Gauss Map Harmonicity and Mean Curvature of a Hypersurface in a Homogeneous Manifold

Paciﬁc Journal of Mathematics

GAUSS MAP HARMONICITY AND MEAN CURVATURE OF A HYPERSURFACE IN A HOMOGENEOUS MANIFOLD

FIDELIS BITTENCOURT AND JAIME RIPOLL

Volume 224 No. 1 March 2006

PACIFIC JOURNAL OF MATHEMATICS Vol. 224, No. 1, 2006

GAUSS MAP HARMONICITY AND MEAN CURVATURE OF A HYPERSURFACE IN A HOMOGENEOUS MANIFOLD

FIDELIS BITTENCOURT AND JAIME RIPOLL

We deﬁne a Gauss map of an orientable hypersurface in a homogeneous manifold with an invariant Riemannian metric. Our main objective is to extend to this setting some results on the Gauss map of a constant mean curvature hypersurface of an Euclidean space, namely the Ruh–Vilm theorem relating the harmonicity of the Gauss map and the constancy of the mean curvature, and the Hoffman–Osserman–Schoen theorem characteriz- ing the plane and the circular cylinder as the only complete constant mean curvature surfaces whose Gauss image is contained in a closed hemisphere of the sphere.

1. Introduction

In this article we deﬁne a Gauss map of an orientable hypersurface in a homogeneous manifold with an invariant Riemannian metric. Our main objective is to extend to this setting some results on the Gauss map of a constant mean curvature hypersurface of an Euclidean space. We focus our investigation on the extension of two well known theorems: Theorem (Ruh–Vilm). A hypersurface in ޒn has constant mean curvature if and only if the Gauss map of the hypersurface is harmonic. Theorem (Hoffman–Osserman–Schoen). If the Gauss map of a complete constant mean curvature surface in ޒ3 is contained in a closed hemisphere of the sphere, the surface is a plane or a circular cylinder. Extensions of these results were already given in [Esp´ırito-Santo et al. 2003] in the case that the ambient space is a Lie group with a bi-invariant metric; the main theorems in the present work extend those of that reference to the case of homogeneous spaces. We also obtain certain other results. Section 2 introduces notations and basic facts about homogeneous manifolds which will be used throughout the paper. We introduce some of these notations

MSC2000: 53A10. Keywords: Gauss map, harmonic map, mean curvature.

45 46 FIDELISBITTENCOURTANDJAIMERIPOLL now. Let އ be a Lie group with a fixed bi-invariant Riemannian metric, ވ a compact Lie subgroup of އ. Set k = dim ވ and n +k +1 = dim އ. In the quotient އ/ވ of left residue classes we consider a homogeneous Riemannian metric in such a way that the projection π : އ → އ/ވ becomes a Riemannian submersion. Denote by Ᏻ the Lie algebra of އ, by Ᏼ the Lie algebra of ވ, and let ޓn+k be the unit sphere centered at the origin of Ᏻ. Let M be an orientable hypersurface in އ/ވ oriented by a unit normal vector field η. In Section 3 we define, prove some general properties of, and give examples of the Gauss map N : M → ޓn+k of M; this map is defined by taking the horizontal lifting η˜ of η to އ followed by the right translation of η˜ to the identity of އ, that is, the right translation of η˜ to Ᏻ. In the Euclidean space, taking އ = ޒn and ވ = {e}, N coincides with the usual Gauss map of M. In Section 4 we study the Gauss map of constant mean curvature hypersurfaces of އ/ވ. Our main result asserts that M has constant mean curvature if and only if N is harmonic (Corollary 4.4). This fact has many implications and, to state some of them, we first introduce a definition. A codimension one vector subspace of Ᏻ divides Ᏻ in two connected components; the closure of any of these components is called a half-space of Ᏻ and its n+k n+k l intersection with ޓ is a half-sphere, denoted by ޓ+ . Given l ∈ ގ, a (1/2 )- sphere ޓn+k,l of ޓn+k is the intersection of l linearly independent half spheres n+k n+k n+k (ޓ1 )+,... , (ޓl )+ of ޓ (by “linearly independent” we mean that the normal vectors to the bounding half-spaces are linearly independent). We prove: Let M be a compact constant mean curvature hypersurface of އ/ވ and N : M → ޓn+k the Gauss map of M. Assume that either M is not totally geodesic or the Ricci curvature of އ is positive. Given l ∈ ގ, let

l n+k,l \ n+k ޓ = (ޓi )+ i=1 be a (1/2l )-sphere of ޓn+k. Then there is equivalence between: (1) N(M) ⊂ ޓn+k,l . T l n+k (2) N(M) ⊂ i=1 ∂(ޓi )+. T l n+k ⊥ (3) ᏷ := i=1 ∂(ޓi )+ is a Lie subalgebra of Ᏻ and M is invariant under the Lie subgroup ދ of އ whose Lie algebra is ᏷. It follows under the same general hypotheses that if l ≥ n, then M is an extrinsically homogeneous manifold (that is, there is a Lie subgroup of isometries of އ/ވ acting transitively on M). If M is assumed only to be complete, a similar result holds when dim(އ/ވ) = 3. In this case, more explicit applications are given in Corollaries 4.13 and 4.14. In GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 47 particular, we obtain the following result related to a conjecture of B. Lawson (which asserts that an embedded minimal torus in ޓ3 has to be a Clifford torus, that is, the Riemannian product of two circles): A complete minimal surface M in ޓ3 = SO(4)/ SO(3) is a Clifford torus if and only if the Gauss image N(M) of M lies in a certain hemisphere of the unit sphere through the origin in so(4) (Corollary 4.14). The harmonicity of the Gauss map of a constant mean curvature surface can also be used to establish a stability criterion: If D is a domain of a constant mean curvature surface M of އ/ވ (dim(އ/ވ) = 3) such that N(D) is contained in a half space of Ᏻ, then D is stable (Theorem 4.10).

