Totally umbilical hypersurfaces of manifolds admitting a unit Killing field

We prove that a Riemannian product of type M x R (where R denotes the Euclidean line) admits totally umbilical hypersurfaces if and only if M has locally the structure of a warped product and we give a complete description of the totally umbilical hypersurfaces in this case. Moreover, we give a necessary and sufficient condition under which a Riemannian three-manifold carrying a unit Killing field admits totally geodesic surfaces and we study local and global properties of three-manifolds satisfying this condition.


Introduction
The starting point of the research leading to this paper was the classification of totally umbilical surfaces in three-dimensional homogeneous spaces with a fourdimensional isometry group, which can be found in [11] and [12]. As is well-known, these three-spaces admit Riemannian submersions onto surfaces of constant Gaussian curvature and the unit vector field tangent to the fibers is Killing. It turns out that such a space admits totally umbilical surfaces if and only if it is a Riemannian product of the base surface and the fibers, i.e., if and only if its universal covering is either S 2 (κ) × R or H 2 (κ) × R. Moreover, the obtained classification of totally umbilical surfaces was extended to a classification of totally umbilical hypersurfaces of the conformally flat symmetric manifolds S n (κ) × R and H n (κ) × R in [13] and [3].
Two questions for further generalizations come up naturally now.
(1) When does a Riemannian product of type M n × R admit totally umbilical hypersurfaces and what are they? (2) When does a Riemannian three-space with a unit Killing field admit totally umbilical surfaces and what are they? In this paper, we give a complete answer to the first question. Our Theorem 1 states that a necessary and sufficient condition for M n × R to admit totally umbilical hypersurfaces is that M n itself has (locally) the structure of a warped product. Moreover, we give a full description of all totally umbilical hypersurfaces of such a manifold and we remark that our results are still valid if we start with a warped product instead of with a Riemannian product as ambient space.
For the second question we give a partial answer. We find a necessary and sufficient condition for a three-manifold with a unit Killing field to admit totally Key words and phrases. totally umbilical, totally geodesic, product manifold, Killing field, warped product.
The geodesic surfaces. Remark that it is not necessary that the three-manifold reduces to a Riemannian product, a fact which is already illustrated by the standard threesphere. We describe the totally geodesic surfaces and we study the local and global properties of three-spaces satisfying our condition.
We are grateful to Eric Toubiana for valuable remarks on a first version of this work.

Preliminaries
Let (M n , g) → (M n+1 ,g) be an isometric immersion between Riemannian manifolds. If N is a unit normal vector field along the immersion and ∇ and∇ are the Levi-Civita connections of (M n , g) and (M n+1 ,g), then the second fundamental form h and the shape operator S associated to N are defined by the formulas of Gauss and Weingarten: for any vector fields X, Y on M n one has It is easy to check that S is a symmetric (1, 1)-tensor field on M n , which is related to h by h(X, Y ) = g(SX, Y ). We call the immersion totally umbilical if S is a multiple of the identity at every point and we call it totally geodesic if S vanishes identically.
In our results, some special types of vector fields on Riemannian manifolds will occur. Let (M, g) be a Riemannian manifold and let ξ be a vector field on M . Then ξ is said to be Killing if and only if L ξ g = 0, where L is the Lie derivative. This condition means that the flow of ξ consists of isometries, and in terms of the Levi-Civita connection ∇ one can reformulate it as g(∇ X ξ, Y ) + g(∇ Y ξ, X) = 0 for all p ∈ M and X, Y ∈ T p M .
More generally, ξ is said to be conformal if and only if L ξ g = 2φg for some function φ. This means that the flow of ξ consists of conformal maps.
Finally, we say that ξ is closed conformal if and only if it is conformal and its dual one-form is closed. It can be checked by a straightforward computation that ξ is closed conformal if and only if for all p ∈ M and all X ∈ T p M , where φ is as above.
In all what follows the manifolds will be assumed of class C ∞ .
3. Totally umbilical hypersurfaces of manifolds of type M n × I Denote by M n × I the Riemannian product of a Riemannian manifold (M n , g M n ) and an open interval I of the Euclidean line and let π : M n × I → M n be the canonical projection. We shall denote by ξ a unit vector field on M n × I, tangent to the fibres of π. Remark that ξ is a unit Killing field.
There are two natural families of examples of totally geodesic hypersurfaces of M n × I, namely the slices M n × {t 0 }, t 0 ∈ I and the inverse images under π of totally geodesic hypersurfaces of M n , if they exist. We are thus interested in totally umbilical hypersurfaces which are at some point neither orthogonal nor tangent to ξ.
