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Constructions of harmonic maps between Hadamard manifolds

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64 CHAPTER 3. DAMEK-RICCI SPACES<br />

where y, ȳ ∈ R + and n = v + z (resp. n ′ = v ′ + z ′ ) with dim v = n (resp. dim v ′ = n ′ ),<br />

dim z = m (resp. dim z ′ = m ′ ). Take an orthonormal basis {v 1 ,... ,v n ,z 1 ,... ,z m } (resp.<br />

{¯v 1 ,... ,¯v n ′, ¯z 1 ,... ,¯z m ′}) <strong>of</strong>n (resp. n ′ ) as above, and denote their left invariant extensions<br />

on S by the same letters. Let<br />

⎧<br />

e 0 = ∂ ⎧<br />

∂y ,<br />

f 0 = ∂ ∂ȳ ,<br />

⎪⎨<br />

⎪⎨<br />

e i = v i (1 ≤ i ≤ n),<br />

f α =¯v α (1 ≤ α ≤ n ′ ),<br />

⎪⎩ e i = z i−n (n +1≤ i ≤ n + m),<br />

⎪⎩ f α =¯z α−n ′ (n ′ +1≤ α ≤ n ′ + m ′ ),<br />

and<br />

u α j = f ∗ α (du(e i)), u α ij = e j · u α i , τα (u) =f ∗ α (τ(u)),<br />

where f ∗ α is the dual frame <strong>of</strong> f α . Then, since the tension field <strong>of</strong> u is given by<br />

we get<br />

(3.2.2)<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

τ 0 (u) =<br />

τ α (u) =<br />

τ α (u) =<br />

τ(u) :=<br />

n+m<br />

∑<br />

i=0<br />

n+m<br />

∑<br />

i=0<br />

n∑<br />

( ˜∇ ei du(e i ) − du(∇ ei e i ))=0,<br />

i=1<br />

n+m<br />

∑<br />

g ii u 0 ii +(1− n − 2m)u 0 0y − (u 0 ) −1<br />

n+m<br />

∑<br />

+(u 0 ) −1<br />

i=0<br />

g ii n ′<br />

i=0<br />

∑<br />

n+m<br />

∑<br />

(u β i )2 +2(u 0 ) −3<br />

β=1<br />

i=0<br />

g ii (u 0 i ) 2<br />

∑<br />

g ii n ′ +m ′<br />

β=n ′ +1<br />

n+m<br />

g ii u α ii +(1− n − 2m)uα 0 y − ∑<br />

2(u0 ) −1 g ii u 0 i uα i<br />

n+m<br />

∑ ∑n ′<br />

+(u 0 ) −2 g ii<br />

n+m<br />

∑<br />

i=0<br />

i=0<br />

n<br />

∑<br />

′ +m ′<br />

β=1 γ=n ′ +1<br />

i=0<br />

(u β i )2 ,<br />

u γ i uβ i Γ γ−n′<br />

αβ<br />

(1 ≤ α ≤ n ′ ),<br />

n+m<br />

g ii u α ii +(1− n − 2m)u0 0 y − ∑<br />

4(u0 ) −1 g ii u 0 i uα i<br />

i=0<br />

(n ′ +1≤ α ≤ n ′ + m ′ ),<br />

where u 0 =ȳ(u), and (g ij ) denotes the matrix component <strong>of</strong> the metric g M , (g ij ) its inverse<br />

matrix. Note that u α ij ≠ uα ji because [v i,v j ] ≠0.

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