M If S 2 = 0, then τ(φ) constant and M K M v g = 0 or S 2 = τ(φ) 2 2 g. If τ(φ) 2 constant and K M 0, then S 2 is parallel and M is flat or S 2 = τ(φ) 2 2 g.
Biharmonic stress-energy tensor Riemannian immersions Geometric version Let φ : (M 2, g) (N n, h) be a proper-biharmonic Riemannian immersion, from a compact oriented surface. If A H = 0, then M is topologically a torus or pseudo-umbilical. If CMC and K M 0. Then A H = 0 and M is flat or pseudo-umbilical.
Biharmonic stress-energy tensor Local description Let φ : (M 2, g) (N n, h) be a CMC proper-biharmonic Riemannian immersion. λ 1, λ 2 principal curvatures of M corresponding to A H, λ 1 λ 2, and let µ = λ 1 λ 2. Let p M such that µ(p) > 0, i.e. p is a non pseudo-umbilical point. Then, around p, there is a local chart (U; x, y) which is both isothermal and a line of curvature coordinate system for A H.
Biharmonic stress-energy tensor Local description g = 1 µ (dx 2 + dy 2 ), A H (.),. = 1 µ (λ 1dx 2 + λ 2 dy 2 ), 2 λ 1,2 = H 2 ± 2 R 2 N (X i, H, X i, H) H 2 2 H 4, i=1 with X 1 = µ x, X 2 = µ y.
Biharmonic stress-energy tensor Local description ( 2 ln R N (X i, H, X i, H) H 2 2 H 4) = 4K M, i=1 Importance of pseudo-umbilical points wrt topology.
Biharmonic stress-energy tensor 3-dim ambient spaces Let φ : (M 2, g) N 3 (c) be a CMC proper-biharmonic Riemannian immersion in a three dimensional real space form. Then it is umbilical. Let φ : (M 2, g) (N 3, h) be a CMC proper-biharmonic Riemannian immersion. Assume that there exists c > 0 such that Ricci N (U, U) c U 2 with H 2 (0, c 2 ). Then M2 has no pseudo-umbilical point.
Biharmonic stress-energy tensor Hopf differential Let φ : (M 2, g) (N n, h) be a proper-biharmonic Riemannian immersion with mean curvature vector field H, M 2 oriented. Let z be a complex coordinate on M 2 then the function B( z, z), H is holomorphic if and only if the norm of H is constant.
Biharmonic stress-energy tensor Pseudo-umbilical points Let φ : (M 2, g) (N n, h) be a CMC proper-biharmonic Riemannian immersion, M 2 oriented. If M 2 is not pseudo-umbilical then its pseudo-umbilical points are isolated. Let φ : (M 2, g) (N n, h) be a CMC proper-biharmonic Riemannian immersion. If M 2 is a topological sphere S 2 then M is pseudo-umbilical. Let φ : (M 2, g) (N n, h) be a CMC proper-biharmonic Riemannian immersion. Assume M is compact, oriented and has no pseudo-umbilical point, then M is topologically a torus.
CMC biharmonic into S n Constraint on H Let φ : M m S n be a CMC proper-biharmonic immersion. Then H (0, 1] H = 1 if and only if φ induces a minimal immersion of M m into a small hypersphere S n 1 ( 1 ) S n. 2
CMC biharmonic into S n Type decomposition (B.Y. Chen) A Riemannian immersion ψ : (M m, g) R n+1 is called of finite type if it can be expressed as a finite sum of R n+1 -valued eigenmaps of the Laplacian of (M, g), i.e. ψ = ψ 0 + ψ t1 + + ψ tk, (1) where ψ 0 R n+1 is a constant vector and ψ ti : M R n+1 are non-constant maps satisfying ψ ti = λ ti ψ ti, for i = 1,..., k. If, in particular, all eigenvalues λ ti are mutually distinct, the submanifold is said to be of k-type and (1) is called the spectral decomposition of ψ.
CMC biharmonic into S n Types for biharmonic Let φ : (M m, g) S n be a proper-biharmonic immersion. Denote by ψ = i φ : M R n+1 the immersion of M in R n+1, where i : S n R n+1 is the canonical inclusion map. Then The map ψ is of 1-type if and only if H = 1. In this case, ψ = ψ 0 + ψ t1, with ψ t1 = 2mψ t1, ψ 0 is a constant vector. Moreover, ψ 0, ψ t1 = 0 at any point, ψ 0 = ψ t1 = 1 and ( ) 2 φ t1 : M S n 1 2 1 is a minimal immersion.
CMC biharmonic into S n Types for biharmonic The map ψ is of 2-type if and only if H is constant, H (0, 1). In this case ψ = ψ t1 + ψ t2, with ψ t1 = m(1 H )ψ t1, ψ t2 = m(1 + H )ψ t2 and ψ t1 = 1 2 ψ + 1 2 H H ψ t2 = 1 2 ψ 1 2 H H. Moreover, ψ t1, ψ t2 = 0, ψ t1 = ψ t2 = 1 2 and ( ) φ ti : (M, g) S n 2 1, i = 1, 2, are harmonic maps with constant density energy.
