or (
0 αδ−βγ
−αδ+βγ 0
)
=
(
0 1
−1 0
)
so
det(
α β
γ δ)
=αδ−βγ= 1This says that we can have any linear transformation with unit determinant.
In other words, we find thatSp(2,R) =SL(2,R). This isomorphism with a
special linear group occurs only ford= 1.
Now turning to the Lie algebra, for group elementsg∈GL(2,R) near the
identity,gcan be written in the formg=etLwhereLis in the Lie algebra
gl(2,R). The condition thatgacts onMpreserving Ω implies that (differenti-
ating 16.4)
d
dt(
(etL)T(
0 1
−1 0
)
etL)
= (etL)T(
LT
(
0 1
−1 0
)
+
(
0 1
−1 0
)
L
)
etL= 0Settingt= 0, the condition onLis
LT
(
0 1
−1 0
)
+
(
0 1
−1 0
)
L= 0 (16.5)
This requires thatLmust be of the form
L=
(
a b
c −a)
(16.6)
which is what one expects:Lis in the Lie algebrasl(2,R) of 2 by 2 real matrices
with zero trace.
The homogeneous degree two polynomials inpandqform a three dimen-
sional sub-Lie algebra of the Lie algebra of functions on phase space, since the
non-zero Poisson bracket relations on a basisq
2
2 ,p^2
2 ,qpare{
q^2
2,
p^2
2}=qp {qp,p^2 }= 2p^2 {qp,q^2 }=− 2 q^2We haveTheorem 16.1. The Lie algebra of degree two homogeneous polynomials on
M=R^2 is isomorphic to the Lie algebrasp(2,R) =sl(2,R), with the isomor-
phism given explicitly by
−aqp+bq^2
2−
cp^2
2=
1
2
(
q p)
L
(
0 − 1
1 0
)(
q
p)
↔L=
(
a b
c −a)
(16.7)
Proof.One can identify basis elements as follows:
q^2
2↔E=
(
0 1
0 0
)
−
p^2
2↔F=
(
0 0
1 0
)
−qp↔G=