2. Preliminaries

Let އ be an (n+k+1)-dimensional connected Lie group, for n ≥ 2, k ≥ 0, with a bi-invariant Riemannian metric h , i. Denote by ∇e the Riemannian connection on އ determined by h , i. Let Ᏻ be the Lie algebra of އ. Given x ∈ އ, let L x and Rx denote the left and right translations on އ, that is, L x (y) = xy and Rx (y) = yx. As usual, exp : Ᏻ → އ denotes the Lie exponential map. Given g ∈ އ and X ∈ Ᏻ, we have the maps Adg, adX : Ᏻ → Ᏻ deﬁned by

−1 Adg = d(Rg ◦ Lg)e and adX (Y ) = [X, Y ]. We note that d adX (Y ) = Adexp t X (Y ) , dt t=0 and that Adg is a Lie group isomorphism and an isometry. We recall that if X, Y , Z are both left (or both right) invariant vector ﬁelds of އ, then ∇e and the curvature tensor R on އ are given by ∇ = 1 [ ] = 1 [[ ] ] eX Y 2 X, Y and R(X, Y )Z 4 X, Y , Z . We have h[X, Y ], Zi = h[Z, Y ], Xi. The Ricci tensor of އ is given by

n+k+1 X Ric(u, v) = trace X 7→ R(u, X)v = R(u, E j )v, E j , i=1 where {E j } is an orthonormal basis of tangent vectors of އ. The Ricci curvature of އ on the u-direction is Ric(u) = Ric(u, u). Throughout this article, Ric will always mean the Ricci curvature of އ. Let ވ be a k-dimensional closed Lie subgroup of އ (recall that k ≥ 0), with Lie algebra Ᏼ, and let އ/ވ be the smooth (n+1)-dimensional manifold of left residue classes of ވ: އ/ވ = {xވ : x ∈ އ}. 48 FIDELISBITTENCOURTANDJAIMERIPOLL

We consider on އ/ވ a Riemannian metric, also denoted by h , i, induced by the projection π : އ → އ/ވ, π(x) = xވ, as follows. Given z ∈ އ/ވ, take x ∈ π −1(z) and deﬁne an inner product h , i in Tz(އ/ވ) in such a way that

−1 ⊥ lx := dπx −1 ⊥ : Tx (π (z)) → Tz(އ/ ވ) Tx (π (z)) ⊥ is a linear isometry, where denotes the orthogonal complement in Tx (އ). It is easy to see that h , i is well defined in Tz(އ/ވ) (that is, it does not depend on the choice of x ∈ π −1(z)) and defines a Riemannian homogeneous (bi-invariant) metric on އ/ވ. Denote by ∇ the Riemannian connection associated to h , i. Denote by ᐄ(އ/ވ) and by ᐄ(އ) the spaces of smooth vector fields on އ/ވ and ⊥ އ. A vector field eX in އ is called horizontal if eX(x) ∈ Tx (xވ) for x ∈ އ, and is called vertical if eX(x) ∈ Tx (xވ) for x ∈ އ. Given eX ∈ ᐄ(އ), let eX v be the vertical vector field on އ given as the orthogonal h projection of eX(x) on Tx (xވ), and let eX be the horizontal vector field on އ given ⊥ as the orthogonal projection of eX(x) on Tx (xވ) , x ∈ އ. It follows forthwith from the definition of h , i that the projection π : އ → އ/ވ is a Riemannian submersion. It is not difficult to see that given X, Y ∈ ᐄ(އ/ވ), we have ∇ = −1∇ + 1 [ ]v eeX Ye(x) lx X Y (π(x)) 2 eX, Ye (π(x)), −1 −1 where eX, Ye∈ ᐄ(އ) are defined by eX(x) =lx (X (π(x))) and Ye(x) =lx (Y (π(x))) for x ∈ އ, and [ , ] is the Lie bracket in އ. Any element of އ acts as an isometry on އ/ވ via g(xވ) = gxވ for x ∈ އ, g ∈ އ, or

(2–1) g(π(x)) = π(Rx (g)).

We may then consider އ as a Lie subgroup of the isometry group Iso(އ/ވ) of އ/ވ. It follows that any vector w ∈ Ᏻ acts on އ/ވ as a Killing vector ﬁeld, which we denote by ζ(w): d ζ(w)(z) = (exp tw)(z) for z ∈ އ/ވ. dt t=0 Given z ∈ އ/ވ and x ∈ π −1(z), we have, using (2–1) d d ζ(w)(z) = (exp tw)(z) = (exp tw)(π(x)) dt t=0 dt t=0 d = π(Rx (exp tw)) = dπx (d(Rx )e(w)), dt t=0 GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 49 that is

ζ(w)(z) = dπx (d(Rx )e(w)). Note that ζ : Ᏻ → ᐄ(އ/ވ) is a Lie algebra monomorphism. From now on, unless otherwise stated, we keep the assumptions and notations of this section concerning the Lie groups އ and ވ.

3. The Gauss map of a hypersurface in އ/ވ

We begin by introducing a “translation” 0 in އ/ވ. Given z ∈ އ/ވ, deﬁne a map

0z : Tz(އ/ވ) → Ᏻ by choosing x ∈ π −1(z) and setting

−1 −1 (3–1) 0z(u) := d(Rx )x (lx (u)).