If Σ is a hypersurface of M n × I with unit normal N , one can define a vector field T and a real-valued function ν on Σ by the following orthogonal decomposition of ξ: Then T and ν satisfy the following equations.
Lemma 1. Let Σ be a hypersurface in M n × I and denote by ∇ the Levi-Civita connection of Σ, by S the shape operator of the immersion and by h the second fundamental form. Then for any vector X tangent to Σ: Proof. Denote by ∇ the Levi-Civita connection of M n ×I. Since ξ is a parallel vector field on M n × I, one has ∇ X ξ = 0. Now let X be tangent to Σ. By the definitions of T and ν and by using the formulas of Gauss and Weingarten, it follows that The result follows by considering the tangent, resp. normal, component of the above equation.
If Σ is totally umbilical in M n × I, say S = λ id, then T is a closed conformal field on Σ. Indeed, in this case it follows from Lemma 1 above that ∇ X T = νλX for all vector fields X tangent to Σ. We shall now prove that if Σ is non-vertical and non-horizontal at some point p, we can use T to construct a local non-vanishing conformal field on M n . Proposition 1. Let Σ be a totally umbilical hypersurface of M n ×I, which is neither vertical nor horizontal at some point. Then the canonical projection π : M n × I → M n is locally a diffeomorphism between an open neighborhood U of this point in Σ and the open subset πU of M n . Let T be as above and denote by T 0 be the projection of T to πU , rescaled such that it has the same length as T again. Then T 0 is a closed conformal field on πU .
Proof. Let Σ be a totally umbilical hypersurface of M n × I. Let ξ, N , T and ν be as above and assume that the shape operator associated to N is S = λ id. Suppose Σ is non-vertical and non-horizontal at some point, then it is clear that there is an open neighborhood U of this point in Σ where ν does not vanish and such that π is a local diffeomorphism between U and its image πU in M n . First, extend the vector fields T , N and the functions ν, λ to the whole of πU × I by using the one-parameter group of translations corresponding to the Killing field ξ and denote these again by T , N , ν and λ. Since ν and λ are constant on fibres of π, one can also view them as functions on πU . Using these notations, the vector field T 0 on πU is and its horizontal lift to πU × I is Remark that T 0 is, up to the sign, the projection of N to πU . Now let X be a vector field on πU and denote by X its horizontal lift. Then Let Y be a local vector field on Σ such that (dπ)(Y ) = X. Denote the extension to πU × I, using the flow of ξ, again by Y . Then X = Y − Y, ξ ξ and Here we used that [ξ, N ] = 0 implies ∇ ξ N = ∇ N ξ = 0. From (2) and (3), we obtain ∇ M n X T 0 = λX, which proves that T 0 is indeed closed conformal. The fact that M n admits a local closed conformal field, determines locally its Riemannian structure, as shown by the following known result (see [6,9] and the references therein).
Proposition 2. Let V be a local closed conformal field without zeros on a Riemannian manifold M n , say ∇ M n X V = f X for some non-vanishing function f and for all vector fields X on M n . Then M n has locally the structure of a warped product of an interval of the Euclidean line with some (n − 1)-dimensional Riemannian manifold.
Proof. One can check that the distribution orthogonal to V is integrable and hence one can find a local coordinate system (x 1 , . . . , x n ) on M n such that ∂ x1 = V and ∂ xj is orthogonal to ∂ x1 for j ≥ 2. With respect to these coordinates, the metric on M n takes the form It follows from a straightforward computation that ∂ xj g 11 = 0 for j ≥ 2 and that ∂ x1 g ij = 2f g ij for i, j ≥ 2. Hence, one has To conclude, we prove that ∂ xj f = 0 for j ≥ 2, such that, after a change of the x 1 -coordinate, the metric above is indeed a warped product metric. To see this, let R be the curvature tensor of M n , then Remark 1. The converse to Proposition 2 is also true. In a warped product I × f M = (I × M, dt 2 + f (t) 2 g M ), the field f (t)∂ t is closed conformal and vanishes nowhere.
We can now prove our main result in this section. Theorem 1. A Riemannian product space M n × I admits a totally umbilical hypersurface Σ, which is neither vertical nor horizontal at some point (p, t) ∈ M n × I, if and only if M n has in a neighborhood of p the structure of a warped product of an interval of the Euclidean line with some (n − 1)-dimensional Riemannian manifold.
