68
S. Goto and H. Hino
One is interested in how the (1, 1)-tensor field φ discussed in Sect. 4.2.1 plays a
role for the contact Hamiltonian vector field (4.9). To give an answer, one needs the
following.
Lemma 1 ([28]). Let { ˙
x a }, { ˙
y a }, ˙
z be some functions, and X 0 the vector field
X 0 =
m
a=1
˙
x
a ∂
∂ x a + ˙
y a
∂
∂ y a
+ ˙
z
∂
∂z
.
Then, φ(X 0 ) and φ
2
(X 0 ) are calculated as
φ
μ
(X 0 ) =
m
a=1
(−1)
μ
˙
x
a
∂
∂ x a + y a
∂
∂z
+ ˙
y a
∂
∂ y a
, μ = 1, 2.
Proof With the local expressions shown in Sect. 4.2.1, one has
θ
a
± (X 0 ) =
√ y a
2
˙
x
a
±
˙
y a
2
√ y a
,
e
+
a + e
−
a =
2
√ y a
∂
∂ x a + y a
∂
∂z
, and e
+
a − e
−
a = 2
√ y a
∂
∂ y a
.
Combining these, one has
φ(X 0 ) =
m
a=1
− θ
a
− (X 0 ) e
+
a + θ
a
+ (X 0 ) e
−
a
=
m
a=1
−
√ y a
2
˙
x
a
( e
+
a + e
−
a ) +
˙
y a
2
√ y a
( e
+
a − e
−
a )
=
m
a=1
− ˙
x
a
∂
∂ x a + y a
∂
∂z
+ ˙
y a
∂
∂ y a
.
For φ
2
(X 0 ), substituting λ(X 0 ) = ˙
z −
m
a=1 y a ˙
x
a into φ
2
(X 0 ) = X 0 − λ(X 0 )ξ ,
one has the desired expression.
Applying Lemma 1, one has the following.
Proposition 5 (Roles of φ for the contact Hamiltonian system, [28]). Let X be
the contact Hamiltonian vector field in Proposition 2. Then
L φ(X ) h = L φ 2 (X ) h = 0.