Representations of linear Lie
groups and applications
Salem Ben Said
(i) This course requires knowledge of the courses done by J.-P. Anker and L.
Kamoun.
(ii) These lecture notes are written in english for the students to get use to
this language.
As one can see from the works of Euler and Gauss, mathematicians of previous
centuries did not only publish general theorems but also examples which they
might enjoyed and which might have educated the reader.
From Siegel [Gesammelte Abhandlungen IV, 1979]
Contents
Part
1.
2.
3.
1. Basic Representation Theory of Linear Lie Groups
Lie algebras of linear Lie groups: A brief review
Representations of linear Lie groups
An application of Nelson’s theorem
5
6
10
22
Part
4.
5.
6.
7.
2. Representations of SL(2, R), and Beyond
Basic properties of SL(2, R) and SU (1, 1)
Finite dimensional representation of SL(2, R)
An application: The wave equation
The discrete series representation of SL(2, R)
29
30
38
42
48
Bibliography
57
3
Part 1
Basic Representation Theory of
Linear Lie Groups
1. Lie algebras of linear Lie groups: A brief review
Definition 1.1. A topological group G is a set that is simultaneously a
group and a topological space such that:
(i) the map (a, b) !→ ab from G × G to G is continuous,
(ii) the map a !→ a−1 from G to G is continuous.
Let H be a subgroup of G. By inheriting the topology from G, H is
a topological group. If H is an open in G, then H is closed. Indeed, the
complement of H in G is the union of the orbits xH, with x ∈ G and x %∈ H.
This is due to the fact that if x %∈ H then xH ∩ H = ∅ and, therefore
G \ H = ∪x"∈H xH. On the other hand, the translation Lx : y !→ xy is
continuous from G to G. Thus xH is an open set in G, and the union of
open sets is an open. Hence the complement of H in G is an open and,
therefore H is a closed subgroup in G.
On M (n, R) we consider the norm
!
)A, B* := tr(tAB).
(A( := )A, A*,
Clearly we have (AB( ≤ (A( (B(. The group
"
#
GL(n, R) := A ∈ M (n, R) | det(A) %= 0 ,
inheriting the topology from M (n, R), is a topological group. Moreover,
GL(n, R) is an open set in M (n, R) as det : M (n, R) −→ R is a continuous
map and GL(n, R) is the complement of the inverse image of the closed
subset {0} ⊂ R.
Definition 1.2. A group G is called a linear Lie group if there is an
n ∈ N such that G is isomorphism to a closed subgroup of GL(n, R).
Example 1.3.
Its kernel
(i) The map det : GL(n, R) −→ R∗ is continuous.
"
#
SL(n, R) = g ∈ GL(n, R) | det(g) = 1
is a closed subgroup of GL(n, R). It is a non compact group unless
n = 1. $
%
0
In
(ii) Let J =
. The symplectic group is by definition
−In 0
"
#
Sp(2n, R) = g ∈ GL(2n, R) | tgJg = J .
In particular Sp(2, R) = SL(2, R). We can show that for every
g ∈ Sp(2n, R) we have det(g) = 1. Thus Sp(2n, R) is a closed
subgroup of GL(2n, R).
(iii) We set
"
#
GL(n, C)“ = ” g ∈ GL(2n, R) | g −1 Jg = J .
1. LIE ALGEBRAS OF LINEAR LIE GROUPS: A BRIEF REVIEW
7
It is the inverse image of the closed subset {0} ⊂ M (2n, R) by the
continuous map g !→ g −1 Jg − J from GL(2n, R) to M (2n, R). Thus
GL(n, C) is a closed subgroup of GL(2n, R).
(iv) The orthogonal group O(n) is the set of g ∈ GL(n, R) such that
tgg = g tg = I . That is | det(g)| = 1 for all g ∈ O(n). This is an
n
example of a compact linear group since (g( = (g −1 ( = 1. The
subgroup SO(n) of O(n) defined by SO(n) = SL(n, R) ∩ O(n) is a
closed subgroup of GL(n, R).
Let K be R or C. A K-vector space g provided with a K-bilinear map
g × g −→ g,
is called a Lie algebra if
(x, y) !→ [x, y]
[x, y] = −[y, x],
and the Jacobi identity
∀x, y ∈ g,
[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0
holds. The dimension of g as a K-vector space is called the dimension of
the Lie algebra g. Henceforth we are mainly interested in Lie algebras over
K = R.
Example 1.4.
(i) Every associative algebra g is a Lie algebra with
[x, y] := xy − yx for all x, y ∈ g.
(ii) The set M (n, K) is an associative non-commutative algebra with
composition XY of two elements X, Y ∈ M (n, K) given by matrix multiplication. In view of (i), M (n, K) is a Lie algebra with
[X, Y ] := XY − Y X, for X, Y ∈ M (n, K). Henceforth we will write
gl(n, K) = M (n, K).
(iii) If g is a Lie algebra, and g0 is a subspace with [x, y] ∈ g0 for all
x, y ∈ g0 , then g0 is a Lie algebra. For instance
"
#
sl(n, R) :=
X ∈ gl(n, R) | tr X = 0 ,
"
#
so(n) :=
X ∈ gl(n, R) | X = −t X
are examples of such Lie algebras. The dimension of the Lie algebra
sl(n, R) is n2 − 1, and the dimension of so(n) is n(n − 1)/2.
Exercise 1.5. Verify that the matrices
$
%
$
%
0 1
0 0
E=
,
F =
,
0 0
1 0
H=
$
1 0
0 −1
form a basis of sl(2, R) with the relations
[E, H] = −2E,
[F, H] = 2F,
[E, F ] = H.
%
8
Exercise 1.6. Show that g = R3 provided with the composition
((p, q, r), (p% , q % , r% )) !→ (0, 0, (pq % − p% q))
is a Lie algebra, called the Heisenberg algebra heis(R), with basis
P = (1, 0, 0),
Q = (0, 1, 0),
R = (0, 0, 1)
and relations
[P, R] = [Q, R] = 0,
[P, Q] = R.
Definition 1.7. Let G be a linear Lie group contained in GL(n, R).
Henceforth we will write
"
#
g = Lie(G) := X ∈ gl(n, R) | exp(tX) ∈ G for all t ∈ R .
Theorem 1.8. The set g provided with the bilinear map
[·, ·] : g × g −→ g,
(X, Y ) !→ [X, Y ] = XY − Y X
is a real Lie algebra, called the Lie algebra of G.
Remark 1.9.
(i) Clearly we have Lie(GL(n, R)) = M (n, R), which
we already denoted by gl(n, R).
(ii) If G ⊂ G% , then Lie(G) ⊂ Lie(G% ).
(iii) Let G1 and G2 be two closed connexe subgroups of GL(n, R) with
the same Lie algebra. Then G1 and G2 are identical.
"
Example 1.10.
(i) Recall that SL(n, R) = X ∈ GL(n, R) | det(g) =
#
1 . Its Lie algebra sl(n, R) is the space of matrices X such that
det(exp(tX)) = 1, for all t ∈ R. Using the fact that det(exp(tX)) =
exp(t tr(X)), it follows that
"
#
sl(n, R) = Lie(SL(n, R)) = X ∈ gl(n, R) | tr(X) = 0 ,
which coincides with
" the definition of sl(n, R) given #previously.
(ii) Recall that O(n) = g ∈ GL(n, R) | tg g = g tg = In . Its Lie alge-
bra o(n) is the set of matrices X such that t exp(sX) = exp(−sX)
for all s ∈ R. Since t exp(A) = exp(tA), the Taylor series expansion at s = 0 gives tX = −X. Conversely, the condition tX = −X
implies that exp(sX) ∈ O(n) for all s ∈ R. Thus
"
#
o(n) = Lie(O(n)) = X ∈ gl(n, R) | tX = −X .
(iii) Recall that Sp(2n, R) denotes
$ the Lie
% group of g ∈ GL(2n, R) such
0
I
n
that tgJg = J, where J =
. Since
−In 0
(exp(sX))J exp(sX) = J + s(tXJ + JX) + o(t),
t
1. LIE ALGEBRAS OF LINEAR LIE GROUPS: A BRIEF REVIEW
9
thus the Lie algebra sp(2n, R) = Lie(Sp(2n, R)) is the set of matrices X ∈ gl(2n, R) such that tXJ + JX = 0. Conversely, if we
choose X such that tXJ + JX = 0, then
∞ i
∞ i
&
&
s t i
s
t
(exp(sX))J =
XJ=
(−1)i JX i = J exp(−sX).
i!
i!
i=0
i=1
Definition 1.11. A one-parameter subgroup γ of a topological group G
is a continuous homomorphism γ : R −→ G, t !→ γ(t). If the one-parameter
subgroup γ = γ(t) is differentiable in t, we assign to it an infinitesimal
generator Xγ by
d
Xγ := γ(t)|t=0 .
dt
Remark 1.12. Let g be the Lie algebra of a linear Lie group G. For
every X ∈ g, the following map γ : R −→ G, t !→ exp(tX) is a one
parameter subgroup. This is due to the fact that the map t !→ exp(tX)
is continuous (even differentiable) and exp(A + B) = exp(A) exp(B) for
commuting A, B ∈ M (n, R). In this example we have X as infinitesimal
generator.
$
%
a b
Example 1.13. Every element g =
∈ SL(2, R) can be written
c d
as g = n(x)t(y)r(θ), where
$
%
$ 1/2
%
$
%
1 x
y
0
cos θ sin θ
n(x) =
,
t(y) =
,
r(θ)
=
,
0 1
− sin θ cos θ
0
y −1/2
with x ∈ R, y ∈ R∗+ and θ ∈ [0, 2π[. The above decomposition of SL(2, R) is
called the Iwasawa decomposition. See Part 2 for more details. We set
$
%
$ t
%
1 t
e
0
γ1 (t) :=
,
γ2 (t) :=
,
γ3 (t) := r(t).
0 1
0 e−t
Thus, by differentiating and then putting t = 0, we obtain the following three
infinitesimal generators
$
%
$
%
$
%
0 1
1 0
0 1
X1 =
,
X2 =
,
X3 =
.
0 0
0 −1
−1 0
We deduce that the Lie algebra sl(2, R), which is of dimension 3, is generated
by {X1 , X2 , X3 }. That is {X1 , X2 , X3 } forms a basis for sl(2, R) with the
relations
[X1 , X2 ] = −2X1 ,
Example

1

γ1 (t) = 0
0
[X2 , X3 ] = −2X2 ,
[X1 , X3 ] = −X2 .


" 1 x z
#
1.14. Let G = Heis(R) := 0 1 y  | x, y, z ∈ R . Take
0 0 1





t 0
1 0 0
1 0 t
1 0 ,
γ2 (t) = 0 1 t  ,
γ3 (t) = 0 1 0 .
0 1
0 0 1
0 0 1
10
Thus the three