In particular, when n = 2, there exists a totally umbilical surface in M 2 × I, which is neither vertical nor horizontal at some point (p, t) ∈ M 2 × I, if and only if M 2 admits a non zero Killing field in a neighborhood of p. Moreover any such surface is invariant by a local one-parameter group of local isometries of M 2 × I keeping the factor I pointwise fixed.
Proof. It follows from Propositions 1 and 2 above that M n having the structure of a warped product in a neighborhood of p is a necessary condition for M n × I to admit a totally umbilical hypersurface which is non-vertical and non-horizontal at (p, t).
We shall now prove that this condition is also sufficient. Assume that M n = J × f M n−1 , i.e., that the metric on M n can be written as . . , x n ). Then the metric on M n × I can be written as . . , x n ). We know from above that a non-vertical and non-horizontal totally umbilical hypersurface Σ of M n × I should be tangent to the distribution orthogonal to the vector fields ∂ x0 and ∂ x1 at any of its points. This means that Σ is generated by a curve in the (x 0 , x 1 )-plane, say α(s) = (x 0 (s), x 1 (s)). Assume that α is parametrized by arc length, then there exists a function θ such that In this case, the tangent space to Σ is spanned by X 1 = sin θ(s)∂ x0 + cos θ(s)∂ x1 , X 2 = ∂ x2 , . . . , X n = ∂ xn and a unit normal to Σ is given by One can compute the Levi-Civita connection ∇ of M n × I from the metric above to verify One can now use this equation to determine the functions x 0 (s) and x 1 (s). Indeed, we have which yields after a first integration The function x 0 (s) is then determined by Thus there does always exist a non-vertical and non-horizontal totally umbilical hypersurface of M n × I if M n is locally isometric to the warped product described above. A parametrization for such a totally umbilical hypersurface Σ is ϕ(s, u 1 , . . . , u n−1 ) = (x 0 (s), x 1 (s), u 1 , . . . , u n−1 ). In the particular case when n = 2, observe the following general fact that can be checked straightforwardly. Let J denote the rotation over 90 degrees of an oriented Riemannian surface M 2 , which is locally well-defined on any Riemannian surface M 2 . Then a vector field X on M 2 is closed conformal if and only if JX is Killing. Hence, if Σ is a totally umbilical surface in M 2 × I, then JT is a Killing field on Σ. Moreover, JT is orthogonal to the fibers of π and (dπ)(JT ) = JT 0 is a Killing field on M 2 . This implies the result.
In particular, from Theorem 1, we recover the classification of totally umbilic surfaces in S 2 × R and H 2 × R obtained in [11] and [12].
Remark 2. Theorem 1 also gives a classification of totally umbilical hypersurfaces in Riemannian product spaces of type M n × I.
Remark 3. As a further particular case, it is interesting to observe that the results above provide a new proof for the classification of totally umbilic hypersurfaces in the Euclidean space R n+1 since the latter can be viewed (in various ways) as a product R n × R. The present proof has the advantage to work assuming only a C 2 regularity for the hypersurfaces. It can indeed easily be checked that C 2 regularity for the hypersurfaces is enough in the above results. The standard proof of the classification of totally umbilic hypersurfaces in R n+1 works for hypersurfaces with at least C 3 regularity. It is known this classification also holds for C 2 -hypersurfaces (see [7] and [10]).

Corollary 1.
A Riemannian warped product space I × f M n admits a totally umbilical hypersurface Σ, which is neither vertical nor horizontal at some point (t, p) ∈ I × f M n , if and only if M n has in a neighborhood of p the structure of a warped product of an interval of the Euclidean line with some (n − 1)-dimensional Riemannian manifold.
Proof. First observe that a Riemannian warped product space I × f M n is conformal to a Riemannian product J × M n , for some open interval J of R. Indeed, with some abuse of notation, the metric on I × f M n writes Choosing a new parameter s = ψ(t) = dt f (t) and introducing the function h(s) = f (ψ −1 (s)), we can write: as desired. Now the claim follows from Theorem 1 and the known fact that totally umbilic hypersurfaces are preserved under conformal diffeomorphisms between the ambient manifolds.

Totally geodesic surfaces in a three-dimensional space admitting a unit Killing field
In this section we will characterize locally the Riemannian three-manifolds admitting a unit Killing field which possess totally geodesic surfaces. A good reference on manifolds admitting a Killing field of constant length is given by [1].
