The Harmonic Mapping Whose Hopf Differential Is a Constant

: Suppose that h ( z ) is a harmonic mapping from the unit disk D to itself with respect to the hyperbolic metric. If the Hopf differential of h ( z ) is a constant c > 0, the Beltrami coefﬁcient µ ( z ) of h ( z ) is radially symmetric and takes the maximum at z = 0. Furthermore, the mapping γ : c → µ ( 0 ) is increasing and gives a homeomorphism from ( 0, + ∞ ) to ( 0, 1 )


Introduction
Let h(z) be an orientation preserving diffeomorphism from the unit disk D to itself. The mapping h(z) is harmonic with respect to the hyperbolic metric on D if it satisfies the Euler-Lagrange equation where ρ 2 (z)|dz| 2 = 4 (1−|z| 2 ) 2 |dz| 2 is the hyperbolic metric on D. The harmonic mapping is closed related to the quasiconformal mapping and Teichmüller theory. For example, the Schoen conjecture, proved by Li and Tam in [1,2] and Markovic in [3], says that for every quasisymmetric homeomorphism on the unit circle T, there exists a unique harmonic extension to the interior D.
The Hopf differential φ(z) = ρ 2 (h)h z hz of a harmonic mapping h(z) is always holomorphic by (1). Given a Hopf differential, the existence and uniqueness of the harmonic mapping is established by Wan in [4]. Also in [4], it is proved that a harmonic mapping is quasiconformal if and only if the Bers norm of the Hopf differential is bounded. Especially if the Hopf differential is constant, the harmonic mapping is quasiconformal. In the paper, we consider such mappings and prove the following result. Theorem 1. Suppose h(z) is a harmonic mapping whose Hopf differential is a constant c > 0. The Beltrami coefficient µ(z) of h(z) is a radially symmetric real function and takes the maximum at z = 0. The mapping γ : c → µ(0) is increasing and gives a homeomorphism from (0, +∞) to (0, 1).
Let h c (z) be a harmonic mapping with the Hopf differential cφ(z) between compact surfaces. In [5], Wolf proved that the Beltrami coefficient µ c (z) of h c (z) is an increasing function of c > 0, which is used to give a compactification of Teichmüller space. But the method can not be generalized to the harmonic mapping on the unit disk. In the proof of Theorem 1, we use the similar method as in [4,6], i.e., the generalized maximum principle of Cheng and Yau [7].
In the following, the Laplace-Beltrami operator is denoted by ∆ g on a Riemannian manifold (M, g). For a conformal metric σ|dz| 2 on the unit disk D, the Laplace-Beltrami operator has the following form where ∆ = ∂ 2 ∂x 2 + ∂ 2 ∂y 2 is the Laplace operator.

The Absolute Value of the Beltrami Coefficient Is Radially Symmetric
First we give some known facts about harmonic mappings. Let h(z) be a harmonic mapping with respect to the hyperbolic metric on the unit disk D. Let φ(z) be the Hopf differential and µ(z) be the Beltrami coefficient of h(z). The Bochner formula gives: By the first equality, we have ∆v = e 2v − |φ| 2 e −2v and ∆v > 0.
where e 2v = ρ 2 (h(z))|h z | 2 . The Beltrami coefficient µ(z) satisfies that On the other hand, the following theorem was proved by Wan in [4].
Theorem 2 (Theorem 12, [4]). For any holomorphic function φ(z) on D, there exists a unique smooth function v(z) such that Theorem 2 gives a way to construct a harmonic mapping from a Hopf differential. From the Hopf differential φ one has the Beltrami coefficient µ(z), then solve the Beltrami equation to obtain the harmonic mapping.
In the following, we discuss the harmonic mappings which has special Hopf differentials. First we show for a class of Hopf differentials, the metric e 2v |dz| 2 is radially symmetric. Lemma 1. Suppose the Hopf differential φ(z) is cz n where c ∈ C and n = 0, 1, 2, · · · . The metric e 2v |dz| 2 is radially symmetric, i.e., v(e iθ z) = v(z).
Proof. For the Hopf differential φ = cz n , by Theorem 2, there is a unique smooth function v(z) such that we have ∆v 1 = e 2v 1 − |cz n | 2 e −2v 1 .
By the uniqueness of the solution, the function v 1 (z) must be v(z). So v(z) is radially symmetric.
In particular, if the Hopf differential is a constant c ∈ C, the metric e 2v |dz| 2 is radially symmetric. If φ 1 (z) and φ 2 (z) are two holomorphic functions which satisfy that φ 1 (z) = e iα φ 2 (z), the corresponding v 1 (z) and v 2 (z) satisfy v 1 (z) = v 2 (z) by the uniqueness. As a result, the corresponding Beltrami coefficients satisfies that µ 1 (z) = e −iα µ 2 (z). So we only need consider the Hopf differential is a positive number c > 0. In these cases, the Beltrami coefficients are real functions.
In the following lemma, we consider the maximum of |µ(z)|.