0

X1 = 0
0
infinitesimal generators associated with γi , 1 ≤ i ≤ 3, are





1 0
0 0 0
0 0 1
0 0 ,
X2 = 0 0 1 ,
X3 = 0 0 0 .
0 0
0 0 0
0 0 0
Thus, the set {X1 , X2 , X3 } spans the 3-dimensional Lie algebra heis(R) =
Lie(Heis(R)) and, therefore forms a basis for heis(R). An easy computation
shows that
[X1 , X2 ] = X3 ,
[X1 , X3 ] = [X2 , X3 ] = 0,
called the Heisenberg commutator relations. The Lie algebra heis(R) is the
same as in Exercise 1.6.
2. Representations of linear Lie groups
Let G be a linear Lie group and H a complex Hilbert space.
Definition 2.1. A representation (π, H) of G (with identity e) is a
group homomorphism from G to the set GL(H) of linear bounded operators
in H, that is
π(g1 g2 ) = π(g1 )π(g2 ),
such that for all v ∈ H the map
π(e) = identity transformation,
g !→ π(g)v
is continuous. This means that for any g0 ∈ G we have
(π(g)v − π(g0 )v( −→ 0
as g → g0 ,
for all v ∈ H. The above condition is called the strong continuity property.
The condition π(e) = idH guarantees that the representation g !→ π(g) is
in terms of invertible operators. Indeed π(g)π(g −1 ) = π(g −1 )π(g) = π(e) =
idH , and thus π(g −1 ) = π(g)−1 .
A representation (π, H) is called trivial if π(g) = idH for all g ∈ G.
A representation is said to be unitary if π(G) consists of unitary operators. Recall that a unitary operator T on a Hilbert space is an operator
such that T T ∗ = T ∗ T = I, where T ∗ is the Hilbert space adjoint operator
of T.
A G-invariant subspace for a representation (π, H) of G is a closed subspace H0 ⊂ H such that
In this case, the map
where
π(g)H0 ⊂ H0 ,
for all g ∈ G.
π|H0 : G −→ GL(H0 ),
π|H0 (g) = π(g)|H0 ,
for all g ∈ G,
defined by restriction to H0 is a representation of G on H0 . We call (π|H0 , H0 )
a subrepresentation of (π, H).
2. REPRESENTATIONS OF LINEAR LIE GROUPS
11
We say that π is irreducible if there are no nontrivial closed G-invariant
subspaces. Thus an irreducible representation (π, H) only has {0} and H as
closed G-invariant subspaces of H.
H0⊥
Lemma 2.2. For a unitary representation, the orthogonal complement
of a closed invariant subspace H0 is closed invariant subspace.
Proof. This is due to the fact that
)π(g)u⊥ , u* = )u⊥ , π(g)−1 u* ∈ )u⊥ , H0 * = 0,
for u⊥ ∈ H0⊥ , u ∈ H0 .
!
A representation (π, H) of G is called to be weakly continuous if for
arbitrary u, v ∈ H
)π(g)u, v* −→ )π(g0 )u, v*,
as g → g0 .
Proposition 2.3. Let (π, H) be a unitary representation of a group G
in a complex Hilbert space. Then the following statements are equivalent:
(i) π is strongly continuous,
(ii) π is weakly continuous,
(iii) the function g !→ )π(g)u, u* is continuous at e for all u ∈ H.
Proof. Clearly we have i) ⇒ ii) ⇒ iii). We will show that iii) ⇒ i).
(π(g)u − π(g0 )u(2 = )π(g)u − π(g0 )u, π(g)u − π(g0 )u*
= )π(g)u, π(g)u* − )π(g)u, π(g0 )u*
−)π(g0 )u, π(g)u* + )π(g0 )u, π(g0 )u*
+
,
= 2(u(2 − 2 Rel )π(g)u, π(g0 )u*
+
,
= 2(u(2 − 2 Rel )π(g0−1 g)u, u*
≤ 2 -(u(2 − )π(g0−1 g)u, u*−→ 0 as g → g0 .
!
Example 2.4. Let G be a linear Lie group with left Haar measure dg.
Allowing G to act on L2 (G) by left translations
πl (g)f (g̃) = f (g −1 g̃),
g, g̃ ∈ G,
implies that (πl , L2 (G)) is a unitary representation of G. Proving the continuity of g !→ πl (g)f requires observing by dominated convergence theorem
that
.
G
|f (g −1 g̃) − f (g0−1 g̃)|2 dg̃ −→ 0
as g −→ g0 for f ∈ Cc (G) and knowing that Cc (G) is dense in L2 (G). This
is the so-called left regular representation of G. One could also define the
right regular representation by allowing G to act by right translations.
12
Example 2.5. Let G be a topological transformation group which acts
continuously on locally compact measure space S and leaves the measure
invariant. Set H = L2 (S, dµ). Define the map g !→ π(g) by
π(g)f (s) = f (g −1 s),
f ∈ H, g ∈ G, s ∈ S.
It follows that π(g) is a linear operator, for all g ∈ G. Moreover,
π(g1 )π(g2 )f (s) = (π(g2 )f )(g1−1 s) = f (g2−1 g1−1 s) = π(g1 g2 )f (s)
and π(e) = idH . The G-invariance of the measure µ implies
.
.
2
−1
2
(π(g)f ( =
|f (g s)| dµ(s) =
|f (s)|2 dµ(s) = (f (2 .
S
S
Thus π is a unitary operator.
Let f ∈ Cc (S) be a continuous function on S with compact support.
Then
sup |f (g −1 s) − f (s)| −→ 0
s∈S
as g → e. Moreover, there exists a fixed compact set K ⊂ S, supporting f
and π(g)f for g sufficiently close to e, and we have
.
2
(π(g)f − f ( =
|f (g −1 s) − f (s)|2 dµ(s)
S
≤ sup |f (g −1 s) − f (s)|2 Vol(K)
s∈K
−→ 0
as g → e.
The continuity property as g → g0 follows if one replaces f by π(g0 )f and
uses the general recipe
(π(g)f − π(g0 )f ( ≤ (π(g0 )( (π(g0−1 g)f − f (.
Now if f ∈ L2 (S, dµ) and $ > 0, there exists an f˜ ∈ Cc (S) such that (f −
f˜( < $. Then
(π(g)f − f ( ≤ (π(g)[f − f˜]( + (π(g)f˜ − f˜( + (f˜ − f (
< 2$ + (π(g)f˜ − f˜(.
For g sufficiently close to e, we deduce that (π(g)f − f ( < 3$. Thus the map
g !→ π(g)f defines a unitary representation of G in L2 (S, dµ).
Example 2.6. Let G = GL(n, R) and define
1
π(g)f (y) = | det g|− 2 f (g −1 y),
where y ∈ Rn , g ∈ G and f ∈ L2 (Rn ). This gives a representation (π, L2 (Rn ))
1
of G. The purpose of the factor | det g|− 2 is to make π(g) unitary, that is
.
.
π(g)f1 (y)π(g)f2 (y)dy =
f1 (y)f2 (y)dy.
Rn
Rn
2. REPRESENTATIONS OF LINEAR LIE GROUPS
13
Note that if we restrict π to the subgroup SL(n, R) of GL(n, R), we get a
unitary representation of SL(n, R). In this case the representation is simpler
since | det(g)| = 1.
Let (π1 , V1 ) be a representation of G acting on a vector space V1 and
(π2 , V2 ) be a representations of G acting on a vector space V2 . A continuous
linear map A : V1 −→ V2 is called an intertwining operator from V1 to V2 if
Aπ1 (g) = π2 (g)A
for all g ∈ G.
Two representations are called equivalent if A is an isomorphism.
Example 2.7. Let (πl , L2 (G)) and (πr , L2 (G)) be the left and the right
regular representation of a unimodular group G. The involution A : u(g) !→
u(g −1 ) defines a unitary map of L2 (G) onto itself. We have
(Aπr (g1 )u)(g2 ) = (πr (g1 )u)(g2−1 ) = u(g2−1 g1 ) = (Au)(g1−1 g2 ) = (πl (g1 )Au)(g2 ).
Hence
Aπr (g) = πl (g)A,
for all g ∈ G.
Proposition 2.8. Let (π1 , V1 ) and (π2 , V2 ) two representations of G
and let A : V1 −→ V2 an intertwining operator. Then:
(i) The kernel of A is a stable vector subspace in V1 .
(ii) The rang of A is a stable vector subspace in V2 .
In particular if A %= 0, then:
(iii) If π1 is irreducible, A is one to one.
(iv) If π2 is irreducible, A is onto.
(v) If π1 and π2 are irreducible, then A is an isomorphism.
Proof. Since π2 (g)A(v) = A(π1 (g)v), for all g ∈ G and v ∈ V1 . Then
Ker A = {v ∈ V1 | A(v) = 0}
is stable by π1 . Let v2 ∈ Im(A) ⊂ V2 and let v1 ∈ V1 such that v2 =
A(v1 ). For all g ∈ G, we have π2 (g)v2 = π2 (g)A(v1 ) = A(π1 (g)v1 ). Thus
π2 (g)v2 ∈ Im(A), which implies that Im(A) is stable by π2 . In particular,
if π1 is irreducible then Ker(A) = {0} or Ker(A) = V1 . Since A =
% 0, then
Ker(A) = {0}, and A is one to one. Further, since π2 is irreducible and
A %= 0, then Im(A) = V2 . The statement (v) follows from (iii) and (iv). !
Definition 2.9. A finite dimensional representation of a topological
group G on a vector space V over R or C is a Lie group homomorphism
π : G −→ GL(n, K),
K = R or C,
where dim(V ) = n.
An n-dimensional representation of a group G is a prescription π associating to each g ∈ G a matrix π(g) = A(g) ∈ GL(n, K) such that
A(gg % ) = A(g)A(g % )
holds for all g, g % ∈ G.
14
Example 2.10. Let (π, Cn ) be the following representation π(g) = g,
where π(g) will act by matrix multiplication on the vector space Cn .
Theorem 2.11. Let π be a finite dimensional irreducible representation
of a topological group on a vector space V over C. The only operators which
commute with all π(g) are scalar multiples of the identity.
Proof. Let A : V −→ V be an intertwining operator of π. The A
possesses an eigenvalue λ ∈ C. This is due to the fact that the characteristic
polynomial p(λ) = det(A − λ idV ) possesses at least one complex solution.
Thus A% := A − λ idV is a linear application with Ker(A% ) %= {0}. Since π
is irreducible we deduce from Proposition 2.7 that A% = 0, i.e. A = λ idV
where λ ∈ C.
!
An direct consequence of the above (Schur) statement is:
Proposition 2.12. Every finite dimensional irreducible representation
of an abelian group G on a vector space over C is one-dimensional.
Proof. Since G is abelian, then for all g, h ∈ G we have gh = hg
and π(g)π(h) = π(gh) = π(h)π(g). Thus for fixed h ∈ G, the operator
π(h) commutes with every operator π(g), and, using the irreducibility of
π, we deduce that π(h) = λ(h) idV for all h ∈ G. Hence, every subspace
in V is then stable by π. Using again the irreducibility of π it follows that
dim(V ) = 1.
!
Proposition 2.13. A unitary representation (π, H) of a linear Lie
group G in a complex Hilbert space H is irreducible if and only if the only
operators commuting wit all the π(g) are scalar multiples of the identity.
Proof. See Knapp, page 12. The proof uses functional analysis and
mainly a spectral theorem.
!
Let g be a Lie algebra on a field K, and V a K-vector space. A representation (ω, V ) of g is a homomorphism of g into End(V ), the space of
endomorphisms of V, i.e.
ω(αX + βY ) = αω(X) + βω(Y ),
α, β ∈ K,
ω([X, Y ]) = [ω(X), ω(Y )] = ω(X)ω(Y ) − ω(Y )ω(X).
In analogy with group representations, a representation (ω, V ) of a Lie
algebra g is called irreducible if there is no nontrivial g-invariant subspace
in V.
Example 2.14.
(i) Each Lie algebra has the trivial representation
ω(X) = 0 for all X ∈ g. This is trivially an irreducible representation.
(ii) Let g be an arbitrary Lie algebra. For x ∈ g, let ad(x) act on g via
ad(x)y := [x, y],
y ∈ g.
2. REPRESENTATIONS OF LINEAR LIE GROUPS
15
Using Jacobi identity we have
ad([x, y])(z) = [[x, y], z] =
=
=
=
[x, [y, z]] − [y, [x, z]]
ad(x)(ad(y)(z)) − ad(y)(ad(x)(z))
(ad(x) ad(y) − ad(y) ad(x))(z)
[ad(x), ad(y)](z).
Thus (ad, g) is a representation of g; it is called the adjoint representation of the Lie algebra g.
