Example of Extrinsically Homogeneous Real Hypersurface in H_{3}(C)

Hyang Sook Kim

Abstract

The purpose of this paper is to give an example of extrinsically homogeneous real hypersurface in a complex hyperbolic space H_{3}(C), which is an orbit under a solvable Lie subgroup of the isometry group of H_{3}(C) and not a Hopf hypersurface.

Mathematics Subject Classification: 53C40, 53C15 Key words: extrinsically homogeneous, Hopf hypersurface, principal structure vector, hyperbolic complex space form, symmetric space

Introduction

Let H_{n}(C) be the hyperbolic complex space form of complex dimension n(≥ 2) endowed with the metric of constant holomorphic sectional curvature c. A submanifold in H_{n}(C) is said to be extrinsically homogeneous if it is an orbit under a closed subgroup of the group of isometries on H_{n}(C). If the structure vector field of a real hypersurface M in H_{n}(C) is principal, then M is called a Hopf hypersurface.

As proposed also in R. Niebergall and P. J. Ryan ([3]), the following is an open problem : Classify all extrinsically homogeneous real hypersurfaces in H_{n}(C).

As a partial answer of this problem, there is a theorem of J. Berndt ([1]) to the effect that if M is an extrinsically homogeneous real hypersurface in H_{n}(C) and M is a Hopf hypersurface, then M is congruent to one of well-known homogeneous model spaces of A_{0}, A_{1}, A_{2} and B type.

In this paper, we shall give an example of extrinsically homogeneous real hypersurface in H_{3}(C) which is not a Hopf hypersurface.

1 A constuction of an example

At first we construct an example of extrinsically homogeneous real hypersurface in H_{3}(C). Basically we shall adopt the notations in S. Helgason ([2]).

Balkan Journal of Geometry and Its Applications, Vol.6, No.1, 2001, pp. 23-28 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

Let GL(4, C) be the general linear group of degree 4 over C, and E_{jk} the element (δ_{ja}δ_{kb})_{1≤a, b≤4} of GL(4, C), where 1 ≤ j, k ≤ 4. For I = E_{11} − E_{22} − E_{33} − E_{44}, we put G = {σ ∈ GL(4, C)|^{t}σ I σ̄ = I, det σ = 1} and

K = \{( \begin{matrix}σ & 0 \\ 0 & τ \end{matrix} )| σ ∈ U(1), τ ∈ U(3), det σ det τ = 1\} .

Then K is a closed subgroup of G, and the homogeneous space G/K is just the hyperbolic complex space form of complex dimension 3, which is denoted by H_{3}. The Riemannian metric and the complex structure on H_{3} will be stated later.

In the following, given a Lie group (e.g. G), we denote the associated Lie algebra of G by the corresponding bold character(e.g. g). Conversely, given a subalgebra (e.g.

l) of g, we denote by the corresponding Roman character(e.g. L) the connected Lie subgroup of G whose Lie algebra is l.

We put

\begin{matrix}A_{1} = iE_{11} − iE_{33}, & A2 = iE_{11} − iE_{22}, & A_{3} = iE_{11} − iE_{44}, \\ Y_{1} = iE_{23} + iE_{32}, & Y_{2} = E_{23} − E_{32}, & Y_{3} = iE_{24} + iE_{42}, \\ Y_{4} = E_{24} − E_{42}, & Y_{5} = iE_{34} + iE_{43}, & Y_{6} = E_{34} − E_{43}, \\ X_{1} = iE_{12} − iE_{21}, & X_{2} = iE_{13} − iE_{31}, & X_{3} = E_{13} + E_{31}, \\ X_{4} = E_{12} + E_{21}, & X_{5} = iE_{14} − iE_{41}, & X_{6} = E_{14} + E_{41}. \end{matrix}

Then the set of the above eight vectors (resp. the set {A_{1}, A_{2}, Y_{1}, Y_{2}}) forms basis for g (resp. k). By a simple computation of bracket product operation in g, we have the following table :

