Rabah Souam

Eric Toubiana

Totally umbilic surfaces in homogeneous 3-manifolds

Keywords: Mathematics Subject Classification. 53C30, 53B25. Key words and phrases. totally umbilic, totally geodesic, homogeneous 3-manifolds . 1

come

Introduction

During the recent years, there has been a rapidly growing interest in the geometry of surfaces in S 2 × R and H 2 × R focusing on minimal and constant mean curvature surfaces. This was initiated by H. Rosenberg, [START_REF] Rosenberg | Minimal surfaces in M 2 × R[END_REF]. More generally many works are devoted to studying the geometry of surfaces in homogeneous 3-manifolds. See for example [START_REF] Meeks | The theory of minimal surfaces in M × R, Comment[END_REF], [START_REF] Daniel | Isometric immersions into S n ×R and H n ×R and applications to minimal surfaces[END_REF], [START_REF] Daniel | Isometric immersions into 3-dimensional homogeneous manifolds[END_REF], [START_REF] Sanini | Gauss map of a surface of the Heisenberg group[END_REF], [START_REF] Nelli | Minimal surfaces in H 2 × R, Bull[END_REF], [START_REF] Hauswirth | Minimal surfaces of Riemann type in three-dimensional product manifolds[END_REF], [START_REF] Hauswirth | Associate and conjugate minimal immersions in M × R, to appear in Tohoku Math[END_REF], [START_REF] Figueroa | Invariant surfaces of the Heisenberg groups[END_REF], [START_REF] Sa Earp | Screw motion surfaces in H 2 × R and S 2 × R[END_REF], [START_REF] Aledo | Complete surfaces of constant curvature in H 2 × R and S 2 × R, Calc[END_REF], [START_REF] Abresch | A Hopf differential for constant mean curvature surfaces in S 2 × R and H 2 × R[END_REF], [START_REF] Fernandez | Harmonic maps and constant mean curvature surfaces in H 2 × R, to appear in[END_REF], [START_REF] Caddeo | SO(2)-invariant minimal and constant mean curvature surfaces in 3-dimensional homogeneous spaces[END_REF] and [START_REF] Earp | Parabolic and hyperbolic screw motion surfaces in H 2 × R[END_REF].

In the space forms the classification of totally umbilic surfaces is well known and very useful, see [START_REF] Spivak | A comprehensive introduction to differential geometry[END_REF]. In R 3 they are planes and round spheres and in S 3 they are round spheres. In H 3 they are totally geodesic planes and their equidistants, horospheres and round spheres. In particular they all have constant mean curvature.

A natural question is to understand the totally umbilic surfaces in the remaining homogeneous 3-manifolds. Untill now the only known result in this direction was the non-existence of totally umbilic surfaces in the Heisenberg space due to A. Sanini, see [START_REF] Sanini | Gauss map of a surface of the Heisenberg group[END_REF]. In this paper we study totally umbilic surfaces in simply connected and homogeneous 3-manifolds. More precisely we first consider the manifolds having a 4-dimensional isometry group, denoted by M 3 (κ, τ ) (see section 2). Namely these manifolds are S 2 (κ) × R (κ > 0, τ = 0), H 2 (κ) × R (κ < 0, τ = 0), the Berger spheres (κ > 0, τ = 0) and the manifolds having the isometry group of either the Heisenberg space (κ = 0, τ = 0) or PSL 2 (R) (κ < 0, τ = 0), see [START_REF] Bonahon | Geometric structures on 3-manifolds, Handbook of geometric topology[END_REF], [START_REF] Scott | The geometries of 3-manifolds[END_REF] or [START_REF] Thurston | Three-Dimensional Geometry and Topology[END_REF]. Except for the Berger spheres, these manifolds are four of the eight model geometries of Thurston [START_REF] Thurston | Three-Dimensional Geometry and Topology[END_REF]. The remaining model geometries are the three space forms and the Sol geometry which has a 3-dimensional isometry group. As a matter of fact we also consider the Sol geometry.

In section 2 we prove (Theorem 1) the non-existence of totally umbilic surfaces, in particular the non-existence of totally geodesic ones, in the homogeneous manifolds M 3 (κ, τ ) for τ = 0, that is those which are not Riemannian products. This extends the result of Sanini, [START_REF] Sanini | Gauss map of a surface of the Heisenberg group[END_REF].

In section 3 we construct and classify all rotational and totally umbilic surfaces in S 2 (κ)×R. In section 4 we construct and classify all totally umbilic surfaces in H 2 (κ)× R which are invariant under a one-parameter group of ambient isometries. Except for the totally geodesic ones, these surfaces do not have constant mean curvature. In section 5, we prove that the surfaces obtained in sections 3 and 4 are the only totally umbilic surfaces in respectively S 2 (κ) × R and H 2 (κ) × R.

In section 6 we show that there exist, up to ambient isometries, only two totally umbilic surfaces in Sol, one of them being totally geodesic.

Finally, in section 7 we apply our results to prove that any conformal diffeomorphism of H 2 × R, S 2 × R and Sol is an isometry.

Throughout this paper all the surfaces are assumed of class C 3 , see however the remark [START_REF] Souam | On the classification and regularity of umbilic surfaces in homogeneous 3-manifolds[END_REF].

We are grateful to H. Rosenberg for valuable comments and to the referee for his observations which improved the paper.

Non-existence of totally umbilic surfaces in some homogeneous 3-manifolds

In this section we consider the connected and simply connected homogeneous Riemannian 3-manifolds, whose isometry group has dimension 4 and which are not Riemannian products. We recall that such a manifold is a fibration over some complete and simply connected surface, M 2 (κ), of constant curvature κ ∈ R, with geodesic fibers. Actually, for each κ, there is a one-parameter family M 3 (κ, τ ) of such fibrations, parametrized by the bundle curvature τ ∈ R * . The unit vector field ξ tangent to the fibers is a Killing field and satisfies :

∇ X ξ = τ (X ∧ ξ), (1)
for any tangent vector X in T M 3 (κ, τ ), where ∇ is the connection on M 3 (κ, τ ). The field ξ defines the vertical direction of the Riemannian submersion M 3 (κ, τ ) → M 2 (κ). As a matter of fact, the bundle curvature τ can be zero, but in this case M 3 (κ, 0) is just a Riemannian product M 2 (κ) × R. These product manifolds will be considered in the following sections. Moreover we assume κ -4τ 2 = 0, otherwise the manifold is a space form and its isometry group has dimension 6. These manifolds are of three types : when κ > 0 they are the Berger spheres, for κ = 0 they have the isometry group of the Heisenberg space, Nil 3 , and for κ < 0 they have the isometry group of PSL 2 (R).

For more details we refer to [START_REF] Bonahon | Geometric structures on 3-manifolds, Handbook of geometric topology[END_REF], [START_REF] Scott | The geometries of 3-manifolds[END_REF] and [START_REF] Thurston | Three-Dimensional Geometry and Topology[END_REF].

We can now state our result.

Theorem 1. There exist no totally umbilic surfaces (even non complete) in the 3manifolds M 3 (κ, τ ), with τ = 0 and κ -4τ 2 = 0. In particular, there are no totally geodesic surfaces.

For the special case of the Heisenberg space ( κ = 0, τ = 1/2), we recover the result proved by A. Sanini, see [START_REF] Sanini | Gauss map of a surface of the Heisenberg group[END_REF].

Proof. Let S be an immersed totally umbilic surface in M 3 (κ, τ ). Locally S is the image of an embedding X : Ω → M 3 (κ, τ ), where Ω is an open disk in R 2 . Call (u, v) the coordinates on Ω and consider a unit normal field N on X(Ω). As X is totally umbilic, there exists a function λ : Ω → R such that

∇ Xu N = λX u ∇ Xv N = λX v Therefore ∇ Xv (∇ Xu N ) = λ v X u + λ∇ Xv X u ∇ Xu (∇ Xv N ) = λ u X v + λ∇ Xu X v
Substracting the second equation from the first one we get

∇ Xv (∇ Xu N ) -∇ Xu (∇ Xv N ) = λ v X u -λ u X v . That is R(X u , X v )N = λ v X u -λ u X v , (2)
where R denotes the curvature tensor of M 3 (κ, τ ).

We define the function ν on Ω setting ν := N, ξ . We denote by T the projection of ξ on S, that is T = ξ -νN .

As the projection M 3 (κ, τ ) → M 2 (κ) is a Riemannian submersion, we have the following formula derived by Daniel, see [START_REF] Daniel | Isometric immersions into 3-dimensional homogeneous manifolds[END_REF]:

R(X u , X v )N = (κ -4τ 2 )ν ( X v , T X u -X u , T X v ) .
Taking into account the relation (2) we get

∇λ = (κ -4τ 2 )νT, (3)
where ∇ denotes the gradient on S.

Observe that if T = 0 on a nonempty open set, then we can take N = ξ on this set and we deduce from (1) that this surface cannot be umbilic. We can thus assume that T does not vanish on Ω.

Set JT = N ∧ T , thus JT is tangent to S and horizontal.

Claim:

We have [T, JT ] ≡ 0. We need to show that:

∇ T JT = ∇ JT T . Since JT = N ∧T = N ∧(ξ -νN ) = N ∧ξ, we have using (1) ∇ T JT = ∇ T (N ∧ ξ) = ∇ T N ∧ ξ + N ∧ ∇ T ξ = λ(T ∧ ξ) + τ (N ∧ (T ∧ ξ)). As T ∧ ξ = T ∧ (T + νN ) = ν(T ∧ N ) = -νJT , we deduce that ∇ T JT = -λνJT + τ νT.
On the other hand

∇ JT T = ∇ JT (ξ -νN ) = τ (JT ∧ ξ) -JT.(ν)N -λνJT. (4)
We have

JT ∧ ξ = JT ∧ (T + νN ) = JT ∧ T + νJT ∧ N = -|T | 2 N + νT, (5)
and

JT.(ν) = JT. N, ξ = ∇ JT N, ξ + N, ∇ JT ξ = λ JT, ξ + τ N, JT ∧ ξ = τ N, JT ∧ ξ .
Using (5) we obtain

JT.(ν) = -τ |T | 2 . (6)
Inserting ( 5) and ( 6) in (4) we end with:

∇ JT T = τ νT -λνJT = ∇ T JT, which proves the claim. Now (3) implies JT.(λ) = 0 T.(λ) = (κ -4τ 2 )ν|T | 2 = (κ -4τ 2 )ν(1 -ν 2 ) Since [T, JT ] = 0, we get (κ -4τ 2 )JT.(ν -ν 3 ) = 0. As κ -4τ 2 = 0 we infer that (1 -3ν 2 )JT.(ν) = 0. This implies easily JT.(ν) = 0. As JT.(ν) = -τ |T | 2 , see (6)
, and τ = 0, we deduce that T ≡ 0, which is a contradiction. This concludes the proof.

3. Symmetric totally umbilic surfaces in S 2 (κ) × R

In this section we classify totally umbilic surfaces which are rotationally invariant in S 2 × R. The classification in M 2 (κ) × R, for any κ > 0 is completely analogous. We will see that besides the obvious totally geodesic ones, up to isometries of S 2 × R, there are two one-parameter families of complete totally umbilic rotationally invariant surfaces homeomorphic to the two-sphere and a unique complete surface which has the topology of R 2 . The surfaces of the first family are homologous to zero and those of the second family are not. Moreover these surfaces are embedded, analytic and any totally umbilic rotationally invariant surface is a part of one of these complete surfaces.

