Zu einer gegebenen Darstellung
ρ
:
G
→
GL
(
V
)
{\displaystyle \rho \colon G\to \operatorname {GL} (V)}
kann man die duale Darstellung
ρ
∗
:
G
→
GL
(
V
∗
)
{\displaystyle \rho ^{*}\colon G\to \operatorname {GL} (V^{*})}
in den dualen Vektorraum
V
∗
{\displaystyle V^{*}}
(
ρ
∗
(
s
)
ν
)
(
v
)
=
ν
(
ρ
(
s
−
1
)
v
)
{\displaystyle \left(\rho ^{*}(s)\nu \right)(v)=\nu \left(\rho (s^{-1})v\right)}
für alle
s
∈
G
,
v
∈
V
{\displaystyle s\in G,v\in V}
ν
∈
V
∗
.
{\displaystyle \nu \in V^{*}.}
Mit dieser Definition gilt für die natürliche Paarung
⟨
ν
,
v
⟩
:=
ν
(
v
)
{\displaystyle \langle \nu ,v\rangle :=\nu (v)}
V
∗
{\displaystyle V^{*}}
V
:
{\displaystyle V:}
⟨
ρ
∗
(
s
)
(
ν
)
,
ρ
(
s
)
(
v
)
⟩
=
⟨
ν
,
v
⟩
{\displaystyle \langle \rho ^{*}(s)(\nu ),\rho (s)(v)\rangle =\langle \nu ,v\rangle }
s
∈
G
,
v
∈
V
,
ν
∈
V
∗
.
{\displaystyle s\in G,v\in V,\nu \in V^{*}.}
Nach Wahl einer Basis und der kanonischen dualen Basis wird
ρ
(
g
)
{\displaystyle \rho (g)}
Matrix
A
{\displaystyle A}
ρ
∗
(
g
)
{\displaystyle \rho ^{*}(g)}
Transponierte der inversen Matrix beschrieben, also
ρ
∗
(
g
)
=
(
A
T
)
−
1
{\displaystyle \rho ^{*}(g)=(A^{T})^{-1}}
Beweis : Sei
v
1
,
…
,
v
n
{\displaystyle v_{1},\ldots ,v_{n}}
V
{\displaystyle V}
v
1
∗
,
…
,
v
n
∗
{\displaystyle v_{1}^{*},\ldots ,v_{n}^{*}}
V
∗
{\displaystyle V^{*}}
g
v
i
=
∑
j
a
i
j
(
g
)
v
j
∈
V
{\displaystyle gv_{i}=\sum _{j}a_{ij}(g)v_{j}\in V}
und
g
v
j
∗
=
∑
i
b
i
j
(
g
)
v
i
∗
∈
V
∗
{\displaystyle gv_{j}^{*}=\sum _{i}b_{ij}(g)v_{i}^{*}\in V^{*}}
dann ist
b
j
i
(
g
)
=
(
g
v
j
)
∗
v
i
=
v
j
∗
(
g
−
1
v
i
)
=
v
j
∗
(
∑
k
a
i
k
(
g
−
1
)
v
k
)
=
a
i
j
(
g
−
1
)
{\displaystyle b_{ji}(g)=(gv_{j})^{*}v_{i}=v_{j}^{*}(g^{-1}v_{i})=v_{j}^{*}\left(\sum _{k}a_{ik}(g^{-1})v_{k}\right)=a_{ij}(g^{-1})}
Sei
G
=
Z
/
3
Z
{\displaystyle G=\mathbb {Z} /3\mathbb {Z} }
ρ
:
Z
/
3
Z
→
GL
2
(
C
)
{\displaystyle \rho \colon \mathbb {Z} /3\mathbb {Z} \to \operatorname {GL} _{2}(\mathbb {C} )}
Z
/
3
Z
{\displaystyle \mathbb {Z} /3\mathbb {Z} }
ρ
(
0
¯
)
=
Id
,
ρ
(
1
¯
)
=
(
cos
(
2
π
3
)
−
sin
(
2
π
3
)
sin
(
2
π
3
)
cos
(
2
π
3
)
)
,
und
ρ
(
2
¯
)
=
(
cos
(
4
π
3
)
−
sin
(
4
π
3
)
sin
(
4
π
3
)
cos
(
4
π
3
)
)
.
{\displaystyle \rho ({\overline {0}})={\text{Id}},\,\,\rho ({\overline {1}})=\left({\begin{array}{cc}\cos({\frac {2\pi }{3}})&-\sin({\frac {2\pi }{3}})\\\sin({\frac {2\pi }{3}})&\cos({\frac {2\pi }{3}})\end{array}}\right),\,\,{\text{ und }}\,\,\rho ({\overline {2}})=\left({\begin{array}{cc}\cos({\frac {4\pi }{3}})&-\sin({\frac {4\pi }{3}})\\\sin({\frac {4\pi }{3}})&\cos({\frac {4\pi }{3}})\end{array}}\right).}
Dann ist die duale Darstellung
ρ
∗
:
Z
/
3
Z
→
GL
(
(
C
2
)
∗
)
{\displaystyle \rho ^{*}\colon \mathbb {Z} /3\mathbb {Z} \to \operatorname {GL} ((\mathbb {C} ^{2})^{*})}
ρ
∗
(
0
¯
)
=
Id
,
ρ
∗
(
1
¯
)
=
(
cos
(
4
π
3
)
sin
(
4
π
3
)
−
sin
(
4
π
3
)
cos
(
4
π
3
)
)
,
und
ρ
∗
(
2
¯
)
=
(
cos
(
2
π
3
)
sin
(
2
π
3
)
−
sin
(
2
π
3
)
cos
(
2
π
3
)
)
.
{\displaystyle \rho ^{*}({\overline {0}})={\text{Id}},\,\,\rho ^{*}({\overline {1}})=\left({\begin{array}{cc}\cos({\frac {4\pi }{3}})&\sin({\frac {4\pi }{3}})\\-\sin({\frac {4\pi }{3}})&\cos({\frac {4\pi }{3}})\end{array}}\right),\,\,{\text{ und }}\,\,\rho ^{*}({\overline {2}})=\left({\begin{array}{cc}\cos({\frac {2\pi }{3}})&\sin({\frac {2\pi }{3}})\\-\sin({\frac {2\pi }{3}})&\cos({\frac {2\pi }{3}})\end{array}}\right).}
Bröcker, Theodor; tom Dieck, Tammo: Representations of compact Lie groups. Graduate Texts in Mathematics, 98. Springer-Verlag, New York, 1985. ISBN 0-387-13678-9
Fulton, William; Harris, Joe: Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. 129. New York: Springer-Verlag, 1991. ISBN 978-0-387-97495-8 
