P
-spaces of Iwasawa type 61
Using the Bianchi identity and the connection formulas we compute RY, Z H + RY, H Z
= 2 RY, H Z − RZ , H Y
= 2∇
[H,Y ]
Z − ∇
[H,Z ]
Y =
− j
Z
ad
H
Y + 1
2 λ
j
Z
Y, that substituted in the above expression, gives i as stated since R
H
= − ad
2 H
. ii-iii Assume that X ∈
v is an eigenvector of ad
H
with eigenvalue µ, and let Y ⊥ X in
v. Using the expression of γ
′ X
t, in the same way as i we get RZ t, γ
′ X
tγ
′ X
t = 1
cosh
2
µ t
dL
γ
X
t e
· R
X
Z + sinh
2
µ t R
H
Z − sinh µt RZ , X H + RZ , H X .
Hence, the expression of R
γ
′ X
t
Z t follows as claimed since RZ , X H + RZ , H X
= 2∇
[H,Z ]
X − ∇
[H,X ]
Z = 2λ∇
Z
X − µ∇
X
Z =
−λ − 1
2 µ
j
Z
X. Finally, we have
RY t, γ
′ X
tγ
′ X
t = 1
cosh
2
µ t
dL
γ
X
t e
· R
X
Y + sinh
2
µ t R
H
Y − sinh µt RY, X H + RY, H X .
In the same way as above, in the case of 2-step nilpotent n, we compute
RY, X H + RY, H X =
2∇
[H,Y ]
X − ∇
[H,X ]
Y =
− [H, [X, Y ]] +
1 2
µ [X, Y ] hX, Y i = 0,
which completes the proof of the proposition.
3. Eigenvectors and eigenvalues along Jacobi operators
In this section we assume that n = z ⊕ v is non-abelian and j
Z
is non-singular on v for
all Z ∈ z. Note that if λ is an eigenvalue of ad
H
|
z
and Z ∈ z
λ
then j
Z
|
v
µ
: v
µ
→ v
λ−µ
for any eigenvalue µ of ad
H
|
v
. In fact, for X ∈
v
µ
and Y ∈ v
µ
′
h j
Z
X, Y i = h[X, Y ], Z i 6= 0 ⇒ [X, Y ] 6= 0, and thus µ + µ
′
= λ since [X, Y ] ∈
n
µ+µ
′
and has non-zero component in z
λ
.
62 M. J. Druetta
Hence, for any eigenvalue µ of ad
H
|
v
, λ − µ is also an eigenvalue of ad
H
|
v
, λ µ and the symmetric operator j
2 Z
preserves v
µ
. Moreover, the map Z → [X, j
Z
X ] de- fines a symmetric operator on
z h[X, j
Z
X ], Z
∗
i = h j
Z
∗
X, j
Z
X i = h[X, j
Z
∗
X ], Z i such that [X, j
Z
X ] ∈ z
λ
for all X ∈ v
µ
since h[X, j
Z
X ], Z i = | j
Z
X |
2
6= 0,
[X, j
Z
X ] ∈ n
λ
the Jacobi identity and λ µ. Next, we describe the eigenvalues of the operators R
γ
′ Z
t
, R
γ
′ X
t
and the parallel vector fields along the geodesics γ
Z
t and γ
X
t for some Z ∈ z
λ
and X ∈ v
µ
. L
EMMA
2. Let Z ∈ z
λ
and X ∈ v
µ
be unit vectors. We set Y =
j
Z
X | j
Z
X |
. i If X is an eigenvector of j
2 Z
, then R
γ
′ Z
t
has an eigenvector U t = x tX t + ytY t with x t
2
+ yt
2
= 1 and associated eigenvalue a
Z
t satisfying a
Z
t cosh
2
λ t = hR
Z
Y, Y i − λ − µ
2
sinh
2
λ t −
1 2
λ − µ | j
Z
X | x t
yt sinh λt
whenever yt 6= 0. ii Assume that
n is 2-step nilpotent. If X satisfies [X, j
Z
X ] = | j
Z
X |
2
Z , then R
γ
′ X
t
has an eigenvector U t = x tZ t + ytY t, x t
2
+ yt
2
= 1, whose
associated eigenvalue a
X
t is given by a
X
t cosh
2
µ t
= 1
4 | j
Z
X |
2
− λµ − λ
2
sinh
2
µ t + λ −
1 2
µ | j
Z
X | yt
x t sinh µt or
a
X
t cosh
2
µ t
= −
3 4
| j
Z
X |
2
− λ − µµ + λ − µ sinh
2
µ t + λ −
1 2
µ | j
Z
X | x t
yt sinh µt
according to x t 6= 0 or yt 6= 0, respectively. Proof. Note that by the properties of Z and X , the spaces span{X t, Y t} and