Proposition 3.1. The map 0z is well-deﬁned, that is, (3–1) does not depend on −1 x ∈ π (z). Moreover, 0z : Tz(އ/ވ) → Ᏻ is linear and preserves the metric. −1 Proof. Consider x, y ∈ π (z) and u ∈ Tz(އ/ވ). Let h ∈ ވ be such that x = yh. −1 −1 We obviously have d(Rh)y(Tyπ (z)) = Tx π (z) and, since Rh is an isometry, −1 ⊥ −1 ⊥ −1 −1 ⊥ d(Rh)y(Tyπ (z) )= Tx π (z) . It follows that dg(Rh)y(lx (u))∈(Tx π (z)) . Observing that

−1 −1 −1 −1 lx (d(Rh)y(ly (u))) = d(π ◦ Rh)y(lx (u)) = dπy(ly (u)) = u = dπx (lx (u)),

−1 −1 so that dg(Rh)y(ly (u)) = lx (u), we have

−1 −1 −1 dg(Rx )x (lx (u)) = d(Rx−1 )x (dg(Rh)y(ly (u))) −1 −1 = dg(Rx−1 ◦ Rh)y(ly (u)) = dg(Ry−1 )y(ly (u)), proving that 0z is well defined. Moreover, 0z is obviously linear and, since lx and −1 Rx are isometries, 0z preserves the metric. The generalized Gauss map. Let M be an immersed, orientable hypersurface of އ/ވ. Let η be a unit normal vector field to M in އ/ވ. We define the Gauss map N of M as the map

n n+k N : M → ޓ ⊂ Ᏻ given by N(z) = 0z(η(z)).

A family of examples. Let ދ be a compact Lie subgroup of އ and ᏷ its Lie algebra. Consider the action of ދ on އ/ވ by isometries, k(xވ) = kxވ. We assume that this is a cohomogeneity-one action, that is, the orbits of ދ of highest dimension have codimension 1 in އ/ވ. 50 FIDELISBITTENCOURTANDJAIMERIPOLL

Let g ∈ އ be such that

Mg := ދ(ވ) := {kgވ | k ∈ ދ} is an orbit of ދ of codimension 1. Then Mg is a compact orientable embedded hypersurface of އ/ވ.

Lemma 3.2. The vector subspace Adkg(Ᏼ) + ᏷ has codimension 1 in Ᏻ, for any k ∈ ދ. −1 Proof. Set Meg := π (Mg). Then Meg is an embedded hypersurface of އ (possibly not connected if ވ is not connected). Note that

Meg = {kgh | k ∈ ދ, h ∈ ވ} and it follows that Tx (Meg) = d(L x )e(Ᏼ) + d(Rx )e(᏷) for any x ∈ Meg, so that −1 (3–2) Adx (Ᏼ) + ᏷ = d(Rx )x (Tx (Meg)) is a codimension-one vector subspace of Ᏻ. Since a generic element x of Meg is of the form x = kgh, we have Adx (Ᏼ) = (Adk ◦ Adg ◦ Adh)(Ᏼ) = Adkg(Ᏼ), so that −1 Adkg(Ᏼ) + ᏷ = d(Rx )x (Tx (Meg)) is a codimension-1 vector subspace of Ᏻ. n+k Proposition 3.3. Set N0 = N(gވ), where N : Mg → ޓ is the Gauss map of Mg. Then N(kgވ) = Adk(N0) for all k ∈ ދ.

Proof. Let η be a unit normal vector field to Mg defining N. Given z = kgވ ∈ Mg, −1 −1 choose x ∈ Meg = π (M), x = kgh. Then note that lx (η(z)) is orthogonal to Tx (Meg) so that, by definition of N, −1 −1 N(kgވ) = N(z) = d(Rx )x (lx (η(z))) is orthogonal to −1 d(Rx )x (Tx (Meg)) = Adkg(Ᏼ) + ᏷.

In particular, N0 = N(gވ) is orthogonal to Adg(Ᏼ)+᏷. Then Adk(N0) is orthogonal to Adk(Adg(Ᏼ) + ᏷) = Adkg(Ᏼ) + Adk(᏷) = Adkg(Ᏼ) + ᏷ and N(kgވ) and Adk(N0) are both orthogonal to Adkg(Ᏼ) + ᏷. Since

dim(Adkg(Ᏼ) + ᏷) = dim Ᏻ − 1, we have N(z) = N(kgވ) = ± Adk(N0).

By continuity, N(kgވ) = Adk(N0), since N(gވ) = N0 = Ade(N0) at k = e. GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 51

Remark. As we see from the proof, the vector N0 quoted on the statement of Proposition 3.3 can be easily determined, since it is orthogonal to the codimension- 1 subspace Adg(Ᏼ) + ᏷. The next result gives a characterization of the homogeneous hypersurfaces of އ/ވ in terms of the Gauss map. This proposition is important for the characterization of the constant mean curvature hypersurfaces given in Section 4. Proposition 3.4. Let M be an orientable hypersurface of އ/ވ and let N : M → ޓn+k be the Gauss map of M. Then

᏷ := N(M)⊥ = {w ∈ Ᏻ | hw, vi = 0 for all v ∈ N(M)} is a Lie subalgebra of Ᏻ and M is invariant under the Lie subgroup of އ whose Lie algebra is ᏷. In particular, if dim ᏷ = n = dim(އ/ވ) − 1, then M is an extrinsically homogeneous submanifold of އ/ވ. Conversely, if M is invariant under a Lie subgroup ދ of އ, then ᏷ ⊂ N(M)⊥, where ᏷ is the Lie algebra of ދ. ⊥ Lemma 3.5. If w ∈ N(z) then ζ(w)(z) ∈ Tz M. Proof. If w ∈ N(z)⊥ then

−1 0 = hw, N(z)i = hd(Rx )ew, lx (η)i −1 = hdπx (d(Rx )e)w, dπx (lx (η))i = hζ(w)(z), η(x)i so that ζ(w)(z) ∈ Tz M, proving the lemma. Proof of Proposition 3.4. Let v, w ∈ N(M)⊥ be given. It follows by Lemma 3.5 that ζ(v) and ζ(w) are vector ﬁelds on M so that [ζ(v), ζ(w)] is a vector ﬁeld on M. Since ζ is a Lie monomorphism, [ζ(v), ζ(w)] = ζ([v, w]). Then, given z ∈ M,