A rotational surface in S 2 × R is by definition a surface obtained by rotating a curve in a totally geodesic cylinder C := Γ × R, where Γ ⊂ S 2 is a geodesic, around an axis R = {p} × R where p is a fixed point of Γ.

In the coordinates (x, y, t) given by the stereographic projection with respect to the north pole, the metric on S 2 × R reads as follows:

d s2 = 2 1 + x 2 + y 2 2 (dx 2 + dy 2 ) + dt 2 ,
where x, y, t ∈ R.

Up to an ambient isometry we can assume that Γ ⊂ S 2 corresponds to the complete geodesic defined by y = 0 and that p = (0, 0, 0) ∈ Γ is the south pole of S 2 . Therefore the axis is R = {(0, 0, t), t ∈ R}.

Remark 2. Let us remark that for any given curve α in the cylinder C, the surface generated by rotating α around the axis through the south pole, that is R, is the same as the one generated by rotating α around the axis through the north pole.

We consider the vertical (noncomplete) geodesic plane P = {y = 0} ⊂ C, that is C = P ∪ ({N } × R), where N ∈ S 2 is the north pole.

Let ρ ∈ ]-π, π[ denote the signed distance to the origin (0, 0) on Γ ∩ P . Thus we have x = tan(ρ/2). In the coordinates (ρ, t) the metric on the plane P writes

ds 2 = dρ 2 + dt 2 .
Consider now a smooth curve α(s) = (ρ(s), t(s)) parametrized by arclength in P . Let θ(s) be the oriented angle between the ρ-axis and α ′ (s). Therefore, we have:

ρ ′ (s) = cos θ(s) t ′ (s) = sin θ(s) (7)
In the plane P we consider the unit normal N to the curve α so that the basis (α ′ (s), N (s)) is positively oriented for each s. We orient by N the symmetric surface generated by α. The principal curvatures computed with respect to this orientation are as follows:

     λ 1 (s) = θ ′ (s) λ 2 (s) = sin θ(s) tan ρ(s) .
Thus, the umbilicity condition is:

θ ′ (s) = sin θ(s) tan ρ(s) • (8)
A priori the equation ( 8) makes sense only for ρ = 0, but as we will see later the surfaces we obtain are regular even at such points.

Differentiating the first equation in [START_REF] Daniel | Isometric immersions into 3-dimensional homogeneous manifolds[END_REF] and using equation [START_REF] Earp | Parabolic and hyperbolic screw motion surfaces in H 2 × R[END_REF] we get

ρ ′′ = (ρ ′2 -1) tan ρ • (9)
Assume that ρ ′ (s 0 ) = 1 for some s 0 where ρ(s 0 ) = 0. Note that the function f (s) = ρ(s 0 ) + s -s 0 is a solution of (9) with the same initial conditions at s 0 than ρ. Therefore ρ ≡ f and t ′ ≡ 0 and the surface is part of a slice S 2 × {t 0 }. The same happens in case where ρ ′ (s 0 ) = -1. Henceforth we will assume that ρ ′2 (s) = 1 for all s and ( 9) is equivalent to ρ ′′ (ρ ′2 -1) = cos ρ sin ρ •

Multiplying both sides by 2ρ ′ and integrating we get

ρ ′2 -1 = λ sin 2 ρ,
for some nonzero real constant λ. Since the curve α is parametrized by arclength we must have ρ ′2 < 1. Thus λ = -a 2 for some a > 0.

Conversely, any solution ρ of (9) satisfying ρ ′2 ≤ 1 defines a function θ(s) setting cos θ(s) = ρ ′ (s). Consider the function t defined by setting t ′ (s) = a sin ρ(s) and t(s 0 ) = t 0 for some s 0 in the domain of ρ and some real number t 0 . Then t satisfies the second equation of [START_REF] Daniel | Isometric immersions into 3-dimensional homogeneous manifolds[END_REF] and therefore the curve α(s) = (ρ(s), t(s)) ∈ C generates a rotational totally umbilic surface in S 2 × R.

Let ρ be a solution of (9) satisfying ρ ′2 < 1. Observe that equations ( 7) and [START_REF] Earp | Parabolic and hyperbolic screw motion surfaces in H 2 × R[END_REF] show that ρ ′ cannot be identically zero on an open interval unless the generated surface is part of the totally geodesic cylinder ρ ≡ π/2. Henceforth we assume that ρ is not this trivial solution and so, up to restricting the domain of ρ, we can suppose that ρ takes its values in ]0, π/2[ or ]π/2, π[. So, we can consider an interval on which ρ ′ never vanishes. Changing s into -s if needed we can suppose that ρ ′ > 0. Therefore we get

ρ ′ = 1 -a 2 sin 2 ρ. (10)
It is interesting to note that, when a = 1, the function ρ is the Jacobi amplitude function : ρ(s) = am(s, a 2 ) and, up to the sign, we have t ′ (s) = a sn(s, a 2 ) and θ ′ (s) = a cn(s, a 2 ), where sn(s, a 2 ) = sin am(s, a 2 ) and cn(s, a 2 ) = cos am(s, a 2 ) are respectively the sinus and cosinus amplitudinis elliptic Jacobi functions, see for instance [START_REF] Abramowitz | Handbook of mathematical functions[END_REF]Chapter 16] and [18, pp 286-307]. However, for reader's convenience and to be self-contained, we will treat in a direct and elementary way this ODE. Now observe that the transformation (ρ, t) → (π-ρ, t) is an isometry which changes rotations around the axis through the south pole into rotations around the axis through the north pole. Therefore, taking into account Remark 2, we can assume that, up to an ambient isometry, ρ takes its values in ]0, π/2[.

Let us call ρ a the maximal solution of (10) extending ρ without restrictions on its values, that is for the moment we do not require that ρ a takes its values in ]-π, π[. Lemma 3. Up to a reparametrization of the form s → s + c for some real constant c, the maximal solution ρ a is defined on an interval ]-δ, δ[, where δ ∈ ]0, +∞]. Furthermore ρ a is odd and so satisfies ρ a (0) = 0 and ρ ′ a (0

) = 1. Proof. Let us call ]u, v[ the domain of ρ a where -∞ ≤ u < v ≤ +∞.
We first show that ρ a vanishes at some point. Since ρ ′ a ≤ 1 such a point clearly exists if u = -∞. In case u is finite, ρ a has a limit l ∈ [-∞, +∞[ at u as it is nondecreasing.

If l < 0 then ρ a vanishes at some point since ρ is positive. Consider now the case where l ≥ 0 and call I ⊂ ]u, v[ the domain of ρ. Suppose that, as s decreases starting from I, ρ a never vanishes. Then the function ρ ′ a increases (since ρ(I) ⊂ ]0, π/2[) and thus has a positive limit at u. But then we could extend the solution ρ a of equation [START_REF] Fernandez | Harmonic maps and constant mean curvature surfaces in H 2 × R, to appear in[END_REF] beyond u, which contradicts the maximality of ρ a .

Therefore ρ a (s 0 ) = 0 for some point s 0 . Changing s into s -s 0 we can assume that ρ a (0) = 0. The function f (s) := -ρ a (-s) is then also a solution of (10) satisfying f (0) = 0. We conclude that f ≡ ρ a , which means ρ a is odd. Lemma 4. Suppose a ∈ ]0, 1[. Then ρ a is defined on the whole of R and it gives rise to a unique, up to an ambient isometry, curve α a generating a rotational totally umbilic surface. This curve is an analytic Jordan curve in the cylinder C, it is nonhomologous to zero and symmetric with respect to the axis of rotation R. The rotational totally umbilic surface, S 1 (a), generated by α a is analytic, embedded and homeomorphic to the sphere. Moreover S 1 (a) is nonhomologous to zero in S 2 × R.

Proof. With the notations of the lemma 3, if δ < +∞ then, since ρ ′ a takes its values in ]0, 1], ρ a would have a finite limit l at δ. Since a ∈ ]0, 1[, we have 1 -a 2 sin 2 l > 0 which allows to extend ρ a beyond δ, contradicting the maximality of ρ a . This shows that δ = +∞.

Since

ρ ′ a ≥ √ 1 -a 2 > 0 and ρ a (0) = 0, there is a smallest s 1 > 0 such that ρ a (s 1 ) = π. Now let us consider the function f (s) := 2π -ρ a (2s 1 -s), s ∈ R.
We observe that f is also a solution of equation ( 10) and satisfies f (s

1 ) = 2π -ρ a (s 1 ) = π = ρ a (s 1 ). Consequently f ≡ ρ a , that is: ρ a (2s 1 -s) = 2π -ρ a (s), ∀s ∈ R. (11)
As we are interested in curves generating rotational totally umbilic surfaces, we look for a function t satisfying t ′2 = 1 -ρ ′2 a = a 2 sin 2 ρ a . Let t a be the function defined on R by setting t ′ a (s) = a sin ρ a (s) and t a (0) = 0. Since ρ a is an odd function and t a (0) = 0 we deduce that t a is an even function. Observe that the function g(s) := t a (2s 1 -s) satisfies g ′ (s) = t ′ a (s) (using equation ( 11)) and g(s 1 ) = t a (s 1 ). Thus g ≡ t a , that is t a (2s 1 -s) = t a (s) for any s ∈ R. Using the evenness of t a we get

t a (s + 2s 1 ) = t a (s), ∀s ∈ R. (12)
Using equation [START_REF] Figueroa | Invariant surfaces of the Heisenberg groups[END_REF] and the oddness of ρ a we obtain for any s ∈ R

ρ a (s) = -ρ a (-s) = -(2π -ρ a (2s 1 + s)),
and so

ρ a (s + 2s 1 ) = 2π + ρ a (s), ∀s ∈ R. ( 13
)
Now the curve α a (s) = (ρ a (s), t a (s)), s ∈ R, is a curve in the Riemannian universal cover C of C. Observe that the equations ( 12) and ( 13) show that restricting s to

rho 3 2 t 2.5 1 2 1.5 0 1 0.5 -1 0 -2 -3 a=0.9 rho 3 2 t 1 1 0.8 0 0.6 -2 0.4 -1 0.2 0 -3 a=0.5 [-s 1 , s 1 ], α a gives rise to an analytic closed curve α a in C. Since ρ a is increasing on [-s 1 , s 1 ]
and ρ a (-s 1 ) = -ρ a (s 1 ) = -π we deduce that α a is embedded and nonhomologous to zero.

As ρ a is odd and t a is even, the curve α a has the desired symmetry.

It is clear that the other choice t ′ a (s) = -a sin ρ a (s) leads to the curve deduced from α a by the isometry (ρ, t) → (ρ, -t).