(1.1)  \begin{matrix}[A_{1}, A_{2}] = 0, & [A_{1}, A_{3}] = 0, & [A_{1}, Y_{1}] = −Y_{2}, \\ [A_{1}, Y_{2}] = Y_{1}, & [A_{1}, Y_{3}] = 0, & [A_{1}, Y_{4}] = 0, \\ [A_{1}, Y_{5}] = Y_{6}, & [A_{1}, Y_{6}] = −Y_{5}, & [A_{1}, X_{1}] = −X_{4}, \\ [A_{1}, X_{2}] = −2X_{3}, & [A_{1}, X_{3}] = 2X_{2}, & [A_{1}, X_{4}] = X_{1}, \\ [A_{1}, X_{5}] = −X_{6}, & [A_{1}, X_{6}] = X_{5}, & [A_{2}, A_{3}] = 0, \\ [A_{2}, Y_{1}] = Y_{2}, & [A_{2}, Y_{2}] = −Y_{1}, & [A_{2}, Y_{3}] = Y_{4}, \\ [A_{2}, Y_{4}] = −Y_{3}, & [A_{2}, Y_{5}] = 0, & [A_{2}, Y_{6}] = 0, \\ [A_{2}, X_{1}] = −2X_{4}, & [A_{2}, X_{2}] = −X_{3}, & [A_{2}, X_{3}] = X_{2}, \\ [A_{2}, X_{4}] = 2X_{1}, & [A_{2}, X_{5}] = −X_{6}, & [A_{2}, X_{6}] = X_{5}, \\ [A_{3}, Y_{1}] = 0, & [A_{3}, Y_{2}] = 0, & [A_{3}, Y_{3}] = −Y_{4}, \\ [A_{3}, Y_{4}] = Y_{3}, & [A_{3}, Y_{5}] = −Y_{6}, & [A_{3}, Y_{6}] = Y_{5}, \\ [A_{3}, X_{1}] = −X_{4}, & [A_{3}, X_{2}] = −X_{3}, & [A_{3}, X_{3}] = X_{2}, \\ [A_{3}, X_{4}] = X_{1}, & [A_{3}, X_{5}] = −2X_{6}, & [A_{3}, X_{6}] = 2X_{5}, \\ [Y_{1}, Y_{2}] = 2A_{2} − 2A_{1}, & [Y_{1}, Y_{3}] = −Y_{6}, & [Y_{1}, Y_{4}] = Y_{5}, \\ [Y_{1}, Y_{5}] = −Y_{4}, & [Y_{1}, Y_{6}] = Y_{3}, & [Y_{1}, X_{1}] = X_{3}, \\ [Y_{1}, X_{2}] = X_{4}, & [Y_{1}, X_{3}] = −X_{1}, & [Y_{1}, X_{4}] = −X_{2}, \\ [Y_{1}, X_{5}] = 0, & [Y_{1}, X_{6}] = 0, & [Y_{2}, Y_{3}] = −Y_{5}, \\ [Y_{2}, Y_{4}] = −Y_{6}, & [Y_{2}, Y_{5}] = Y_{3}, & [Y_{2}, Y_{6}] = Y_{4}, \\ [Y_{2}, X_{1}] = −X_{2}, & [Y_{2}, X_{2}] = X_{1}, & [Y_{2}, X_{3}] = X_{4}, \\ [Y_{2}, X_{4}] = −X_{3}, & [Y_{2}, X_{5}] = 0, & [Y_{2}, X_{6}] = 0, \\ [Y_{3}, Y_{4}] = 2A_{2} − 2A_{3}, & [Y_{3}, Y_{5}] = −Y_{2}, & [Y_{3}, Y_{6}] = −Y_{1}, \\ [Y_{3}, X_{1}] = X_{6}, & [Y_{3}, X_{2}] = 0, & [Y_{3}, X_{3}] = 0, \\ [Y_{3}, X_{4}] = −X_{5}, & [Y_{3}, X_{5}] = X_{4}, & [Y_{3}, X_{6}] = −X_{1}, \\ [Y_{4}, Y_{5}] = Y_{1}, & [Y_{4}, Y_{6}] = −Y_{2}, & [Y_{4}, X_{1}] = −X_{5}, \end{matrix}

