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We need:
(180)
Ag(Ag'(y)) = Ag"(y)
Ag(y) = y × g
Ag'(y) = y × g'
Ag(Ag'(y)) = y × g' × g
But:
The product of two matrices is not, in general, commutative. Consequently:
(181) Ag(y) = y × g
is not a group action: it does not satisfy the aforementioned axioms. However, it corresponds to an "anti-action":
(182)
For matrices:
(183)
We continue our search for actions and anti-actions. Starting from the vector x, we can construct its transpose and try:
(184)
Is this an action? Let's proceed.
g" = g × g'
(185)
(186)
Here we use a theorem from linear algebra:
(187) M⁻¹ × N = (N × M)⁻¹
where M and N are arbitrary (n,n) matrices. Hence:
(188) g'⁻¹ × g⁻¹ = (g × g')⁻¹ = g"⁻¹
and:
(189)
which indeed constitutes a group action. Now consider:
(190)
Ag(m) = g × m × g⁻¹
Show that this is an action. We will examine the following three matrices:
(191)
g
g'
g" = g × g'
Ag(m) = g × m × g⁻¹
Ag'(m) = g' × m × g'⁻¹
Ag"(m) = g" × m × g"⁻¹
We must verify:
(192) Ag(Ag'(m)) = Ag"(m)
Compute the left-hand side:
(193) g × (g' × m × g'⁻¹) × g⁻¹
or equivalently:
(194) g × g' × m × g'⁻¹ × g⁻¹
that is:
(195) (g × g') × m × (g × g')⁻¹ = g" × m × g"⁻¹
This is indeed a group action. Following Souriau, we shall call it the
adjoint action:
(193)
We will now consider an anti-action of the group on a matrix m.
(194) AAg(m) = g⁻¹ × m × g
Show that it satisfies:
(195) AAg'(AAg(m)) = AAg"(m)
Compute the left-hand side:
(196) g'⁻¹ × (g⁻¹ × m × g) × g
or equivalently:
(197) g'⁻¹ × g⁻¹ × m × g × g'
that is:
(198) (g × g')⁻¹ × m × (g × g')
or:
(199) g"⁻¹ × m × g"