| 15 |
We need:
(180)
Ag
( Ag'(y)) = Ag"(y)
Ag(y)
= y x
g
Ag'(y)
= y x
g'
Ag
( Ag'(y)) = y x
g' x g
But:
The product of two matrices
is not commutative, in general. In conclusion:
(181)
Ag(y)
= y x g
is not a group action
: It does not satisfy the above axioms. But it satisfies another one, corresponding to
an "anti-action".
(182)
For matrices:
(183
We are still looking for
actions and anti-actions. From the x vector we can construct the transposed
one and try:
(184)
is it an action
? Let us proceed.
g" = g
x g'
(185)
(186)
Here we use a theorem of linear calculus:
(187)
M -1
x N
= ( N x
M ) -1
where M and N
are arbitrary (n,n) matrices. Whence:
(188)
g' -1
x g-1 = ( g x
g' ) -1
= g" -1
and:
(189)
which is a group
action. Consider
now:
(190)
Ag(m)
= g x mx g-1
Show that it is an action.
We shall consider the three following matrices.
(191)
g
g'
g" = g
x g'
Ag(m)
= g x
m x g-1
Ag'(m)
= **g'**x
m x
g' -1
Ag"(m)
= g" x
m x
g" -1
We need:
(192)
Ag(Ag'(m))
= Ag"(m)
Form the first member:
(193)
g x (g'
x m
x g'
-1) x g-1
or:
(194)
g x
g'x
m x
g' -1
x
g -1
i.e.:
(195)
(g x
g') x
m x ( g
x g')-1 = g" x
m x
g" -1
It is a group action.
We will call it, after Souriau:
Adjoint action
:
(193)
We are now going to consider an anti-action of the group on a matrix m.
(194)
AAg(m)
= g -1 x
m x
g
Show that it satisfies:
(195)
AAg'(Ag(m))
= Ag"(m)
Form the first member:
(196)
g'-1 x
(g -1 x
m x
g) x
g'
or:
(197)
g' -1
x
g -1
x
m x
g x
g'
i.e.:
(198)
(g x
g') -1
x
m x
( g x
g')
or:
(199)
g" -1
x
m x
g"