Lemma 2.
Let φ(z) be Hopf differential of a harmonic mapping h(z) and µ(z) be the Beltrami coefficient.
If φ(z) has no zeros, the function |µ(z)| has no minimum in D.
which implies that ∆ log |µ(z)| < 0 holds for every z ∈ ∆. So log |µ(z)| is a superharmonic function on D, which implies that log |µ(z)| can not obtain the minimum in D. Then the function |µ(z)| has no minimum in D.
If the Hopf differential of a harmonic mapping h(z) is a constant c ∈ C, the absolute value of the Beltrami coefficient µ is radially symmetric by Lemma 1. Since lim |z|→1 |µ(z)| = 0, the function |µ(z)| has an absolute maximum in D. If |µ(z)| obtains a maximum at z 0 = 0, |µ(z)| also obtains a maximum at e iθ z 0 . As a result, |µ(z)| have a minimum in |z| < |z 0 |, which contradicts to Lemma 2. So |µ(z)| takes the absolute maximum only at z = 0. It is the context of the following lemma. Here we give a direct proof. Lemma 3. Suppose the Hopf differential of a harmonic mapping h(z) is a constant c ∈ C. The maximum of |µ(z)| is obtained at z = 0.
Recall that the Laplace operator in polar coordinates is given by Since τ(z) is radially symmetric, it is a function of r = |z| and satisfies The left side is equal to (rτ (r)) r , so (rτ (r)) > 0, which implies rτ (r) is a strictly increasing function of r > 0. The function τ(x) on the line (−1, 1) is an even function of x. It satisfies that τ (0) = 0.
Finally we give a geometric description of a harmonic mapping whose Hopf differential is a constant. (1). Suppose the mapping h(z) satisfies that h(0) = 0, h(1) = 1, then h(z) maps the x-axis onto x-axis and the y-axis onto y-axis.
(2). Let γ 1 and γ 2 be a pair of intersecting lines in the disk D which are horizontal and vertical respectively. Then the curves h(γ 1 ) and h(γ 2 ) are orthogonal.
Proof. (1). The mapping H(z) = h(z) is also harmonic and has the same Hopf differential with h(z). Since H(z) keeps 0, 1 fixed, we have H(z) = h(z). So h(z) maps the x-axis onto x-axis. Similarly we have h(z) = −h(−z), which implies that h(z) maps the y-axis onto y-axis. Now we prove (2). First assume that the Hopf differential of a harmonic mapping h(z) = u(x, y) + iv(x, y) is a real number. By the following calculations, we find u x u y + v x v y = 0, which implies that the curves h(γ 1 ) and h(γ 2 ) are orthogonal.

The Value µ c (0) Is Strictly Increasing
Let's recall the following lemma of Cheng and Yau [8].
Lemma 5. Let (M, g) be a complete manifold with the Ricci curvature bounded below by some constant.
where f (x) is a continuous function which is positive and non-decreasing near ∞ and satisfies that for some constants p > q. Then α is bounded from above.
The following maximum principle given by Omori [9] and Yau [7] is well-known.
Lemma 6 (Omori-Yau maximum principle). Let (M, g) be a complete manifold with the Ricci curvature of M bounded below by some constant. If α ∈ C 2 (M) is bounded from above, then for any ε > 0, there is a point Let φ be a Hopf differential and e 2v |dz| 2 be the corresponding complete metric in Theorem 2. The Gauss(Ricci) curvature of e 2v |dz| 2 is Now we can prove the following lemma.
Since K(e 2v 1 ) = |µ 1 | 2 − 1 has a lower bound, the Gauss curvature of e 2τ 1 also has a lower bound. Then we can use Lemma 5 to conclude that η is bounded from above. By Lemma 6, there is a sequence of points x k in D, such that Since the sequence {e −4τ 1 (x k ) } is bounded, we can take a subsequence such that lim k→∞ e −4τ 1 (x k ) = ξ(≤ 1). By taking limits on two sides of (3), we get 0 ≥ c(e 2η − ξe −2η − 1 + ξ). Let It is an increasing function of t and g(0) = 0. Then g(η) ≤ 0 implies thatη ≤ 0. So we have η(z) ≤ 0 and τ 2 ≤ τ 1 .
To prove that γ is strictly increasing, we use the following lemma of Heinz [10].
Lemma 8 has the following corollary. It can also be proved directly. We sketch a proof for the sake of completeness.
When the Hopf differential is equal to 0, the harmonic mapping is conformal on D. So the function γ can be defined at c = 0 with γ(0) = 0.

Lemma 10. The function γ(x) is continuous on [0, +∞).
Proof. Let φ 1 and φ 1 be two Hopf differentials whose Bers norms are bounded. Let µ i = φ i e 2v i (i = 1, 2) be the corresponding Beltrami coefficient. It is known that For its proof one can see Proposition 14 of [4]. So the function γ(x) is continuous on [0, ∞). Proof. Let φ(z) be the Hopf differential of a quasiconformal harmonic mapping and µ(z) be the Beltrami coefficient. It is known (see [4]) that if µ ∞ ≤ k < 1, the Hopf differential satisfies that Thus for φ = c, we have |φ| ρ 2 ≤ 1 1−γ(c) 2 . In particular, Proof of Theorem 1. Let h(z) be a harmonic mapping whose Hopf differential is a constant c > 0. By Lemmas 7 and 9, the function γ(x) : c → µ(0), is strictly increasing from (0, +∞) to (0, 1). To prove it is a homeomorphism, we only need show γ is continuous and surjective by Brouwer invariance of domain theorem. These are finished in Lemmas 10 and 11.

Conclusions
We have discussed a class of harmonic mappings whose Hopf differentials are constants. Let h c (z) be a harmonic mapping whose Hopf differential is a constant c ∈ (0, +∞). When c is increasing from 0 to +∞, it is proved that the vaule µ c (0) is strictly increasing. Since the Beltrami coefficient µ c (z) of h c (z) is radially symmetric and takes the maximum at z = 0, the quasiconformal distortion K c of h c (z) is also strictly increasing.