CMC biharmonic into S n Type decomposition uniqueness Let φ 1, φ 2 : (M m, g) S n be two CMC proper-biharmonic immersions. If φ 1 and φ 2 agree on an open subset of M, then they agree everywhere.
CMC biharmonic into S n Positive curvature Diagonal sums of Boruvka spheres. A Riemannian immersion φ is a CMC proper-biharmonic map from a surface with positive Gaussian curvature in S n if and only if it is (the restriction to an open subset of) the diagonal sum φ = (αφ 1, βφ 2 ), where φ 1 : S 2 (r) S 2n 1 (r 1 ), φ 2 : S 2 (r) S 2n 2 (r 2 ) are Boruvka minimal immersions
CMC biharmonic into S n Positive curvature Parameters α 2 = q 1 and β 2 = q 2 q 1 + q 2 q 1 + q 2 q1 + q 2 q1 + q 2 r 1 = and r 2 = 2q 1 2q 2 r = 1 2 q1 + q 2, with q 1 = n 1 (n 1 + 1) and q 2 = n 2 (n 2 + 1), and n 1 n 2. Moreover, H 2 = (q 1 q 2 ) 2 (q 1 +q 2 ) 2 and φ is pseudo-umbilical.
CMC biharmonic into S n Negative curvature There can be no CMC proper-biharmonic surface in a sphere with K M < 0 (Bryant). Alternative proof? Bochner formula?
CMC biharmonic into S n Flat curvature Local description : D small disk about the origin in R 2. φ : D S n CMC proper-biharmonic immersion with H (0, 1). Then n is odd, n 5. φ extends uniquely to a CMC proper-biharmonic immersion of R 2 into S n. ψ = i φ : R 2 R n+1
CMC biharmonic into S n Flat curvature ψ(z) = 1 2 m k=1 m + 1 2 j=1 ( λ1 Rk e 2 (µ k z µ z) λ1 ) k Z k + e 2 ( µ k z+ µ k z) Zk ( λ2 R j e 2 (η j z η z) λ2 j W j + e 2 ( η j z+ η z) j W j ), (*)
CMC biharmonic into S n Flat curvature where ) 1 Z k = 1 2 (E 2k 1 ie 2k, k = 1,..., m, ) 2 W j = 1 2 (E 2(m+j) 1 ie 2(m+j), j = 1,..., m, 3 {E 1,..., E 2m+2m } is an orthonormal basis of R n+1, n = 2m + 2m 1, 4 λ 1 = 2(1 H ), λ 2 = 2(1 + H ), H constant, H (0, 1) 5 k R k = 1, j R j = 1, R k > 0, R j > 0, 6 (1 H ) k µ2 k R k + (1 + H ) j η2 j R j = 0. 7 {±µ k } m k=1 are 2m distinct complex numbers of norm 1, 8 {±η j } m j=1 are 2m distinct complex numbers of norm 1.
CMC biharmonic into S n Flat curvature (1 h) k µ 2 k R k + (1 + h) j η 2 j R j = 0 h (0, 1) k R k = j R j = 1, R k, R j > 0 {±µ k } and {±η j } distinct (sets of) unit complex numbers.
Flat CMC biharmonic into S 5 Structure theorem Take h (0, 1) Then one-parameter family of CMC proper-biharmonic surfaces φ h,ρ = φ ρ : R 2 S 5 with mean curvature h, ρ [0, 1 2 arccos h 1 1+h ].
CMC biharmonic into S n Flat curvature ψ ρ = i φ ρ : R 2 R 6 can be written as ψ ρ (z) = 1 λ1 (e 2 (z z) λ1 ) Z 1 + e 2 ( z+ z) Z1 2 + 1 2 2 j=1 ( R j e λ2 2 (η j z η z) j W j + R j e λ2 2 ( η j z+ η j z) Wj ), where ) 1 Z 1 = 1 2 (E 1 ie 2, ) 2 W j = 1 2 (E 2(1+j) 1 ie 2(1+j), j = 1, 2, 3 {E 1,..., E 6 } is an orthonormal basis of R 6, 4 λ 1 = 2(1 h), λ 2 = 2(1 + h),
CMC biharmonic into S n Flat curvature R 1, R 2, η 1 = e iρ and η 2 = e i ρ are given by R 1 = ( ) 2 1 h 1 h+1 ( ), 2 1 + h 1 h+1 cos 2ρ if ρ (0, 1 2 R 2 = 1 ( ) 2 1 h 1 h+1 ( ), 2 1 + h 1 h+1 cos 2ρ ρ = arctan ( 1 h tan ρ ), h 1 arccos 1+h ], and
CMC biharmonic into S n Flat curvature if ρ = 0. R 1 = h 1 + h, R 2 = 1 1 + h, ρ = π 2,
CMC biharmonic into S n Flat curvature Conversely, assume that φ : R 2 S 5 is a CMC proper-biharmonic surface with mean curvature h (0, 1). Then, up to isometries of R 2 and R 6, ψ = i φ : R 2 R 6 is one of the above maps.