0 = ζ([v, w]), η = 0z(ζ([v, w])), 0z(η) −1 −1 = d(Rx )x lx (dπx (d(Rx )e([v, w]))) , N(z) −1 h −1 −1 = d(Rx )x (d(Rx )e([v, w]) ), d(Rx )x (lx (η)) h −1 = (d(Rx )e([v, w])) , lx (η) −1 −1 −1 = d(Rx )e([v, w]), lx (η) = [v, w], d(Rx )xlx (η) = [v, w], N(z) . Since z is arbitrary, we get [v, w] ∈ N(M)⊥, proving that N(M)⊥ is a Lie subalgebra of Ᏻ. Conversely, if M is ދ-invariant then, given w ∈ ᏷, it follows that ζ(w) is a vector ﬁeld on M, so that hζ(w), ηi = 0 on M. We then have, as above,

0 = hζ(w), ηi = 0z(ζ(w)), 0z(η) = hw, N(z)i ⊥ proving that ᏷ ⊂ N(M) . 52 FIDELISBITTENCOURTANDJAIMERIPOLL

4. The harmonicity of the Gauss map and the mean curvature

It is well known that a hypersurface of ޒn+1 has constant mean curvature if and only if its Gauss map is harmonic. Our main purpose in this section is to extend this result to hypersurfaces of a homogeneous manifold. We recall that in ޒn+1, this equivalence between constant mean curvature hypersurfaces and the harmonicity of the Gauss map is an immediate corollary of the well known formula

(4–1) 1N = − grad H − kBk2 N.

This formula was extended to hypersurfaces in a Lie group in [Esp´ırito-Santo et al. 2003] and, more generally, using Killing vector ﬁelds, to Killing parallelizable Riemannian manifolds; see [Fornari and Ripoll 2004]. Our objective now is to extend (4–1) to homogeneous manifolds using the Gauss map deﬁned here. Let އ, ވ and އ/ވ be as before and let M be an immersed orientable hypersurface in އ/ވ. Consider the Gauss map N : M → ޓn+k ⊂ Ᏻ. Let

{e1, e2,..., en+k+1} ⊂ Ᏻ be an orthonormal basis of Ᏻ, and deﬁne functions Ni : M → ޒ by setting N(z) = Pn+k+1 i=1 Ni (z)ei for z ∈ M. The Laplacian of N is deﬁned by

n+k+1 X 1N := 1Ni ei , i=1 where 1Ni is the usual Laplacian of Ni with respect to the metric on M induced by the immersion of M in އ/ވ. Then N is harmonic if and only if

> 1ޓn+k N = (1N) = 0, where > denotes the orthogonal projection of Ᏻ on T ޓn+k [Eells and Sampson 1964]. To get a proper extension of (4–1) to އ/ވ, we need to introduce a new algebraic- geometric invariant of M on އ/ވ, a kind of Lie invariant second fundamental form (to appear later in Lemma 4.2). Given z ∈ M, deﬁne

i 1 v Bz(u) = √ adN(z)(0z(u)) for u ∈ Tz M. 2

− Lemma 4.1. Let M be a hypersurface in އ/ވ and set Me =π 1(M). Let f : M →ޒ ˜ be a differentiable function and set f = f ◦ π : Me → ޒ. Then ˜ = ◦ 1Me f (1M f ) π. GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 53

Proof. Choose x0 ∈ Me and set z0 =π(x0). Let {Ei }, i =1,..., n, be an orthonormal −1 frame on a neighborhood V of z0 in M. Set W = π (V ) and deﬁne Eei on W, by −1 Eei (x) = lx (Ei (π(x)).

Let ei , i = n+1,..., n+k, be an orthonormal basis of Ᏼ and let Eei be the left- invariant vector fields on އ such that Eei (e) = ei . We have ∇ E = 1 [E , E ] = 0 eEei ei 2 ei ei = + + { }n+k for i n 1,..., n k, and Eei i=1 is an orthonormal basis of Tx Me for x in W. We have n+k X 1 f˜(x ) = ∇ grad f˜, E . Me 0 eEei ei i=1 Since ˜ Eei ( f )(x) = Eei ( f ◦ π)(x) = dg( f ◦ π)x (Eei (x)) = d fπ(x)(dπx (Eei (x))) ˜ and, since dπx (Eei (x)) = 0 for i = n+1,..., n+k, we have Eei ( f )(x) = 0 for i = n+1,..., n+k. Moreover, dπx (Eei (x))) = Ei (π(x)) for i = 1,..., n so that ˜ Eei ( f )(x) = Ei ( f )(π(x)), i = 1,..., n, and we have n+k n ˜ X ˜ X grad f (x) = Eei ( f )(x)Eei (x) = Ei ( f )(π(x))Eei (x) i=1 i=1 ˜ −1 so that grad f (x) = lx (grad f (π(x))). Moreover, for i = n+1,..., n+k, ∇ grad f˜, E = E grad f˜, E − grad f˜, ∇ E = 0, eEei ei ei ei eEei ei so that n X 1 f˜ (x ) = ∇ grad f˜, E Me 0 eEei ei i=1 n n 1 = X −1 ∇ + X [ ˜]v l ( Ei grad f ), Eei Eei , grad f , Eei x 2 i=1 i=1 n = X h∇ i = Ei grad f, Ei 1M f (π(x0)), i=1 proving the lemma. − Lemma 4.2. Let M be a hypersurface in އ/ވ and set Me := π 1(M). Denote by kBk (keBk) the norm of the second fundamental form B (eB) of M (Me) in އ/ވ (އ), and by H (He) the mean curvature of M (Me) with respect to a normal vector field −1 η (η˜ = lx η). Suppose dim އ = n + k + 1 and dim ވ = k. 54 FIDELISBITTENCOURTANDJAIMERIPOLL n (i) H(x) = H (z) for all z ∈ M and x ∈ π −1(z). e n+k n (ii) grad H(x) = l−1(grad H(z)) for all z ∈ M and x ∈ π −1(z). e n+k x p (iii) keBk = kBk2 + kBi k2. −1 Proof. Choose z ∈ M and x ∈ π (z). Let Ei (z), i = 1,..., n, be an ortho- −1 normal basis of Tz M and define Eei (x) = lx (Ei (z)) for i = 1,..., n. Let Eei , i = n+2,..., n+k+1, be left-invariant vector fields on އ such that Eei (x) ∈ Tx Me and {Eei }, i = 1,..., n+k+1 is an orthonormal basis of Tx Me. Note that since Me is ވ-invariant, Eei is a vector field on Me for i =n+2,..., n+k+1. It is also easy to see that η˜ is a unit normal vector field to Me. It follows that, for i, j =n+2,..., n+k+1, h∇ η(˜ x), E (x)i = −h∇ E (x), η(˜ x)i = − 1 hη( ˜ x), [E , E ]i = 0 eEei ej eEei ej 2 ei ej since [Eei , Eej ] is a vector field on Me. Proof of (i). By the definition of the mean curvature, we have:

n+k+1 n+1 1 X 1 X He(x) = ∇e Eei (x), η(˜ x) = ∇e Eei (x), η(˜ x) n + k Eei n + k Eei i=1 i=1 n+ 1 1 = X −1 ∇ + 1 [ ] v ˜ l ( Ei Ei (z)) ( Eei , Eei (x)) , η(x) n + k x 2 i=1 n+1 1 X = l−1(∇ E (z)), l−1(η(x)) n + k x Ei i x i=1 n 1 X n = ∇ E (z), η(z) = H(z). n + k Ei i n + k i=1

Proof of (ii). Choose v˜ ∈ Tx Me. We may then write

−1 v˜ = lx (v) + w, where v ∈ Tz M and w is vertical. Then

−1 −1 −1 −1 −1 lx (grad H (z)), v˜ = lx (grad H(z)), lx (v) + w = lx (grad H(z)), lx (v)

= hgrad H(z), vi = d(H)z(v) = dg(H)z(dπx (v))˜ n+k n+k = d(H ◦ π)x (v)˜ = dg He (v)˜ = d Hex (v).˜ n x n Therefore, n grad H(x) = l−1(grad H (z)). e n+k x GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 55

Proof of (iii). As we have seen, ∇ η(˜ x), E (x) = 0 if i, j ∈ {n+2,..., n+k+1}. eEei ej Moreover,

n+1 n+1 X 2 X 2 ∇ η(˜ x), E (x) = l−1(∇ η(z)) + 1 ([E , η˜](x))v, E (x) eEei ej x Ei 2 ei ej i, j=1 i, j=1 n+1 = X −1 ∇ −1 2 lx ( Ei η(z)), lx (E j (z)) i, j=1 n+1 = X ∇ 2 = k k2 Ei η, E j B . i, j=1

Assume that i ∈ {1,..., n+1} and j ∈ {n+2,..., n+k+1}. We have, since Eej is a vector ﬁeld on Me, ∇ η(˜ x), E (x) = − η(˜ x), ∇ E (x) . eEei ej eEei ej −1 Set Fl = d(L x )x (Eel ), l = 1,..., n + k + 1. We then have ∇ η(˜ x), E (x) = − η(˜ x), ∇ d(L ) (F ) eEei ej ed(L x )e(Fi ) x e j = − 1 ˜ [ ] 2 η(x), d(L x )e(Fi ), d(L x )e(Fj ) (x) = − 1 −1 [ ] 2 N(π(x)), d(Rx )x ( d(L x )e(Fj ), d(L x )e(Fj ) (x)) = − 1 [ ] 2 N(π(x)), Adx (Fi ), Adx (Fj ) (x)) = − 1 [ ] 2 N(π(x), Adx (Fi ) , Adx (Fj ) = − 1 2 adN(π(x)(Adx (Fi )), Adx (Fj ) . Note that

−1 −1 −1 Adx (Fl ) = d(Rx )x d(L x )e(d(L x )x (Eel ) = d(Rx )x (Eel ), so that, if l = i,

Adx (Fl ) = Adx (Fi ) = 0z(Ei ). We then have

∇ η(˜ x), E (x) = − 1 ad (0 (E )), Ad (F ) . eEei ej 2 N(π(x) z i x j } = { −1 Moreover, Adx (Fj ) d(Rx )x (fE j) j=n+2,...,n+k+1, is an orthonormal basis of Ᏼ. In particular, √ 2 i −1 ∇e η(˜ x), Eej (x) = − B (Ei ), d(R )x (fE j) , Eei 2 z x 56 FIDELISBITTENCOURTANDJAIMERIPOLL and thus X 2 ∇ η(˜ x), E (x) = 1 kBi k2. eEei ej 2 z i=1,...,n+1 j=n+2,...,n+k+1 Then n+k+1 X 2 kB(x)k2 = ∇ η(˜ x), E (x) e eEei ej i, j=1 n+1 n+k+1 X 2 X 2 = ∇ η(˜ x), E (x) + ∇ η(˜ x), E (x) eEei ej eEei ej i, j=1 i, j=n+2 X 2 +2 ∇ η(˜ x), E (x) eEei ej i=1,...,n+1 j=n+2,...,n+k+1 2 i 2 = kBk + kB k . Keeping the same meanings for B, Bi , H and grad H as above, we have: Theorem 4.3. Let M be an immersed orientable hypersurface of އ/ވ, and let N : M → ޓn+k be the Gauss map of M. Then