span{Z t, Y t} are invariant under the symmetric operators R
γ
′ Z
t
and R
γ
′ X
t
, re-
spectively. The assertion of the lemma follows from the expression of R
γ
′ Z
t
and R
γ
′ X
t
given in Proposition 1 and using in each case the equalities R
γ
′ Z
t
U t = a
Z
tU t and R
γ
′ X
t
U t = a
X
tU t. Note that in the last case,
R
X
Z = 1
4 | j
Z
X |
2
Z − λµZ and R
X
Y = − 3
4 | j
Z
X |
2
Y − µλ − µY since
n is 2-step nilpotent. P
ROPOSITION
2. Let Z ∈ z
λ
be a unit vector. If X ∈ v
µ
, | X | = 1, is an eigenvec-
tor of j
2 Z
, then the parallel vector field U along the geodesic γ
Z
with U 0 = x X +y
Y
P
-spaces of Iwasawa type 63
Y =
j
Z
X | j
Z
X |
, x
2
+ y
2
= 1, is given by U t = x tX t + ytY t where
x t = x cos st − y
sin st , yt = x sin st + y
cos st, and st =
1 2
| j
Z
X | Z
t
du cosh λu
. Proof. We first note that dL
γ
Z
t e
span{X, j
Z
X } is invariant under ∇
γ
′ Z
t
since γ
′ Z
t =
1 cosh λt
Z t − tanh λt H t, ∇
Z
X = −
1 2
j
Z
X, ∇
Z
j
Z
X =
1 2
| j
Z
X |
2
X and ∇
H
= 0.
Hence the parallel vector field U along γ
Z
with U 0 = x X + y
Y is given by U t = x tX t + ytY t satisfying the equation ∇
γ
′ Z
t
U t = 0, which gives x
′
tX + x t 1
cosh λt ∇
Z
X + y
′
tY + yt 1
cosh λt ∇
Z
Y = 0 for all real t since X and Y are left invariant. Thus,
x
′
tX − x t 1
2 cosh λt j
Z
X + y
′
tY − yt 1
2 cosh λt j
Z
Y = 0, and x t, yt are solutions of the differential equations
x
′
t + | j
Z
X | 2 cosh λt
yt = 0, y
′
t − | j
Z
X | 2 cosh λt
x t = 0 since j
Z
Y = − | j
Z
X | X. By expressing these equations in the matrix form as x
′
t y
′
t =
1 2
| j
Z
X | cosh λt
− 1
1 x t
yt ,
the solutions are given by x t
yt =
Expst J x
y ,
where J = −
1 1
, st =
1 2
| j
Z
X | R
t du
cosh λu
Exp denotes the exponential map of matrices. The assertion of the proposition follows since Exps J =
cos s −
sin s sin s
cos s .
P
ROPOSITION
3. Assume that n is 2-step nilpotent. Let X ∈ v
µ
and Z ∈ z
λ
be unit vectors satisfying [X, j
Z
X ] = | j
Z
X |
2
Z . Then the parallel vector field U along the geodesic γ
X
with U 0 = x Z + y
Y Y =
j
Z
X | j
Z
X |
, x
2
+ y
2
= 1, is given by
U t = x tZ t + ytY t where x t = x
cos st − y sin st , yt = x
sin st + y cos st and
st = 1
2 | j
Z
X | Z
t
du cosh µu
.
64 M. J. Druetta
Proof. Note that the parallel displacement U t of U 0 along γ
X
t is expressed as U t = x tZ t+ ytY t, since dL
γ
X
t e
span{Z , j
Z
X } is invariant under ∇
γ
′ X
t
. In the same way as in Proposition 2, the equation ∇
γ
′ X
t
U t = 0 gives x
′
tZ − x t | j
Z
X | 2 cosh µt
Y + y
′
tY + yt | j
Z
X | 2 cosh µt
Z = 0 for all real t. Hence, x t and yt are solutions of the differential equations
x
′
t + | j
Z
X | 2 cosh µt
yt = 0, y
′
t − | j
Z
X | 2 cosh µt
x t = 0, which are given by
x t = x cos st − y
sin st, yt = x sin st + y
cos st with st =
1 2
| j
Z
X | R
t du
cosh µu
.
4. Condition P on Lie algebras of Iwasawa type and rank one
