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2. mathematics
3. geometry

TOPOLOGICAL GROUPS - PART 3/3
T.K.SUBRAHMONIAN MOOTHATHU
Contents
1.
Dual groups
1
2.
Sample results about the structure of LCSC abelian groups
7
3.
Some Major Theorems (without proof) and their consequences
10
4.
Abstract Fourier Transform
21
1. Dual groups
For the rest of our discussion about characters, we assume X is a LCSC group, for simplicity.
However, the theory is valid in most places without the assumption of second countability.
b of a (LCSC) topological group X is defined as X
b = {all characters of X}. It is
The dual group X
b ⊂ C(X, S 1 ) ⊂ C(X, C),
a group with respect to the operation of pointwise multiplication. Since X
b with the compact-open topology coming from C(X, S 1 ) (or equivalently, from
we may equip X
b itself is a topological group.
C(X, C)). Then, it may be checked that X
b converging pointwise to a map
Observation: Let X be a LCSC group and (fn ) be a sequence in X
f : X → S 1 . Then f (xy −1 ) = limn→∞ fn (xy −1 ) = limn→∞ fn (x)[limn→∞ fn (y)]−1 = f (x)f (y)−1 so
that f is a homomorphism. If in addition the convergence (fn ) → f is uniform on compact subsets
b
of X, then f is continuous and thus f ∈ X.
b is closed in C(X, S 1 ). Therefore, by [113], [114] and [115], we
This observation implies that X
have:
b is completely metrizable and second
[156] Let X be a LCSC group. Then the dual group X
countable.
b
Remark : If X is a compact metric group, then the supremum metric is an admissible metric on X.
b is a LCSC abelian group.
[157] Let X be a LCSC group. Then X
b is abelian since S 1 is abelian, and we already know X
b is second countable. To show X
b is
Proof. X
b with Γ compact.
locally compact, it suffices to produce a neighborhood Γ of the identity 1X ∈ X
b is closed in C(X, S 1 ), for a subset Γ ∈ X,
b Γ is compact iff Γ is pointwise
Note that since X
equicontinuous, by Arzela-Ascoli Theorem, [1170 ]. Let P ⊂ X be a neighborhood of e with P
compact, Q = {z ∈ S 1 : Re(z) > 3/4}, and let Γ = S(P , Q). Then Γ is a neighborhood of 1X .
1
2
T.K.SUBRAHMONIAN MOOTHATHU
First we show Γ is equicontinuous at e. Let ² > 0. We need to find a neighborhood U ⊂ X
of e such that |1 − f (x)| < ² for every x ∈ U and every f ∈ Γ. Choose r ∈ (3/4, 1) such that
V := {z ∈ S 1 : Re(z) > r} ⊂ B(1, ²). For each k ∈ N, let Qk = {z ∈ Q : z j ∈ Q for 1 ≤ j ≤ k}.
T
Then Q = Q1 ⊃ Q2 ⊃ · · · and ∞
k=1 Qk = {1}. So there is k ∈ N such that Qk ⊂ V . Now,
let U ⊂ X be a neighborhood of e with U k ⊂ P . Then for x ∈ U ⊂ U k ⊂ P and f ∈ Γ, we
have f (x) ∈ Q. If f (x) ∈ Q \ V , then f (x) ∈
/ Qk and therefore there is j ∈ {1, . . . , k} such that
f (x)j = f (xj ) ∈
/ Q, which is a contradiction since xj = xj ek−j ∈ U . Hence f (x) ∈ V ⊂ B(1, ²).
To prove equicontinuity of Γ at an arbitrary point a ∈ X, consider ² > 0. If U is as above, then
aU is a neighborhood of a, and for any b ∈ aU we have a−1 b ∈ U so that ² > |1 − f (a−1 b)| =
|1 − f (a)−1 f (b)| = |f (a)−1 ||f (a) − f (b)| = |f (a) − f (b)| for every f ∈ Γ.
