Constructions of harmonic maps between Hadamard manifolds

80 CHAPTER 4. NON-EXISTENCE OF PROPER HARMONIC MAPS
It is then easy to see that u is explicitly given by
u(z 1 ,z 2 )=
1
(1 −|z| 2 +2|1 − z 2 | 2 ) 2 +8(I(z 2 )) 2
× (−8 √ 2|1 − z 2 | 2 I(z 2 ), (1 −|z| 2 ) 2 − 4|1 − z 2 | 4 +8(I(z 2 )) 2 ),
where z =(z 1 ,z 2 ). Now, by a long but straightforward calculation, we can verify that u has
the following properties:
(1) u(0, 1) = (0, 1).
(2) u is Lipschitz continuous at (0, 1).
(3) u is not C 1 at (0, 1).
(4) u is not a constant map at the ideal boundary.
Therefore the C 1 -regularity up to the ideal boundary is an optimal condition for our theorem.
Now, we prove that u satisfies the properties mentioned above. Since the boundary map
of ũ : CH 2 → RH 2 is given by ũ(0,η,ξ,x)=(0, 2 √ 2x), it is clear that u : B 2 → D 2 is not
a constant map at the ideal boundary. Also it is easily verified that
(1 −|z| 2 +2|1 − z 2 | 2 ) 2 +8(Iz 2 ) 2 =0, |z| ≤1 ⇐⇒ z 1 =0,z 2 =1.
Claim 1.
Let
lim u(z 1 ,z 2 )=(0, 1).
(z 1 ,z 2 )→(0,1)
z 1 = r(θ 1 + √ −1θ 2 ), z 2 =1+r(θ 3 + √ −1θ 4 ),
where (θ 1 ) 2 +(θ 2 ) 2 +(θ 3 ) 2 +(θ 4 ) 2 =1. Note that
(z 1 ,z 2 ) ≠(0, 1) ⇐⇒ |θ| 2 := (θ 3 ) 2 +(θ 4 ) 2 ≠0.
Then it is easy to see that
1 −|z| 2 = −r(2θ 3 + r), and |1 − z 2 | 2 = r 2 |θ| 2 .