Flat CMC biharmonic into S 5 Existence on R 2 Let h (0, 1), then there exist proper-biharmonic immersions of R 2 into S 5 with CMC equal to h. Infinitely in fact, one for each ρ.
Flat CMC biharmonic into S 5 Existence on cylinders Let h (0, 1), then there exist proper-biharmonic cylinders in S 5 with CMC equal to h. Periodicity condition : F(ρ) = ( sin ρ K 1 sin ρ λ 2 λ 1 K 0 cos ρ ) + λ 2 λ 1 K 0 cos ρ Z
Flat CMC biharmonic into S 5 Existence on cylinders lim ρ 0 ( F(ρ) = sgn lim F(ρ) = ρ π 2 λ 2 λ 1 K 0. K 1 λ 2 λ 1 K 0 ) Choice of K 0, K 1 Z, so infinitely many solutions.
Flat CMC biharmonic into S 5 Existence on tori φ h,ρ : R 2 S 5, CMC proper-biharmonic immersion ρ (0, π 2 ), quotients to a torus if and only if h = 1 (a b) 2 1 + (a b) 2 + 2(a + b), where a = p 2 /q 2 and b = r 2 /t 2 with p, q, r, t N, (0 (b a) 2 < 1).
Flat CMC biharmonic into S 5 Existence on tori Λ ψh,ρ = {mv 2 + nv 1 : m, n Z s.t. m q p n qr pt Z}, where v 1 = ( π ) (a b) 2 + a + b, 0, a and v 2 = ( π ) b/a(1 (a b)) (a b) 2 + a + b, π (a b) 2 + 2(a + b) + 1 (a b) 2. + a + b
Flat CMC biharmonic into S 5 Existence on tori F (a, b) = is not injective on Q 2 1 (a b) 2 1 + (a b) 2 + 2(a + b) F(0, 1/9) = F(1/16, 1/16) = 4/5 and F(0, 1/(24) 2 ) = F(1/36, 1/36) = 144/145 Non-isometric proper-biharmonic tori with the same mean curvature.
Flat CMC biharmonic into S 5 Example Take R 1 = 1, R 1 = R 2 = 1 2, µ 1 = 1 and h (0, 1). Then η 2 1 + η2 2 = 2(h 1) h + 1. As s = 1 2, η 1 = h 1 h+1 + i h+1 and η 2 = h 1 h+1 i h+1 then {±η 1, ±η 2 } has 4 distinct elements. Then ψ = 1 2 (e i 2(1 h)y, 0, 0) + 1 2 (0, ei 2(x+ hy), 0) + 1 2 (0, 0, ei 2( x+ hy) ).
Flat CMC biharmonic into S 5 Example Periodicity condition : i λ 1, v i λ 2 η 1, v i λ 2 η 2, v 0 (mod 2π) Lattice Λ ψ = { h nv 1 + mv 2 : m, n Z s.t. 2m 1 h }, Z v 1 = 2π(1, 0), and v 2 = ( 2π h 1 h, 1 1 h ) rank Λ ψ = 2 if and only if h = 1 4b+1, where b = r 2 t 2, r, t N..
Flat CMC biharmonic into S 5 Example If b = 1 4, then h = 1 2 and Λ ψ = { 2π(n, 2m) : m, n Z }. This is the CMC proper-biharmonic immersion originally found by Sasahara.
Flat CMC biharmonic into S n Existence on R 2 Any (odd) dimension of S n, n 5. Let h (0, 1). Then CMC proper-biharmonic Riemannian immersion φ : R 2 S 2n+1, n 7, n odd, with mean curvature h.
Flat CMC biharmonic into S n Existence on R 2 Solution on S 7 from solution on S 5 if then sη 2 1 + (1 s)η2 2 = 1 h 1+h. hs(iη 1 ) 2 + h(1 s)(iη 2 ) 2 + (1 h)i 2 = 1 h 1+h
Flat CMC biharmonic into S n Existence on R 2 General case if solution on S n n j=1 R j η2 j = 1 h 1+h, n = 2m + 1.
Flat CMC biharmonic into S n Existence on R 2 then solution on S n+4 from with sη 2 m +1 + (1 s)η2 m +2 = 1 h 1+h, because η m +1 ±η j, η m +2 ±η j, j = 1,... n. (1/2) R j η2 j + s 2 η2 m +1 + 1 s 2 η2 m +2 = 1 h 1+h, m j=1 Only case of R 2, no periodicity result (cylinder or torus).