Remark 5. We observe that the curve α a is globally invariant under the isometry (ρ, t) → (π -ρ mod(2π), t a (2s 0 ) -t). Lemma 6. Assume a = 1. Then ρ 1 (s) = π/2 -2 arctan e -s , s ∈ R. This gives rise to a unique, up to an ambient isometry, curve α 1 in P ⊂ C generating a rotational totally umbilic surface. The curve α 1 is complete, open, embedded and symmetric with respect to the axis of rotation R. The rotational totally umbilic surface, S 1 , generated by α 1 in S 2 × R is complete, properly embedded, analytic and homeomorphic to R 2 .

Proof. Since ρ 1 is the maximal solution of ρ ′ = 1 -sin 2 ρ satisfying ρ(0) = 0, we deduce that ρ 1 is solution of ρ ′ = cos ρ. A straightforward computation shows that the maximal solution of this last equation with the initial condition ρ(0) = 0 is

ρ 1 (s) = π 2 -2 arctan e -s , ∀s ∈ R. Note that ρ 1 takes its values in ]-π/2, π/2[.
As in the proof of the lemma 4, we can assume that t satisfies t ′ (s) = sin ρ 1 (s), up to an ambient isometry. We consider the function t 1 defined by setting t ′ 1 (s) = sin ρ 1 (s) and t 1 (0) = 0. It is straightforward to check that sin ρ 1 (s) = tanh(s) and then t 1 (s) = log cosh(s).

As ρ 1 is odd and t 1 is even, the curve α 1 has the desired symmetry. As a matter of fact, α 1 is the graph of the function t(ρ) = -log cos ρ, ρ ∈ ]-π/2, π/2[. This concludes the proof. Then the maximal solution ρ a is defined on a bounded interval ]-δ a , δ a [ where δ a is a positive number. It gives rise to a unique, up to an ambient isometry, Jordan curve α a in P ⊂ C generating a rotational totally umbilic surface. The curve α a is analytic and symmetric with respect to the axis of rotation R. The rotational totally umbilic surface, S 2 (a), generated by α a is analytic, embedded and homeomorphic to the sphere. Furthermore, S 2 (a) is homologous to zero in S 2 ×R.

Proof. Since a > 1 we deduce that ρ a (s) ∈ ]-arcsin 1/a, arcsin 1/a[. Recall that ρ a is defined on an open interval ]-δ a , δ a [, see the lemma 3. We first show that δ a is finite. Assume by contradiction δ a = +∞. Since ρ a is nondecreasing it admits a limit l ∈ ]0, arcsin 1/a] as s → +∞. Necessarily l = arcsin 1/a since otherwise it would follow from the equation ( 10) that ρ ′ a > 1 -a 2 sin 2 l > 0 for all s > 0. Therefore ρ a would not be bounded.

Using the equation ( 9) we see that for s big enough : ρ ′′ a (s) ≤ -1/(2 tan l) < 0. Consequently ρ ′ a would be negative for s big enough which is a contradiction. This proves that δ a is finite.

Let us call again l the limit of ρ a as s → δ a . If l < arcsin 1/a then we could extend the solution ρ a , which is maximal, beyond δ a , which is absurd. So l = arcsin 1/a and ρ ′ a → 0 as s → δ a .

Observe that, since the function ρ a satisfies equations ( 9) and [START_REF] Fernandez | Harmonic maps and constant mean curvature surfaces in H 2 × R, to appear in[END_REF], it satisfies also the following equation

ρ ′′ = -a 2 sin ρ cos ρ. ( 14
)
As the second member of ( 14) is bounded, its maximal solutions are defined on the whole of R. Call ρ a the maximal solution of ( 14) extending ρ a . Set f (s) := ρ a (2δ a -s). It is clear that f and ρ a satisfy equation ( 14) with the same initial conditions at δ a . Thus we have

ρ a (2δ a -s) = ρ a (s), ∀s ∈ R. (15)
As we are interested in curves generating rotational totally umbilic surfaces, we look for a function t satisfying t ′2 = 1 -ρ ′2 a = a 2 sin 2 ρ a . Let t a be the function defined by t ′ a = a sin ρ a and t a (0) = 0. As ρ a is an odd function we deduce that t a is even. Observe that the function g(s) := 2t a (δ a ) -t a (2δ a -s) satisfies g ′ (s) = t ′ a (s) (using equation ( 15)) and g(δ a ) = t a (δ a ). Thus g ≡ t a , that is

t a (2δ a -s) = 2t a (δ a ) -t a (s), ∀s ∈ R. ( 16
)
It follows from [START_REF] Nelli | Minimal surfaces in H 2 × R, Bull[END_REF] and the oddness of ρ a that ρ a (s + 4δ a ) = ρ a (s), ∀s ∈ R.

In the same way, using [START_REF] Rosenberg | Minimal surfaces in M 2 × R[END_REF] and the evenness of t a , we get

t a (s + 4δ a ) = t a (s), ∀s ∈ R.
Now the curve α a (s) = ( ρ a (s), t a (s)), s ∈ R, parametrizes a closed analytic curve in P ⊂ C. Taking into account [START_REF] Nelli | Minimal surfaces in H 2 × R, Bull[END_REF], the oddnes of ρ a , ( 16) and the evenness of t a , we deduce that the curve α a is symmetric with respect to the axis R.

Considering the fact that t a is increasing on [0, 2δ a ] and the symmetry of α a , we infer that α a defines a Jordan curve in P .

To conclude the proof observe that the other choice for t a , that is t ′ a = -a sin ρ a , leads to the curve deduced from α a by the isometry (ρ, t) → (ρ, -t). 15) and [START_REF] Rosenberg | Minimal surfaces in M 2 × R[END_REF] show that the curve α a in the lemma 7 is symmetric with respect to the horizontal reflection (ρ, t) → (ρ, 2t a (δ a ) -t) in P . Consequently, the surface S 3 (a) is symmetric with respect to the slice S 2 × {t a (δ a )}.

Furthermore, we observe that each surface S 3 (a) is contained in S 2 -× R where S 2 is the south hemispshere.

Summarizing we can state the following result.

Theorem 9. Besides the totally geodesic slices S 2 × {t 0 } and the vertical cylinder Γ × R where Γ ⊂ S 2 is a geodesic, the surfaces introduced in the lemmas 4, 6 and 7 are, up to ambient isometries, the only complete totally umbilic rotational surfaces in S 2 × R. In particular they are all embedded and homeomorphic either to R 2 or to S 2 .

Among the surfaces homeomorphic to S 2 some are homologous to zero and some are not.

Remark 10. It is interesting to observe that unlike in the case of space forms, the totally umbilic surfaces we obtained do not have constant mean curvature, except for the totally geodesic ones.

Symmetric totally umbilic surfaces in

H 2 (κ) × R
In this section, we classify the totally umbilic surfaces in H 2 × R which are invariant under a one-parameter group of isometries. The case of M 2 (κ) × R, κ < 0, is completely similar.

We recall that in H 2 there are three kinds of one-parameter families of positive isometries: the rotations around a fixed point (elliptic isometries), the translations along a fixed geodesic (hyperbolic isometries) and the "translations" along the horocycles sharing the same point at infinity (parabolic isometries). An isometry of H 2 obviously induces an isometry of H 2 × R fixing the factor R pointwise. Such an isometry of H 2 × R obtained from an elliptic (resp. parabolic, hyperbolic) isometry of H 2 will thus be called elliptic (resp. parabolic, hyperbolic).

We will see that for each of the associated families of isometries of H 2 × R there exist complete and globally invariant totally umbilic surfaces. In fact, we are going to classify all of them. More precisely, we prove they are all embedded, those which are invariant under elliptic isometries are either totally geodesic slices H 2 × {t 0 } or homeomorphic to the sphere and the remaining ones are all homeomorphic to the plane. In particular the only totally geodesic ones are the slices and the products Γ × R where Γ ⊂ H 2 is a geodesic.

We will work with the disk model for H 2 , so that

H 2 = {(x, y) ∈ R 2 , x 2 + y 2 < 1},
and the metric is

ds 2 H = 2 1 -(x 2 + y 2 ) 2 (dx 2 + dy 2 ).
Therefore the product metric on H 2 × R reads as follows:

d s2 = 2 1 -(x 2 + y 2 ) 2 (dx 2 + dy 2 ) + dt 2 ,
where (x, y) ∈ H 2 and t ∈ R. We consider the following particular geodesics of

H 2 : Γ = {(x, 0), x ∈ ]-1, 1[ } ⊂ H 2 L = {(0, y), y ∈ ]-1, 1[ } ⊂ H 2
Up to ambient isometries, we can assume that the symmetric surfaces are generated by curves in the geodesic plane

P := Γ × R ⊂ H 2 × R.
On the geodesic Γ we denote by ρ ∈ R the signed distance to the origin (0, 0), thus x = tanh ρ/2. Therefore the metric on P is

ds 2 = 2 1 -x 2 2 dx 2 + dt 2 = dρ 2 + dt 2 .
Given a curve α(s) = (ρ(s), t(s)) parametrized by arclength in P , we let θ(s) be the oriented angle between the ρ-axis and α ′ (s). Therefore, we have:

ρ ′ (s) = cos θ(s) t ′ (s) = sin θ(s) (17)
In the elliptic case, the isometries of H 2 × R under consideration are the rotations around the vertical axis R := {(0, 0)}×R. In the parabolic case, the isometries are the ones corresponding to the point at infinity (-1, 0) ∈ ∂ ∞ H 2 . Finally, the hyperbolic isometries correspond to translations along L in H 2 .

In the plane P we consider the unit normal N to the curve α so that the basis (α ′ (s), N (s)) is positively oriented for each s. In the three cases we orient by N the symmetric surface generated by α. The principal curvatures computed with respect to this orientation are as follows:

λ 1 (s) = θ ′ (s) and λ 2 (s) =        sin θ(s) tanh ρ(s) (elliptic case) sin θ(s) (parabolic case) sin θ(s) tanh ρ(s) (hyperbolic case) 4.1. Elliptic case.
The umbilicity condition is

θ ′ (s) = sin θ(s) tanh ρ(s) •
This case is similar to the case a > 1 in S 2 × R, so we will omit the details. Differentiating the first equation in [START_REF] Sanini | Gauss map of a surface of the Heisenberg group[END_REF] and using the umbilicity condition we get

ρ ′′ = (ρ ′2 -1) tanh ρ • (18)
Discarding the trivial totally geodesic surfaces H 2 × {t 0 }, we can show as in the case of S 2 × R that ρ ′2 (s) = 1 for any s such that ρ(s) = 0. Therefore we may assume that ρ ′2 < 1.

We can state the following.

Proposition 11. Any local solution of (18) satisfying ρ ′2 < 1 gives rise to a unique, up to ambient isometries, complete rotational totally umbilic and nongeodesic surface in H 2 × R. Moreover, there exists a one-parameter family of such surfaces and all of them are analytic, embedded and homeomorphic to the sphere. Besides the totally geodesic slices H 2 × {t}, these surfaces are the only complete rotational and totally umbilic surfaces in H 2 × R.

Furthermore, any rotational umbilic (including geodesic) surface in H 2 × R is, up to an ambient isometry, part of one of the above surfaces.