¤
Exercise-41 : Let X be a LCSC group. Then, {Γ(K, ²) : K ⊂ X compact with e ∈ int[K], ² > 0}
b where Γ(K, ²) = {f ∈ X
b : |1 − f (x)| < ² ∀x ∈ K}.
is an open base at 1X in X,
c1 = Z.
b = S1, R
b = R, and S
[158] We have the following topological isomorphisms: Z
Proof. The group isomorphims are clear by [155]. We have to verify the topological part also. We
b be φ(y) = fy , where fy = e2πixy . As observed already,
do this just for the second one. Let φ : R → R
φ is a group isomorphism. To check φ is continuous, consider a convergent sequence (yn ) → y and
M > 0. We need to show that (fyn ) → fy uniformly on [−M, M ]. Given ² > 0 choose δ > 0 so that
|β| < δ implies |1 − e2πiβ | < ². Let n0 ∈ N be such that |y − yn | < δ/M for every n ≥ n0 . Then
we see that |fy (x) − fyn (x)| = |e2πixy ||1 − e2πix(yn −y) | < ² for every n ≥ n0 and every x ∈ [−M, M ].
The continuity of φ−1 follows from [157] and Open Mapping Theorem.
¤
cn = Tn , R
cn = Rn , and T
cn = Zn , where Tn is the n-torus
b × Yb . Hence, Z
\
[159] X
×Y = X
1
S
· · × S }1 .
| × ·{z
n−times
b = X. [Hint: First assume X is finite
Exercise-42 : If X is a finite discrete abelian group, then X
cyclic of order n generated by a ∈ X. If f ∈ X, then f (a) is an nth root of unity and f (a) uniquely
b∼
determines f . Deduce X
= {nth roots of unity} ∼
= X. In the general case use the fact that X is a
finite product of cyclic groups, and [159].]
[160] Let X be a LCSC group. Then:
b is discrete and countable.
(i) If X is compact, then X
b is compact.
(ii) If X is discrete, then X
b is finite.
(iii) If X is finite, then X
b is topologized by the supremum metric d. If f ∈ X
b \{1}, then there is x ∈ X
Proof. (i) Note that X
such that f (x) 6= 1 and therefore there is n ∈ N such that 1/4 < |1−f (x)n | = |1−f (xn )| ≤ d(1X , f ).
b is discrete. Being second countable, X
b is countable then.
Hence Bd (1X , 1/4) = {1X } and so X
(ii) If X is discrete, then compact subsets are finite subsets and therefore the compact-open topology
b Thus X
b may be thought of as a subspace
and the topology of pointwise convergence coincide on X.
TOPOLOGICAL GROUPS - PART 3/3
3
Q
of the compact product space x∈X Sx , where Sx = S 1 . Also, the Observation given just before
b is closed in the compact space Q
[156] says that X
x∈X Sx .
b
(iii) X is compact and discrete, so X is discrete and compact, hence finite.
¤
Q∞
Q∞
Exercise-43 : It is FALSE to say \
n=1 Xn even for compact metric abelian groups Xn .
n=1 Xn =
N
cn = Xn but X
b 6= X since X
b is discrete and countable.]
[Hint: If Xn = {0, 1}, X = {0, 1} , then X
bb
Remark : Later we will see that when X is a locally compact abelian group, then X
= X. So in the
case of compact metric abelian X, all the information about X is contained in the purely algebraic
b (∵ X
b is discrete).
properties of X
[161] Let X be a LCSC group. Then,
b has enough characters to separate members of X.
b
(i) X
(ii) If X is discrete, then X has enough characters to separate points of X.
bb
b → S 1 as αx (f= f (x). Then αx ∈ X.
b are distinct,
If f, g ∈ X
Proof. (i) For x ∈ X, define αx : X
then there is x ∈ X such that f (x) 6= g(x), i.e., αx (f ) 6= αx (g).
b Let a ∈ X \ {e} and let H = {an :
(ii) Note that any group homomorphism f : X → S 1 belongs X.
n ∈ Z}, the subgroup generated by a. If an 6= e for every n ∈ N, let b ∈ S 1 \ {1}, and if am = e for
some m ∈ N, choose the smallest such m (clearly m ≥ 2 since a 6= e), and let b ∈ S 1 \ {1} be an mth
root of unity. Define f : H → S 1 as f (an ) = bn for n ∈ Z. Then f is a homomorphism with f (a) 6= 1.
