Constructions of harmonic maps between Hadamard manifolds
Constructions of harmonic maps between Hadamard manifolds
Constructions of harmonic maps between Hadamard manifolds
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54 CHAPTER 2. EQUIVARIANT HARMONIC MAPS<br />
The equation (2.2.3) is a special case <strong>of</strong> the equation (1.1.1), where q = m, f 1 (t) =<br />
cosh t, f 2 (t) = sinh t, and h 1 (r) =h 2 (r) = sinh r. Since these functions satisfy the conditions<br />
(F-1) through (F-4) and (H-1) through (H-3), respectively, a solution r = r(t) to the<br />
equation (2.2.3) exists globally. Moreover, since<br />
∫ ∞<br />
dr<br />
sinh r < ∞,<br />
we have a global, strictly monotone increasing, and unbounded solution r = r(t) to the<br />
equation (2.2.3) (see Theorem A in Chapter 1).<br />
In particular, let p = q and ψ : RH p → RH p the identity map. Then, from Theorem<br />
2.1.4, we have the following<br />
Theorem 2.3.1. There exists an equivariant <strong>harmonic</strong> map u : RH m+p+1 → RH n(m)+p+1 .<br />
Here n(m) is the integer for which an eigenmap ϕ : S m → S n(m) exists.<br />
On the other hand, when m =1, there exists a family <strong>of</strong> eigen<strong>maps</strong> S 1 ∋ z ↦→ z k ∈ S 1 ,<br />
where k ∈ Z and z ∈ C. Thus we obtain<br />
Theorem 2.3.2.<br />
itself, which is parameterized by Z.<br />
Pro<strong>of</strong>.<br />
Let m ≥ 3. Then there exists a family <strong>of</strong> <strong>harmonic</strong> <strong>maps</strong> from RH m onto<br />
Since the equation (2.2.3) has a global, strictly monotone increasing, and unbounded<br />
solution r = r(t), if we define u : RH p+2 → RH p+2 by<br />
u : RH p+2 ∋ (t, x, z) ↦→ (r(t),x,z k ) ∈ RH p+2 (x ∈ RH p ,z ∈ S 1 ),<br />
then u is a surjective <strong>harmonic</strong> map.<br />
2.3.2 Warped product <strong>manifolds</strong><br />
Let M = ([0, ∞) × S m ,dt 2 + f(t) 2 g S m) and N = ([0, ∞) × S n ,dr 2 + h(r) 2 g S n), where t, r ∈<br />
[0, ∞). Let f = f(t) and h = h(r) be smooth functions on [0, ∞) satisfying f(t) > 0 and<br />
h(r) > 0 for t>0 and r>0. If f and h satisfy<br />
f(0) = 0,<br />
˙ f(0) = 1, h(0) = 0 and h ′ (0) = 1,