Remark 2.15. One of the principal difficulties in the general representation theory of Lie algebras is due to the fact that in many cases representatives ω(X), for X ∈ g, are given by unbounded operators. Hence we have
to consider the problem of the selection of proper common domain D for
the set of unbounded operators. The domain D cannot be the whole space,
because on such domain only bounded operators would be defined. Moreover, we want to define the commutator ω(X)ω(Y ) − ω(Y )ω(X), then the
Rang(ω(X)) has to be in D for all X ∈ g. Hence D must be g-invariant.
Example 2.16. The poincaré group in 4 dimension is defined by
$
%
"
#
A a
G(4) = g = (A, a) =
, A ∈ SO(3, 1), a ∈ R4 .
0 1
The dimension of its Lie algebra g(4) is 10 and the generators satisfy the
commutation relations
[Mµ,ν , Mρ,σ ]
Mµ,ν
[Pµ , Pν ]
[Mµ,ν , Pσ ]
=
=
=
=
gµ,ρ Mν,σ + gν,σ Mµ,ρ − gν,ρ Mµσ − gµ,σ Mν,ρ ,
−Mν,µ
0
gν,σ Pµ − gµ,σ Pν ,
where µ, ν, ρ, σ = 0, 1, 2, 3, g0,0 = −g1,1 = −g2,2 = −g3,3 = 1, and gµ,ν = 0
for ν %= µ.
Let H = L2 (Ω) where Ω is the forth-dimensional Minkowski space. We
can verify that the above commutation relations are satisfied by the following
differential operators
Eµ,ν = xµ ∂ν − xν ∂µ ,
Fµ = ∂µ ,
µ, ν = 0, 1, 2, 3,
(2.1)
where ∂µ = ∂/∂xµ . In order to obtain a representation ω of g(4), we have
to determine a common dense invariant domain D for Eµ,ν and Fµ with
µ, ν = 0, 1, 2, 3. We can take either one of the following two dense subspaces
of L2 (Ω) :
(i) D = Cc∞ (Ω),
(ii) D = S(Ω),
where S(Ω) is the Schwartz space of C ∞ (Ω)-functions ϕ(x) with
sup |xα ∂ β ϕ(x)| < ∞
x∈Ω
16
where
xα = xα0 0 xα1 1 xα2 2 xα3 3 ,
∂ β = ∂ |β| /∂xβ0 0 ∂xβ1 1 ∂xβ2 2 ∂xβ3 3 ,
αi , βi = 0, 1, 2, . . . ,
|β| = β0 + β1 + β2 + β3 ,
0 ≤ i ≤ 3.
Clearly the domains i) and ii) are invariant under the operators (2.1).
Representations of a Lie group G give rise to representations of its Lie
algebra g in a natural way.
Let (π, H) be a representation of G on a Hilbert space H. We call a vector
v ∈ H smooth for the representation π if the map g !→ π(g)v is a smooth
function from G to H. The set of smooth vectors for π form a subspace H∞
of H.
Let v ∈ H∞ and X ∈ g, we define
f (X) = π(exp(X))v.
Then f is of class C ∞ . Put
dπ(X)v := df (0)X.
(2.2)
It follows that
π(exp(tX))v − v
= dπ(X)v.
(2.3)
t
By (2.2) and (2.3) we have dπ(X) as a linear mapping of H∞ into H and it
depends linearly on X.
lim
t→0
Theorem 2.17. Let π be a representation of a linear Lie group G on a
Hilbert space H. For X ∈ g, define a linear mapping from H∞ into H by
π(exp(tX))v − v
.
(2.4)
t→0
t
For every X ∈ g, dπ(X) leaves H∞ stable, and dπ is a representation of g
on H∞ . We call it the derived (or the infinitesimal) representation of π.
dπ(X)v = lim
Proof. For v ∈ H∞ , define the map f˜ : G 1 g !→ π(g)v. Applying π(g)
to (2.4), we obtain
π(g exp(tX))v − π(g)v
t→0
t
d
=
π(g exp(tX))v.
dt |t=0
π(g)dπ(X)v = lim
Since the map g !→ π(g)v is a C ∞ -map it follows that g !→ π(g)dπ(X)v is
smooth and, therefore dπ(X)v ∈ H∞ . Hence H∞ is stable by dπ(X), for all
X ∈ g.
Next we will show that dπ([X, Y ]) = [dπ(X), dπ(Y )] for all X, Y ∈ g. Let
c(t) := exp((− sgn t)|t|1/2 X) exp(−|t|1/2 Y ) exp((sgn t)|t|1/2 X) exp(|t|1/2 Y )
=
exp((sgn t)|t|[X, Y ] + o(|t|3/2 )).
2. REPRESENTATIONS OF LINEAR LIE GROUPS
17
The function t !→ c(t) is a C 1 -curve from R to G, with c% (0) = [X, Y ]. For
v ∈ H∞ the map t !→ π(c(t))v has differential df˜(e)([X, Y ]) at t = 0, where
f˜(g) = π(g)v. That is
π(c(t))v − v
= dπ([X, Y ])v.
t→0
t
lim
Hence
π(c(t2 ))v − v
= dπ([X, Y ])v.
t→0
t2
Using the strong continuity of the representation (π, H) of G, we obtain
lim
π(exp(tX) exp(tY ))v − π(exp(tY ) exp(tX))v
t→0
t2
π(c(t2 ))v − v
= lim π(exp(tY ))π(exp(tX))
t→0
t2
= dπ([X, Y ])v.
lim
On the other hand, the map
(s, t) !→ π(exp(sX) exp(tY ))v
(2.5)
is of class C ∞ . This is due to the fact that (2.5) is the composition of the
following C ∞ maps
(s, t) !→ (exp sX, exp tY ),
(g1 , g2 ) !→ g1 g2 ,
g !→ π(g)v.
In particular, for each v ∈ H, the map (s, t) !→ )π(exp(sX) exp(tY ))v, v* is
of class C ∞ . Hence
)dπ(X)dπ(Y )v, w*
∂ ∂
=
)π(exp sX exp tY )v, w*|s=t=0
∂s ∂t "
#
= lim)t−2 π(exp(tX) exp(tY )) − π(exp(tX)) − π(exp(tY )) + I v, w*.
t→0
Replace X by Y and vice versa, we obtain
)dπ(Y )dπ(X)v, w*
"
#
= lim)t−2 π(exp(tY ) exp(tX)) − π(exp(tX)) − π(exp(tY )) + I v, w*.
t→0
Thus
"
#
) dπ(X)dπ(Y ) − dπ(Y )dπ(X) v, w*
"
#
= lim)t−2 π(exp(tX) exp(tY )) − π(exp(tY ) exp(tX)) v, w*.
t→0
In conclusion dπ([X, Y ]) = dπ(X)dπ(Y ) − dπ(Y )dπ(X).
!
When (π, V ) is a finite dimensional representation of a Lie group, there
is a much easier proof for the existence of its infinitesimal representation.
Let V be a K-vector space, where K = R or C, and dim(V ) = n. Let
π : G −→ GL(V ) 2 GL(n, K) be a representation of a linear Lie group G.
18
For all X ∈ g, the map t −→ π(exp(tX)) from R to G% = GL(n, K) is a oneparameter subgroup of G% . Thus there exists a unique element X % ∈ gl(n, K)
which depends only π and on X, such that
π(expG (tX)) = expG! (tX % ).
Put X % = dπ(X). One can verify that
dπ(αX + βY ) = αdπ(X) + βdπ(Y ),
dπ[X, Y ] = [dπ(X), dπ(Y )],
∀α, β ∈ K, ∀X, Y ∈ g,
∀X, Y ∈ g.
Notice that dπ is also characterized by the relation
π(expG (X)) = expG! (dπ(X)),
∀X ∈ g.
(2.6)
Proposition 2.18. If (π, H) is a representation of a linear Lie group G,
then H∞ is stable under each π(g). Further π(g)dπ(X)π(g −1 ) = dπ(Ad(g)X)
for all g ∈ G and X ∈ g, where Ad(g)X = gXg −1 .
Proof. The map g !→ π(g)π(g0 )v is the composition of g !→ gg0 and
the map g !→ π(g)v. Hence H∞ is stable under π(g0 ). Let X ∈ g, g ∈ G,
and v ∈ H∞ . Then
π(exp tX)π(g)−1 v − π(g)−1 v
π(exp tX) − I
π(g)−1 v =
t
t
−1 π(exp t Ad(g)X)v − v
= π(g)
.
t
Taking the limit t → 0, we obtain dπ(X)π(g)−1 v = π(g)−1 dπ(Ad(g)X)v.
That is π(g)dπ(X)π(g)−1 = dπ(Ad(g)X).
!
Proposition 2.19. Let g !→ π(g) be a unitary representation of a linear
Lie group G on a Hilbert space H. Then the operators idπ(X), for all X ∈ g,
are symmetric.
Proof. Let v, w ∈ H∞ . Then
)idπ(X)v, w* = lim t−1 )i[π(exp(tX)) − I]v, w*
t→0
= lim t−1 )v, −i[π(exp(tX))∗ − I]w*.
t→0
Since π is unitary, then π(exp(tX))∗ = π(exp(tX))−1 = π(exp(−tX)). Thus
)idπ(X)v, w* = lim t−1 )v, −i[π(exp(−tX)) − I]w*
t→0
= lim s−1 )v, i[π(exp(sX)) − I]w*
s→0
= )v, idπ(X)w*.
!
It is very important to know at this stage whether H∞ is significantly
large. Indeed:
Theorem 2.20. (Garding). The space H∞ is dense in H.
2. REPRESENTATIONS OF LINEAR LIE GROUPS
Proof. See Section 4 of Chapter III of Knapp.
19
!
Corollary 2.21. When (π, V ) is a finite dimensional representation of
a linear Lie group G, then V ∞ = V. Hence π is smooth as a mapping of G
into GL(V ). This matches well with the observation made above about the
existence of dπ in the finite dimensional case.
Proof. This is due to the fact that V ∞ denses in the finite dimensional
vector space V.
!
Proposition 2.22. Assume that G is a connected linear Lie group. Let
(π, V ) be a finite dimensional representation, and W a vector subspace of V.
Then W is stable by π if and only if it is stable by dπ.
Proof. Let f be an element of the dual space V ∗ of V such that f
vanishes on W, and recall the identity (2.6). Since W is stable by π, then
for all t ∈ R, X ∈ g and w ∈ W, we have
0 = f (π(exp(tX))w) = tf (dπ(X)w) + o(t2 ),
thus f (dπ(X)w) = 0, and W is stable by dπ. The other direction is clear
since G is connected (recall that in this case the group G is generated by
exp(g)).
!
Corollary 2.23. Let G be a connected linear Lie group. A finite dimensional representation π of G is irreducible if and only if the infinitesimal
representation dπ of g is irreducible.
Example 2.24. Recall the adjoint representation Ad : G −→ GL(g)
given by Ad(g)X = gXg −1 , for X ∈ g and g ∈ G. The infinitesimal
representation associated with Ad is given by ad : g −→ End(g), where
ad(X)Y = [X, Y ]. Indeed:
(i) first method:
ad(X)Y
d
Ad(exp(tX))Y|t=0
dt
d
=
exp(tX)Y exp(−tX)|t=0
dt
= XY − Y X = [X, Y ].
=
(ii) second method: We have expGL(g) ◦ ad = Ad ◦ expG .
Example 2.25. Let G = R, H = L2 (R), and define the map π : R −→
GL(H) by
π(t)f (x) = f (x − t).
$
%
1 b
We may identify G with the closed subgroup of GL(2, R) of matrices
,
0 1
$
%
0 1
with b ∈ R. Hence the matrix X0 =
consists a basis of its Lie algebra.
0 0
20
The infinitesimal representation of π is given by
dπ(X0 )f (x) =
d
f (x − s) = −f % (x).
ds |s=0
Thus dπ is defined on the space of functions f ∈ L2 (R) such that f % ∈ L2 (R).
The subspace L2 (R)∞ consists of smooth functions f ∈ L2 (R) such that
f (n) ∈ L2 (R) for all n = 1, 2, . . . . In particular, the space Cc∞ (R) is a
subspace of L2 (R)∞ .
Example 2.26. Let
$
%
"
#
a b
G = SU (2) = g =
| |a|2 + |b|2 = 1 ,
−b̄ ā
and V = C[x, y]2 be the space of homogeneous polynomials of degree 2 in
two variables. Note that dim(V ) = 3. Choose as a basis of V the monomials
fp given by
xp+1
fp (x, y) = !
(p + 1)!
!
y 1−p
(1 − p)!
,
p = −1, 0, 1.
Let π be a representation of SU (2) on V given by
π(g)fp (x, y) = fp (āx − by, b̄x + ax),
with p = −1, 0, 1.
On the other hand, recall that the Lie algebra su(2) of SU (2) is given by
"
#
su(2) = X ∈ gl(2, C) | X = −t X̄, tr(X) = 0 .
We can show that every X ∈ su(2) can be written as
$
%
it
−r − is
X=
,
t, r, s ∈ R.
r − is
−it
Hence, the matrices
$
%
1 0 −i
X1 =
,
2 −i 0
1
X2 =
2
$
%
0 −1
,
1 0
1
X3 =
2
$
%
i 0
,
0 −i
consist a basis for su(2). Consider the following one-parameters subgroups
$
%
$
%
cos(t/2)
−i sin(t/2)
cos(t/2) − sin(t/2)
exp(tX1 ) =
, exp(tX2 ) =
,
−i sin(t/2)
cos(t/2)
sin(t/2) cos(t/2)
exp(tX3 ) =
$
%
eit/2
0
.
0
e−it/2
2. REPRESENTATIONS OF LINEAR LIE GROUPS
21
Since dim(V ) < ∞ then V ∞ = V. The infinitesimal representation is given
by
√
√
d
π(exp(tX1 ))f−1 = (i/ 2)xy = (i/ 2)f0 ,
dπ(X1 )f−1 =
(2.7)
dt |t=0
√
d
π(exp(tX1 ))f0 = (i/2)(x2 + y 2 ) = (i/ 2)(f−1 + f1 ),
dπ(X1 )f0 =
dt |t=0
(2.8)
√
√
d
dπ(X1 )f1 =
π(exp(tX1 ))f1 = (i/ 2)xy = (i/ 2)f0 .
(2.9)
dt |t=0
Similarly we get
√
dπ(X2 )f−1 = −(1/ 2)f0 ,
dπ(X3 )f−1 = if−1 ,
√
dπ(X2 )f0 = (1/ 2)(f−1 − f1 ),
dπ(X3 )f0 = 0,
√
dπ(X2 )f1 = (1/ 2)f0 ,
dπ(X3 )f1 = −if1 .
The matrices representations of dπ are of the form
√
√