Let F = {(Y, g) : Y is a subgroup of X containing H, g : Y → S 1 is a homomorphism and g|H =
f }. Define a partial order on F by saying (Y1 , g1 ) ≤ (Y2 , g2 ) if Y1 ⊂ Y2 and g2 |Y1 = g1 . Now,
(H, f ) ∈ F and it is clear that every chain in F has an upper bound. So by Zorn’s Lemma, F has
a maximal element, say (Y, g). We have to show Y = X.
If possible, let x ∈ X \ Y and let Y 0 = {xn y : n ∈ Z, y ∈ Y } be the subgroup generated by Y and
x. Define a homomorphism h : Y 0 → S 1 as follows. If xm ∈
/ Y for every m ∈ N, let h(xn y) = g(y)
for every n ∈ Z. If xm ∈ Y for some m ∈ N, choose the smallest such m (≥ 2), let b ∈ S 1 be
such that bm = g(xm ) (here we use the fact that S 1 is divisible, i.e., every element has an mth root
for every m ∈ N), and let h(xn y) = bn g(y). Then (Y 0 , h) ∈ F, contradicting the maximality of
(Y, g).
¤
cn , R
cn , and T
cn have enough characters (verify this directly first for n = 1; then use the
Example: Z
b h ∈ Yb ).
\
fact that any f ∈ X
× Y is of the form f (x, y) = g(x)h(y) for some g ∈ X,
bb
as αx (f ) = f (x).
Definition and Observations: Let X be a LCSC abelian group and let α : X → X
We say X satisfies Pontryagin duality if α is a topological isomorphism. In general, we have the
following easy observations:
(i) α is always a continuous homomorphism.
(ii) α is one-one iff X has enough characters.
(iii) If α is a bijective continuous homomorphism, then α−1 is continuous by Open Mapping Theorem.
4
T.K.SUBRAHMONIAN MOOTHATHU
bb
b satisfies Pontryagin duality [∵ if Y = X,
(iv) If X satisfies Pontryagin duality, then X
then
b → Yb is one-one by [161]. If p ∈ Yb , define f (x) = p(αx ). Then f ∈ X
b and αf = p.]
α:X
[162] (i) If X, Y are LCSC abelian groups satisfying Pontryagin duality, then X × Y satisfies
Pontryagin duality.
(ii) If X = Zn , Tn , Rn , a finite discrete abelian group, or a finite product of these, then X satisfies
Pontryagin duality.
b × Yb .
\
×Y =X
Proof. (i) Make use of the topological isomorphism X
(ii) First assume that X = Z, S 1 , R, or a finite cyclic group. Then by the Observation above α is a
continuous one-one homomorphism. By [155] and [158], α is onto, and then α−1 is continuous by
Open Mapping Theorem. The general case follows by (i).
¤
Remark : If a countable abelian group is not finitely generated (eg: Q), then it cannot be written
in the form Zn × F for any finite abelian group F . Therefore, from [162] we cannot conclude that
all countable discrete abelian groups satisfy Pontryagin duality.
Q
Example: Let X = {0, 1}N = ∞
n=1 {0, 1} with coordinatewise addition modulo 2, and let A =
L∞
1
n=1 {0, 1}, which is a dense subgroup of X. For a ∈ A, define fa : X → S as fa (x) =
Qna
an xn where n ∈ N is such that a = 0 for every n ≥ n . Then, f ∈ X
b and a →
7 fa is a
a
n
a
a
n=1 (−1)
group homomorphism. If e(n) ∈ X is the sequence with 1 at the nth place and 0’s elsewhere, we
b then f (e(n)) = ±1
see that fa (e(n)) = (−1)an and it follows that a 7→ fa is one-one. Now if f ∈ X,
since e(n) has order two in X. Also (f (e(n))) → f ((0, 0, . . .)) = 1 since (e(n)) → (0, 0, . . .). Hence
f (e(n)) = 1 for all large n. Define a = (a1 , a2 , . . .) ∈ A by putting an = 0 if f (e(n)) = 1 and an = 1
if f (e(n)) = −1. Then f (e(n)) = (−1)an = fa (e(n)) and hence fa = f since e(n)’s generate A
and A is dense in X. Thus a 7→ fa is onto also, and therefore is a topological isomorphism from
b (we know by [160] that X
b is discrete).