Proof. Let ρ be a local solution of [START_REF] Sansone | Lectures on the theory of functions of a complex variable[END_REF]. Proceeding as in the case of S 2 × R, we can suppose that ρ ′ > 0 and so

ρ ′ = 1 -b 2 sinh 2 ρ, (19)
for some real number b > 0. As in the case of S 2 × R, the functions ρ(s) and t(s) are related to the Jacobi elliptic functions as follows: ρ(s) = -iam(is, -b 2 ) and, up to the sign, t ′ (s) = ib sn(is, -b 2 ), see [START_REF] Abramowitz | Handbook of mathematical functions[END_REF]Chapter 16]. Again, we prefer to give direct and elementary arguments. Let ρ b be the maximal solution of ( 19) extending ρ. As in the proof of the lemma 3, we can prove that ρ b vanishes at some point. Thus, up to a reparametrization we can assume that ρ b (0) = 0, consequently we prove as in lemma 3 that ρ b is an odd function. Therefore ρ b is defined on an interval ]-δ b , δ b [. As in lemma 7, it can be shown that δ b is a finite positive number, that ρ b has a finite limit l = argsinh 1/b at δ b and that ρ ′ b (s) → 0 as s → δ b .

Elliptic case: b=10

From equations ( 18) and [START_REF] Scott | The geometries of 3-manifolds[END_REF] we deduce that ρ b satisfies Taking into account that t b is increasing on [0, 2δ b ], we deduce that the curve α b (s) = ( ρ b (s), t(s)), s ∈ R, parametrizes an analytic Jordan curve in P , symmetric with respect to the axis R.

ρ ′′ = -b 2 cosh ρ sinh ρ (20
To conclude the proof we just observe that the other choice for t b , that is t ′ b = -b sinh ρ b , leads to the curve deduced from α b by the isometry (ρ, t) → (ρ, -t).

Parabolic case.

The umbilicity condition is

θ ′ (s) = sin θ(s)
Integrating this equation we get

θ(s) = 2 arctan λe s , ∀s ∈ R,
for some real constant λ.

First observe that λ = 0 leads to the curve Γ which generates a slice H 2 × {t 0 }. Now if λ < 0 then θ is a negative function. Note that the symmetry (ρ, t) → (ρ, -t) changes θ into -θ. Therefore, up to an ambient isometry, we can assume that θ is positive and then λ > 0. Finally observe that up to the reparametrization s → s-log λ we can assume λ = 1 and then:

θ(s) = 2 arctan e s , ∀s ∈ R, (21)
Taking into account the first equation in ( 17) and ( 21) we obtain ρ ′ (s) = cos 2 arctan e s = -tanh s, ∀s ∈ R.

Thus ρ(s) = -log cosh s + µ, ∀s ∈ R,
for some real constant µ. Note that the isometries of H 2 × R obtained from the hyperbolic translations along the geodesic Γ in H 2 send any surface invariant under the parabolic isometries fixing the point at infinity (-1, 0) ∈ ∂ ∞ H 2 to a surface of the same type. Consequently, up to an ambient isometry, we can assume µ = 0. Thus

ρ(s) = -log cosh s, ∀s ∈ R,
As for the function t, taking into account the second equation in ( 17) and ( 21) we obtain t ′ (s) = sin 2 arctan e s = 2e s 1 + e 2s , ∀s ∈ R.

Integrating we get t(s) = 2 arctan e s + β, ∀s ∈ R, for some real constant β. Up to a vertical translation we can take β = -π/2 so that t(0) = 0 and t(s) = 2 arctan e sπ 2 , ∀s ∈ R.

Note that t(-s) = -t(s) and ρ(-s) = ρ(s) for any s ∈ R so that the curve L par parametrized by (ρ(s), t(s)), s ∈ R, is symmetric with respect to Γ. Summarizing we state the following proposition.

Proposition 12. Besides the slices H 2 × {t}, up to ambient isometries, there exists a unique complete totally umbilic surface S P in H 2 × R invariant under parabolic isometries. This surface is analytic, properly embedded, homeomorphic to a plane and is invariant under reflection with respect to a horizontal slice. Moreover, any totally umbilic surface invariant under parabolic isometries is either part of a slice or, up to an ambient isometry, part of this surface. Remark 13. Consider the surface M in H 2 ×R generated by the same curve L par ⊂ P under parabolic isometries fixing, now, the point at infinity (1, 0) ∈ ∂ ∞ H 2 (and not (-1, 0) as before). Observe that at each point of M the principal curvatures λ 1 , λ 2 of M are given by λ 1 = λ 1 and λ 2 = -λ 2 . Therefore we get

λ 1 + λ 2 = λ 1 -λ 2 ≡ 0.
We deduce that M is a complete minimal surface of H 2 × R, embedded and invariant under parabolic isometries. Consequently, M is foliated by horocycles. This minimal surface was considered by B. Daniel [START_REF] Daniel | Isometric immersions into S n ×R and H n ×R and applications to minimal surfaces[END_REF], R. Sa Earp [START_REF] Earp | Parabolic and hyperbolic screw motion surfaces in H 2 × R[END_REF] and L. Hauswirth [START_REF] Hauswirth | Minimal surfaces of Riemann type in three-dimensional product manifolds[END_REF].

Hyperbolic case.

The umbilicity condition is

θ ′ (s) = sin θ(s) tanh ρ(s)
Proceeding as in the elliptic case, we discard the totally geodesic surfaces H 2 × {t} and therefore we can we assume that ρ satisfies ρ ′2 < 1 and:

ρ ′′ = (ρ ′2 -1) tanh ρ, ρ ′2 -1 = -c 2 cosh 2 ρ, (22)
for some real constant c ∈ ]0, 1[. Again, invoking the Jacobi elliptic functions, it can be shown that, up to the sign, we have ρ(s) = iam(is + K, c 2 ) -iπ/2 and

t ′ (s) = c sn(is + K, c 2 ) where K = π/2 0 dt √ 1-c 2 sin 2 t
, see [START_REF] Abramowitz | Handbook of mathematical functions[END_REF]Chapter 16]. Nevertheless, as in the previuous cases, we prefer to give direct and elementary arguments. We deduce from ( 22) that

ρ ′′ = -c 2 cosh ρ sinh ρ. (23)
If ρ ≡ 0, then the generated surface is the vertical totally geodesic plane {(0, y, t), -1 < y < 1, t ∈ R}. Discarding this case, we consider only the nontrivial solutions of (23). Again, as in the elliptic case, it can be shown that any maximal solution of the last equation is defined on the whole of R, is periodic, vanishes at some point and, up to a reparametrization, is odd. We can therefore assume that there exists a unique maximal solution ρ c satisfying ρ(0) = 0 and ρ ′ (0) = √ 1 -c 2 . As we are interested in curves generating totally umbilic surfaces, we look for a function t satisfying t ′2 = 1 -ρ ′2 c = c 2 cosh 2 ρ c . Let t c be the function defined by t ′ c = c cosh ρ c and t c (0) = 0. The function t c is odd. Moreover consider any s 0 ∈ R such that ρ ′ c (s 0 ) = 0, then it can be shown that ρ c (2s 0 -s) = ρ c (s). Set T (s) := -t c (2s 0 -s) + 2t c (s 0 ), s ∈ R. We have T ′ ≡ t ′ c and T (s 0 ) = t c (s 0 ), therefore T = t c . As the function t c is odd, we have t c (s + 4s 0 ) = 4t c (s 0 ) + t c (s) for every s ∈ R. We can deduce that t c is an increasing and nonbounded function on R. This shows that the curve α c = (ρ c , t c ) is properly embedded.

Observe that the other choice t ′ = -c cosh ρ c changes the curve α c = (ρ c , t c ) into the symmetric curve (ρ c , -t c ) with respect to Γ.

We summarize stating the following.

Proposition 14. Any non identically zero local solution of (23) satisfying ρ ′2 < 1 gives rise to a unique, up to ambient isometries, complete totally umbilic and nongeodesic surface S c , c > 0, in H 2 × R invariant under hyperbolic isometries. Moreover, there exists a one-parameter family of such surfaces and all of them are analytic, properly embedded and homeomorphic to the plane. These surfaces are periodic in the vertical direction and symmetric with respect to a discrete set of horizontal slices. Furthermore, any umbilic surface in H 2 × R invariant under hyperbolic isometries is either part of a vertical totally geodesic plane, a slice H 2 × {t} or, up to an ambient isometry, part of one of the surfaces S c .

Unicity of totally umbilic surfaces in

H 2 (κ) × R and S 2 (κ) × R
In this section M 2 stands for H 2 or S 2 . The cases M 2 = M 2 (κ) for κ < 0 and κ > 0 are completely analogous.

We will need the following result which is of independent interest.

Proposition 15. Let S ⊂ M 2 × R be an orientable surface transversal to each slice M 2 × {t}. We suppose the following:

(1) The geodesic curvature of each horizontal curve S t := S ∩ (M 2 × {t}) in M 2 is constant (depending on t).

(2) The angle between S and M 2 × {t} is constant along S t for each t.

Then:

In case M 2 = S 2 the surface S is part of rotational surface.

In case M 2 = H 2 the surface S is part of either

(1) a rotational surface, (2) or a surface invariant by a family of parabolic isometries having the same fixed point at infinity, (3) or a surface invariant by a family of hyperbolic isometries along the same fixed geodesic of H 2 .

Proof. Let N be a unit normal field along S. We define the function ν on S setting ν := N, ∂ ∂t . We denote by T the projection of ∂ ∂t on S, that is T = ∂ ∂t -νN . As the angle between S and M 2 × {t} is constant along S t , we deduce that S t is a line of curvature. Indeed let γ : s ∈ I ⊂ R → γ(s) ∈ S t be a regular parametrization of S t , then taking into account that ∂ ∂t is a parallel field and the definition of T :

0 = d ds N, ∂ ∂t = ∇ γ ′ (s) N, ∂ ∂t = ∇ γ ′ (s) N, T ,
where ∇ is the connection on M 2 × R. It follows that T is a principal direction on S. Let c : u ∈ I ⊂ R → c(u) ∈ S be a line of curvature associated to the field T . We are going to show that c(I) is contained in a vertical totally geodesic plane. This is equivalent to showing that the horizontal projection c h :

I ⊂ R → M 2 of c is a geodesic.
Assume first that c is never vertical, that is ν = 0 along c. Thus c ′ h does not vanish. Let ∇ be the connection on

M 2 . It is sufficient to show that ∇ c ′ h c ′ h is always parallel to c ′ h . As c ′ = c ′ h + (1 -ν 2 ) ∂ ∂t we have ∇ T T = ∇ c ′ c ′ = ∇ c ′ h c ′ h + d du (1 -ν 2 ) ∂ ∂t = ∇ c ′ h c ′ h -2νν ′ ∂ ∂t
As T is a principal direction there exists a function λ such that ∇ T N = λT . Therefore

ν ′ = d du N, ∂ ∂t = ∇ T N, ∂ ∂t = λ(1 -ν 2 ).
Thus we obtain:

∇ T T = ∇ c ′ h c ′ h -2λν(1 -ν 2 ) ∂ ∂t •
Moreover we have

∇ T T = ∇ T ( ∂ ∂t -νN ) = -∇ T νN = -ν ′ N -λνT = ( ν′ ν -λν)c ′ h + (-ν ′ ν -λν(1 -ν 2 )) ∂ ∂t = ( ν′ ν -λν)c ′ h -2λν(1 -ν 2 ) ∂ ∂t .
Thus we get

∇ c ′ h c ′ h = ( ν′ ν -λν)c ′ h , which shows that c h (I) is a geodesic in M 2 .
We denote by w a unit horizontal field along c tangent to S and for each u ∈ I we let P (u) be the vertical totally geodesic plane containing c(u) and orthogonal at c(u) to w(u).

