This is legitimate if Gis a group of matrices.
So γ[r, s] = 0. But {γ(r), γ(s)} 6= 0, since
γ(r) = p1
(momentum generates translations), and
γ(s) = mq1
(mass time position generates boosts), so
{γ(r), γ(s)}=
n
X
i=1
∂
∂pi
p1
∂
∂qi
mq1−∂
∂qi
p1
∂
∂pi
mq1
=m
Here m∈C∞(X) is really the constant function equal to mat all points of X. But what vector
field on Xdoes this observable generate? What is vm={m, ·}? It is zero! It generates this flow:
φ:R×X→X
(t, x)7→ x
All constant functions, or indeed, all locally constant functions fon any Poisson manifold give
vf= 0.
picture of phase space with two connected components
The problem is that this diagram:
g
γ""
E
E
E
E
E
E
E
E
E
α//Vect(X)
C∞(X)
β
99
s
s
s
s
s
s
s
s
s
s
βis not 1-1; it sends all locally constant functions to 0, so we can have
βγ[x, y] = [βγ (x), βγ (y)]
even though
γ[x, y]6= [γ(x), γ(y)].
This also means that different choices of γcan make this diagram commute.
Definition 2 If Gis a Lie group acting on a Poisson manifold X:
A:G×X→X
we say Ais weakly Hamiltonian if there exists a linear map
γ:g→C∞(X)
such that
α=βγ
with γnot necessarily a Lie algebra homomorphism.
2
docsity.com