(A, discrete topology) to X
b must be compact. We see that x 7→ gx , where
Now consider (A, discrete topology). Then A
Qna
b Note that since A is discrete, the
gx (a) = n=1 (−1)an xn , is a group isomorphism from X to A.
b coincides with the topology of pointwise convergence. If (x(k)) is a
compact-open topology on A
sequence in X converging to 0∞ = (0, 0, . . .) and a ∈ A, then there is k0 ∈ N such that [x(k)]i = 0 for
1 ≤ i ≤ na for every k ≥ k0 and then gx(k) (a) = 1 = g0∞ (a) for every k ≥ k0 . Therefore x 7→ gx is
continuous at 0∞ and hence continuous on X. Its inverse is continuous by Open Mapping Theorem.
b Making use of the two topological isomorphisms discussed
Thus X is topologically isomorphic to A.
Q∞
L
above, we may see that n=1 {0, 1} and ( ∞
n=1 {0, 1}, discrete topology) satisfy Pontryagin duality.
We generalize:
L
and A = ∞
n=1 Xn . Equip X with
b = A and A
b = X. Moreover, X, A satisfy
product topology and A with discrete topology. Then X
[163] Let Xn ’s be finite discrete abelian groups, X =
Pontryagin duality.
Q∞
n=1 Xn
TOPOLOGICAL GROUPS - PART 3/3
5
Qna
1
cn = Xn . For x = Q∞ X
c
Proof. Recall that X
n=1 n = X, define gx : A → S as gx (a) =
n=1 xn (an ),
b
where an = 0 = identity of Xn for n ≥ na . Then we see that gx ∈ A and x → gx is a oneb define xn : Xn → S 1 as xn (z) = g(0, . . . , 0, z, 0, 0, . . .).
one group homomorphism. If g ∈ A,
Q
c
Then x = (xn ) ∈ ∞
n=1 Xn = X and gx = g. Thus x 7→ gx is a group isomorphism. Continuity is
established as in the Example above, and the inverse map is continuous by Open Mapping Theorem.
L∞ c
Q
If a = (an ) ∈
Xn = A, define fa : X → S 1 as fa (x) = na an (xn ), where an = 0 =
n=1
n=1
cn for n ≥ na . Then a 7→ fa is a one-one group homomorphism. Now, let f ∈ X.
b
identity of X
If there is xn ∈ Xn such that f ((0, . . . , 0, xn , 0, 0, . . .)) 6= 1, then there is yn ∈ Xn (take yn = xkn
for some k) such that |1 − f ((0, . . . , 0, yn , 0, 0, . . .))| ≥ 1/2. Since f is continuous at (0, 0, . . .), this
implies that f restricted to {0} × · · · × {0} × Xn × {0} × {0} × · · · must be constant function
1 for all large n. Hence if we define an : Xn → S 1 as an (xn ) = f ((0, . . . , 0, xn , 0, 0, . . .)), then
L∞ c
a = (an ) ∈
n=1 Xn = A. Also fa = f . Thus a 7→ fa is a topological isomorphism from
b since X
b is discrete.
(A, discrete topology) to X,
Make use of the maps x 7→ gx and a 7→ fa to show Pontryagin duality.
¤
L∞ c
Q
Exercise-44 : (i) Let X = ∞
n=1 Xn , where Xn ’s are compact metric abelian
n=1 Xn and A =
b = A.
groups with enough characters. If A is given discrete topology, then X
Q∞ c
L∞
(ii) Let A = n=1 An and X = n=1 An where An ’s are countable discrete abelian groups. If A is
b = X. [Hint: Imitate the proof of [163]; use [161] for injectivity in
given discrete topology, then A
(ii).]
Projective limit (Inverse limit): We discuss a special case of this construction in the category of
topological groups. Suppose that Xn ’s are topological groups and fn : Xn+1 → Xn be continuous
Q
Q∞
open homomorphisms. We know that ∞
n=1 Xn is a topological group. Let X = {(xn ) ∈
n=1 Xn :
Q∞
fn (xn+1 ) = xn for every n}. Then X is a closed subgroup of n=1 Xn (so X is compact if Xn ’s
are compact). We call X the projective limit (inverse limit) of the system {Xn , fn } and we write
X = lim Xn or X = lim{Xn , fn }.