Suppose now that that ν vanishes on an open interval J ⊂ I. Let u 0 ∈ J. Observe that along the horizontal curve of S through c(u 0 ) the vector field N is horizontal. This means that an open set of S, including c(J), is part of a cylinder γ × R where γ ⊂ M 2 is some horizontal curve. Clearly this implies that w is constant along J, and thus so is P .

Combining those two arguments we see that P is locally constant on an open and dense subset of I. As P (u) depends in a differentiable way on u, we conclude that P is constant.

Let us now consider a horizontal curve γ : I → S t parametrized by arclength. Let s 1 , s 2 ∈ I and call c : ]-ε, ε[→ S the integral curve of T such that c(0) = γ(s 1 ) and c : ]-ε, ε[→ S the integral curve of T such that c(0) = γ(s 2 ).

Let us call c 3 (resp. c 3 ) the vertical coordinate of c (resp. c). Calling again u the parameter in ]-ε, ε[, we have

c ′ 3 (u) = c ′ (u), ∂ ∂t = T (c(u)), ∂ ∂t = 1 -ν 2 (c(u)) = 1 -ν 2 (c 3 (u)).
Thus c 3 and c 3 verify the same first order differential equation with the same initial condition at u = 0. We conclude that c 3 ≡ c 3 .

Recall that c and c are contained in vertical totally geodesic planes P and P . Let us call Γ ⊂ M 2 the complete constant geodesic curvature line defined by γ, that is γ ⊂ Γ.

Observe that there is a unique positive isometry ϕ of M 2 such that ϕ(Γ) = Γ, ϕ(c(0)) = c(0) and preserving the orientation of Γ. Therefore the isometry Φ(z, t) = (ϕ(z), t) of M 2 × R sends P to P . Note that the curves c and Φ • c in the vertical plane P have the same vertical component and make the same angle with the horizontal for each u ∈ ]-ε, ε[. We deduce that these curves coincide: Φ • c = c. This concludes the proof.

We now state the main result of this section.

Theorem 16. Let S ⊂ M 2 × R be an immersed totally umbilic surface. Then S is part of a complete and embedded totally umbilic surface S which is invariant by a one-parameter group of isometries of M 2 × R. More precisely, up to an ambient isometry, in case M 2 = S 2 , then S is one of the examples described in the section 3, and in case M 2 = H 2 then S is one of the examples described in the section 4.

In particular, any totally geodesic surface is part of a slice M 2 × R or part of a product Γ × R where Γ ⊂ M 2 is a geodesic.

Proof. Locally S is the image of an embedding X : Ω → M 2 × R, where Ω is an open disk in R 2 . As X is totally umbilic, there exists a function λ : Ω → R such that

∇ w N = λw,
for any vector w tangent to S.

Proceeding as in the proof of Theorem 1, as τ = 0 we obtain:

∇λ = κνT, (24)
where κ is the Gaussian curvature of M 2 , that is

κ = 1 if M 2 = S 2 and κ = -1 if M 2 = H 2 .
Assume for the moment that λ has no critical point. In particular each level curve of λ is orthogonal to T and is therefore horizontal, that is belongs to some M 2 × {t 0 }. Let γ : I ⊂ M 2 × {t 0 } be such a curve parametrized by arclength. We have

dν ds = ∇ γ ′ (s) N, ∂ ∂t + N, ∇ γ ′ (s) ∂ ∂t = λ(γ(s)) γ ′ (s), ∂ ∂t = 0, therefore ν is constant along γ.
We now call n the unit normal field in T (M 2 × {t 0 }) along γ with the orientation induced by N . Let θ be the oriented angle between n and N , hence N (γ(s)) = cos θ(s)n(γ(s)) + sin θ(s) ∂ ∂t (γ(s)), we deduce that θ is constant along γ. On the other hand:

λ(γ(s)) = ∇ γ ′ (s) N, γ ′ (s) = ∇ γ ′ (s) (cos θ n + sin θ ∂ ∂t ), γ ′ (s) = cos θ ∇ γ ′ (s) n, γ ′ (s) + sin θ ∇ γ ′ (s) ∂ ∂t , γ ′ (s) = cos θ ∇ γ ′ (s) n, γ ′ (s) . Now observe that ∇ γ ′ (s) n, γ ′ (s)
is the geodesic curvature of γ in M 2 × {t 0 }. Since λ and θ are constant along γ we deduce that γ has constant geodesic curvature. We conclude using the proposition 15 and results in the sections 3 and 4 that S is as stated.

Suppose now that λ has some critical points. Let U ⊂ S be a connected component, if any, of the interior of the set of critical points of λ. The formula (24) shows that N is either always vertical or always horizontal in U . In the former case U is part of a slice M 2 × {t 0 } and in the latter case U is part of a cylinder, that is part of a product Γ × R where Γ is some curve in M 2 . As S is totally umbilic, Γ has to be a geodesic and so U is totally geodesic.

Let now V ⊂ S be a connected component, if any, of the set of regular points of λ. From the first part of the proof, we know that V is part of one the symmetric examples given in the sections 3 and 4.

Therefore S is obtained by gluing pieces of totally geodesic surfaces and pieces of the symmetric examples constructed in the sections 3 and 4. A closer look at these different types of surfaces shows that the whole of S is either totally geodesic or part of one of the complete symmetric examples, which concludes the proof.

Remark 17. The local existence of totally umbilic surfaces in S 2 × R and H 2 × R can be seen in an alternate way. Indeed, it is known that umbilicity is preserved by conformal diffeomorphisms, see [START_REF] Spivak | A comprehensive introduction to differential geometry[END_REF] (Vol. 4). It can be shown that S 2 × R is conformally diffeomorphic to R 3 \ {(0, 0, 0)}, see Section 7. This implies the umbilic surfaces in S 2 × R correspond through this conformal diffeomorphism to those of R 3 \ {(0, 0, 0)}. However to classify them in S 2 × R up to congruences, and to understand their geometry requires a nontrivial work. Regarding H 2 × R, it can be shown that H 2 × ]0, π[ is conformally diffeomorphic to H 3 , see Section 7. Nonetheless this is not enough to understand the global geometry and topology of the umbilic surfaces in H 2 × R.

Totally umbilic surfaces in Sol

The Sol geometry is the eighth model geometry of Thurston, see [START_REF] Thurston | Three-Dimensional Geometry and Topology[END_REF]. It is a Lie group endowed with a left-invariant metric, it is a homogeneous simply connected 3-manifold with a 3-dimensional isometry group, see [START_REF] Bonahon | Geometric structures on 3-manifolds, Handbook of geometric topology[END_REF]. It is isometric to R 3 equipped with the metric:

ds 2 = e 2z dx 2 + e -2z dy 2 + dz 2 .
The group structure of Sol is given by

(x ′ , y ′ , z ′ ) ⋆ (x, y, z) = (e -z ′ x + x ′ , e z ′ y + y ′ , z + z ′ ).
The isometries are: (x, y, z) → (±e -c x + a, ±e c y + b, z + c) and (x, y, z) → (±e -c y + a, ±e c x + b, -z + c), where a, b and c are any real numbers. We set

E 1 = e -z ∂ ∂x , E 2 = e z ∂ ∂y and E 3 = ∂ ∂z . Thus (E 1 , E 2 , E 3
) is a global orthonormal frame. A straightforward computation gives:

∇ E 1 E 1 = -E 3 ∇ E 2 E 1 = 0 ∇ E 3 E 1 = 0 ∇ E 1 E 2 = 0 ∇ E 2 E 2 = E 3 ∇ E 3 E 2 = 0 ∇ E 1 E 3 = E 1 ∇ E 2 E 3 = -E 2 ∇ E 3 E 3 = 0 (25)
We deduce from (25) that the vertical planes {x = x 0 } and {y = y 0 } are totally geodesic complete surfaces and that the horizontal planes {z = z 0 } are not totally umbilic surfaces (in fact they are minimal surfaces).

We now look for totally umbilic surfaces which are invariant under the one parameter group of isometries (x, y, z) → (x + c, y, z). Clearly, such a surface is generated by a curve γ in the totally geodesic plane {x = 0}. Discarding the trivial case of a vertical plane {y = y 0 }, we can assume that γ locally is a graph over the y-axis. Thus γ is given by γ(y) = (0, y, z(y)). Therefore the generated surface is parametrized by: X(t, y) := (t, y, z(y)).

We have X t = (1, 0, 0) = e z E 1 and X y = (0, 1, z ′ ) = e -z E 2 + z ′ E 3 . As a unit normal field we can take

N = e z z ′ 1 + e 2z z ′ 2 E 2 - 1 1 + e 2z z ′ 2 E 3 .
We have:

∇ Xt N = - 1 1 + e 2z z ′ 2 X t ∇ Xy N = e -z (1 + e 2z z ′ 2 ) 3/2 (1 + 2e 2z z ′ 2 + e 2z z ′′ )E 2 + z ′ (1 + e 2z z ′ 2 ) 3/2 (1 + 2e 2z z ′ 2 + e 2z z ′′ )E 3
So that X is a totally umbilic immersion if and only if

∇ Xy N = - 1 1 + e 2z z ′ 2 X y ,
that is if and only if

z ′′ + 3z ′ 2 + 2e -2z = 0. ( 26
)
A first integral of ( 26) is

z ′ 2 = ae -6z -e -2z ,
where a is any positive real number. Assume z ′ (y 0 ) = 0 for some y 0 . Considering the function f (y) = z(2y 0 -y), we can see that the curve γ is symmetric with respect to the vertical geodesic {y = y 0 }. Therefore, up to the isometry (x, y, z) → (x, -y, z) and restricting the domain of z if needed, we can assume z ′ > 0. Therefore z satisfies

z ′ = e -z ae -4z -1. (27)
We consider the maximal solution of (27) defined by z and we call it again z, it is defined on an open interval ]y 1 , y 2 [, -∞ ≤ y 1 < y 2 ≤ +∞. By (27) the function z is bounded above and is increasing we deduce using (26) that z ′′ has a negative limit at y 2 . Taking into account the fact that z ′ is positive, we deduce that y 2 is finite, y 2 < +∞. Moreover, since z ′ is a positive and decreasing function, it has a nonnegative limit at y 2 . If this limit were positive, we could extend the solution z of (27) beyond y 2 which contradicts the maximality of z. Thus we have lim y→y 2 z ′ (y) = 0 and consequently lim y→y 2 z(y) = 1 4 log a. Consider now the maximal solution of (26) defined by z and call it z a . As z ′ a (y 2 ) = 0 we have z a (2y 2 -y) = z a (y). Up to the horizontal translation (x, y, z) → (x, y -y 2 , z), which is an ambient isometry, we can assume that y 2 = 0 and therefore z a is an even fonction and is defined on an interval ]-y a , y a [ where 0 < y a ≤ +∞. Observe that there exist A > 0 and y 3 > 0 such that z ′′ a (y) < -A < 0 for any y > y 3 . Therefore, if y a = +∞ we have lim y→+∞ z a (y) = -∞. Suppose now that y a is finite. If z a had a finite limit at y a then z ′ a would have also a finite limite but then we could extend the solution z a beyond y a , which is absurd. We deduce that in both cases, that is y a < +∞ and y a = +∞, we have lim y→ya z a (y) = -∞. Now we show that y a < +∞. Indeed, as z a is a solution of (26) satisfying z a (0) = 0 we have

z ′ (y) = - √ ae -3z 1 - e 4z a ,
for any y > 0. Since lim y→ya z a (y) = -∞, we deduce that for some y 4 > 0 we have

1 √ a z ′ e 3z < - 1 2
for any y > y 4 . Therefore we have

1 3 √ a e 3z < - y 2 + c
for some real constant c and for any y > y 4 . This implies that y a < +∞. Call γ a the graph of the function z a : γ a := {(0, y, z a (y)), -y a < y < y a } and call F a the totally umbilic complete surface generated by γ a :

F a := {(x, y, z a (y)), x ∈ R, -y a < y < y a }
Let z a and z b be two maximal solutions of (26) where a and b are any real numbers.