Example of Extrinsically Homogeneous Real Hypersurface in H_{3}(C) 25

\begin{matrix}[Y_{4}, X_{2}] = 0, & [Y_{4}, X_{3}] = 0, & [Y_{4}, X_{4}] = −X_{6}, \\ [Y_{4}, X_{5}] = X_{1}, & [Y_{4}, X_{6}] = X_{4}, & [Y_{5}, Y_{6}] = 2A_{1} − 2A_{3}, \\ [Y_{5}, X_{1}] = 0, & [Y_{5}, X_{2}] = X_{6}, & [Y_{5}, X_{3}] = −X_{5}, \\ [Y_{5}, X_{4}] = 0, & [Y_{5}, X_{5}] = X_{3}, & [Y_{5}, X_{6}] = −X_{2}, \\ [Y_{6}, X_{1}] = 0, & [Y_{6}, X_{2}] = −X_{5}, & [Y_{6}, X_{3}] = −X_{6}, \\ [Y_{6}, X_{4}] = 0, & [Y_{6}, X_{5}] = X_{2}, & [Y_{6}, X_{6}] = X_{3}, \\ [X_{1}, X_{2}] = Y_{2}, & [X_{1}, X_{3}] = −Y_{1}, & [X_{1}, X_{4}] = 2A_{2}, \\ [X_{1}, X_{5}] = Y_{4}, & [X_{1}, X_{6}] = −Y_{3}, & [X_{2}, X_{3}] = 2A_{1}, \\ [X_{2}, X_{4}] = −Y_{1}, & [X_{2}, X_{5}] = Y_{6}, & [X_{2}, X_{6}] = −Y_{5}, \\ [X_{3}, X_{4}] = −Y_{2}, & [X_{3}, X_{5}] = Y_{5}, & [X_{3}, X_{6}] = Y_{6}, \\ [X_{4}, X_{5}] = Y_{3}, & [X_{4}, X_{6}] = Y_{4}, & [X_{5}, X_{6}] = 2A_{3}. \end{matrix}

We put p = RX_{1} + RX_{2} + RX_{3} + RX_{4} + RX_{5} + RX_{6}. Then we have a Cartan decomposition of g

g = k + p.

For any element X of g, we denote the k (resp. p)-component of X by Xk (resp. Xp).

We shall identify p with the tangent space T_{o}(H_{3}) of H_{3} at the origin o. For a constant c(< 0), we give on H_{3}, regarded as a symmetric space, a Riemannian metric ⟨ , ⟩ in such a way that

⟨\sqrt{−c} X_{i}, \sqrt{−c} X_{j}⟩ = δ_{ij} (1 ≤ i, j ≤ 6)

at o. Such an H_{3} is the hyperbolic complex space form of constant holomorphic sectional curvature 4c of complex dimension 3, which is denoted by H_{3}(C). Then G acts on H_{3}(C) as a group of isometries. The complex structure J on H_{3}(C) is given by (cf. Helgason [2], p. 393)

(1.2)  J(X_{1}) = X_{4},  J(X_{2}) = X_{3}  and  J(X_{5}) = X_{6},

where we have J = ad (−\frac{1}{4}(A_{1} + A_{2} + A_{3})).

For any element σ of G and for any 4 × 4 matrix X over C, we put

Ad(σ) X = σ X σ^{−1}.

Then Ad(σ) (σ ∈ G) is an isomorphism of G as well as an inner automorphism of g. The exponential map exp of g into G is given by

exp X = ∑_{n=0}^{∞} \frac{1}{n!}X^{n}  for  X ∈ g.

Then the followings are well-known, or can be easily checked :

(1.3)  Ad(σ) p ⊂ p for σ ∈ K,

(1.4)  \frac{d}{dt}|_{o} Ad(exp tX) Y = [X, Y] for X, Y ∈ g,

(1.5)  σ (exp X) σ^{−1} = exp(σ X σ^{−1}) for σ ∈ G, X ∈ g,

(1.6)  \begin{matrix}exp(sA_{1} + tA_{2} + uA_{3}) = e^{i(s+t+u)}E_{11} + e^{−it}E_{22} + e^{−is}E_{33} + e^{−iu}E_{44} \\ for s, t, u ∈ R, \end{matrix}

(1.7)

The group Ad(K) acts on any hypersphere of p centered at the origin transitively.

Remark 1.1. The statement (1.7) is a property of a symmetric space of rank 1. For two subalgebras l_{1} and l_{2} of g, if there is an element σ of K such that

Ad(σ) l_{1} = l_{2},

then the corresponding two orbits L_{1}(o) and L_{2}(o) are congruent in H_{3}(C) since σ(L_{1}(o)) = L_{2}(o) by (1.5).