We start with a result which is valid in all dimensions. Let M denote a Riemannian manifold which admits a unit Killing field ξ. The product manifolds M n × I considered in Section 3 are a particular case. Denote by ∇ the Levi-Civita connection of M. Let Σ be a hypersurface in M with unit normal N . Then we can, as in the previous section, define a vector field T and a real-valued function ν on Σ by the orthogonal decomposition The following result is a key fact for our purposes: Proposition 3. Let Σ be a totally geodesic hypersurface in a Riemannian manifold M admitting a unit Killing field ξ. Suppose ξ is not tangent to Σ at some point. Then one can extend the vector field T to a neighborhood of this point in M using the local flow of ξ. If one denotes the resulting vector field again by T , then T is a local Killing field on M .
Proof. Since ξ is transversal to Σ in a neighborhood of the point, using the local flow (φ t ) t∈I of ξ we obtain a foliation F of an open subset of M by the totally geodesic hypersurfaces φ t (Σ). In this way we have local extensions of the fields T and N and the function ν. Denote these extensions again by T , N and ν. Note that (4) is again valid for these extensions.
We have to verify that ∇ X T, Y + X, ∇ Y T = 0 for all vector fields X and Y . First we note that Indeed, since the hypersurfaces φ t (Σ) are totally geodesic and ξ is a unit Killing field, we have Consider now a vector field X which is tangent to the leaves of the foliation F . Then Furthermore, Using (6) and (5) we get Using (6) and (7), it is easy to check that T is Killing.
We now particularize to the case where the ambient manifold is three-dimensional. Well-known examples of such manifolds are Riemannian products of type M 2 × R and also the unit three-sphere S 3 , Berger spheres and the Thurston spaces SL(2, R) and Nil 3 . In the next section we will describe more such spaces. We first prove an important basic formula.
Lemma 2. Let M 3 be an oriented Riemannian manifold carrying a unit Killing field ξ. Denote by ∇ the Levi-Civita connection of M 3 and by × its cross product. Then there exists a real-valued function τ on M 3 , with ξ(τ ) = 0, such that for all vector fields X on M 3 .
Proof. It is clear that ∇ X ξ is perpendicular to ξ and X since ξ is a unit Killing field. Because the space is three-dimensional, we obtain that ∇ X ξ = τ (X)(X × ξ) for some real number τ (X).
Since the mapping T p M 3 → T p M 3 : X → ∇ X ξ = τ (X)(X × ξ) must be linear for every point p, it is easily seen that τ can only depend on the choice of p ∈ M 3 and not on the choice of X ∈ T p M 3 . Hence τ is a real-valued function on M 3 .
To see that this function satisfies ξ(τ ) = 0, let (φ t ) t∈I be the local flow of ξ as above. Then for every parameter t and for every p ∈ M 3 and X ∈ T p M 3 . We conclude that τ (φ t (p)) = τ (p) and hence that ξ(τ ) = 0.
The first main result in this section is the following. (i) M admits a totally geodesic surface passing through p which is neither orthogonal nor tangent to ξ at p. (ii) There is in a neighborhood of p in M an orthogonal decomposition ξ = X 1 + X 2 of ξ, where X 1 and X 2 are Killing fields without zeros that commute. (iii) There exist local coordinates (x, y, z) around p in M with ξ = ∂ y + ∂ z , such that the metric takes the form g = dx 2 + sin 2 θ(x)dy 2 + cos 2 θ(x)dz 2 . Proof. We denote by (φ t ) t∈I the local flow of ξ.
(1) Suppose M admits a totally geodesic surface Σ, passing through p, which is everywhere orthogonal to ξ. Restricting Σ and the interval I if necessary, the mapping (x, t) ∈ Σ × I → φ t (x) ∈ M parametrizes an open subset of M and is easily checked to be an isometry between the Riemannian product manifold Σ × I and that open set.
Conversely it is clear that if M has locally a product structure of some surface Σ and an interval I to which ξ is tangent then the surfaces Σ× {t} are totally geodesic and orthogonal to ξ at each point.
The mapping (s, t) → φ t (γ(s)) parametrizes a surface Σ in M and τ | Σ = 0. We now check that Σ is totally geodesic. A unit normal field to Σ is the field N (s, t) = (dφ t )(γ ′ (s)) × ξ. Note that ξ commutes with (dφ t )(γ ′ (s), so that where we used that τ | Σ = 0. Therefore Since s → φ t (γ(s)) is a geodesic in M for each t, we have It follows that Σ is totally geodesic.
Conversely, suppose M admits a totally geodesic surface Σ passing through p which is everywhere tangent to ξ. As geodesics on Σ are also geodesics on M, it is enough to check that τ | Σ = 0. This is indeed the case: take an arbitrary point q of Σ and a vector X tangent to Σ and linearly independent of ξ. Since Σ is totally geodesic, the vector ∇ X ξ = τ (q)(X × ξ) has to be tangent to Σ, which is only possible if τ (q) = 0.