0√ i/ 2
0√
1/ 2
0√
0√
dπ(X1 ) = i/ 2
0√ i/ 2 , dπ(X2 ) = −1/ 2
0√ 1/ 2 ,
0
i/ 2
0
0
−1/ 2
0


i 0 0
dπ(X3 ) = 0 0 0  .
0 0 −i
Example 2.27. Consider the Heisenberg

1 0
"
λ 1
G = Heis(R) = g = (λ, µ, k)1 := 
0 0
0 0
group

0 µ
#
µ k 
 | λ, µ, k ∈ R .
1 −λ
0 1
For m ∈ R∗ , define the representation πm of G on L2 (R) by
πm (g)f (x) := e2πim [k + (λ + 2x)µ]f (x + λ),
where g = (λ, µ, k)1 ∈ G, x ∈ R, f ∈
algebra heis(R) is given by

0
"
λ
heis(R) = (λ, µ, k)0 := 
0
0
L2 (R), and em [u] = e2πimu . The Lie
P = (1, 0, 0)0 ,
R = (0, 0, 1)0 ,
0
0
0
0

0 µ
#
µ k 
 | λ, µ, k ∈ R .
0 −λ
0 0
Clearly the Lie algebra g = heis(R) is generated by
Q = (0, 1, 0)0 ,
with the relations [P, Q] = R and [R, P] = [R, Q] = 0. In particular
exp(tP) = (t, 0, 0)1 ,
exp(tQ) = (0, t, 0)1 ,
exp(tR) = (0, 0, t)1 .
22
For differentiable f we compute
d
d
πm (exp(tP))f (x) =
f (x + t) = f % (x),
dπm (P)f (x) =
dt |t=0
dt |t=0
d
d
πm (exp(tQ))f (x) =
em [2xt]f (x) = 4πimxf (x),
dt |t=0
dt |t=0
d
d
dπm (R)f (x) =
πm (exp(tR))f (x) =
em [t]f (x) = 2πimf (x).
dt |t=0
dt |t=0
dπm (Q)f (x) =
Assume that V is a vector space of finite dimension, and G a linear
Lie group connected and simply connected such that dim(g) = m, where
g = Lie(G). If (ω, V ) is a representation of the Lie algebra g, then in order to
reproduce a representation π of G it is sufficient to know m basis infinitesimal
operators ω(X1 ), . . . , ω(Xm ). Namely, at first we set
m
m
&
&
π(exp(
µi Xi )) = exp(
µi ω(Xi ))
i=1
i=1
and then extend this representation of the part exp(g) of G to the whole
group. This extension is possible and unique since G is connected and simply
connected. See also the construction above of the infinitesimal representation dπ in the finite dimensional case.
In the case where V = H is a Hilbert space, the situation is less obvious
since the infinitesimal operators are not in general bounded and therefore
are not defined on the entire H.
Next we will give Nelson’s fundamental theorem which establishes when
a representation of a Lie algebra g, given in terms of skew-symmetric operators, can be seen as the infinitisimal representation of a unitary representation of a simply connected Lie group G having g as a Lie algebra.
Theorem 2.28. (Nelson). Let g be a real Lie algebra of dimension m
and H be a Hilbert space. Let {X1 , . . . , Xm } be a basis of the Lie algebra
g. Assume that ω(X1 ), . . . , ω(Xm ) are skew-symmetric operators which have
2)
a common invariant dense domain D. If the operator ω(X12 + · · · + Xm
is essentially self-adjoint, then there is on H a unique unitary representation π of the universal covering Lie group G with Lie algebra g such that
d
ω(X) = dt
π(exp(tX)), for all X ∈ g. A symmetric operator O is said to
|t=0
be essentially self adjoint if its closure is an auto-adjoint operator.
3. An application of Nelson’s theorem
Let
h=
so that
denote the standard basis of sl(2, R) where
%
$
%
$
%
0 −1
0 0
1 0
+
−
,
e =
,
e =
0 0
−1 0
0 −1
{e+ , e− , h}
$
[e+ , h] = −2e+ ,
[e− , h] = 2e− ,
[e+ , e− ] = h.
3. AN APPLICATION OF NELSON’S THEOREM
23
Theorem 3.1. Denote by ∆ the usual Laplacian on RN , and (x(2 =
1N 2
N
i=1 xi the usual norm on R . The map
e+ !−→ multiplication by i
e− !−→ application of i
∆
2
(x(2
2
N
h !−→ application of
N &
+
xi ∂i
2
i=1
is a Lie algebra homomorphism from sl(2, R) to operators on functions
and/or distributions on RN .
Proof. Define the operators E, F and H on the Schwartz function f,
(x(2
Ef = i
f (x),
2
N
&
N
Hf (x) = f (x) +
xi fi (x)
2
i
F f = ∆f (x),
2
i=1
where fi (x) = ∂i f (x). Then
[E, F ]f = (EF − F E)f
1
1
= − (x(2 ∆f (x) + ∆((x(2 f )
4
4
N
,
&
1
1+
2
= − (x( ∆f +
2N f + 4
xi fi + (x(2 ∆f
4
4
i=1
N
f (x) +
2
=
N
&
xi fi .
i=1
Now we compute [E, H] = EH − HE. For f in S(RN ) we have
,
&
i+
N
N &
(x(2 { f +
xi fi } −{
+
xi ∂i }{(x(2 f }
2
2
2
N
[E, H]f = =
N
i=1
=
i+
2
(x(2
N
&
i=1
xi fi −
N
&
i=1
i=1
,
xi ∂i {(x(2 f } .
Using the fact that (xi ∂i )(x2j f ) − x2j (xi ∂i )f = 0 for i %= j, we deduce that
[E, H]f = −i
N
&
i=1
x2i f = −i(x(2 f = −2Ef.
24
Finally
[F, H]f = (F H − HF )f
#
#
i "N &
i "N &
∆
+
xi ∂i f −
+
xi ∂i ∆f
2
2
2 2
=
i
2
=
= i
N
&
j=1
N
&
N
N
i=1
i=1
∂j2 (xj fj ) − xj ∂j (fjj )
fjj = 2F f.
j=1
!
Define the following representation (ω, S(RN )) of sl(2, R) by
i
ω(e ) = (x(2 ,
2
+
i
ω(e ) = ∆,
2
−
N
N &
ω(h) =
+
xi ∂i .
2
(3.1)
i=1
We shall use Nelson’s Theorem 2.28 to prove that ω is the infinitesimal
representation of a unitary representation of a simply connected Lie group
with Lie algebra sl(2, R) on L2 (RN ).
Define the group SU (1, 1) as SU (1, 1) = c−1 SL(2, R)c, where
√ %
$ √
1/√ 2 i/ √2
c=
∈ SL(2, C).
i/ 2 1/ 2
In particular, if we take g =
c
−1
1
gc =
2
$
$
a b
c d
%
∈ SL(2, R), we obtain
%
$
%
(a + ib) − i(c + id)
(ai + b) − i(ic + d)
α β̄
:=
,
(−ia + b) + i(−ic + d) (a − ib) + i(c − id)
β ᾱ
with
|α|2 − |β|2 = ad − bc = 1.
That is
SU (1, 1) =
"$
α β̄
β ᾱ
%
#
| α, β ∈ C, |α|2 − |β|2 = 1 .
In part II of this course we will study the groups SL(2, R) and SU (1, 1) in
more details.
Hence the triple
{k, n+ , n− } := c−1 {h, e+ , e− }c
(3.2)
3. AN APPLICATION OF NELSON’S THEOREM
25
is a basis of the Lie algebra su(1, 1) which is isomorphic to sl(2, R). More
precisely
√ %$
√ %
$ √
%$ √
1/ √2 −i/√ 2
1 0
1/√ 2 i/ √2
k =
0 −1
−i/ 2 1/ 2
i/ 2 1/ 2
$
%
0 i
=
−i 0
= i(e− − e+ ).
Similarly we obtain n+ = 12 (−ih + e+ + e− ) and n− = 12 (ih + e+ + e− ). By
(3.1) we have a representation of su(1, 1) on S(RN ), which we will denote
by the same Greek letter ω,
(x(2 − ∆
2
(=: H),
2
H − ∆/2 − (x(2 /2
2
ω(n+ ) = −i
(=: E),
2
H + ∆/2 + (x(2 /2
ω(n− ) = i
(=: F2).
2
Lemma 3.2. For all ν ∈ R we have
ω(k) =
eν)x)
2 /2
Proof. check that
eν)x)
2 /2
◦ ∆eν)x)
◦ ∆eν)x)
2 /2
2 /2
= ν 2 (x(2 + ∆ − 2νH.
= ν 2 (x(2 + ∆ − ν
N
&
(xi ∂i + ∂i xi ).
(3.3)
(3.4)
i=1
Using the fact that [AB, C] = A[B, C] + [A, C]B, with A = B = ∂i and
1
C = (x(2 we deduce that [∆, (x(2 ] = 2 N
i=1 xi ∂i + ∂i xi . In view of the
proof of Theorem 3.1, equation (3.4) becomes
eν)x)
2 /2
◦ ∆e−ν)x)
2 /2
= ν 2 (x(2 + ∆ − 2νH.
Corollary 3.3. The above lemma implies
2 = i e)x)2 /2 ◦ ∆e−)x)2 /2 ,
E
4
i
2
2
F2 = − e−)x) /2 ◦ ∆e)x) /2 ,
4
+
,
2 = e−)x)2 /2 H − ∆ e)x)2 /2 .
H
2
!
Denote by L2 (RN ) the space of square integrable functions on RN . For
m ∈ Z, let Hm (RN ) := Ker(∆) : Pm (RN ) −→ Pm−2 (RN ) be the space of
harmonic polynomials of degree m. Notice that
$
% $
%
m+N −1
m+N −3
dim Hm (RN ) =
−
,
if N > 1,
N −1
N −1
26
and dim Hm (R) = 0 for m ≥ 2. For ., m ∈ N and p ∈ Hm (RN ) set
(λ)
Φ( (p, x) := p(x)L( ((x(2 )e−)x)
where λ = m +
given by
N
2
(α)
− 1 and L(
(α)
L( (x)
2 /2
,
x ∈ RN ,
denotes the classical Laguerre polynomial
(
(α + 1)( & (−.)j xj
=
.
.!
(α + 1)j j!
j=0
Proposition 3.4. For ., s, m, n ∈ N, p ∈ Hm (RN ) and q ∈ Hn (RN ) we
have
.
.
Γ(λ + . + 1)
Φ( (p, x)Φs (q, x)dx = δmn δsl
p(ω)q(ω)dσ(ω),
2Γ(. + 1) S N −1
RN
where dσ is the Lebesgue measure on the sphere S N −1 .
Proof. Using the polar coordinates x = rω, with r > 0 and ω ∈ S N −1 ,
the left hand side becomes
3.
4
. ∞
(λ) 2
(λ) 2 −r 2 m+n+N −1
p(ω)q(ω)dσ(ω) dr.
L( (r )Ls (r )e r
S N −1
0
If m %= n, it follows that
.
p(ω)q(ω)dσ(ω) = 0.
S N −1
Now assume that m = n and the well known integral formula
. ∞
Γ(α + . + 1)
(α)
α −t
L( (t)L(α)
,
α > −1,
s (t)t e dt = δ(,s
Γ(. + 1)
0
the proposition holds.
(m)
Choose an orthonormal basis {hj
and j ∈ Jm , define
(m)
Φ(,m,j (x) := c(,m hj
where
!
}j∈Jm of Hm (RN ). For m ∈ N, . ∈ N,
(λ)
(x)L( ((x(2 )e−)x)
2 /2
,
%1/2
Γ(N/2).!
.
π N/2 Γ(λ + m + .)
By proposition 3.4 the functions {Φ(,m,j | . ∈ N, m ∈ N, j ∈ Jm } form an
orthonormal basis of L2 (RN )
c(,m =
$
Lemma 3.5. For ψ ∈ S(R+ ) and p ∈ Hm (RN ), we have
5
6 5
6
H p(x)ψ((x(2 ) = (λ + 1)ψ((x(2 ) + 2(x(2 ψ % ((x(2 ) p(x)
5
6 5
6
∆ p(x)ψ((x(2 ) = 4(λ + 1)ψ % ((x(2 ) + 4(x(2 ψ %% ((x(2 ) p(x).
Proof. Exercise
One of the main results is the following:
!
3. AN APPLICATION OF NELSON’S THEOREM
27
Theorem 3.6. For {k, n+ , n− } is as in (3.2), we have
ω(k)Φ(,m,j (x) = (m + λ + 2.)Φ(,m,j (x),
(3.5)
ω(n+ )Φ(,m,j (x) = −i(. + 1)1/2 (λ + m + .)1/2 Φ(+1,m,j (x),
(3.6)
ω(n− )Φ(,m,mj (x) = −i.1/2 (m + λ + . + 1)1/2 Φ(−1,m,j (x),
with Φ−1,m,j = 0 and λ = m +
N
2
(3.7)
− 1.
Proof. We recall that the Laguerre polynomials satisfy
(α)
(α)
(α)
t(d2 /dt2 )L( (t) + (α + . + 1)(d/dt)L( (t) = −.L( (t),
(α)
(α)
(α)
t(d/dt)L( (t) = .L( (t) − (. + α)L(−1 (t),
(α)
(α)
(α)
t(d/dt)L( (t) = (. + 1)L(+1 (t) − (. + α + 1 − t)L( (t).
We will show the identity (3.5). Using lemma 3.5, we have
ω(k)Φ(,m,j (x)
2 (,m,j (x)
= HΦ
+
,
∆ ,+ (m) λ)
2
= c(,m e−)x) /2 H −
hj L( ((x(2 )
2
2
(m)
= (λ + 1)Φ(,m,j (x) − 2c(,m e−)x) /2 hj (x)
"
#
(λ)
(λ)!
(x(2 L( ((x(2 ) + (λ + 1 − (x(2 )L( ((x(2 )
= (λ + 1 + 2.)Φ(,m,j (x).
In the same way we can check the other two identities (3.6) and (3.7).
!
Notice that the operators E, F and H are skew symmetric with respect
to the inner product
.
)f, g* =
f (x)g(x)dx.
RN
Thus ω(X), for any X ∈ g, regarded as an operator, is skew-symmetric.
Definition 3.7. If a symmetric operator O is closable and its closure