Put Z_{1} = X_{1} + Y_{3}, Z_{2} = X_{2} + Y_{5}, Z_{3} = X_{5} − A_{3}, Z_{4} = X_{6} and Z_{5} = X_{3} − Y_{6}. Then it follows from (1.1) that

(1.8)  \begin{matrix}[Z_{1}, Z_{2}] = [Z_{1}, Z_{3}] = [Z_{1}, Z_{5}] = [Z_{2}, Z_{3}] = [Z_{3}, Z_{5}] = 0, \\ [Z_{1}, Z_{4}] = −Z_{1}, [Z_{2}, Z_{4}] = −Z_{2}, [Z_{4}, Z_{5}] = Z_{5}, \\ [Z_{3}, Z_{4}] = [Z_{2}, Z_{5}] = −2Z_{3}. \end{matrix}

If we define a subspace l of g by

(1.9)  l = RZ_{1} + RZ_{2} + RZ_{3} + RZ_{4} + RZ_{5},

then we see from (1.8) that l is a solvable Lie subalgebra of g.

Now we can state our example as follows. Theorem 1.1. Let L be the connected Lie subgroup of G whose associated Lie algebra is l given in (1.9), and denote by σ_{t} the 1-parameter subgroup exp tX_{4} of G. Then, for any t ∈ R, the orbit L(σ_{t}(o)) of the point σ_{t}(o) under L is an extrinsically homogeneous real hypersurface in H_{3}(C) whose structure vector is not principal.

In order to prove Theorem 1.1, we must make some preperations. Let ∇ be the Riemannian connection of H_{3}(C) with respect to the Riemannian metric ⟨ , ⟩. Each element X of g induces a differentiable vector field X^{∗} on H_{3}(C) such as

X_{p}^{∗} = \frac{d}{dt}|_{o}(exp tX)(p), p ∈ H_{3}(C).

Then the following results are well-known in the theory of a symmetric spaces

(1.10)  For any X ∈ g, X^{∗} is a Killing vector field on H_{3}(C),

(1.11)  [X^{∗}, Y^{∗}] = −[X, Y]^{∗} for any X, Y ∈ g,

(1.12)  ∇_{X_{o}^{∗}}Y^{∗} = 0 for any X, Y ∈ p.

It is clear that, for any X ∈ g, we have X_{o}^{∗} = Xp ∈ p ≡ T_{o}(H_{3}(C)). In particular, we see that

X_{o}^{∗} = \{ \begin{matrix}0 & if & X ∈ k \\ X & if & X ∈ p. \end{matrix}

To find out the shape operator of an orbit under G, we shall prove Proposition 1.2. Let m be any Lie subalgbra of g, and M be the corresponding analytic Lie subgroup of G such that dim M(o) ≤ 5. Let ν ∈ p be a normal vector of the orbit M(o) at o. Then the shape operator T_{ν} of M(o) in the direction ν is given by T_{ν}(Xp) = [Xk, ν]_{M} for X ∈ m, where [Xk, ν]_{M} indicates the T_{o}(M)-component of [Xk, ν]. Proof. First we assert that

(1.13)  ∇_{X_{o}^{∗}}Y^{∗} = −[X, Y] for any X ∈ p and Y ∈ k.

In fact, we have

∇_{X^{∗}}Y^{∗} = ∇_{Y^{∗}}X^{∗} + [X^{∗}, Y^{∗}] = ∇_{Y^{∗}}X^{∗} − [X, Y]^{∗}

by (1.11). Evaluating this equation at o, we obtain (1.13).

Next, by use of the result in R.Takagi and T.Takahashi ([4], 471p) and the equations (1.12) and (1.13), we get

T_{ν}(Xp) = −∇_{ν}X^{∗} = −∇_{ν_{o}^{∗}}(Xk^{∗} + Xp^{∗}) = [Xk, ν]_{M},

and this completes the proof. □ Remark 1.2. As seen from the above proof, Proposition 1.2 holds for any symmetric space and the isometry group of it. Proof of Theorem 1.1. It is clear by definition that the orbit L(σ_{t}(o)) is an extrinsically homogeneous real hypersurface in H_{3}(C).