(3) First, we prove that (i) implies (iii). Let Σ be totally geodesic in M such that ξ is not tangent to Σ at p. Extend T , N , JT = N × T and ν to a neighborhood of this p in M using the local flow of ξ. Using equations (6), (5) and (4), we have: Hence, we can take local coordinates (x, y, z) on M such that ∂ x = JT , ∂ y = T and ∂ z = νN . With respect to these coordinates, the metric takes the form From (5) one has ∂ y ν = T (ν) = 0 and ∂ z ν = νN (ν) = (ξ − T )(ν) = 0. After a change of the x-coordinate, we obtain the form for g given in the theorem.
To see that (iii) implies (ii), it suffices to take X 1 = ∂ y and X 2 = ∂ z . It remains to prove that (ii) implies (i). Let u be a unit vector field perpendicular to X 1 and X 2 . We shall first prove that u commutes with X 1 and X 2 . By using that u is perpendicular to X 1 and that X 1 is Killing, we obtain Furthermore, by using that u is perpendicular to X 2 and that X 1 is Killing, we find Finally, since u = 1 and X 1 is Killing, We conclude that [X 1 , u] = 0. Analogously, we can prove that [X 2 , u] = 0. Now consider an integral surface of the distribution spanned by u and X 1 . Of course, this surface is nowhere tangent or orthogonal to ξ = X 1 + X 2 . We shall prove that it is totally geodesic. It is sufficient to verify that ∇ u u, ∇ u X 1 and ∇ X1 X 1 are all perpendicular to X 2 . Using Koszul's formula and the facts that [X 1 , X 2 ] = [X 1 , u] = [X 2 , u] = 0, X 1 , X 2 = X 1 , u = X 2 , u = 0 and u, u = 1, gives immediately that ∇ u u, X 2 = 0 and ∇ u X 1 , X 2 = 0. Finally, using the facts that X 1 and X 2 are orthogonal and that X 2 is Killing, gives ∇ X1 X 1 , X 2 = − X 1 , ∇ X1 X 2 = 0.
Remark 5. For later use, we note that in the coordinates where the metric takes the given form in case (iii) we have from the proof of the theorem: ∂ x = JT / T , ∂ y = T , cos θ(x) = ξ, N and T = sin θ(x).
We now study further the case (3) in Theorem 2. We are able to determine all the totally geodesic surfaces of M in a neighborhood of p in this case. We will need the following result which can be verified by straightforward computations.
Our second main result in this section decribes the totally geodesic surfaces in M in case (3) in Theorem 2. It characterizes in particular the flat and spherical metrics. It can be compared to a result of E. Cartan (see [2] p. 233). In the three-dimensional case, Cartan's theorem asserts a three-dimensional Riemannian manifold with a totally geodesic surface passing through any point with any specified plane as tangent plane must be a space form. When the manifold admits a unit Killing field, our result says that the existence of very few totally geodesic surfaces suffices to characterize the space forms of non-negative curvature.
Theorem 3. Let M be a Riemannian three-manifold carrying a unit Killing field ξ and let p ∈ M . Suppose there is a totally geodesic surface Σ 1 passing through p which is neither orthogonal nor tangent to ξ at p. Then: (1) There is a second totally geodesic surface passing through p which is orthogonal to Σ 1 . Proof. From case (3) in Theorem 2 we can find local coordinates (x, y, z) in a neighborhood W of p where the metric takes the form (8), the point p corresponding to the origin. Restricting Σ 1 if necessay we can assume that Σ 1 is given by the equation z = 0.
(1) From the above local expression (8) for the metric we see that the surface Σ 0 , defined by the equation y = 0, is totally geodesic and is orthogonal to Σ 1 .
(2) Suppose there is a third totally geodesic surface Σ 2 containing p which is not tangent to ξ at p.
We first treat the case when Σ 2 is not orthogonal to ξ at p. We will show that the function τ is constant in a neighborhood of p. This will conclude the proof as this means that, in the coordinates introduced above, θ(x) = αx + β for some constants α and β. If α = 0, the metric g is flat. If α = 0, then it is straightforward to check that g has constant sectional curvature α 2 .