coincides with its adjoint, i.e. Ō∗ = Ō, then O is said to be essentially
self-adjoint.
Lemma 3.8. (Davis, Ch. 1) Let O be a symmetric operator on a Hilbert
space H with domain D(O), and let {fn }n be an orthonormal basis in H. If
each fn ∈ D(O) and there exists λn ∈ R such that Ofn = λn fn , for every
n, then O is essentially self-adjoint.
Now we arrive to the following conclusion.
28
Theorem 3.9. The representation ω described by (3.1) exponentiates to
a unitary representation on the Hilbert space L2 (RN ) of universal covering
Lie group with Lie algebra sl(2, R).
Proof. Choose u1 := e+ − e− , u2 := e+ + e− and u3 := h, so that
{u1 , u2 , u3 } is a basis of g = sl(2, R). Moreover, we have u21 + u22 + u23 =
(e+ − e− )2 + (e+ + e− )2 + h2 = −k2 + 2n+ n− + 2n− n+ . Furthermore, recall
that {Φ(,m,j } is an orthonormal basis of L2 (RN ). By Theorem 3.6 we deduce that Φ(,m,j are eigenvectors for ω(u21 + u22 + u23 ) with real eigenvalues.
Moreover, for all X ∈ sl(2, R), ω(X) is a skew symmetric operator. That
is ω(u21 + u22 + u23 ) is symmetric operator. By Lemma 3.8 it follows that
ω(u21 + u22 + u23 ) is an essentially self adjoint operator and, therefore the conditions of Nelson’s theorem are satisfied. Hence there exists a unique unitary
representation (π, L2 (RN )) of a simply connected Lie group with Lie algebra
d
π(exp(tX)).
!
sl(2, R), such that ω(X) = dt
|t=0
Part 2
Representations of SL(2, R), and
Beyond
4. Basic properties of SL(2, R) and SU (1, 1)
Consider the group
%
"$
#
a b
GL(2, C) :=
| a, b, c, d, ∈ C with ad − bc %= 0 .
c d
Each element g ∈ GL(2, C) defines a fractional linear transformation on C
via
$
%
7 az+b
a b
cz+d for cz + d %= 0,
·z =
c d
∞
for cz + d = 0.
A straightforward calculation shows that
Set
(g, z) :=
(g1 g2 ) · z = g1 · (g2 · z).
a(cz + d) − c(az + d)
ad − bc
det(g)
d(g · z)
=
=
=
. (4.1)
2
2
dz
(cz + d)
(cz + d)
(cz + d)2
Then  : GL(2, C) × C∗ −→ C∗ satisfies the following cocycle condition
(g1 g2 , z) = (g1 , g2 · z)(g2 , z),
and (I, z) = 1 where I is the identity matrix.
Let D ⊂ C be the open unit disk
D := {z ∈ C | |z| < 1} ⊂ C.
The condition defining D can be written as
$ %∗ $
%$ %
#
"
z
−1 0
z
>0 ,
D= z∈C |
1
0 1
1
where T ∗ = t T . Define the so-called standard unitary group U (1, 1) by
%#
$
%
$
"
−1 0
∗ −1 0
g=
.
U (1, 1) := g ∈ GL(2, C) | g
0 1
0 1
We check
that
$
% the group U (1, 1) stabilizes the unit disk D : let z ∈ D and
a b
g=
∈ U (1, 1). Then
c d
$
%∗ $
%$
%
g·z
−1 0
g·z
2
1 − |g · z| =
1
0 1
1
$
%∗ $
%$
%
1
1
az + b
−1 0
az + b
=
cz
+
d
0
1
cz
+
d
cz + d
cz + d
% $ %
$ %∗ $
1
z
z
−1 0
g
=
g∗
1
0 1
|cz + d|2 1
$ %∗ $
%$ %
1
z
−1 0
z
=
0 1
1
|cz + d|2 1
1
=
(1 − |z|2 ).
|cz + d|2
4. BASIC PROPERTIES OF SL(2, R) AND SU (1, 1)
31
Thus |cz + d|2 (1 − |g · z|2 ) = (1 − |z|2 ). Since 1 − |z|2 > 0 and |cz + d| ≥ 0,
it follows that necessarily |cz + d|2 > 0 and 1 − |g · z|2 > 0.
The complex upper half plane is by definition
Note that
T := {z = x + iy ∈ C | Im z > 0}.
1
Im z =
2i
$ %∗ $
%$ %
z
0 −1
z
.
1
1 0
1
This suggests defining the group
$
%
$
%#
"
0 −1
∗ 0 −1
G := g ∈ GL(2, C) | g
g=
.
1 0
1 0
For z ∈ T and g ∈ G we compute
$
%∗ $
%$
%
g·z
0 −1
g·z
2i Im (g · z) =
1
1 0
1
%∗ $
%$
%
$
1
1
0 −1
az + b
az + b
=
1 0
cz + d cz + d
cz + d cz + d
$ %∗ $
% $ %
1
z
0 −1
z
=
g∗
g
2
1 0
1
|cz + d| 1
$ %∗ $
%$ %
1
z
0 −1
z
=
.
2
1
1
0
1
|cz + d|
This gives
Im z = |cz + d|2 Im (g · z).
Since z ∈ T and |cz + d|2 ≥ 0, it follows that necessarily |cz + d|2 > 0 and
Im (g · z) > 0.
Consider the group
$
%
$
%#
"
0 −1
0 −1
t
SL(2, C) := g ∈ GL(2, C) | g
g=
.
1 0
1 0
The subgroup in SL(2, C) stabilizing the upper half plane is
$
%
$
%
$
%
$
%#
"
0 −1
0 −1
0 −1
%
t
∗ 0 −1
G = g ∈ GL(2, C) | g
g=
and g
g=
.
1 0
1 0
1 0
1 0
Thus, for g ∈ G% we have tg = g ∗ . That is the entries of g are real. We will
denote G% by SL(2, R) and we have shown that
$
%
$
%#
"
0 −1
0 −1
t
SL(2, R) = g ∈ GL(2, R) | g
g=
,
1 0
1 0
which is already by definition the set
{g ∈ SL(2, C) | g · T = T}.
32
w+i
We now define the Cayley transform as c · w = −i
, where
w−i
$
%
1 1 i
c= √
∈ SL(2, C).
2 i 1
Since det c %= 0, this fractional linear transformation in invertible. A direct
z−i
calculation shows the inverse Cayley transform is c−1 · z = i
.
z+i
Proposition 4.1. The Cayley transform is a biholomorphic equivalence
between D and T.
Proof. Since the only singularities
of 9c occurs at w = i, c is holo8
1−|w|2
w+i
morphic on D. Now, for w ∈ D, Im −i w−i = |1+iw|
2 > 0 and, therefore
z−i
c(D) ⊂ T. Now letting z ∈ T. Then z %= −i and i z+i
∈ D. To see this, just recall that if z ∈ T then z is closer to i than to −i, and then |z −i| < |z +i|. !
Thus the disk and the upper half plane are the same thing apart from
coordinates. In particular
c·0
c·1
c · (−1)
c · (−i)
c·i
=
=
=
=
=
i,
1,
(−1),
0,
∞.
If we use the following group
$
%
$
%#
"
−1 0
∗ −1 0
SU (1, 1) := g ∈ SL(2, C) | g
g=
0 1
0 1
instead of U (1, 1) as the group stabilizing the disk D, then it is easy to check
that
c−1 SL(2, R)c = SU (1, 1).
$
%
a b
Thus, if we take g =
∈ SL(2, R), we obtain
c d
$
%
$
%
1
(a + ib) − i(c + id)
(ai + b) − i(ic + d)
α β̄
−1
:=
,
c gc =
β ᾱ
2 (−ia + b) + i(−ic + d) (a − ib) + i(c − id)
with
|α|2 − |β|2 = ad − bc = 1.
That is
SU (1, 1) =
"$
α β̄
β ᾱ
%
#
| α, β ∈ C, |α|2 − |β|2 = 1 .
On the upper half plane, the subgroup
%$
%
"$
#
a 0
1 b
∗
|
b
∈
R,
a
∈
R
⊂ SL(2, R)
0 1
0 a−1
4. BASIC PROPERTIES OF SL(2, R) AND SU (1, 1)
33
acts by affine transformations
$
%$
%
$
%
a 0
1 b
1 b
·
z
=
· a2 z = a2 z + b.
0 1
0 1
0 a−1
In particular, this shows that SL(2, R) is transitive on T. Since SL(2, R)
acts transitively on the upper half plane, certainly SU (1, 1) acts transitively
on the disk. Also, this can be seen directly by verifying that for |w| < 1 the
matrix
$
%
1
1 w
!
1 − |w|2 w̄ 1
lies in SU (1, 1) and maps 0 to w.
In the action of SU (1, 1) on the disk D, the subgroup
%
"$
#
α
0
| |α| = 1
−1
0 α
acts by affine transformations
$
%
α
0
· z = α2 z.
0 α−1
This is the subgroup of SU (1, 1) fixing the point 0 ∈ D. When the latter
subgroup is conjugated by the Cayley element c to obtain the corresponding
subgroup of SL(2, R) fixing c(0) = i ∈ T, consisting of elements
:
; $
$
%
%
α+α−1
α−α−1
α
0
cos θ sin θ
−1
2
2i
c
c =
=
−1
α+α−1
− sin θ cos θ
0 α−1
− α−α
2i
2
where α = eiθ with θ ∈ R. Thus, the isotropy group in SL(2, R) of the point
i ∈ T is the special orthogonal group
%
"
"$
#
cos θ sin θ
t
SO(2) = g ∈ SL(2, R) | gg = I2 } =
| θ∈R .
− sin θ cos θ
The map
SL(2, R)/SO(2) −→ T
g · SO(2) !→ g · i
is then a diffeomorphism. The disk D is an instance of Harish-Chandra’s
realization of quotients analogous to SL(2, R)/SO(2) as bounded subsets of
Cn . The Cayley map from a bounded to unbounded model was studied in
general circumstances by Piatetski-Shapiro for the classical domains, and
later by Koranyi and Wolf intrinsically.
Theorem 4.2.
(i) The measure
µ(dx, dy) =
dxdy
y2
is an SL(2, R)-invariant measure on the tube T.
34
(ii) The measure
ν(dx, dy) =
dxdy
(1 − |z|2 )2
is an SU (1, 1)-invariant measure on the disk D.
(iii) ν = 4c−1 (µ).
Proof. (i) For g ∈ SL(2, R) such that
Im (g
−1
g −1
=
$
%
a b
, we have
c d
7
<
ac|z|2 + bd + bcz̄ + adz
· z) = Im
|cz + d|2
7
<
ac|z|2 + bd + bc(z + z̄) + det(g)z
= Im
|cz + d|2
Im z
=
.
|cz + d|2
In view of (6.1) we obtain
.
dxdy
f (g · z)
(Im z)2
T
.
1
dz
4
|cz + d| (Im g −1 · z)2
.T
dz
=
f (z)
.
(Im z)2
T
=
f (z)
(ii) Similarly to the statement (i).
(iii) Let z ∈ T and w ∈ D such that z = c · w. We have Im z =
Further, the Jacobian of the map w !→ c · w is 4/|1 + iw|4 . Thus
.
T
f (c−1 · z)
dz
(Im z)2
=
.
f (w)
D
= 4
.
D
$
|1 + iw|2
1 − |w|2
%2
1 − |w|2
.
|1 + iw|2
4
dw
|1 + iw|4
f (w)(1 − |w|2 )−2 dw.
!
Consider the subgroups K, A, N, and N of SL(2, R) with infinitesimal
generators
k=
$
%
0 1
,
−1 0
h=
$
%
1 0
,
0 −1
e+ =
$
%
0 1
,
0 0
e− =
$
%
0 0
,
1 0
4. BASIC PROPERTIES OF SL(2, R) AND SU (1, 1)
respectively. That is
%
<
cos θ sin θ
K= k=
| θ ∈ [0, 2π[ ,
− sin θ cos θ
7
$
%
<
a 0
A = ã =
|
a
>
0
,
0 a−1
7
$
%
<
1 x
N = n=
|x∈R ,
0 1
7
$
%
<
1 0
N = n̄ =
|x∈R .
x 1
$
%
α β
Every matrix
in SL(2, R) can be written as
γ δ
$
%
α β
= exp(θk) exp((log a)h) exp(xe+ )
γ δ
$
%$
%$
%
a 0
cos θ sin θ
1 x
,
=
− sin θ cos θ
0 1
0 a−1
where
a=
!
7
35
$
α2 + γ 2 ,
cos θ = !
α
α2 + γ 2
,
x=
αβ + γδ
.
α2 + γ 2
The above decomposition is called the Iwasawa decomposition of SL(2, R).
Thus the map
F : K × A × N −→ SL(2, R)
is an isomorphism. If (dF )g is the differential of F at the point
then
g = kãn
$
%$
%$
%
a 0
cos θ sin θ
1 x
=
,
− sin θ cos θ
0 1
0 a−1
(dF )g (h) =
8
9
(cos 2θ)h − (sin 2θ)k − 2(sin 2θ)e+ g,
(dF )g (k) = kg,
8 1
9
(dF )g (e+ ) = ( a2 sin 2θ)h − (a2 sin2 θ)k + (a2 cos 2θ)e+ g,
2
and so the Jacobian is given by a2 .
Let dg, dk, da/a, and dn be the Haar measures on SL(2, R), K, A, and
N, respectively. For an integrable function f on SL(2, R) we have
.
. . .
f (g)dg =
f (kãn)dk ada dn.
SL(2,R)
N
A
K
36
The same arguments in the Iwasawa decomposition allow us to write
SL(2, R) = N AN ∪ AN.
%
α β
More precisely, every
in SL(2, R) can be written
γ δ
%$
%$
%
 $
β
α
0
1
0
1