Set c = 1 4 log b a and consider the ambient isometry (x, y, z) → (e -c x, e c y, z + c). Observe that this isometry maps the planar curve γ a onto the planar curve γ b and maps any Euclidean line parallel to the x-axis onto a line of the same type. Therefore this isometry maps the totally umbilic surface F a onto the totally umbilic surface F b .

Summarizing, we state the following result.

Proposition 18. Up to ambient isometries, there exist only two complete totally umbilic surfaces in the Sol group invariant under the horizontal translations (x, y, z) → (x + t, y, z), t ∈ R. The first one is the totally geodesic plane {y = 0}. The second one is nongeodesic, is contained in a slab delimited by two totally geodesic planes {y = ±y 0 }, y 0 > 0, and is asymptotic to these planes. Moreover it is symmetric with respect to the totally geodesic plane {y = 0}. As a matter of fact we have the following.

Theorem 19. Up to ambient isometries, any totally umbilic surface in the Sol group is part of one of the two complete totally umbilic surfaces given in the proposition 18. In particular, up to ambient isometries, there exists a unique complete totally geodesic surface in the Sol group.

Proof. Let S be an immersed totally umbilic surface in the Sol group. Locally S is the image of an embedding X : Ω → Sol, where Ω is an open disk in R 2 . Call (u, v) the coordinates on Ω and consider a unit normal field N on X(Ω). As X is totally umbilic, there exists a function λ : Ω → R such that

∇ Xu N = λX u ∇ Xv N = λX v
where ∇ is the Riemannian connection of the Sol group. As in the proof of the theorem 1 we find

R(X u , X v )N = λ v X u -λ u X v , (28)
where R denotes the curvature tensor of the Sol group. Let us express the later. let X, Y, Z and W be any vector fields. Proceeding as in [START_REF] Daniel | Isometric immersions into 3-dimensional homogeneous manifolds[END_REF], Proposition 2.1, after some computations and using (25) we obtain the following:

R(X, Y )Z, W = ( X, Z Y, W -X, W Y, Z ) + 2 X, W Y, ∂ ∂z Z, ∂ ∂z + Y, Z X, ∂ ∂z W, ∂ ∂z -X, Z Y, ∂ ∂z W, ∂ ∂z -Y, W X, ∂ ∂z Z, ∂ ∂z .
We define the function ν on Ω setting ν := N, ∂ ∂z . We denote by T the projection of ∂ ∂z on S, that is T = ∂ ∂z -νN . We then have:

R(X u , X v )N = 2ν( X v , ∂ ∂z X u -X u , ∂ ∂z X v ) = 2ν( X v , T X u -X u , T X v )
From what we deduce using (28):

∇λ = 2νT. (29)
Assume first that ν and T do not vanish on Ω. Thus T is of the form

T = αE 1 + βE 2 + γE 3 ,
where α and β do not vanish simultaneously. Since

|T | 2 = 1 -ν 2 we have α 2 + β 2 = ν 2 (1 -ν 2 ). We thus have N = - α ν E 1 - β ν E 2 + νE 3 .
We set

JT = - β ν E 1 + α ν E 2
therefore JT is tangent to S, orthogonal to T and E 3 and satisfies |JT | 2 = |T | 2 . Furthermore we have

N ∧ T = JT, T ∧ JT = (1 -ν 2 )N, JT ∧ N = T.
We now compute the derivative [T, JT ](λ) in two different ways. We first compute [T, JT ] = ∇ T JT -∇ JT T . We have

∇ T JT = ∇ T N ∧ T + N ∧ ∇ T T = N ∧ ∇ T T (since ∇ T N = λN ) = N ∧ ∇ T (E 3 -νN ) = N ∧ ∇ T E 3 -λνJT
Furthermore, using (25) we obtain

∇ T E 3 = α∇ E 1 E 3 + β∇ E 2 E 3 + (1 -ν 2 )∇ E 3 E 3 = αE 1 -βE 2
from what we deduce after some straightforward computations

∇ T E 3 = α 2 -β 2 1 -ν 2 T -2 αβ ν(1 -ν 2 ) JT -ν(1 -ν 2 )N.
Consequently:

∇ T JT = 2 αβ ν(1 -ν 2 ) T + α 2 -β 2 1 -ν 2 JT -λνJT.
In the same way we obtain:

∇ JT T = ∇ JT E 3 -∇ JT νN = -2 αβ ν(1 -ν 2 ) T + νJT -λνJT We deduce that [T, JT ] = 4 αβ ν(1 -ν 2 ) T + ( α 2 -β 2 1 -ν 2 -ν)JT.
Using this last expression and (29), we find

[T, JT ](λ) = 8αβ. ( 30
)
On the other hand, using again (29) we have

[T, JT ](λ) = T (JT (λ)) -JT (T (λ)) = -JT ( ∇λ, T ) = 2(-1 + 3ν 2 )JT (ν) = 4 αβ ν (-1 + 3ν 2 ) (31)
since an easy computation gives JT (ν) = 2αβ/ν. From (30) and (31) we deduce

αβ(3ν 2 -2ν -1) = 0.
Observe that if ν is constant on an open subset then JT (ν) ≡ 0 which implies αβ ≡ 0. So in all cases we have αβ ≡ 0.

Recall that α and β do not vanish simultaneously since by our assumption ν = 0. Therefore we have either α ≡ 0 or β ≡ 0.

Considering the isometry (x, y, z) → (y, x, -z) we can assume that α ≡ 0. Then the surface is part of a product R×Γ where Γ is a curve in the geodesic plane {x = 0}. This case is considered in the proposition [START_REF] Sansone | Lectures on the theory of functions of a complex variable[END_REF].

Let us suppose now that T ≡ 0 on an open set. Then this open set is part of a horizontal plane {z = z 0 }, but this contradicts the assumption of umbilicity.

To finish the proof we consider the case where ν ≡ 0 on an open subset. Therefore T ≡ E 3 and so this piece of the surface is part of a product L × R where L is a curve in the horizontal plane {z = 0}. If L is contained in a line parallel to the y-axis, then the surface is contained in a totally geodesic plane {x = x 0 }. Discarding this trivial case, we can assume that L is a graph over the x-axis. Consequently, the embedding X is given by X(x, t) = (x, y(x), t).

As a unit normal we take N = 1 e -2t y ′2 + e 2t (e -2t y ′ , -e 2t , 0)

= y ′ y ′2 + e 4t E 1 - 1 e -4t y ′2 + 1 E 2
As X t = E 3 , using (25) we obtain:

∇ Xt N = -2
y ′ e 4t (y ′2 + e 4t ) 3/2 E 1 -2 e -4t y ′2 (e -4t y ′2 + 1) 3/2 E 2 .

The condition ∇ Xt N = λX t is therefore equivalent to λ ≡ 0 and y ′ ≡ 0. So L is part of a line parallel to the x-axis and the surface is part of a geodesic plane {y = y 0 }. This concludes the proof.

Remark 20. It can be proved that any twice differentiable totally umbilic surface in a space form, in S 2 (κ) × R or in H 2 (κ) × R, is in fact C 3 and then analytic by the previous discussions, see [START_REF] Souam | On the classification and regularity of umbilic surfaces in homogeneous 3-manifolds[END_REF].

An application

As an application of the classification of totally umbilic surfaces obtained in the previous sections, we can prove the following result: Theorem 21. Any conformal diffeomorphism of H 2 × R, S 2 × R and Sol is an isometry.

Proof. The result for S 2 × R is a consequence of the fact that the mapping: S 2 × R → R 3 -{(0, 0, 0)} (p, t) → e t p is a conformal diffeomorphism -here S 2 is viewed as the unit sphere of R 3 centered at the origin. Indeed the conformal diffeomorphisms of R 3 -{(0, 0, 0)} are the Möbius transformations fixing (0, 0, 0) or sending (0, 0, 0) to the point at infinity and these transformations correspond through the above conformal diffeomorphism to isometries of S 2 × R. We leave the details to the reader.

We now prove the result for the space H 2 × R. We claim that, except for the slices H 2 × {t 0 }, all the non-compact maximal (for the inclusion) totally umbilic surfaces in H 2 × R are conformal to C. This is clear for the products γ × R, where γ ⊂ H 2 is a geodesic. As for the surfaces invariant under a one parameter group of parabolic transformations and which are all congruent to the surface S P described in 4.2, this is seen as follows. Consider in H 3 a totally geodesic plane which we call H 2 and denote by N a unit normal along it. We let exp denote the exponential map in H 3 . Then the map:

H 2 × ]0, π[ → H 3 (p, t) → exp p (ln(tan( t 2
))N (p)) is a conformal diffeomorphism (cf. [START_REF] Souam | On the classification and regularity of umbilic surfaces in homogeneous 3-manifolds[END_REF] for the details) which sends S P onto a totally umbilic surface of H 3 with one point at infinity, that is a horosphere. It remains to consider the case of the surfaces invariant under a one parameter group of hyperbolic transformations. Consider such a surface Σ. We know that Σ is invariant under a set of reflections of H 2 × R through horizontal slices H 2 × {t = a + nb}, for all n ∈ Z, and a, b ∈ R depending on Σ. Suppose by contradiction that Σ is conformal to H 2 . The isometries of Σ induced by those reflections correspond then to conformal diffeomorphisms of H 2 and so to isometries of H 2 . In particular all the horizontal curves Σ ∩ {t = an + b} correspond to geodesics of H 2 . Now observe that all these curves are invariant by the hyperbolic isometries leaving Σ invariant. We thus get isometries of H 2 which leave (globally) invariant more than one geodesic. This is a contradiction as only the identity of H 2 has this property. Take now a conformal diffeomorphism φ of H 2 × R. Then φ sends any horizontal slice H 2 × {t 0 } to a maximal totally umbilic non-compact surface which is conformal to H 2 . From what preceeds it follows that φ sends H 2 × {t 0 } conformally to some horizontal slice H 2 × {t 1 } and so isometrically. As φ is conformal this implies that for any x ∈ H 2 × {t 0 }, the tangent map D x φ is an isometry. So φ is an isometry of H 2 × R.