We still denote by Σ 2 the component of W ∩ Σ 2 containing p. Restricting W and replacing Σ 2 by an open subset of it if necessary, we can assume the intersection Σ 1 ∩ Σ 2 is connected. For i = 1, 2, denote by N i a unit normal to Σ i . As before we introduce the vector field T i tangent to Σ i and the real valued function ν i on Σ i by the orthogonal decomposition We again use the same notations to denote the extensions of N i , T i and ν i to W using the flow of ξ. Note that along Σ 1 ∩Σ 2 the vectors N 1 and N 2 are independent, so, up to restricting W if necessary, we can assume that their extensions are also pointwise independent. In the same way, as Σ 1 and Σ 2 are distinct, T 1 and T 2 are independent along Σ 1 ∩ Σ 2 and we can assume their extensions are independent in W.
Suppose that ξ, T 1 and T 2 are linearly independent in an open set U ⊂ W. It follows from Proposition 4 and Remark 5 that τ does not depend on T 1 , that is, T 1 (τ ) = 0. In the same way T 2 (τ ) = 0. As ξ(τ ) = 0, see Lemma 2, we conclude that grad τ = 0 in U, where grad denotes the gradient on M.
Suppose now that ξ, T 1 and T 2 are (pointwise) linearly dependent in some connected open set V ⊂ W. Let S denote the surface tangent to the distribution spanned by ξ and T 1 which passes through p. From the expression of the metric (8) obtained in Theorem 2 using the surface Σ 1 , we see that the coordinate x is the signed distance function to the surface S. As we are assuming ξ, T 1 and T 2 are dependent in V, we conclude that we obtain the same coordinate function x when we use Σ 2 in Theorem 2. The Killing fields T 1 and T 2 are tangent to the integral surfaces of the distribution spanned by ξ and T 1 , that is, the level surfaces of the coordinate function x, and so are Killing fields on each of them. Moreover their norms depend only on x (see Remark 5). For each x, the level surface corresponding to x is flat, that is, locally euclidean. The Killing fields ξ, T 1 and T 2 on such a surface correspond to constant fields under an isometry with an open subset of the Euclidean plane since they have constant norms. Therefore we have in V a relation of the form We next show that the functions α 1 and α 2 are actually constant. As ξ, T 1 and T 2 are Killing fields, we have for any vector field Y : where γ i = α i /(α 1 + α 2 − 1), i = 1, 2. We have: Taking the inner product of both sides with ξ we obtain: That is, τ {γ 1 sin 2θ 1 (x) + γ 2 sin 2θ 2 (x)} = 0.
Summarizing we have shown that grad τ = 0 on an open dense set in a neighborhood of p in M 3 . Therefore τ is constant near p. This concludes the proof of (2) when Σ 2 is not orthogonal to ξ at p.
Suppose now that Σ 2 is orthogonal to ξ at p. Let S 0 ⊂ Σ 2 denote the subset where Σ 2 is not orthogonal to ξ. By the above argument, the function τ is locally constant on S 0 . Suppose now that Σ 2 is orthogonal to ξ in an open set S 1 ⊂ Σ 2 . So ξ is the unit normal to Σ 2 on S 1 . As Σ 2 is totally geodesic, by the formula in Lemma 2, we have τ ≡ 0 on S 1 . Consequently, denoting by grad Σ2 the gradient on Σ 2 , we have grad Σ2 τ = 0 on an open dense subset of Σ 2 and so τ is constant on Σ 2 . As ξ is transversal to Σ 2 and ξ(τ ) = 0, we conclude again that τ is constant in a neighborhood of p in M 3 . This concludes the proof of (2).
(3) Suppose τ (p) = 0, that is, θ ′ (0) = 0. Then from Proposition 4 the surface given by x = 0 is totally geodesic. Conversely, suppose there is a connected totally geodesic surface Σ through p which is tangent to ξ at p. We may assume Σ is contained in the coordinate neighborhood where the metric on M takes the form (8). By the same arguments as in (2), we get that τ is constant on any connected open subset of Σ where ξ is not tangent to Σ. Moreover τ vanishes on any open subset of Σ where ξ is tangent to Σ (see the proof of (2) in Theorem 2). As previously, we conclude that τ is constant on Σ.
Denote by π the projection on the x−axis. We consider three cases: -First case: I := π(Σ) contains an open interval containing 0. It follows that τ (which depends only on x) is constant in a neighborhood of p. So M is, near p, flat or has constant positive curvature. The second possibility is excluded by hypothesis and consequently τ is identically zero near p.
-Third case: π(Σ) ⊂ [0, +∞) or π(Σ) ⊂ (−∞, 0]. This means the surface Σ is on one side of the surface {x = 0}. The extrinsic curvature of the surface {x = 0} is K ext = −(θ ′ (0)) 2 = −τ (p) 2 as is seen from Proposition 4. It is therefore a saddle surface if τ (p) = 0 and we are led in this case to a contradiction since it is a general fact that a totally geodesic surface tangent to a saddle surface at a point cannot lie on one side of it near the tangency point. So necessarily τ (p) = 0.