2 i 2 −1 (4–2) 1N(z)=−n0z(grad H(z))− kBk +kB k +(n+k) Ric(lx (η(z))) N(z) for all z ∈ M and x ∈ π −1(z), where η is a unit normal vector ﬁeld to M and Ric(w) the Ricci curvature of އ with respect to w ∈ T އ. −1 Proof. As before, set η˜ = lx (η). As mentioned, η˜ is a unitary normal vector ﬁeld − + to Me = π 1(M). Let Ne : Me → ޓn k be the Gauss map of Me: −1 Ne(x) = d(Rx )x (η(˜ x)). We claim that, if z = π(x), = (4–3) 1M N (z) 1Me Ne (x). −1 To see this, note that N(z) = 0z(η(z)) = d(Rx )x (η(˜ x)) = Ne(x), so writing

n+k+1 n+k+1 X X N (z) = Ni (z)ei and Ne(x) = Nei (x)ei i=1 i=1 in some orthonormal basis {ei }i=1 ..., n+k+1 of Ᏻ, we have

Nei (x) = Ni (z) = (Ni ◦ π)(x), where x ∈ π −1(z). By Lemma 4.1, we have = ◦ = 1Me Nei (x) (1M Ni ) π (x) 1M Ni (z), GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 57 proving (4–3). It follows by Theorem 1 of [Esp´ırito-Santo et al. 2003] that = − + −1 − k k2 + + ˜ 1Me Ne(x) (n k)d(Rx )x (grad He(x)) eB (n k) Ric η(x) Ne(x) so that, applying Lemma 4.2, we obtain n 1N (z) = −(n + k)d(R−1) l−1(grad H (z)) x x n+k x 2 − (keBk + (n + k) Ric(η(˜ x)))N(z) 2 i 2 −1 = −n0z(grad H(z)) − kBk + kB k + (n + k) Ric(lx (η(z))) N(z), proving the theorem. Theorem 4.3 has an application to constant mean curvature immersions: Corollary 4.4. Let M be an orientable hypersurface in އ/ވ and N : M → ޓn+k the Gauss map of M. There is equivalence between: (i) M has constant mean curvature. (ii) The Gauss map N : Mn → ޓn+k is harmonic. (iii) N satisﬁes the equation 2 i 2 −1 (4–4) 1N (z) = − kBk + kB k + (n + k) Ric(lx (η(z))) N(z). We apply the deﬁnition of a (1/2l )-sphere (page 46) to a result that extends Corollary 2 of [Esp´ırito-Santo et al. 2003]: Theorem 4.5. Let M be a compact constant mean curvature hypersurface of އ/ވ and N : M → ޓn+k the Gauss map of M. Assume that M is not totally geodesic, or that kBi k > 0, or that Ric > 0. Given l ∈ ގ, let l n+k,l \ n+k ޓ = (ޓi )+ i=1 be a (1/2l )-sphere of ޓn+k. There is equivalence between: (i) N(M) ⊂ ޓn+k,l . Tl n+k (ii) N(M) ⊂ i=1 ∂(ޓi )+. Tl n+k ⊥ (iii) ᏷ := i=1 ∂(ޓi )+ is a Lie subalgebra of Ᏻ and M is invariant under the Lie subgroup ދ of އ whose Lie algebra is ᏷.

Proof. We ﬁrst prove that (i) implies (ii). Given i, we have N(M) ⊂ Hi , where Hi n+k n+k n+k is the half-space of Ᏻ such that ޓi = Hi ∩ ޓ . Therefore, there is v ∈ ޓ orthogonal to Hi such that hN(z), vi≤0 for all z ∈ M. Using (4–4), and considering that Ric ≥ 0, we obtain 2 i −1 1hN, vi = h1N, vi = −(kBk + kB k + (n + k) Ric(lx (η(z))))hN(z), vi ≥ 0 58 FIDELISBITTENCOURTANDJAIMERIPOLL so that hN, vi is a subharmonic function on the compact manifold M. It follows that hN, vi is constant and then 1hN, vi = 0. Hence, since either kBk > 0, kBi k > 0, −1 or Ric(lx (η(z))) > 0, we have hN, vi = 0, so that N(M) ⊂ ∂ Hi . This proves that (i) implies (ii). The implication (iii) ⇒ (ii) follows from Proposition 3.4, and (ii) ⇒ (i) is obvious. i −1 Remark. If M is admitted to be totally geodesic, kB k=0, and Ric(lx (η(z)))=0, the theorem is false. Planes in ޒ3 give counterexamples to the theorem. Corollary 4.6. Assume that M is not totally geodesic, or |Bi | > 0, or Ric > 0. Let M be a compact constant mean curvature hypersurface of އ/ވ and N : M → ޓn+k n+k l n+k the Gauss map of M. Let ޓl be a (1/2 )-sphere of ޓ . Assume that l ≥ n and n+k N(M) ⊂ ޓl . Then M is an extrinsically homogeneous submanifold of އ/ވ (recall that އ/ވ has dimension n + 1). To investigate if a constant mean hypersurface M is invariant under a speciﬁc Lie subgroup of isometries, we shall make use of the following result: Corollary 4.7. Let M be a compact constant mean curvature hypersurface of އ/ވ and N : M → ޓn+k the Gauss map of M. Assume that M is not totally geodesic, or |Bi | > 0, or Ric > 0. Let ދ be a Lie subgroup of އ of dimension l and ᏷ the Lie algebra of ދ. Let ᏹ be the orthogonal complement of ᏷ in Ᏻ. Let n+k n+k n+k ∩ n+k ⊂ n+k (ޓ1 )+,...,(ޓl )+ be l half-spheres of ޓ such that ᏹ ޓ (ޓi )+ for i = 1,..., l, that is, l n+k \ n+k ᏹ ∩ ޓ = ∂(ޓi )+. i=1 Then M is ދ-invariant if and only if

l \ n+k N(M) ⊂ (ޓi )+. i=1 Corollary 4.8. Assume that އ/ވ is a two-point homogeneous manifold (that is, one such that އ = Iso(އ/ވ)). Let M be a compact constant mean curvature hypersurface of އ/ވ and N : M → ޓn+k the Gauss map of M. Let ᏹ be the n+k n+k orthogonal complement of the Lie algebra Ᏼ of ވ in Ᏻ. Let (ޓ1 )+,...,(ޓl )+ n+k n+k n+k be l half-spheres of ޓ such that ᏹ ∩ ޓ ⊂ (ޓi )+ for i = 1,..., l. Then l \ n+k N(M) ⊂ (ޓi )+ i=1 if and only if M is a geodesic sphere. GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 59