α
 γ
$
% 
if α %= 0,
1
0 α−1
0 1 %
α β
α
$
%
$
%
$
=
γ δ
1 γδ
γ
0
0 −1



if α = 0.
1 0
0 γ −1
0 1
$
This is the so-called Bruhat decomposition. The map
is an isomorphism. For
F2 : N × N × A −→ SL(2, R)
g = n̄ãn
$
%$
%$
%
a 0
1 0
1 y
=
,
x 1
0 1
0 a−1
we have
(dF2)g (e− ) = e− g,
8
9
(dF2)g (h) = (1 + 2xy)h − 2ye+ + (2x + 2x2 y)e− g,
8
9
(dF2)g (e+ ) =
− xh + e+ − x2 e− g.
The Jacobian of the map is 1 and
.
. . .
f (g)dg =
f (n̄na)dn̄dnda.
SL(2,R)
N
N
A
For t1 , t2 ∈ R, we have
$
%
a 0
exp(t1 k)
exp(t2 k)
0 a−1
$
%$
%
a 0
0
t1 + t2 a2
∼ exp
0 a−1
−t1 − t2 a−2
0
$
%
a 0
−2 −
2 +
∼ exp(−(t1 + t2 a )e ) exp((t1 + t2 a )e )
,
0 a−1
where ∼ means that the various expressions are equal to first order in t1
and t2 . Thus the Jacobian matrix of the transformation
is given by
K × A × K −→ N × N × A
det
$
−1
1
−a−2 a2
%
= −a2 + a−2 .
4. BASIC PROPERTIES OF SL(2, R) AND SU (1, 1)
The following integral formula holds:
.
. . .
da
f (g)dg =
f (k1 ãk2 )dk1 |a2 − a−2 | dk2
a
SL(2,R)
K A K
37
(4.2)
where dk1 = dk2 is a Haar measure on K. The decomposition KAK of
SL(2, R) is the
Cartan decomposition.
$ so-called
%
α β̄
For g0 =
∈ SU (1, 1), we set
β ᾱ
γ = β/α,
Notice that |γ| < 1, and
α = eiλ (1 − |γ|2 )−1/2 ,
λ = arg α[2π].
β = eiλ γ(1 − |γ|2 )−1/2 .
(4.3)
(4.4)
Thus we can parametrize SU (1, 1) in terms of γ and λ as follows
"
#
SU (1, 1) = (γ, λ) | |γ| < 1, −π < λ ≤ π .
$
%
α β̄
The identities (4.3) and (4.4) show that the correspondence
←→
β ᾱ
(γ, λ) is a bijection. The group multiplication in G0 = SU (1, 1) can be
expressed in terms of γ and λ by multiplying the corresponding matrices: if
g0 = (γ0 , λ0 ) and g0% = (γ0% , λ%0 ) are in SU (1, 1) then g0 g0% = g0%% = (γ0%% , λ%%0 ),
where
γ0%% = (γ0 + γ0% e−2iλ0 )(1 + γ̄0 γ0% e−2iλ0 )−1 ,
(4.5)
#
"
1
λ%%0 = λ0 + λ%0 + ln (1 + γ¯0 γ0% e−2iλ0 )(1 + γ0 γ¯0% e2iλ0 )−1 .
(4.6)
2i
Here ln(z) is defined by its principal value and λ%%0 is taken modulo 2π.
Moreover,
(γ0 , λ0 )−1 = (−e2iλ0 γ0 , −λ0 ),
and the identity element is (0, 0).
(i) The group
"
#
G = (γ, λ) | |γ| < 1, −∞ < λ< ∞
Definition 4.3.
is called the universal covering Lie group of SU (1, 1) (or of SL(2, R)).
The multiplication on G is given by (4.5) and (4.6) except that w%%
is no longer defined modulo 2π. The difference between SU (1, 1)
and G is that on G any two distinct real values of λ give distinct
elements of G.
(ii) The group
"
#
Gd = (γ, λ) | |γ| < 1, −dπ < λ ≤ dπ ,
with the product rules as above, except w%% is taken modulo 2dπ, is
called a d-fold cover of SU (1, 1).
38
The canonical homomorphism Φ : G −→ G0 is defined by
Φ(γ, λ) = (γ, λ mod[2π])
and
Ker Φ = {(0, 2kπ) | k ∈ Z}.
The action of G0 on the open unit disk D induces an action of G on D :
for g ∈ G and z ∈ D, we have
g · z = g0 · z, where Φ(g) = g0 ,
and, since Φ is a homomorphism, we further have
(g1 g2 ) · z = g1 · (g2 · z),
for all g1 , g2 ∈ G.
5. Finite dimensional representation of SL(2, R)
For an integer m ≥ 0, denote by Pm (R) the complex vector space of
polynomials of degree ≤ m in one variable x. Note that dim(Pm (R)) = m+1.
For g ∈ SL(2, R) and p ∈ Pm (R) we set
$
%
a b
−1
m ax + b
.
),
g =
πm (g)p(x) = (cx + d) p(
c d
cx + d
We claim that (πm , Pm ) is a finite dimensional representation of SL(2, R).
Indeed, if we let
$
%
a b
(g −1 , x) = (cx + d)−1 ,
for g −1 =
,
c d
then we have
(g2−1 g1−1 , x) = (g2−1 , g1−1 · x)(g1−1 , x),
(I2 , x) = 1.
Thus πm (I2 )p(x) = p(x), and
πm (g1 g2 )p(x) = (g2−1 g1−1 , x)−m p(g2−1 g1−1 · x)
= (g2−1 , g1−1 · x)(g1−1 , x)p(g2−1 · g1−1 · x)
= (g1−1 , x)(πm (g2 )p)(g1−1 · x)
= πm (g1 )πm (g2 )p(x).
Recall that
"
#
g = sl(2, R) = X ∈ M (2, R) | tr(X) = 0 = )e+ , e− , h*,
where {h, e+ , e− } are given by
%
$
%
$
0 0
0 1
−
+
,
e =
,
e =
1 0
0 0
h=
$
%
1 0
.
0 −1
We shall find the infinitesimal representation dπm of the Lie algebra g.
5. FINITE DIMENSIONAL REPRESENTATION OF SL(2, R)
(i) For h =
$
1 0
0 −1
%
and p ∈ Pm (R), we have
dπm (h)p(x) =
where
πm (exp(th))p(x) = πm
Hence
dπm (h)p(x) =
(ii) For
e+
=
$
39
d
πm (exp(th))f (x)
dt |t=0
$
%
et 0
p(x) = emt p(e−2t x).
0 e−t
d
emt p(e−2t x) = mp(x) − 2xp% (x).
dt |t=0
%
0 1
, we have
0 0
dπm (e+ )p(x) = −p% (x).
$
%
0 0
−
, we have
(iii) For e =
1 0
dπm (e− )p(x) = −mxp(x) + x2 p% (x).
On the space Pm (R) we consider the basis
"
#
(−1)i m−i
pi (x) =
x
|0≤i≤m .
(m − i)!
Thus
dπm (h)pi (x) = (2i − m)pi (x),
dπm (e+ )pi (x) = pi+1 (x),
dπm (e− )pi (x) = i(m − i + 1)pi−1 (x),
with pm+1 = 0 and p−1 = 0.
Theorem 5.1. The representation πm is irreducible.
Proof. Let U be a closed subspace of Pm (R) which is stable by πm .
Thus, by Proposition 2.21, U is stable by dπm . Let f %= 0 be an eigenvector
1
for dπm (h), say dπm (h)f = λf for λ ∈ C. Using the fact that f = m
i=0 αi pi
it follows that
m
&
dπm (h)f =
αi (2i − m)pi
while
i=0
dπm (h)f =
m
&
λαi pi .
i=0
Hence there exists i0 such that λ = (2i0 − m) and αi = 0 for all i %= i0 .
This implies that f = αi0 pi0 . That is pi0 ∈ U. Moreover, since U is stable by
dπm , it follows from pi0 ∈ U that πm (e+ )pi0 = pi0 +1 ∈ U and dπm (e− )pi0 =
40
i0 (m − i0 + 1)pi0 −1 ∈ U. That is pi0 +1 and pi0 −1 belong to U. Now applying
the same argument again to pi0 +1 and pi0 −1 we deduce that pi0 ∈ U for all
0 ≤ i0 ≤ m. Hence U = Pm (R).
!
Theorem 5.2. For every integer m ≥ 1, there exists up to a equivalence
a unique irreducible representation π of SL(2, R) on a complex vector space
V of dimension m + 1. In V, there is a basis {v0 , . . . , vm } such that
dπ(h)vi = (2i − m)vi ,
dπ(e+ )vi = vi+1 ,
dπ(e− )vi = i(m − i + 1)vi−1 ,
with vm+1 = 0 and v−1 = 0.
Proof. Let π be a representation of SL(2, R) on V, where dim(V ) =
m + 1. Let v %= 0 be an eigenvector for dπ(h) with eigenvalue λ ∈ C. Then
dπ(e− )v, dπ(e− )2 v, . . . are also eigenvectors for dπm (h) since
dπ(h)dπ(e− )v = dπ(e− )dπ(h)v + dπ([h, e− ])v
= λdπ(e− )v − 2dπ(e− )v
= (λ − 2)dπ(e− )v.
By induction on i, it follows that dπ(h)dπ(e− )i v = (λ − 2i)dπ(e− )i v. Since
λ, λ − 2, λ − 4, . . . are distincts, the eigenvectors v, dπ(e− )v, dπ(e− )2 v, . . . are
independent (or of course 0). Using the fact that dim(V ) < ∞, we can find
v0 ∈ V such that v0 %= 0, dπ(h)v0 = λv0 and dπ(e− )v0 = 0 (for a redefined
λ). Define vi = dπ(e+ )i v0 . Then
dπ(h)vi = dπ(h)dπ(e+ )i v0
"
#
=
dπ(e+ )dπ(h) + dπ([h, e+ ]) dπ(e+ )i−1 v0
= dπ(e+ )dπ(h)dπ(e+ )i−1 v0 + 2dπ(e+ )dπ(e+ )i−1 v0
= (λ + 2i)vi .
Hence there is a minimum integer n0 such that dπ(e+ )n0 +1 v0 = 0. Therefore,
the eigenvectors {v0 , v1 , . . . , vn0 } are independent, and
dπ(h)vi = (λ + 2i)vi
dπ(e− )v0 = 0
dπ(e+ )vi = vi+1 , with vn0 +1 = 0.
We claim that {v0 , v1 , . . . , vn0 } spans V. It is enough to show that {v0 , . . . , vn0 }
is stable by dπ(e− ). Indeed dπ(e− )v0 = 0,
dπ(e− )v1 =
=
=
=
dπ(e− )dπ(e+ )v0
dπ(e+ )dπ(e− )v0 + dπ([e− , e+ ])v0
−dπ(h)v0
−λv0 .
5. FINITE DIMENSIONAL REPRESENTATION OF SL(2, R)
dπ(e− )v2 =
=
=
=
=
Let us assume that
41
dπ(e− )dπ(e+ )v1
dπ(e+ )dπ(e− )v1 + dπ([e− , e+ ])v1
−λdπ(e+ )v0 − dπ(h)v1
−λv1 − (λ + 2)v1
−2(λ + 1)v1 .
dπ(e− )vi = −i(λ + i − 1)vi−1
with v−1 = 0.
To prove the case i + 1, we write
dπ(e− )vi+1 =
=
=
=
dπ(e− )dπ(e+ )vi
dπ(e+ )dπ(e− )vi + dπ([e− , e+ ])vi
dπ(e+ ){−i(λ + i − 1)vi−1 } − dπ(h)vi
−i(λ + i − 1)vi − (λ + 2i)vi
= {λ(−i − 1) − i2 − i}vi
= −(i + 1)(λ + i)vi .
Thus the set {v0 , v1 , . . . , vn0 } spans V. Using the fact that {v0 , . . . , vn0 } are
independent, it follows that {v0 , . . . , vn0 } forms a basis for V and n0 = m.
To finish the proof we need to prove that λ = −m. Since 0 = dπ(e− )vm+1 =
vm {−(m + 1)(λ + m)}, we obtain λ = −m. In conclusion
dπ(h)vi = (2i − m)vi ,
dπ(e+ )vi = vi+1
with vm+1 = 0,
−
dπ(e )vi = i(m − i + 1)vi−1
with v−1 = 0.
Since π is irreducible it follows from Corollary 2.22 that dπ is also irreducible.
The same holds for dπm . Clearly the linear map A : Pm (R) −→ V defined
by
A(fi ) = vi
for all 0 ≤ i ≤ m,
is an intertwining operator between dπm and dπ. Hence
Aπm (g) = π(g)A,
∀g ∈ SL(2, R).
!
Theorem 5.3. The only finite dimensional unitary irreducible representation of SL(2, R) is the trivial one-dimensional representation.
Proof. Recall that
"
#
X ∈ M (2, R) | tr(X) = 0
%
"$
#
a b
=
| a, b, c ∈ R .
c −a
g = sl(2, R) =
42
Define the involution θ : g −→ g by θ(X) = −t X, and set
"
#
k :=
X ∈ g | θ(X) = X
%
"$
#
0 b
=
|b∈R ,
−b 0
and
p :=
=
"
#
X ∈ g | θ(X) = −X
%
"$
#
a c
| a, c ∈ R .
c −a
Notice that sl(2, R) = k ⊕ p, the so-called Cartan decomposition of sl(2, R).
Further, let su(2) be the Lie algebra of the compact group
%
"$
#
α β
SU (2) =
| α, β ∈ C, |α|2 + |β|2 = 1 .
−β̄ ᾱ
That is
su(2) =
"$
it1
t2 + it3
−t2 + it3
−it1
%
| t1 , t2 , t3 ∈ R} = k ⊕ ip.
Let (dπ, V ) be the infinitesimal representation of sl(2, R) associated with
the unitary irreducible representation (π, V ) of SL(2, R). Let (ω, V ) be the
corresponding representation of su(2) via the direct sums sl(2, R) = k ⊕
p and su(2) = k ⊕ ip. Since SU (2) is compact, then su(2) acts on V by
skew-hermitian operators. It follows that ip acts by diagonable operators
with imaginary eigenvalues. That is p acts by diagonable operators with
real eigenvalues. Moreover, since the representation (π, V ) of SL(2, R) is a
unitary representation, it follow that p acts by diagonable operators with
imaginary eigenvalues. Hence p acts as 0. Using the fact that [p, p] = k
(in general [p, p] ⊂ k) we deduce that dπ(X) = 0 for all X ∈ sl(2, R), i.e.
π(g) = idV for all g ∈ G. In addition, since we assumed that (π, V ) is
irreducible, then dim(V ) = 1.
!
6. An application: The wave equation
The wave equation for N space dimensions
N
&
∂2u
2 (x, t)
∂x
i
i=1
=
∂2u
(x, t),
∂t2
x ∈ RN , t ∈ R,
is one of the most celebrated equations of mathematical physics.
Question 1: Solve the following Cauchy problem
7
∆u(x, t) = ∂tt u(x, t),
u(x, 0) = f (x), ∂t u(x, 0) = g(x),
where f, g ∈ S(RN ).
(6.1)
6. AN APPLICATION: THE WAVE EQUATION
43
For any f in S(RN ) define its euclidean Fourier transform F(f ) by
.
−N/2
F(f )(λ) = (2π)
e−iλ·x f (x)dx,
RN
where dx = dx1 · · · dxN is the Lebesgue measure on RN . We define the
inverse Fourier transform F −1 (f ) by
.
−1
