SOLUTIONS TO SELECTED EXERCISES: SHEET 3
Exercise 1
a: Consider the map Ψ : GL+
n (R) → (0, ∞) × SLn (R) given by:
1 1
Ψ(g) := (det g) n , (det g)− n g
Ψ is continuous and hence Ψ−1 ⊆ (0, 1] × B ⊆ GL+
n (R) is measureable
n2
whenever B ∈ B SLn (R). As GL+
(R)
⊆
R
is
open,
C(B)
= {0} ∪ Ψ−1 ⊆
n
2
(0, 1] × B is measureable in Rn .
Left invariance of the measure:
mSLn (R) (B) := mRn2 C(B)
2
follows immediately from C(gB) = gC(B) and the fact that on Rn an
element g ∈ SLn (R) acts – after an appropriate permutation of the standard
basis – like a block-diagonal n2 × n2 -matrix with n copies of g along the
diagonal.
b: We use the coordinate system φ : Aff 1 (R) 3 (a, b) 7→ (a, a−1 b) ∈ R× × R.
1
On R× × R we define the measure dν(α, β) := |α|
dα dβ and we claim that
−1
(φ )∗ ν is a left-Haar measure
on Aff 1 (R).
Let f ∈ Cc Aff 1 (R) and denote ψ(α) := xα, then for left-translation –
indicated by subscript – follows:
!
Z
Z f x y ◦ φ−1 (α, β)
( 1)
−1
dβ dα
(φ )∗ ν f( x y ) =
|α|
1
R×
R
!
Z
Z
f ◦ φ−1 xα, β + (xα)−1 y
=
dβ dα
|α|
R×
R
Z Z
f ◦ φ−1 (xα, β)
(trans. inv.)
=
dβ dα
|α|
R×
R
!
Z
Z
−1
f ◦φ
ψ(α), β
0
0
(ψ = x)
=
|ψ (α)| dα dβ
|ψ(α)|
R
R×
!
Z
Z
f ◦ φ−1 α, β
×
×
dα dβ = (φ−1 )∗ ν(f )
(ψ(R ) = R )
=
|α|
×
R
R
and thus we have indeed found a left Haar measure. For right translation
– indicated by superscript – follows:
!
Z
Z (x y)
xy f 1 ◦ φ−1 (α, β)
−1
(
)
(φ )∗ ν f 1 =
dβ dα
|α|
R×
R
!
Z
Z
f ◦ φ−1 xα, x−1 β + x−1 y
=
dβ dα
|α|
R×
R
!
Z
Z
f ◦ φ−1 xα, x−1 β
(trans. inv.)
=
dβ dα
|α|
R×
R
1
2
SOLUTIONS TO SELECTED EXERCISES: SHEET 3
Z
Z
(subst. β 7→ xβ)
|x|
=
R×
= |x| (φ
(as above)
R
−1
!
f ◦ φ−1 xα, β
dβ dα
|α|
)∗ ν(f )
−1
Hence follows ∆Aff 1 (R) ( x y1 ) = |x|
First look at
.
Exercise 5
v) = det d0 φAdg−1 v . Fix some δ < δ0 such that:
ρ(Ad−1
g
U := Bδ ∪ Adg−1 Bδ ∪ Adg Bδ ⊆ Bδ0
Note that for v, w ∈ U :
φAdg−1 v (w) = (Adg−1 ◦φv ◦ Adg )(w)
In particular:
−1
ρ(Ad−1
det(d0 Adg )−1 ρ(v)
g v) = det(dv Adg −1 )
As Adg−1 is linear, det(dv Adg−1 ) = det(d0 Adg−1 ) = det(d0 Adg )−1 and thus:
ρ(Ad−1
g v) = ρ(v)
We calculate:
µ(Bg) =µ(g
−1
Bg) =
XZ
k
=
XZ
k
Bδ
1g−1 Bg∩Pk gk exp(v) dv
Bδ
1B∩Pk0 gk0 exp(Adg v) ρ(v) dv
XZ
1
1B∩Pk0 gk0 exp(v) ρ(Adg−1 v) dv
=
|det dAdg |
Adg Bδ
k
Z
X
1B∩Pk0 gk0 exp(v) ρ(v) dv
= det dAdg−1
k
Adg Bδ
= det dAdg−1 µ(B)
where we used that for a fixed measure on g, the Haar measure constructed in the
above manner does not depend on the choice of the neighbourhood U = Bδ of 0.