Proof. In a two-point homogeneous space the isotropy subgroup ވ of އ acts transitively on geodesic spheres; now apply Corollary 4.7. Theorem 4.9. Assume that dim އ/ވ = 3. Let M be a complete constant mean curvature surface of އ/ވ and N : M → ޓ3+k the Gauss map of M. Assume that Ric > 0 or kBi k > 0 or M is not totally geodesic. Then N(M) is contained in a hemisphere of the unit sphere in Ᏻ if and only if M is invariant under a one- parameter subgroup of isometries of އ. Proof. If M is invariant under a one-parameter subgroup of isometries of އ/ވ determined by a X ∈ Ᏻ then N(M) is contained in a hemisphere whose boundary hyperplane is orthogonal to X. The other direction follows from Corollary 2 of [Fornari and Ripoll 2004], since given w ∈ N(M)⊥, ζ(w) is a Killing ﬁeld of އ/ވ such that f (z) := hζ(w)(z), η(z)i does not change sign on M; however, for completeness, we give the details of the proof, which is essentially the same as that of Theorem 1 of [Hoffman et al. 1982]. Let (Mb, π) be the universal covering of M. First assume that Mb is the disk and suppose that f ≤ 0. Then, noting that f (z) = hw, N(z)i, we obtain from (4–4) 2 i 2 −1 (4–5) 1f = hw, 1Ni = − kBk + kB k + (n + k) Ric(lx (η(z))) f ≥ 0, so that f is subharmonic. Therefore, if f vanishes at some point of M, it vanishes identically (by the maximum principle), so M is invariant under ζ(w), proving the theorem in this case. We now show that the case f < 0 leads to a contradiction. By the Gauss equation, we have

2 2 (4–6) kBk = 4H − 2(K − Ke), where K is the Gaussian curvature of M and Ke the sectional curvature of އ/ވ on tangent planes of M. From the second equality in (4–5) and from (4–6) we obtain i 2 −1 2 (4–7) 1f − 2K f + kB k + (n + k) Ric(lx (η(z))) + 2Ke + 4H f = 0. However, (4–7) contradicts [Fischer-Colbrie and Schoen 1980, Corollary 3], which states that when K is the Gaussian curvature of a complete conformal metric on the unit disk there can be no negative solution of equation (4–7) if

i 2 −1 2 kB k + (n + k) Ric(lx (η(z))) + 2Ke + 4H ≥ 0 (note that Ke ≥ 0 and Ric ≥ 0). It follows that Mb is the sphere or the plane, proving the assertion stated above. Thus f ◦ π is a subharmonic and bounded, hence constant, function on ޒ2; thus 2 i 2 −1 f is constant. Then 1f = 0 and kBk + kB k + (n + k) Ric(lx (η(z))) f = 0, and the conclusion of the theorem is now immediate. 60 FIDELISBITTENCOURTANDJAIMERIPOLL

We have the following stability criterion, whose proof is the same as that of [Esp´ırito-Santo et al. 2003, Theorem 5]:

Theorem 4.10. Assume that dim(އ/ވ) = 3 and let M be a constant mean curvature surface in އ/ވ. Let D ⊂ M be a domain such that D lies in the interior of M. If N(D), the spherical image of D, is contained in a open hemisphere of the unit sphere in Ᏻ, then D is stable.

In the most common cases އ has constant Ricci curvature. This always occurs when the Killing form of އ is negative definite, as it is shown below. The Killing form of އ is negative definite, for instance, when އ is compact and has finite center.

Lemma 4.11. Suppose that the Killing form of އ is negative deﬁnite. Then any bi-invariant metric h , i satisﬁes

= 1 h i Ric(u, v) 4 u, v , where u and v are tangent vectors to އ. In particular, އ has constant Ricci curva- 1 ture 4 . Proof. the Killing form B of އ on the Lie algebra Ᏻ is given by

B(u, v) = trace(adu ◦ adv) for u, v ∈ Ᏻ, and we have assumed it to be negative deﬁnite. Consider any bi-invariant metric h , i on އ and let {E j } j=1,...,n+k+1 be an orthonormal basis of Ᏻ with respect to h , i. Given x ∈ އ and u, v ∈ Ᏻ, we have

n+k+1 n+k+1 = X = 1 X [[ ] ] (4–8) Ric(u, v) R(u, E j )v, E j 4 u, E j , v , E j j=1 j=1 n+k+1 n+k+1 = − 1 X [ ] [ ] = 1 X [ ] [ ] 4 E j , v , u, E j 4 E j , v , E j , u j=1 j=1 n+k+1 n+k+1 = 1 X = − 1 X 2 4 adE j (v), adE j (u) 4 (adE j ) (v), u j=1 j=1 n+k+1 = − 1 X 2 4 (adE j ) (v), u , j=1 where in the third and sixth equalities we used the skew-symmetry of adX :

hadX (u), vi = −hadX (v), ui. GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 61

Considering the linear symmetric transformation

n+k+1 := X 2 T (adE j ) , j=1 we obtain, for all u, v ∈ Ᏻ, n+k+1 n+k+1 h i = X 2 = − X T (u), v (adE j ) (u), v adE j (u), adE j (v) j=1 j=1 n+k+1 n+k+1 X X = − adu(E j ), adv(E j ) = E j , adu ◦ adv(E j ) j=1 j=1

= trace( adu ◦ adv) = B(u, v).