The case of the Sol group is treated analogously. There are two maximal totally umbilic surfaces up to congruences. With the notations of Section 6, the first one is the totally geodesic plane {x = x 0 } and is easily seen to be isometric to H 2 . The second one is the surface parametrized by: X(t, y) = (t, y, z(y)), where t ∈ R, y ∈ ]-y 1 , y 1 [ and z is the maximal solution to the equation: z ′ = e -z √ e -4z -1 (we have chosen a = 1 with the notations of Section 6). The metric on this surface writes:

ds 2 = e 2z dt 2 + e -z dy 2 .
Making the change of coordinate ξ = e -4z(y) dy, the metric writes:

ds 2 = e 2z (dt 2 + dξ 2 ).
As the function z is bounded from above (cf. Section 6), the flat metric dt 2 + dξ 2 is complete. It follows that the surface is conformal to C. As the totally geodesic planes {x = const.} fill the whole space Sol, we conclude as before that any conformal diffeomorphism of Sol is an isometry.

Lemma 7 .

 7 Assume a > 1.

Remark 8 .

 8 The relations (

  ) with the initial conditions ρ(0) = 0 and ρ ′ (0) = 1. Therefore we can extend the solution ρ b of (20) beyond δ b . Let ρ b be the maximal solution of (20) extending ρ b . Observe that for any s 0 where ρ b ′ (s 0 ) = 0 we have the symmetry ρ b (2s 0 -s) = ρ b (s). As ρ b ′ (δ b ) = 0 and ρ b is odd, we deduce that ρ b is defined on all of R and that it is 4δ b -periodic. As we are interested in curves generating rotational totally umbilic surfaces, we look for a function t satisfying t ′2 = 1 -ρ b ′2 = b 2 sinh 2 ρ b . Let t b be the function defined by t ′ b = b sinh ρ b and t b (0) = 0, thus t b is an even function. As in the proof of lemma 7, we can show that t b satisfies t b (2δ b -s) = 2t b (δ b ) -t b (s) for any s ∈ R and that it is also 4δ b -periodic.


Peng Zhang

Emmanuel Jacques

Régis Rogel

Laurent Pichon

Olivier Bonnaud

Regis Rogel

Affiliations

Characterization and electrical modeling of polycrystalline silicon vertical thin film transistors

Keywords: vertical thin film transistors, low temperature polycrystalline silicon, density of states, conduction modeling

ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction

Polycrystalline silicon thin-film transistors (TFTs) are generic components of large-surface electronics made on low-temperature substrates such as glass and, more recently, plastics, polymers and even paper. They have met the needs of various fields, for example, by being used in radio frequency identification (RFID) tags [1], random access memories (RAM) [2], flat panel displays [3], and more particularly active matrix flat panel displays involving either liquid crystals (AMLCD [4]) or organic light emitting diodes (AMOLED [5]). For these applications, one of the objectives is to produce a high current density at the pixel level that can be obtained either by shortening the channel length of conventional TFTs with parallel conduction to the substrate surface, or by modifying the architecture with vertical conduction. Thus, vertical TFTs have been proposed, in which the source and drain are positioned on the top and at the bottom of the transistor, respectively, and the channel length is defined by the film thickness between the source and drain regions, regardless of the design rule [6]. As a result, the vertical TFT architecture can provide an ultra-short channel length, which theoretically allows a much higher current density. In our previous work, we designed and manufactured a vertical TFT by stacking three layers of in situ deposited polycrystalline silicon, as shown in figure 1(a), for which a high drain-source current for the on-state (I on ) of about 0.3 mA is obtained, but with a high off-current (I off ) measured at the minimum value (about 0.3 μA), which provides an I on /I off ratio of about 1000. This relatively low ratio is due to the large overlapping area of the source and drain, which limits its potential applications [7]. In order to avoid this phenomenon, a second architecture was proposed with an insulating thin film between source and drain films and with parallel channels deposited on the sidewalls (see figure 1(b)). The length of the channel remains the same but the conduction in the overlapping area of source and drain is considerably decreased. However, in this case there is a modification of the nature of the sidewalls underneath the channel due the stacking of source layer, undoped polysilicon layer, barrier layer, and drain layer. Because the etching rate is not the same for these materials, the interface between channel polycrystalline Si layer and the quasi-vertical sidewalls is not homogeneous with some roughness. The electrical behavior is thus different from the case of a lateral TFT.

Silicon dioxide (SiO 2 ) or silicon nitride (Si 3 N 4 ) insulators were adopted as the barrier layer between source and drain, and considering the high etching selectivity between barrier layer and the polycrystalline silicon layer, a thin SiO 2 /Si 3 N 4 layer is appropriate. On the other hand, in order to minimize the short channel effects, the channel length can be moderately increased with the presence of an undoped polycrystalline Si layer for which the thickness is well controlled. After optimization of the process, by involving SiO 2 , the barrier layer between source and drain incorporates a 100 nm SiO 2 layer with a 1 μm undoped polycrystalline Si (figure 1(c)). are then carried out to define the multi-tooth configuration as well as form the sidewalls, and to discriminate the source and drain regions. Note that, the source and drain regions are discriminated by a partial etching until reaching the bottom N-type heavily-doped polycrystalline Si layer by the aid of a laser interferometer measurement. Afterwards, an undoped polycrystalline Si layer is deposited and patterned by the third RIE step, which forms a channel at each sidewall. Prior to the gate oxide layer deposition, an essential RCA cleaning is carried out to eliminate most of the defects at the surface of the active layer. In sequence, a 70 nm gate oxide layer is deposited by APCVD and densified at 600ºC for 12 hours to eliminate most defects in the gate oxide, and the gate oxide layer is patterned by the fourth mask to show contact holes for source and drain.

By involving

Finally, a thick aluminum layer is deposited by thermal evaporation, patterned by a fifth mask and a wet etching process to define source, drain and gate electrodes. The electrical characterization is carried out by an Agilent B1500 semiconductor parameter analyzer in ambient air.

The fabricated device configuration is shown in figure 1(e), which demonstrates a typical four-tooth structure, and therefore eight channels are formed on the sidewalls. The typical measurement is based on drainon-top configuration, while the source-on-top configuration shows nearly the same electrical characteristics, demonstrating the symmetry of the fabricated structure. Figure 1(c) and 1(d) show the sidewalls based on the 100 nm SiO 2 or 200 nm Si 3 N 4 barrier layer, respectively. Due to the higher etching selectivity between SiO 2 and polycrystalline Si, higher roughness is obtained for the sidewall with SiO 2 barrier layer, in contrast, the lower etching selectivity between Si 3 N 4 and polycrystalline Si enables a much flatter sidewall, even though there is a recess at the barrier layer for both cases. Predicted by the sidewall roughness, which can greatly affect the grain nucleation, the device with Si 3 N 4 barrier enables better electrical characteristics. In the following, a simplified model is expected to explain this behavior.

Results and Discussions

The transfer characteristics of the typical vertical TFTs are shown in figure 2(a) and 2(b), where different drain-source voltages are adopted corresponding to the linear regime. Note that, from output characteristics measurement, a kink region is also observed [8]. The kink effect is due to the short channel length. Indeed, the large electric field at the drain side accelerate the carriers in the channel that collide with the crystal lattice in the pinch-off region to generate extra electron-hole pairs. This adverse effect can be attenuated either by a lightly doped drain structure (LDD) or by an extension of the channel length. Derived from the transfer characteristic curves, the high on/off current ratio reaches about 10 5 for SiO 2 gate insulator, while for the Si 3 N 4 barrier layer device, the current ratio is higher and reaches about 10 6 for lower drain-source voltages (10 mV and 100 mV). When increasing the drain-source voltages, the oncurrent increases proportionally, where the off-current increases superproportionally, which brings about a reduced current ratio. The deduced on/off current ratio I on /I off , transconductance g m , threshold voltage V th , subthreshold slope SS, and field effect mobility μ FE , from each transfer curve are shown in table 1. The threshold voltage V th , and the field effect mobility μ FE are deduced from the transistor current formula in the linear regime shown below:

(1)

2 i FE ds ds WC 1 I = - ] L 2  gs t h ds [ ( V -V )V V
Where W/L are the channel width/length ratio, C i is the gate capacitance per unit area, V gs and V ds are the gate bias and the drain-source voltage, respectively. Note that, in the calculation, the channel length is defined as the total thickness of the barrier layer and the undoped polycrystalline Si layer, which is approximately 1.1 μm for the device with 100 nm SiO 2 layer, and approximately 1.2 μm for the device with 200 nm Si 3 N 4 layer, and the roughness of the sidewall does not greatly affect the channel length. The transconductance g m is calculated from the slope of the transfer characteristics in the linear coordinate, while the subthreshold slope SS is calculated from the reciprocal of the slope in the subthreshold region of the transfer characteristics in the semi-logarithm coordinate. Note that, due to the high stability of the polycrystalline Si active layer in ambient air and the high design rule precision in the fabrication process, the fabricated device is highly reproducible, which can be verified from the relationship between different devices with different channel width/length ratios [8].

Therefore, the electrical characteristics of the adopted devices are typical and representative.

Three main remarks can be made, ( It is also noteworthy that, for the short channel device, the access resistance contributes more to the total resistance, therefore the access resistance needs to be evaluated. Since transistors operate in the linear regime, the access resistance can be deduced from the relationship between R total and 1/(V gs -V th ) as shown in Figure 2c. For the vertical TFT with Si 3 N 4 barrier layer, the deduced access resistance is approximately 225 Ω, and the intrinsic field effect mobility is approximately 15.5 cm 2 /V.s. For reference, the access resistance of a typical lateral TFT is in the order of 100 Ω, and the field effect mobility reaches in the best case 100 cm 2 /V•s [9]. Therefore, the access resistance is not important for the vertical TFT configuration, and comparatively, the lower mobility of vertical TFT is more likely to be due to the low-quality grain nucleation of the active layer on the rugged sidewall, which provides more grain boundaries. In order to further prove that the access resistance does not greatly influence the electrical parameters, such as the threshold voltage v th , a second-derivative method is proposed to extract the threshold voltage v th [10], supposing the ideal condition that the subthreshold current is zero and the drain current increases linearly with the gate bias when it is above the threshold voltage, therefore, a critical transition of the second-derivative curve for the drain current takes place at the threshold voltage. In practice, the critical transition shows as a peak with broadened spreading, with the peak value position corresponding to the threshold voltage v th . Figure 2 The bandgap density of states (DOS) was studied by the simple Suzuki method [11], for which the transfer characteristics of the typical vertical TFTs are studied and analyzed. From Poisson equation, the potential bending of the surface space charge layer u(x) is defined by:

(2)

2 2 d ( ) ( ) a u x x dx    
Where ε a is the dielectric constant of the polycrystalline Si, ρ(x) is the local space charge density at x due to the local Fermi-level shift of eu(x). The local space charge density is related to the gap state density distribution N(E) by:

(3) -eu ( ) (x)= -e ( ) F F E E x N E dE   ( )
where E F is the equilibrium Fermi-level, eu(x) is the band bending, and e is the elementary charge.