Since the orbit L(σ_{t}(o)) is congruent to the orbit (Ad(σ_{t}^{−1})L)(o) under Ad(σ_{t}^{−1})L in H_{3}(C), we shall investigate the shape operator and the structure vector on the latter. For simplicity, we put c_{t} = cosh t and s_{t} = sinh t. Then we see that σ_{t} = c_{t}E_{11} + c_{t}E_{22} + E_{33} + E_{44} + s_{t}E_{12} + s_{t}E_{21}. By a simple calculation, we have

(1.14)  \begin{matrix}Ad(σ_{t})Z_{1} = −2s_{t}c_{t}A_{2} + c_{t}Y_{3} + (c_{t}^{2} + s_{t}^{2})X_{1} + s_{t}X_{5}, \\ Ad(σ_{t})Z_{2} = s_{t}Y_{1} + Y_{5} + c_{t}X_{2}, \\ Ad(σ_{t})Z_{3} = −s_{t}^{2}A_{2} − A_{3} + s_{t}Y_{3} + s_{t}c_{t}X_{1} + c_{t}X_{5}, \\ Ad(σ_{t})Z_{4} = s_{t}Y_{4} + c_{t}X_{6}, \\ Ad(σ_{t})Z_{5} = s_{t}Y_{2} − Y_{6} + c_{t}X_{3}. \end{matrix}

Since ν = X_{4} is the normal vector of (Ad(σ_{t}^{−1})L)(o), it follows from (1.1), Proposition 1.2 and (1.14) that

T_{ν}((c_{t}^{2} + s_{t}^{2})X_{1} + s_{t}X_{5}) = −4s_{t}c_{t}X_{1} − c_{t}X_{5},
T_{ν}(s_{t}c_{t}X_{1} + c_{t}X_{5}) = −(2s_{t}^{2} + 1)X_{1} − s_{t}X_{5},
T_{ν}(c_{t}X_{2}) = −s_{t}X_{2},
T_{ν}(c_{t}X_{3}) = −s_{t}X_{3},
T_{ν}(c_{t}X_{6}) = −s_{t}X_{6}.

Therefore, with respect to a basis {X_{1}, X_{2}, X_{3}} for T_{o}((Ad(σ_{t}^{−1})L)(o)), the shape operator T := T_{\sqrt{−c}X_{4}} of (Ad(σ_{t}^{−1})L)(o) is expressed by

(1.15)

T (X_{1}) = \sqrt{−c}(tanh ^{3}t − 3 tanh t)X_{1} − \sqrt{−c} sech^{3}t X_{5},
T (X_{2}) = −\sqrt{−c}tanht X_{2},
T (X_{3}) = −\sqrt{−c}tanht X_{3},
T (X_{5}) = −\sqrt{−c}sech ^{3}t X_{1} − \sqrt{−c} tanh^{3}t X_{5},
T (X_{6}) = −\sqrt{−c}tanht X_{6}.

Since the structure vector of (Ad(σ_{t}^{−1})L)(o) is X_{1} by (1.2) and sech t ≠ 0 for any t ∈ R, (1.15) shows that X_{1} is not principal. This completes the proof. □ Remark 1.3. By the above computation we see that the vector X_{2}, X_{3} and X_{6} are always principal, and for any t ∈ R the orbit L(σ_{t}(o)) has the following three distinct principal curvatures −\sqrt{−c} tanh t ( of multiplicity 3),

−\sqrt{−c} ( \frac{3}{2} tanh t ± \sqrt{1 − \frac{3}{4} tanh^{2} t} ). In particular, the orbit L(o) is minimal.

References

[1] J. Berndt, Real hypersurfaces with constant principal curvatures in a complex hyperbolic space, J. Reine Angew. Math., 395(1989), 132-141.

[2] S. Helgason, Differential Geometry and Symmetric Spaces, Academic Press, New York, 1978.

[3] R. Niebergall and P. J. Ryan, ”Real hypersurfaces in complex space forms”, pp 233-305 in Tight and Taut Submanifolds, edited by T. E.Cecil et al., MSRI Publications, Cambridge, 1997.

[4] R. Takagi and T. Takahashi, ”On the principal curvatures of homogeneous hypersurfaces in a sphere”, pp 469-481 in Differential Geometry (in honor of K. Yano), edited by S. Kobayashi et al. Kinokuniya, Tokyo, 1972.

Computational Mathematics

School of Computer Aided Science Inje University, Kimhae 621-749, Korea E-mail : mathkim@ijnc.inje.ac.kr