Now take a basis {ui }i=1,...,n+k+1 of Ᏻ orthonormal with respect to −B. Since −B is positive deﬁnite, we obtain hT (ui ), u j i = −δi j , so T = −I , where I denotes the identity in Ᏻ. It follows from (4–8) that = 1 h i Ric(u, v) 4 u, v , concluding the proof of the lemma. We obtain from Theorem 4.3 and this last lemma: Corollary 4.12. Under the hypothesis of Theorem 4.3, assume moreover that the Killing form of އ is negative deﬁnite. Then

2 i 2 1 1N(z) = −n0z(grad H(z)) − (kBk + kB k + (n + k))N(z). 4 We now apply the previous results to the special case of ޓ3 = SO(4)/ SO(3). Recall that the Lie algebra so(4) of SO(4) is given by the 4×4 matrices Z satisfying Z t = −Z. We may consider in so(4) the Killing form h i = − 1 t ∈ X, Y 4 trace(XY ), X, Y so(4), although the result, of course, is independent of the bi-invariant Riemannian metric considered. α 3 Given α ∈ ޒ, let {φt }t∈ޒ be the one-parameter subgroup of isometries of ޓ deﬁned by  cos t sin t 0 0  α −sin t cos t 0 0  φ =   . t  0 0 cos αt sin αt  0 0 −sin αt cos αt 62 FIDELISBITTENCOURTANDJAIMERIPOLL

Corollary 4.13. Let M be an immersed orientable complete surface with constant mean curvature in ޓ3 = SO(4)/ SO(3) with Gauss map N : M → ޓ5. Then N(M) 5 a is contained in ޓ+ if and only if , up to an isometry, M is φt -invariant for some α ∈ ޒ. Proof. One direction is obvious. For the other, note that Theorem 4.9 implies that 3 M is invariant under a one-parameter Lie subgroup φt of isometries of ޓ of the form exp(t X) for some nonzero vector X ∈ so(4). By the maximal torus theorem, there are δ, ε ∈ ޒ and g ∈ SO(4) such that  0 δ 0 0  −δ 0 0 0  Adg X =   .  0 0 0 ε  0 0 −ε 0 We may assume without loss of generality that δ 6= 0 so that the one-dimensional Lie subalgebra generated by Adg X is the same as the one generated by  0 1 0 0  −1 0 0 0  (4–9) Xα =   ,  0 0 0 α  0 0 −α 0 α whose associated Lie subgroup is φt . It follows that M is congruent to g(M) and α g(M) is φt -invariant. An open question on the study of minimal surfaces in the sphere ޓ3 is whether there exist embedded minimal tori in ޓ3 besides the Clifford tori. Lawson con- jectured that the answer is no. We may apply the corollary above to obtain the following characterization of the Clifford torus and of the surfaces of revolution of ޓ3 as well. 0 3 Observe that φt is a rotational one-parameter subgroup of isometries of ޓ and 1 5 5 that φt is the Hopf action. Given α ∈ ޒ, denote by ޓα the half-sphere of ޓ in so(4) whose boundary is orthogonal to Xα. Corollary 4.14. Let M be an immersed orientable complete surface with constant mean curvature in ޓ3 = SO(4)/ SO(3) and let N : M → ޓ5 be the Gauss map of M. ⊂ 5 (1) N(M) ޓ1 if and only if M is a Clifford torus. ⊂ 5 (2) N(M) ޓ0 if and only if M is a surface of revolution. For the proof of (a) we use the fact that a complete cmc surface in ޓ3 which is invariant under the Hopf action is a Clifford torus. This is easy to prove using an argument from the so-called equivariant geometry (see the proof of Theorem 4 of [Esp´ırito-Santo et al. 2003]). GAUSS MAP AND MEAN CURVATURE IN HOMOGENEOUS MANIFOLDS 63

Remark. Surfaces of revolution with cmc in ޓ3 have been described in many places. A description of these surfaces using a special coordinate system is given in [Fornari and Ripoll 2004].

References

[Eells and Sampson 1964] J. Eells, Jr. and J. H. Sampson, “Harmonic mappings of Riemannian manifolds”, Amer. J. Math. 86 (1964), 109–160. MR 29 #1603 Zbl 0122.40102 [Espírito-Santo et al. 2003] N. do Espírito-Santo, S. Fornari, K. Frensel, and J. Ripoll, “Constant mean curvature hypersurfaces in a Lie group with a bi-invariant metric”, Manuscripta Math. 111:4 (2003), 459–470. MR 2004f:53072 [Fischer-Colbrie and Schoen 1980] D. Fischer-Colbrie and R. Schoen, “The structure of complete stable minimal surfaces in 3-manifolds of nonnegative scalar curvature”, Comm. Pure Appl. Math. 33:2 (1980), 199–211. MR 81i:53044 Zbl 0439.53060 [Fornari and Ripoll 2004] S. Fornari and J. Ripoll, “Killing ﬁelds, mean curvature, translation maps”, Illinois J. Math. 48:4 (2004), 1385–1403. MR 2005i:53037 Zbl 02140243 [Hoffman et al. 1982] D. A. Hoffman, R. Osserman, and R. Schoen, “On the Gauss map of complete surfaces of constant mean curvature in ޒ3 and ޒ4”, Comment. Math. Helv. 57:4 (1982), 519–531. MR 84f:53004 Zbl 0512.53008

Received June 1, 2004.

FIDELIS BITTENCOURT UNIVERSIDADE FEDERALDE SANTA MARIA CCNE–DEPARTAMENTO DE MATEMATICA´ CAMPUS CAMOBI 97105-900 SANTA MARIA,RS BRAZIL ﬁ[email protected]

JAIME RIPOLL UNIVERSIDADE FEDERALDO RIO GRANDEDO SUL INSTITUTODE MATEMATICA´ AV.BENTO GONC¸ ALVES 9500 91501-970 PORTO ALEGRE,RS BRAZIL [email protected]