The surface potential bending u a dependence with the gate voltage V GS can be deduced from sheet conductance modulation δG [10]:

i a 0 ( ) / [exp( / ) 1] a GS bi a a a i d G u V u u eu kT G d           (4)
where G 0 corresponds to the conductance at the flat-band condition, d a is the channel thickness of polycrystalline Si (which is set to be 10 nm [12]), ε i and ε a are the relative permittivity of gate dielectric and the active layer of polycrystalline Si respectively, d a and d i are the thickness of channel and the gate dielectric layer respectively, and u bi is approximately the flat-band voltage. By iterative calculation, the relationship between u a and V G can be obtained.

Using this method, the bandgap DOS is calculated from the transfer characteristics shown in Figure 3(a), which have been plotted for both types of vertical transistors and for a lateral transistor as reference. Note that, in order to show the similar drain-source electric field, a lower drainsource voltage of 0.1 V is adopted for the vertical short-channel TFTs (L≈1μm), and a larger drain-source voltage of 1 V is adopted for the lateral TFT (L=20 μm). For the lateral TFT, as shown in table 1, the subthreshold slope SS is approximately 0.87 V/dec, the threshold voltage V th is approximately 3.7 V, the field effect mobility is approximately 81.1 cm 2 /V•s, and the smaller I off in the order of 10 pA is assumed to be due to the high-quality polycrystalline Si, especially in the increase of the average grain size and decrease of the intragrain defect density [13]. The calculated DOS for the vertical TFTs and the lateral TFT are plotted in figure 3(b).

From the DOS curves, four main comments can be made, (1) the larger E F -E shows higher DOS, which means the shallow trap density of states is In order to evaluate the DOS calculated from the Suzuki method shown above, two extra methods are also proposed to show the DOS in the deep trap regime and the shallow trap regime, respectively. The DOS in the deep trap regime is calculated from the simple subthreshold slope method, which adopts the relationship between the subthreshold slope and the deep trap density, and the relationship between the surface potential u(x) and the applied gate voltage V GS is shown below [15]: the subthreshold slope method and the Suzuki method. Indeed, the DOS calculated from the subthreshold slope method is relevant to the subthreshold region that is approximately below the gate voltage of 4 V, and it is found to be consistent with the DOS calculated from the Suzuki method, indicating the accurate DOS distribution calculation. In addition, the subthreshold slope method gives more details of DOS in this subthreshold region. From this method, the DOS is calculated to range from 5.9×10 18 eV -1 .cm -3 to 1.59×10 19 eV -1 .cm -3 in the forward subthreshold region for vertical TFT with SiO 2 barrier layer, and this value is calculated to range from 4.99×10 18 eV -1 .cm -3 to 1.89×10 19 eV -1 .cm -3 for the vertical TFT with Si 3 N 4 barrier layer. Therefore, nearly the same deep trap density is obtained again for the two kinds of vertical TFTs. However, with the further increase of the charge density in the channel layer, the DOS from the subthreshold slope method deviates from the DOS from Suzuki method.

Nevertheless, using the subthreshold slope method, the higher DOS of vertical TFTs are demonstrated again in comparison to the DOS of lateral TFT, as shown in figure 4(c).

In fact, the subthreshold slope is related to the synergistic effect of the bulk traps and the interface traps, which can be expressed as follows [16]:

(6) 2 int SS=ln10 (1 ( )) ln10 (1 )
a bulk i i kT q kT q N qN N q C q C      
Both of the deep bulk traps N bulk and the interface traps N int contribute to the trap density N □ . For vertical TFTs, the larger subthreshold slope SS is due to the calculated trap density for the vertical TFT with SiO 2 barrier layer, the vertical TFT with Si 3 N 4 barrier layer, and the lateral TFT are 4.07×10 12 eV -1 .cm -2 , 3.55×10 12 eV -1 .cm -2 , and 2.97×10 12 eV -1 .cm -2 , respectively. Considering the accumulation layer thickness of approximately 10 nm, the calculated volume trap density can only be regarded as a rough estimate for the traps slightly above the Fermi energy.

The shallow DOS trap calculated from the Suzuki method is also verified by the Grünewald method, for which a different relationship between the surface potential u a and the applied gate voltage U GS is obtained [17]. The surface potential u a as a function of gate voltage V GS is calculated as:

  Si 3 N 4 , the barrier layer incorporates a 200 nm Si 3 N 4 layer with a 1μm undoped polycrystalline Si layer (figure 1(d)). It is noteworthy that a thicker Si 3 N 4 insulating layer in comparison to SiO 2 guarantees the good insulating property of the Si 3 N 4 layer deposited by low temperature chemical vapor deposition (LPCVD) method, which does not greatly affect the electrical characteristics of the fabricated devices. Even though SiO 2 or Si 3 N 4 layers films are very thin, the sidewalls still undergo high roughness and create high density of seed sites for polycrystalline silicon active layer deposition that adversely affects the electrical properties, mainly the transconductance, the on-state and off-state currents. The fabrication process of the vertical TFT is listed as follows. After a basic RCA cleaning of the substrates, a thick SiO 2 buffer layer of 500 nm is deposited by atmospheric pressure chemical vapor deposition (APCVD) at 420ºC. Then, different layers are deposited, in the sequence of a thick N-type in situ heavily-doped polycrystalline Si layer by LPCVD, a 100 nm SiO 2 layer by APCVD or a 200 nm Si 3 N 4 layer by LPCVD, a 1 μm undoped polycrystalline Si layer, and finally another N-type in situ heavily-doped polycrystalline Si layer by LPCVD. Two reactive ion etching (RIE) steps

  (d) and 2 (e) shows the second-derivative curve for the vertical TFTs with SiO 2 and Si 3 N 4 barrier layer, respectively, for both types of vertical TFTs, they show the same threshold voltage v th of approximately 4.4 V for a drain-source voltage of 0.1 V, which is consistent with the calculation from the conventional extrapolating method shown above, indicating the independence of the electrical characteristics on the access resistance for both types of vertical TFTs.

  higher than the deep trap density of states;(2) for smaller E F -E ≤ 0.25 eV, the vertical TFTs with different barrier layers show nearly the same DOS in the order of 10 19 eV -1 .cm -3 , which proves nearly the same subthreshold slope of the two kinds of vertical TFTs;(3) for larger E F -E>0.25 eV, the vertical TFT with Si 3 N 4 barrier layer shows a lower DOS, especially when E F -E = 0.4 eV, the DOS of the Si 3 N 4 barrier layer device decreases by one decade compared to the DOS of the device with SiO 2 barrier layer, which proves the greater field effect mobility of the device with Si 3 N 4 barrier layer reported in table 1; (4) the DOS for the lateral TFT is much smaller than the DOS of the two vertical TFTs, in the smaller E F -E zone, the DOS is approximately 10 18 eV -1 .cm-3 , where for E F -E =0.4 eV, the DOS is ten times smaller than the DOS of vertical TFT with Si 3 N 4 barrier layer, and one hundred times smaller than the DOS of vertical TFT with SiO 2 barrier layer. Indeed, strained and dangling bonds correspond respectively to shallow and deep traps, and two types of density distributions are taken into account: exponential band tail states near the mobility edge and Gaussian distribution with a maximum near the midgap[14]. From our DOS calculation, the vertical TFT with Si 3 N 4 shows smaller DOS in comparison with the vertical TFT with SiO 2 barrier layer, and the lateral TFT shows smaller DOS than vertical TFTs, indicating a smaller density of strained bonds and dangling bonds, which is closely associated with the density of the grain boundaries in polycrystalline Si active layer. In addition, for the same DOS, the larger E F -E for the lateral TFT in comparison with vertical TFT indicates the closer of the tail states to the mobility edge, which is beneficial for enlarging the field effect mobility.

Figure 4

 4 Figure 4(a) and 4(b) shows comparisons of the DOS calculated from

Figure 1 :

 1 Figure 1: Images of the vertical TFT device, (a) the first vertical TFT

Figure 2 :

 2 Figure 2: Electrical characteristics of the vertical TFTs, (a) transfer

Figure 3 :

 3 Figure 3: Transfer characteristics and DOS calculation by Suzuki method

Figure 4 :

 4 Figure 4: Subthreshold slope method for DOS calculation, (a) DOS

Figure 5 :

 5 Figure 5: Grünewald method for DOS calculation, (a) DOS calculated by

Figure 1 Figure 2

 12 Figure 1

Figure 3 Figure 4

 34 Figure 3

Figure 5 Dr 1 . 3 .

 513 Figure 5

  1) for the threshold voltage V th , the devices with SiO 2 barrier layer and with Si 3 N 4 barrier layer have approximately the same threshold voltage of approximately 4 V, with a slightly higher value for the device with Si 3 N 4 insulating layer that does not greatly affect the electrical characteristics; (2) with regard to the subthreshold slope SS, the device with SiO 2 barrier layer has a SS comparable to that of the device with Si 3 N 4 barrier layer. For low drainsource voltages, the SS of the device with the SiO 2 barrier layer is greater than that of the device with the Si 3 N 4 barrier layer, while for high drain source voltage, the SS of the device with SiO 2 barrier layer is smaller than that of the device with Si 3 N 4 barrier layer; (3) the transconductance and the field effect mobility of the device with Si 3 N 4 barrier layer are greater than that of the device with SiO 2 barrier layer.

The first point indicates similar defects or impurities density in the vertical devices, while the second point indicates similar deep trap densities in the two types of vertical TFTs. In contrast, the third point indicates a higher shallow trap density in TFT with SiO 2 barrier layer, in which carriers must fill the shallow traps, knowing that the remaining untrapped carriers contribute to field effect mobility.

Table 1 :

 1 Electrical parameters for vertical TFTs and lateral TFT deduced from transfer characteristics.

	Type	V ds (V)	I on (A)	I off (A)	I on /I off	g m (μS)	V th (V)	SS (V/dec)	μ FE (cm 2 /V•s)
	SiO 2	0.01	1.94x10 -6	7.30 x10 -12	2.66 x10 5	0.104	4.022	1.04	11.6
	barrier	0.1	1.88 x10 -5	9.49 x10 -11	1.95 x10 5	0.984	4.085	1.18	11.0
	layer	1	1.86 x10 -4	2.35 x10 -9	0.79 x10 5	9.855	3.779	1.12	11.0
	Si 3 N 4	0.01	2.44 x10 -6	1.73 x10 -12	1.41 x10 6	0.124	4.369	0.79	15.1
	barrier	0.1	2.44 x10 -5	2.67 x10 -11	0.914 x10 6	1.256	4.354	1.03	15.3
	layer	1	2.67 x10 -4	1.05 x10 -9	2.54 x10 5	14.27	4.133	1.27	17.4
	Lateral TFT	1	1.26 x10 -4	1.25 x10 -11	1.01 x10 7	5.615	3.675	0.87	81.1

Acknowledgement

This work was supported by the Natural Science Foundation of Jiangsu Province (BK20180762), and also sponsored by NUPTSF (Grant No. NY219099). The authors would like to thank Dr J. Lelannic for the realization of the SEM images of the devices. Members of the Microelectronics and Micro Sensors Department are also thanked for their contribution to the process and for their useful discussions.

Declaration of Interest Statement

There are no conflicts to declare.