−N/2
F (f )(λ) = (2π)
eiλ·x f (x)dx.
RN
Since the Fourier transform is linear and continuous in the topology of
S(RN ), we can define the Fourier transform F(T ) of a tempered distributions
T by
F(T )(ψ) = T (F(ψ)),
ψ ∈ S(RN ).
For fixed t ∈ R, let Pt be the following 2 × 2 matrix of tempered distribution on RN
$ 1,1
%
$
%
Pt
P1,2
F −1 (cos(t( ·( ))
F −1 (sin(t( ·( )/( ·( )
t
Pt =
:=
.
F −1 (−( · ( sin(t( ·( ))
F −1 (cos(t( ·( ))
P2,1
P2,2
t
t
$
%
f (x)
Put U (x, 0) :=
. Thus we may define the vector column U (x, t)
g(x)
by
A
B
U (x, t) := Pt ∗ U (·, 0) (x)
% $ %
$ 1,1
f
Pt
P1,2
t
(6.2)
=
2,1
2,2 ∗ g (x),
Pt
Pt
where ∗ indicates convolution on RN . Applying the Fourier transform F to
(6.2) in the x-variable we get
F(U (·, t))(ξ) = etA F(U (·, 0))(ξ),
$
%
0
1
where A =
. That is the function t !→ F(U (·, t)) is a solution
−(ξ(2 0
of the following differential equation
$
%
0
1
∂t F(U (·, t))(ξ) = AF(U (·, t))(ξ) =
F(U (·, t))(ξ).
−(ξ(2 0
Using the fact −(ξ(2 F(f )(ξ) = F(∆f )(ξ), and the injectivity of the euclidean Fourier transform, we deduce that
$
%
0 1
∂t U (x, t) =
U (x, t).
∆ 0
In other words, if we write
U (x, t) =
then
$
$
%
u(x, t)
,
v(x, t)
% $
%$
%
∂t u(x, t)
0 1
u(x, t)
=
.
∂t v(x, t)
∆ 0
v(x, t)
44
That is v(x, t) = ∂t u(x, t), and u(x, t) satisfies the wave equation
∆u(x, t) = ∂tt u(x, t).
Remark 6.1. The solution u(x, t) constructed above is unique. This is
due to the fact that
. "
N
#
&
E(u)(t) :=
|∂t u(x, t)|2 +
|∂i u(x, t)|2 dx
RN
i=1
is independent of t. Thus
E(u)(t) = E(u)(0) =
.
RN
N
#
"
&
2
2
|g(x)| +
|∂i f (x)| dx
(∗)
i=1
for all t ∈ R. Now if u1 and u2 are two solutions to the Cauchy problem
(6.1) with the same initial data (f, g), then (u1 − u2 )(x, t) is a solution to
the Cauchy problem with initial data (0, 0). It follows from (*) that u1 (x, t) =
u2 (x, t). We leave the details for the reader.
by
Theorem 6.2. The unique solution to the Cauchy problem (5.1) is given
1,2
u(x, t) = (P1,1
t ∗ f )(x) + (Pt ∗ g)(x),
(x, t) ∈ RN × R,
= F −1 (cos(t( ·( )) and P1,2
= F −1 (sin(t( ·( )/( ·( )
where, for fixed t, P1,1
t
t
are two tempered distributions, and ∗ indicates convolution on RN .
$ 1,1
%
P
P1,2
N
+1
Define the distribution P on R
by P =
where
P2,1 P2,2
.
i,j
P (ψ1 ⊗ ψ2 ) =
Pi,j
t (ψ1 )ψ2 (t)dt,
R
S(RN )
for ψ1 ∈
and ψ2 ∈ S(R). Here we used the identification S(RN +1 ) 2
N
S(R ) ⊗ S(R).
Question 2: Search the condition(s) on N such that the distributions
P1,1 and P1,2 are supported in the cone
"
#
(x, t) ∈ RN +1 | (x( = |t| .
In physics this means that the wave equation ∆u(x, t)−∂tt u(x, t) = 0 satisfies
the Huygens principle.
Note that
∂ 2 P1,1
∂ 2 P1,2
1,1
=
∆P
,
= ∆P1,2 .
∂t2
∂t2
Corollary 6.3. The distributions P1,1 and P1,2 on RN +1 anuhulated by
the operator ∆ − ∂tt .
For λ > 0 and a function f = f (x, t) on RN +1 , define
Sλx f (x, t) := f (λx, t),
Sλt f (x, t) := f (x, λt).
6. AN APPLICATION: THE WAVE EQUATION
45
Set Sλ := Sλx ◦ Sλt . By duality the operators Sλx , Sλt , and Sλ act on distributions.
Let u(x, t) be a solution to the Cauchy problem (5.1) with initial data
(f, g), then Sλ u(x, t) solves the wave equation with initial data (Sλx f, λSλx g).
Thus, as in (5.2) we have
Sλ U (x, t) :=
$
%
$ x %
Sλ u(x, t)
Sλ f
= Pt ∗
.
∂t Sλ u(x, t)
λSλx g
(6.3)
On the other hand we have
Sλ U (x, t) =
=
=
=
=
=
=
=
$
%
Sλ u(x, t)
∂t Sλ u(x, t)
$
%
u(λx, λt)
λ∂t u(λx, λt)
$
%
u
(λx, λt)
λ∂t u
$
%$ %
1 0
u
(λx, λt)
0 λ
∂t u
$
%"
$ %#
1 0
f
Pλt ∗
(λx)
0 λ
g
$
% "
$ %#
1 0
f
x
Sλ Pλt ∗
(x)
0 λ
g
$
%"
$ x %#
1 0
Sλ f
Sλx Pλt ∗
(x)
0 λ
Sλx g
$
%"
$
%$ x %#
1 0
1 0
Sλ f
x
(x).
Sλ Pλt ∗
−1
0 λ
λSλx g
0 λ
Above we have used the fact that Sλx preserve the convolution of a distribution with a function. Comparing with the identity (6.3) gives
Pt =
$
%
$
%
1 0
1 0
x
Sλ Pλt
0 λ
0 λ−1
and therefore
j−i i.j
Sλx Pi,j
Pt ,
λt = λ
for all 1 ≤ i, j ≤ 2.
46
We claim that Sλ Pi,j = λ1+j−i Pi,j . Indeed, for ψ1 ∈ S(RN ) and ψ2 ∈ S(R)
we have
Sλ (Pi,j )(ψ1 ⊗ ψ2 ) = Pi,j (Sλx−1 ψ1 ⊗ Sλt −1 ψ2 )
.
x
t
=
Pi,j
t (Sλ−1 ψ1 )Sλ−1 ψ2 (t)dt
R
.
x
= λ Pi,j
λt (Sλ−1 ψ1 )ψ2 (t)dt
.R
= λ Sλx Pi,j
λt (ψ1 )ψ2 (t)dt
R
.
= λ1+j−i
Pi,j
t (ψ1 )ψ2 (t)dt
= λ
R
1+j−i i,j
P (ψ1 ⊗ ψ2 ).
Corollary 6.4. The distributions P1,1 and P1,2 satisfy
Sλ P1,1 = λP1,1 ,
Sλ P1,2 = λ2 P1,2 .
Set
EN,1
i
= ((x(2 −t2 ),
2
FN,1
i
= (∆−∂tt ),
2
HN,1
N
N +1 &
=
+
xi ∂i +t∂t .
2
i=1
Recall the standard basis {e+ , e− , h} of the Lie algebra sl(2, R), where
$
%
$
%
$
%
0 −1
0 0
1 0
+
−
e =
,
e =
,
h=
,
0 0
−1 0
0 −1
so that
[e− , h] = 2e− ,
[e+ , h] = −2e+ ,
[e+ , e− ] = h.
Proposition 6.5. The map
i
((x(2 − t2 ),
2
i
e− !→ applying the operator (∆ − ∂tt ),
2
N
N +1 &
h !→ applying the operator
+
xi ∂i + t∂t ,
2
e+ !→ multiplication by
i=1
is a Lie algebra homomorphism from sl(2, R) on functions and/or distributions on RN +1 .
Proof. This follows directly from Theorem 3.1 and the fact that
[t2 ,
1
+ t∂t ] = −2t2 ,
2
[∂tt ,
1
+ t∂t ] = 2∂tt ,
2
[t2 , ∂tt ] = −2 − 4t∂t .
!
6. AN APPLICATION: THE WAVE EQUATION
47
(Howe). If a distribution
" Proposition 6.6. 1
# T is supported on the set
1q
p
p+q
2
2
2
(x, y) ∈ R
| rp,q := i=1 xi − i=1 yi = 0 , then we can find an integer
k ≥ 1 such that
2 k
) f ) = 0,
T ((rp,q
by
∀f ∈ Cc∞ (Rp+q ).
As in Section 3 Part 2, define the representation (ω, S(RN +1 )) of sl(2, R)
i
ω(e ) = ((x(2 −t2 ),
2
+
i
ω(e ) = (∆−∂tt ),
2
−
N
N +1 &
ω(h) =
+
xi ∂i +t∂t .
2
i=1
Recall that our question is to find the condition(s) on N such that P1,1 and
P1,2 are supported on the cone {(x, t) ∈ RN +1 | (x(2 = t2 }. In view of
Proposition 6.6, this is equivalent to the fact that
((x(2 − t2 )k Pi,j = 0
for some k ∈ N∗ ,
with (i, j) = (1, 1) and (1, 2). Recall the homogeneousity of P1,1 and P1,2
and the fact (∆ − ∂tt )P1,1 = 0 and (∆ − ∂tt )P1,2 = 0. Thus we deduce from
the proof of Theorem 5.2 the following:
Theorem 6.7. For (i, j) = (1, 1) or (i, j) = (1, 2), the distribution Pi,j is
supported on {(x, t) ∈ RN +1 | (x(2 = t2 } if and only if Pi,j generates a finite
dimensional representation of sl(2, R) on the space S ∗ (RN +1 ) of tempered
distribution.
Let us recall the following definition:
Definition 6.8. Let (ω, V ) be a representation of a Lie algebra g. Let
V ∗ be the dual space to V, the space of continuous linear functionals on V.
Define an action ω ∗ of g on V ∗ by the recipe
ω ∗ (X)(λ)(v) = −λ(ω(X)v),
X ∈ g, λ ∈ V ∗ , v ∈ V.
We call (ω ∗ , V ∗ ) the contragredient representation to (ω, V ).
In view of Corollary 6.4 we have
N
&
(
xi ∂i + t∂t )Pi,j = (1 + j − i)Pi,j .
i=1
That is
HN,1 Pi,j = −(
N +1
+ i − j − 1)Pi,j .
2
48
Theorem 6.9. Let (i, j) = (1, 1) or (i, j) = (1, 2). The distribution Pi,j
generates a finite dimensional representation of sl(2, R) if and only if
N −1
+ i − j ∈ N.
2
The dimension of the representation generated by Pi,j is
dij =
N +1
+ i − j.
2
The answer to question 2: The two distributions P1,1 and P1,2 generate finite dimensional representations of sl(2, R) simultaneously if and only
if N is odd ≥ 3. That is, the wave equation ∆u(x, t) = ∂tt u(x, t) satisfies
the Huygens principle if and only if N is odd ≥ 3.
7. The discrete series representation of SL(2, R)
In this part we will change slightly
our %previous notations.
$
a11 a12
For an invertible matrix g =
∈ GL(2, C) and z ∈ C, we set
a21 a22
z·g =
We also set
(z, g) =
a11 z + a21
.
a12 z + a22
d(z · g)
det g
=
.
dz
(a12 z + a22 )2
As before, one can check that (z · g1 ) · g2 = z · (g1 g2 ), (z, g1 g2 ) = (z, g1 )(z ·
g1 , g2 ), and (z, I2 ) = 1.
Let
%
" $
#
α β̄
G0 := SU (1, 1) =
| α, β ∈ C, |α|2 − |β|2 = 1 ,
β ᾱ
$
%
1 −1
−1
and consider the group G1 = SL(2, R) = cSU (1, 1)c , where c :=
.
−i −i
$
%
α β̄
Then, if g0 =
∈ SU (1, 1), we have
β ᾱ
$
%
α1 − β1 −α2 + β2
−1
g1 = cg0 c =
(7.1)
α2 + β2 α1 + β1
where
α = α1 + iα2 ,
β = β1 + iβ2 .
(7.2)
The action of c on z ∈ C is given by w = z · c = (z − i)/(−z − i), and
its inverse is z = w · c−1 = −i(w − 1)/(w + 1). The transformation w = z · c
takes the upper half plane T = {z = x + iy | y > 0} onto the open unit disk
D = {w ∈ C | |w| < 1}.
7. THE DISCRETE SERIES REPRESENTATION OF SL(2, R)
49
For an integer m, with m ≥ 2, we define the Hilbert space Hm (D) of
holomorphic functions on D with the inner product
.
m−1
(f, g)m =
f (w)g(w)(1 − |w|2 )m−2 dw.
π
D
The functions
3
41/2
Γ(m + n)
ϕm,n (w) =
wn ,
n ∈ N,
Γ(m)Γ(n + 1)
form an orthonormal basis in Hm (D).
Definition and Proposition 7.1. Let H be a Hilbert space consisting
of complex functions on an open set E.
(i) We call reproducing kernel for H a complex function K : E×E → C
such that, if we put Kw (z) := K(z, w), then the following holds:
(a) for every w ∈ E, the function Kw belongs to H,
(b) for all f ∈ H and w ∈ E, we have f (w) = )f, Kw *.
(ii) For all z, w ∈ E, K(z, w) = K(w, z).
(iii) For every orthonormal basis {ψm } of H, we have
&
ψm (z)ψm (w).
K(z, w) =
m
Theorem 7.2. (Riesz) A Hilbert space H possesses a reproducing kernel
if and only if for all z ∈ E, the map H → C, f !→ f (z) is continuous.
Since the functions
3
Γ(m + n)
ϕm,n (w) =
Γ(m)Γ(n + 1)
41/2
wn ,
n ∈ N,
form an orthonormal basis in Hm (D), for m ≥ 2, using the statement (iii)
above, the following holds.
by
Proposition 7.3. For m ≥ 2, the reproducing kernel of Hm (D) is given
Km (w0 , w) = (1 − w0 w)−m ,
w0 , w ∈ D.
On the upper half plane T, for m ≥ 2, we take the Hilbert space Hm (T)
of holomorphic functions on T with the inner product
.
1
(f, g)m =
f (x + iy)g(x + iy)y m−2 dxdy.
Γ(m − 1) T
For m ≥ 2, define the map Em : O(D) −→ O(T) by
Γ(m)1/2 −1+m
√
2
(1 − iz)−m f (z · c),
π
where z ∈ T and f ∈ O(D). For f˜ ∈ O(T) and w ∈ D we set
√
2 π
−1 ˜
Em f (w) =
(1 + w)−m f˜(w · c−1 ).
Γ(m)1/2
Em f (z) =
50
−1 f˜ are holomorphic on T and D, respectively. A simple
Clearly Em f and Em
−1 = E −1 E = id, the identity transformation
calculation shows that Em Em
m
m
on functions defined on D and T.
Lemma 7.4. For m ≥ 2, Em is a unitary map from Hm (D) to Hm (T).
Proof. For f, g ∈ Hm (D) we have
.
1
Em f (z)Em g(z)y −2+m dxdy
(Em f, Em g)m =
Γ(m − 1) T
.
Γ(m)2−2+2m
=
f (z · c)g(z · c)|1 − iz|−2m y −2+m dxdy
Γ(m − 1)π T
.
m−1