Properties of the three-spaces
In this section we shall discuss some properties of three-dimensional spaces M 3 with a unit Killing field ξ that admit totally geodesic surfaces which are neither orthogonal nor tangent to ξ. From Theorem 2 we know that such a manifold locally admits a metric of type (12) g = dx 2 + sin 2 θ(x)dy 2 + cos 2 θ(x)dz 2 , The following result, which can be checked through straightforward computations, states that these three-spaces admit Riemannian submersions onto a surface with a Killing field. Proposition 5. Given M 3 as above, consider a surface M 2 with local coordinates (u, v) and metric Then the mapping π : M 3 → M 2 : (x, y, z) → (u, v) = (x, y − z) is a Riemannian submersion whose fibers are integral curves of the unit Killing field ξ = ∂ y + ∂ z .
Let us now study some global properties of M 3 . In particular we want to investigate which manifolds admit a smooth metric which, in local coordinates, is given by (12).
We will first recall two lemmas from [8] on a class of more general doubly warped products (13) ( where I ⊆ R is an open interval and g S p and g S q are the standard Riemannian metrics on S p and S q . These results allow us to prove that a smooth metric of type (12) exists on the simply connected manifolds S 3 , S 2 × R and R 3 . Proof. It follows from the assumptions on θ that θ(0, b) = (0, π/2). Hence, the functions ϕ = sin θ and ψ = cos θ are strictly positive on (0, b). Lemma 3 and Lemma 4 yield that (12) then gives rise to a smooth metric on S 3 if and only if the conditions of Lemma 3 are satisfied at x = 0 and the conditions of Lemma 4 are satisfied at x = b, with the roles of ϕ and ψ interchanged.
Remark 6. The function θ(x) = x satisfies the conditions given in Proposition 6. In this case, the metric (12) corresponds to the standard metric on S 3 and the Riemannian submersion of Proposition 5 is the classical Hopf fibration. Proof. Remark that the functions ϕ = sin θ and ψ = cos θ are positive on (0, b). Hence, (12) defines a smooth metric on S 2 × R if and only if the conditions of Lemma 3 and Lemma 4 are satisfied. We can now proceed in an analogous way as in the proof of Proposition 6 to obtain the result.
Proof. It is clear that the metric is smooth under the given assumptions. To prove completeness, we may assume that θ(x) ∈ (0, π/2) for all x ∈ R. Now let γ : [0, T ) → R 3 : t → (γ 1 (t), γ 2 (t), γ 3 (t)) be a curve which diverges to infinity, i.e., such that γ 1 (t) 2 + γ 2 (t) 2 + γ 3 (t) 2 tends to infinity if t tends to T . We have to prove that the length of this curve with respect to the metric (12), is infinite. Therefore, we consider two cases. First, assume that γ 1 is unbounded. In this case, we have Next, assume that γ 1 is bounded. In that case the function θ(γ 1 (t)) is bounded away from 0 and π/2 and hence there exists a real constant c > 0 such that sin(θ(γ 1 (s))) ≥ c and cos(θ(γ 1 (s))) ≥ c. This implies that The last equality is due to the fact that the integral appearing on the left hand side is the Euclidean length of the projection of the curve γ onto the (y, z)-plane.
Since γ diverges to infinity but γ 1 is bounded, this projection must have infinite length.
It is possible to check, using for instance our Theorem 3, that in the examples of Proposition 6 and Proposition 7, through the points where x = 0 or x = b, there is no totally geodesic surface which is not tangent to the unit Killing field unless the function θ ′ is constant in a neighborhood of x = 0 and x = b, respectively. This is not a mere coincidence. We actually have the following global result. Then M is isometric to R 3 endowed with the metric: where θ : R → (0, π/2) is a smooth function whose derivative θ ′ is not constant on any interval. Moreover ξ = ∂ y + ∂ z .
