Constructions of harmonic maps between Hadamard manifolds

66 CHAPTER 3. DAMEK-RICCI SPACES
(2) (n +2m)(u 0 0 )4 −
n∑ ∑n ′
(u β j )2 (u 0 0 )2 − 2
j=0 β=1
(3) (1 + n +2m)u α 0 (u 0 0) 2 −
n∑ ∑n ′
n
∑
′ +m ′
j=0 β=1 γ=n ′ +1
n∑
n
∑
′ +m ′
j=0 β=n ′ +1
(4) (2 + n +2m)u 0 0u α 00 =0 (n ′ +1≤ α ≤ n ′ + m ′ ).
(u β j0 )2 − 2
n+m
∑
n
∑
′ +m ′
j=n+1 β=n ′ +1
Γ γ−n′
α β
u β j uγ j0 =0 (1≤ α ≤ n′ ),
(u β j )2 =0,
Proof. Since u ∈ C 2 ( ¯S, ¯S ′ ) is proper, we have u 0 = O(y) and lim y→0 u 0 y −1 = u 0 0.
Multiplying by (u 0 ) 3 y −2 both sides of the first equation in (3.2.2), we let y → 0. Then, by
virtue of Lemma 3.2.2, the first term on the right hand side tends to 0. Hence we obtain (1).
The rest of the statement follows similarly if we multiply the first, second and third
equations in (3.2.2) by (u 0 ) 3 y −4 , (u 0 ) 2 y −3 and u 0 y −3 , respectively.
Since u γ 0j =0(γ ≥ n′ + 1) at the boundary and ddu =0, we have
∑n ′
(3.2.3) u γ j0 = −
µ,ν=1
Γ γ−n′
µν u µ 0u ν j (0 ≤ j ≤ n, n ′ +1≤ γ ≤ n ′ + m ′ ).
Substituting this into the equation (3) of Lemma 3.2.3 and adding the result in α, we get
∑n ′
(3.2.4) (1 + n +2m)(u 0 0 )2 (u α 0 )2 +
α=1
Now we can deduce the following
n∑
n
∑
′ +m ′
j=0 γ=n ′ +1
( n ′
∑
α,β=1
Γ γ−n′
αβ
u α 0 uβ j
) 2
=0.
Proposition 3.2.4. Let u ∈ C 2 ( ¯S, ¯S ′ ) be a proper harmonic map. If f := u |∂S satisfies
n+m
∑
n
∑
′ +m ′
j=n+1 γ=n ′ +1
(f γ j )2 > 0,
then at the ideal boundary u must satisfy u 0 0 > 0,uα 0 =0(1≤ α ≤ n′ + m ′ ) and u β k0 = uβ 00 =
0(1≤ k ≤ n, n ′ +1≤ β ≤ n ′ + m ′ ).