f (w)g(w)(1 − |w|2 )m−2 dw
=
π
D
= (f, g)m .
!
In view of the above proposition, we can prove that:
by
Proposition 7.5. For m ≥ 2, the reproducing kernel of Hm (T) is given
Km (z0 , z) =
m − 1 + z0 − z ,−m
,
4π
2i
z0 , z ∈ T.
Proof. We already know that the transformation z = w · c−1 takes the
open disk D onto the upper half plane T. It follows that
Im (z) =
1 − |w|2
,
|1 + w|2
and the Jacobian of c−1 is 4/|w + 1|4 . Let z0 = w0 · c−1 ∈ T. We have
.
+ z − z ,−m
m−1
0
f˜(z)
y m−2 dxdy
4π
2i
T
.
+ w · c−1 − w · c−1 ,−m + 1 − |w|2 ,m−2
m−1
0
|1 + w|−4 dw
=
f˜(w · c−1 )
2
π
2i
|1
+
w|
D
.
m−1
(1 − w0 w)−m
(1 − |w|2 )m−2
=
f˜(w · c−1 )
dw
π
(w0 + 1)−m (w + 1)−m |1 + w|2m
D
.
m−1
=
(1 + w0 )m
f˜(w · c−1 )(1 + w)−m (1 − w0 w)−m (1 − |w|2 )m−2 dw
π
D
.
Γ(m)1/2 m − 1
√
=
(1 + w0 )m
f (w)(1 − w0 w)−m (1 − |w|2 )m−2 dw
π
2 π
D
Γ(m)1/2
√ (1 + w0 )m f (w0 )
2 π
= f˜(z0 ).
=
!
7. THE DISCRETE SERIES REPRESENTATION OF SL(2, R)
51
We come now to define the representations of the groups SU (1, 1) and
SL(2, R).
a11 z + a21
The action of SL(2, R) on T is given by z !→ z · g1 =
where
a12 z + a22
$
%
a11 a12
g1 =
∈ SL(2, R). On Hm (T), with m ≥ 2, define
a21 a22
Tm (g1 )f (z) = (a12 z + a22 )−m f (z · g1 ).
For SU (1, 1), on Hm (D), with m ≥ 2, define
Um (g0 )f (w) = (β̄w + ᾱ)−m f (w · g0 ).
Remark 7.6. Assume that m is a real number ν > 1. In this case,
we should replace the group SU (1, 1) by the universal covering Lie group G
defined in a previous section. More precisely, for g = (γ, λ) ∈ G, we set
mν (w, g) := eiλν (1 − |γ|2 )ν/2 (1 + γ̄w)−ν ,
where w ∈ D (recall that |γ| < 1). In particular, if ν = 2, 3, 4, . . . , we have
%
α β̄
∈ G0 . Further, for ν > 1, the map mν : D × G −→ C
β ᾱ
is a continuous function and satisfies the multiplier equation mν (w, g1 g2 ) =
mν (w, g1 )mν (w · g1 , g2 ) with mν (w, e) = 1, where e = (0, 0) is the identity
element in G. In these circumstances, we define the representation g !→
Uν (g) of G in Hν (D) by
where Φ(g) =
$
mν (w, g) = (β̄w + ᾱ)−ν ,
Uν (g)f (w) = mν (w, g)f (w · g)
= eiλν (1 − |γ|2 )ν/2 (1 + γ̄w)−ν f (w · g0 )
where g = (γ, λ) ∈ G and Φ(g) = g0 .
Theorem 7.7. For m ≥ 2, the map g0 !→ Um (g0 ) defines a unitary
representation of G on Hm (D).
$
%
α β̄
Proof. For fixed g0 =
and f ∈ Hm (D), we have
β ᾱ
.
m−1
2
(Um (g0 )f (m =
|Um (g0 )f (w)|2 (1 − |w|2 )m−2 dw
π
D
.
m−1
=
|β̄w + ᾱ|−2m |f (w · g0 )|2 (1 − |w|2 )m−2 dw.
π
D
Making the change of variable w · g0 → w we obtain
.
m−1
(Um (g0 )f (2m =
|f (w)|2 (1 − |w|2 )m−2 dw = (f (2m .
π
D
Thus, in order to prove the continuity of Um , we have to verify that as
g −→ e, we have
(Um (g)f − f (m −→ 0,
f ∈ Hm (D).
52
This is immediate from the dominated convergence theorem.
!
We want to show that Um (g0 ) and Tm (g1 ) are unitarily equivalent where
g0 = c−1 g1 c.
Theorem 7.8. For m ≥ 2, the representations g0 !→ Um (g0 ) of SU (1, 1)
and g1 !→ Tm (g1 ) of SL(2, R) are unitarly equivalent. More precisely
−1
Um (g0 ) = Em
Tm (g1 )Em ,
where g0 = c−1 g1 c.
Proof. Let f ∈ Hm (D). For g0 =
$
α β̄
β ᾱ
%
∈ SU (1, 1) we have
Γ(m)1/2 −1+m
√
2
(1 − iz)−m (β̄(z · c) + ᾱ)−m f (z · cg0 ).
π
$
%
a11 a12
−1
On the other hand, if g1 = cg0 c =
∈ SL(2, R),
a21 a22
Em Um (g0 )f (z) =
Tm (g1 )Em f (z) =
Γ(m)1/2 −1+m
√
2
(a12 z + a22 )−2ν (1 − i(z · g1 ))−m f (z · g1 c).
π
Since cg0 = g1 c and m ∈ N, the result follows from the fact that (1−iz)(β̄(z ·
c) + ᾱ) = (a12 z + a22 )(1 − i(z · g1 )) by means of (7.1) and (7.2).
!
Theorem 7.9. For m ≥ 2, the representation g0 !→ Um (g0 ) of SU (1, 1)
on Hm (D) is irreducible.
Proof. Assume that U is a closed invariant subspace of Hm (D). There
exists then an element f ∈ U and w0 ∈ D such that f (w0 ) %= 0. We can find
g ∈ SU (1, 1) such that w0 · g −1 = 0 (recall that SU (1, 1) acts transitively
on D). Thus Um (g)f (0) = ᾱ−m f (w0 ). Hence we can $
assume that
% f (0) %= 0.
−iθ
1
e
0
Write f (w) = n≥0 an wn , for w ∈ D. Thus for kθ =
we have
0
eiθ
&
Um (kθ )f (w) = e−imθ f (e−2iθ w) = e−imθ
an e−2iθn wn .
n≥0
Since U is closed, it follows that
. 2π
. 2π
1
1
imθ
e Um (kθ )f (w)dθ =
f (e2iθ w)dθ = a0 = f (0)
2π 0
2π 0
belongs to U. That is constants belong to U. On the other hand, the unitarity
of the representation implies that U ⊥ is also stable by Um . Thus, if U is not
every thing, applying the same argument to U ⊥ to show that constants
belong also to U ⊥ , which lead to a contradiction.
!
Let (π, H) be a unitary representation of a linear Lie group G. The
matrix coefficient function attached to u, v ∈ H is defined to be
cu,v (g) = )π(g)u, v*H .
7. THE DISCRETE SERIES REPRESENTATION OF SL(2, R)
53
Theorem 7.10. (Godment). Let (π, H) be an irreducible unitary representation of a unimodular Lie group G. If a matrix coefficient of π is
in L2 (G), then all matrix coefficients of π are in L2 (G). A representation
with this property is called a discrete series representation. Moreover, for
all u, v ∈ H, there exists a positive constant d, which depends on the normalization of the Haar measure, such that
()π(g)u, v*(L2 (G) = d(u(2H (v(2H .
If G is compact, every irreducible representation is in the discrete series.
Moreover, if the Haar measure has total measure 1, then d = (dim π)−1 . For
this reason d−1 is called the formal degree of π.
Proof. Since the map g !→ π(g)v is continuous, each coefficient function
is a continuous C-valued function on G. By the unitarniess of π, for g, x, y ∈
G, the bi-regular representation’s behavior is
cu,v (y −1 gx) = )π(y −1 gx)u, v* = )π(g)π(x)u, π(y −1 )∗ v*
= )π(g)π(x)u, π(y)v* = cπ(x)u,π(y)v (g).
That is, letting L be the left regular representation of G, and R be the right
regular representation of G on the space of functions on G, then
Ly Rx cu,v (g) = cπ(x)u,π(y)v (g).
(7.3)
Let u, v ∈ H such that cu,v ∈ L2 (G). Let
and on W define
W := {x ∈ H | cx,v ∈ L2 (G)} =
% ∅,
T x = cx,v .
Thus T is a linear map T : W −→ L2 (G). The operator T is a closed
operator. Indeed, let (wn ) be a sequence of point in W and suppose that
the sequence of points (wn , T wn ) has a limit (u, f ) ∈ H ⊕ L2 (G). We must
show that f = T u :
(i) Pointwise convergence: Let g ∈ G. As wn → u ∈ H
T wn (g) = cwn ,v (g) = )π(g)wn , v* = )wn , π(g)−1 v*
−→ )u, π(g)−1 v* = )π(g)u, v* = cu,v (g) = T u(g)
as n goes to ∞.
(ii) Uniform convergence: Indeed
|cwn ,v (g) − cu,v (g)| = |)π(g)(wn − u), v*| ≤ (wn − u(H (v(H
by Cauchy-Schwrartz inequality and the unitariness of π.
Now we will show that f = cu,v in the L2 -sense. Let
"
1#
Xn = x ∈ G | |f (x)| ≥
.
n
54
For fixed n, µ(Xn ) < ∞. Indeed,
.
.
1
2
|f (g)|2 dg < ∞.
µ(X
)
≤
|f
(x)|
dx
≤
n
n2
Xn
G
Moreover, since G = ∪n∈N∗ Xn and the sets Xn are two by two non-disjoint,
then
∞
∞
&
&
(cu,v −f (L2 (G) ≤
(cu,v −f (L2 (Xn ) ≤
(cu,v −cwin ,v (L2 (Xn ) +(cwin ,v −f (L2 (Xn )
n=1
n=1
where for every n, the index in is arbitrary. Since the measure of Xn is
finite, we can choose in large enough such that
$
(cu,v − cwin ,v (L2 (Xn ) < n
2
by the uniform convergence, and
$
(cwin ,v − f (L2 (Xn ) ≤ (cwin ,v − f (L2 (G) < n
2
via the convergence in L2 (G). Therefore
(cu,v − f (L2 (G) ≤
∞
&
n=1
$
2n−1
L2 (G).
=$
∞
&
n=1
1
2n−1
= 2$
for every $ > 0. That is cu,v = f ∈
Next we need the following version of the Schur lemma: ”Assume that
H1 and H2 are Hilbert spaces. Let T : H1 −→ H2 be a densely defined,
closed intertwining operators between two unitary representations (π1 , H1 )
and (π2 , H2 ) of a Lie group G. Then, if π1 is irreducible, T is a scalar
multiple of an isometric embedding.“ (See Howe, Theorem 1.2.15).
By (7.3) we have
T (π(g)x) = cπ(g)x,v = Rg cx,v = Rg T (x).
Thus T is an intertwining operator from H to the right regular representation on L2 (G). Moreover, the fact that G is unimodular implies that W is
stable by π(G). Thus, the irreducibility of π implies that W = H. Hence
T is a closed densely defined interwing operator between the two unitary
representations (π, H) and (R, L2 (G)) of the unimodular group G. Using
again the irreducibility of π, and the above version of the Schur lemma, we
deduce that T is a scalar multiple of an isometry. That is
(cx,v (L2 (G) = Cv (x(H
for all x ∈ H. Reversing the role of u and v, we deduce that there exists
d > 0 such that
(cx,y (L2 (G) = d(x(H (y(H
for all x, y ∈ H.
!
7. THE DISCRETE SERIES REPRESENTATION OF SL(2, R)
55
Recall from a previous section the Cartan decomposition KAK of SL(2, R),
where
7$
%
<
cos θ sin θ
K=
| θ ∈ [0, 2π] ,
− sin θ cos θ
7$
%
<
a 0
A=
|a>0 .
0 a−1
The corresponding subgroups of SU (1, 1) are
−1
K̃ = c
7 Kc$ −iθ
%
<
e
0
=
k=
| θ ∈ [0, 2π] ,
0
eiθ
and
−1
à = c
7 Ac $
%
<
ch t −sht
=
ãt =
| t∈R ,
−sht ch t
where a = et . Thus, the corresponding integral formula of (4.2) is
.
. . .
f (g)dg = 2
f (kãt k % )|sh(2t)|dkdtdk % .
SU (1,1)
K
R
K
Theorem 7.11. For m ≥ 2, Um is in the discrete series representation
of SU (1, 1).
Proof. In view of the above theorem, we shall prove that
.
|)Um (g)1, 1*|2 dg
(7.4)
SU (1,1)
is finite, where 1 denotes the constant function 1. The integrand is left and
right K̃-invariant since Um (K̃) acts on 1 by scalars of modulus one. It
follows that (7.4) is, up to a constant, equal to
-2
$
%
. -)Um ch t −sht 1, 1*- |sh(2t)| dt
−sht
ch
t
R
-2
. -.
−m
2 m−2
=
dw-- |sh(2t)| dt
- (−wsht + ch t) (1 − |w| )
R
D
-2
. - . 1
2
m−2
-2π
=
(1 − s )
sds-- (ch t)−2m |sh(2t)| dt
R
. 0∞
= 4(1(2ν
(ch t)−2m+1 sht dt
0
=
2(1(2
,
m−1
which is finite for m > 1.
!
Bibliography
Books.
A.W. Knapp, Representation theory of semisimple groups. An
overview based on examples. Princeton Mathematical Series, 36.
Princeton University Press, Princeton, NJ, 1986
R. Howe and E.-C. Tan, Nonabelian harmonic analysis. Applications of SL(2, R). Universitext. Springer-Verlag, New York, 1992
E.B. Davies, Spectral Theory and Differential Operators, Cambridge University Press (1995).
Papers.
S. Ben Said, On the integrability of a representation of sl(2, R). J.
Funct. Anal. 250 (2007), no. 2, 249–264
S. Ben Said and B. Orsted, The wave equation for Dunkl operators.
Indag. Math. (N.S.) 16 (2005), no. 3-4, 351–391
E. Nelson, Analytic vectors, Ann. of Math. 70 (1959) 572–615
57