Proof. Let p 0 be a fixed point in M. By Theorems 2 and 3, ξ admits an orthogonal decomposition, ξ = X 1 + X 2 , in a neighborhood of p 0 , where X 1 and X 2 are two Killing fields which commute and have no zeros. Moreover this decomposition is unique up to ordering of X 1 and X 2 . By the same theorems, given any point p ∈ M and any continuous path γ joining p 0 to p, we can extend continuously this decomposition along γ till the point p. As M is simply connected, by a standard monodromy argument, the decomposition we obtain at p is independent on the choice of the path γ. We get in this way a global orthogonal decomposition ξ = X 1 + X 2 , where X 1 and X 2 are now global smooth Killing vector fields on M which commute and have no zeros. Denote by u a unit vector field on M orthogonal to X 1 and X 2 . Such a global smooth vector field exists since M is orientable. From the proof of (3)-(ii) of Theorem 2, we know that the distribution spanned by u and X 1 is integrable and its integral surfaces are totally geodesic. Therefore the manifold M admits a foliation F by totally geodesic surfaces. Let F ⊥ be the orthogonal foliation, that is, the foliation by the orbits of X 2 . By a result of Carrière and Ghys [4], (F , F ⊥ ) is a product. This means there is a diffeomorphism between M and the product Σ × R, where Σ is any fixed leaf of F , sending the leaves of F to Σ × { * } and those of F ⊥ to { * } × R. Denote by z the coordinate on the R factor. Under this diffeomorphism, the vector field X 2 therefore corresponds to the field f (z)∂ z for some function f. So, up to reparameterizing the R factor, we can assume that X 2 correponds to the field ∂ z . It is clear that Σ is simply connected and is therefore, topologically, either a plane or a sphere. X 1 is vector field on Σ which has no zeros, so Σ is topologically a plane. It is moreover not difficult to check that Σ is complete.
Fix an orientation on Σ and denote by J the rotation over 90 degrees in T Σ. As in the proof of (3) of Theorem 2, we consider on Σ the fields X 1 and JX 1 which commute, [X 1 , JX 1 ] = 0, and are complete since they have bounded norms and Σ is complete. It follows that we can find a global chart for Σ with domain R 2 and ∂ x = JX 1 and ∂ y = X 1 for the standard coordinates (x, y) on R 2 . We include a proof of this fact for completeness. Let (φ x ) x∈R and (ψ y ) y∈R be the flows of JX 1 and X 1 , respectively. Consider the mapping: (x, y) ∈ R 2 → F (x, y) = (φ x • ψ y )(p 0 ) ∈ Σ.
F is a local diffeomorphism with (dF )(∂ x ) = JX 1 and (dF )(∂ y ) = X 1 . We will show it is a global diffeomeorphism, this will provide the global chart we want.
-F is one-to-one: by contradiction, suppose (x 1 , y 1 ) and (x 2 , y 2 ) are distinct points in R 2 with F (x 1 , y 1 ) = F (x 2 , y 2 ). Assume that x 1 = x 2 and y 1 = y 2 , then the orbit of X 1 through φ x1 (p 0 ) will be closed and will bound a disk in Σ inside which necessarily X 1 will have a zero, which is a contradiction. The case x 1 = x 2 and y 1 = y 2 is similar. Assume now that x 1 = x 2 and y 1 = y 2 . Set p 1 = (φ x1 • ψ y1 )(p 0 ) = (φ x2 • ψ y2 )(p 0 ). The orbit of JX 1 through p 0 and the orbit of X 1 through p 1 intersect at two distinct points, namely φ x1 (p 0 ) and φ x2 (p 0 ). However an orbit of JX 1 can intersect an orbit of X 1 at most once. Indeed let γ 1 and γ 2 be orbits of X 1 and JX 1 , respectively. Assume they intersect more than once. Then there will be a bounded disk Ω in Σ with boundary the union of an arc α 1 ⊂ γ 1 and an arc α 2 ⊂ γ 2 with common endpoints p and q. We assume that along α 2 , the field X 1 points into Ω. The case when X 1 points outside Ω can be treated in a similar way. Consider any point q 1 on α 2 distinct from p and q. The half orbit β := {ψ t (q 1 ), t > 0} of X 1 starting from q 1 will be entirely contained in Ω. It follows from the Poincaré-Bendixon theorem, see for instance [5], that the accumulation set of β must contain a zero or a closed orbit of X 1 , which is again a contradiction.
In the global coordinates (x, y, z), like in the proof of (3) in Theorem 2 and with the same notations, the metric on M writes: where ν(x) = X 2 2 is a function of x alone. We now make the change of coordinatē x(x) = 1 − ν(x) 2 dx. By the completeness of the metric g, the functionx is a bijection from R onto R. Setting ν(x) = cos θ(x) for some smooth function θ : R → (0, π/2), the metric writes in the global coordinates (x, y, z) : ds 2 = dx 2 + sin 2 θ(x)dy 2 + cos 2 θ(x)dz 2 .
The condition (1) in the statement means precisely that θ ′ is not constant on any interval (see the proof of (2) in Theorem 3).