As this is by itself an interesting fact, we give the construction of the Haar measure
for neighbourhoods of 0 which are not necessarily open balls around 0.
Let
V ⊆ g an open neighbourhood of 0 such that exp(V ) ⊆ G is open and
exp V is a diffeomorphism onto its image. Denote U := exp(V ). As exp and
p : G × G → G are continuous, there is some open neighbourhood A ⊆ g of 0 such
that exp(A) exp(A) ⊆ U and hence we obtain a well-defined product on A × A:
∗ :A × A → g
v ∗ w := log exp(v) exp(w)
By the same argument, we can assume that A is chosen so that the above holds
and exp(u) exp(v) exp(w) ∈ U for all u, v, w ∈ A. We note that det(d0 φ0 ) = 1 as
φ0 = id. As v 7→ det(d0 φv ) is continuous, we can assume without loss of generality
that A is chosen so that for v, w ∈ A holds det(dv φw ) > 0.
Lemma 0.1. Let v0 ∈ A, f ∈ Cc (G), then:
Z
Z
f exp(v0 ) exp(v) ρ(v) dv =
∗A
v0 ∗A
f exp(v) ρ(v) dv
SOLUTIONS TO SELECTED EXERCISES: SHEET 3
3
Proof. If v0 , v, w ∈ A, then φφv0 v (w) = (φv0 v)∗w = (v0 ∗v)∗w = v0 ∗(v ∗w) = φv0 ◦
φv (w), hence d0 φφv0 v = dv φv0 d0 φv and thus ρ(φv0 v) = det(dv φv0 )ρ(v), yielding:
Z
Z
f exp(v0 ) exp(v) ρ(v) dv =
f exp(φv0 v) ρ(φv0 v) det(dv φv0 ) dv
A
ZA
=
f exp(v) ρ(v) dv
v0 ∗A
Given this local definition of the measure, we define:
XZ
m(B) :=
1B∩Pk gk exp(v) ρ(v) dv
∀B ∈ BG
A
k≥1
where we chose a symmetric neighbourhood O ⊆ g of 0 such that O ∗ O ∗ O ⊆ A,
(gk )k∈N ∈ GN and (Pk )k∈N ∈ BGN such that:
G
[
G=
Pk =
gk exp(O)
k≥1
k≥1
Pk ⊆gk exp(O) ∀k ≥ 1
We check that this is independent of the choice of the partition and the sequence.
Assume that (Ql )l∈N ∈ BGN and (hl )l∈N ∈ GN are chosen similarly. Assume that
Pk ∩ Ql 6= ∅, then there are vk , wl ∈ O such that:
gk exp(vk ) = hl exp(vl ) ⇒ h−1
l gk ∈ exp(O ∗ O)
Beppo-Levi implies:
m(B) =
XZ
XZ
O
vk,l ∗O
1B∩Pk ∩Ql hl exp(v) ρ(v) dv
A
XZ
l≥1
1B∩Pk ∩Ql hl exp(v) ρ(v) dv
X Z
k,l≥1
=
1B∩Pk ∩Ql hl exp(vk,l ) exp(v) ρ(v) dv
X Z
k,l≥1
=
X Z
k,l≥1
=
1B∩Pk gk exp(v) ρ(v) dv
O
k≥1
=
A
k≥1
=
1B∩Pk gk exp(v) ρ(v) dv
1B∩Ql hl exp(v) ρ(v) dv
A
where in the second to last step we used that for v ∈ A holds hl exp(v) ∈ Pk iff
v = vk,l ∗ w for some w ∈ O, which can be checked easily using the definitions of A
and O. Left-invariance of this measure follows by the same argument as in class:
if (Pk )k∈N is a partition corresponding to O ⊆ A and the sequence (gk )k∈N , then
(g −1 Pk )k∈N is a partition corresponding to O ⊆ A and the sequence (g −1 gk )k∈N .
Finally we note that the measure is independent of the specific choice of O and A
for some fixed choice of measure on g. In order to see this, assume that A1 , O1
and A2 , O2 are two distinct choices as above. Let exp(O1 ) ∩ exp(O2 ) =: U . The
measures defined by (A1 , O1 ) and (A2 , O2 ) agree on U without vanishing which
follows from the fact that we can choose U as an element of both partitions defining
the measures.
4
SOLUTIONS TO SELECTED EXERCISES: SHEET 3
References
1. Alexander S. Kechris, Classical Descriptive Set Theory, Springer, 1995.