Differential Geometry 2
Test 3
Define what is meant by a
Lie group homomorphism
φ
:
G
→
H
and con
struct its derivative
φ
#
:
g
→
h
.
(a) Show that
φ
◦
exp
G
= exp
H
◦
φ
#
and explain briefly why
φ
is uniquely
determined by its derivative
φ
#
when
G
is connected.
(b) Show that if
ϕ
:
g
→
h
is a Lie algebra homomorphism then there need
not exist
φ
:
G
→
H
such that
φ
#
=
ϕ
.
Suggestion
: (b) Let
G
=
H
=
S
1
. Let
ϕ
: Lie(
S
1
)
→
Lie(
S
1
) be given by
ϕ
=
1
2
Id; assume
ϕ
=
φ
#
and take the square.
Solutions
: When
G
and
H
are Lie groups, a
Lie group homomorphism
is a smooth, group homomorphism from
G
to
H
. Being a group homomor
phism,
φ
maps
e
=
e
G
to
e
=
e
H
; being smooth,
φ
has a tangent map
φ
*
:
T
e
G
→
T
e
H
. Now the derivative
φ
#
:
g
→
h
is defined to be the com
posite
g
→
T
e
G
φ
*
→
T
e
H
→
h
in which the first map is evaluation at
e
G
and
the last is the inverse of evaluation at
e
H
.
(a) Let
ξ
∈
g
have integral curve
γ
through
e
G
in
G
: if
t
∈
R
then it
follows that (
φ
◦
γ
) (
t
) =
φ
*
(
γ
(
t
)) =
φ
*
(
ξ
γ
(
t
)
) =
φ
*
(
L
γ
(
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 Spring '09
 Robinson
 Representation theory, Lie group, Functor, General linear group

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