←
←
[164] For a metrizable topological group, the following are equivalent:
(i) X is compact and totally disconnected.
(ii) ∃ finite discrete groups Xn ’s and homomorpisms fn : Xn+1 → Xn such that X = lim{Xn , fn }.
←
Proof. (ii) ⇒ (i) is clear. Now assume (i). By [150], there is a base {Yn : n ∈ N} at e consisting
of compact open normal subgroups. We may assume Yn+1 ⊂ Yn . Since X is compact and Yn
is open, Xn := X/Yn is a finite discrete group. Define fn : Xn+1 → Xn as fn (xYn+1 ) = xYn ,
and let X 0 = lim{Xn , fn }. Define g : X → X 0 as g(x) = (xYn ), which is a homomorphism, and
←
T
ker(g) = {e} since ∞
n=1 Yn = {e}. If x ∈ Yk , then x ∈ Yn for every n ≤ k and therefore g(x) =
(Y1 , . . . , Yk , xYk+1 , xYk+2 , . . .), which shows g is continuous at e; and therefore g is continuous on
X. If z = (z1 Y1 , z2 Y2 , . . .) ∈ X 0 and k ∈ N, then z = (zk Y1 , zk Y2 , . . . , , zk Yk , zk+1 Yk+1 , zk+2 Yk+2 , . . .)
since zj Yj = fj+1 (zj+1 Yj+1 ) = zj+1 Yj , and therefore g(zk ) and z agree in the first k coordinates.
This says that g(X) is dense in X 0 . Since g(X) is compact and hence closed, we get that g(X) =
6
T.K.SUBRAHMONIAN MOOTHATHU
X 0 . And g −1 is continuous by Open Mapping Theorem. Thus g : X → X 0 is a topological
isomorphism.
¤
Remark : A projective limit of finite groups is called a profinite group. The above result says that
totally disconnected compact metric groups are profinite.
Exercise-45 : Let p be a prime and θ ∈ R. If limn→∞ e2πip
nθ
= 1, then θ = m/pk for some integers
k ≥ 0 and m. [Hint: If U = [0, 1/p2 ) ∪ (1 − 1/p2 , 1), then f rac{pn θ} ∈ U for all large n, where
f rac{x} is the fractional part of x. Suppose θ is irrational. If f rac{pn θ} ∈ (0, 1/p2 ), there is
q ∈ Z and k ∈ N such that 1/pk+2 < pn θ − q < 1/pk+1 and then 1/p2 < pn+k θ − pk q < 1/p
which implies f rac{pn+k θ} ∈
/ U . If f rac{pn θ} ∈ (1 − 1/p2 , 1), there is q ∈ Z and k ∈ N such
that 1 − 1/pk+1 < pn θ − q < 1 − 1/pk+2 and then pk − 1/p < pn+k θ − pk q < pk − 1/p2 , which
implies f rac{pn+k θ} ∈
/ U . Thus we must have θ ∈ Q. Now if θ = m/(pk j) with 1 < j < p, then
f rac{pk+n θ} ∈
/ U for every n ∈ N.] [A more elegant argument is to consider the base-p expansion
of θ and to show that the expansion should terminate.]
Exercise-46 : Let X be a LCSC group, Y ⊂ X be a dense subgroup and g : Y → S 1 be a continuous
b such that f |Y = g. [Hint: Let Uj be a neighborhood of e ∈ X
homomorphism. Then there is f ∈ X
such that |1 − g(y)| < 1/j for every y ∈ Y ∩ Uj . Then |g(a) − g(b)| < 1/j for every a, b ∈ Y with
a−1 b ∈ Uj . If x ∈ X, consider (an ) in Y converging to x. Then (g(an )) is Cauchy in S 1 . Define
f (x) = limn→∞ g(an ), and check that f is a well-defined homomorphism. Let Vj be a symmetric
neighborhood of e with Vj2 ⊂ Uj . If z ∈ xVj , there is a, b ∈ Y ∩ xVj with |f (x) − g(a)| < 1/j and
|f (z) − g(b)| < 1/j. Then a−1 b = (x−1 a)−1 (x−1 b) ∈ Y ∩ Vj2 ⊂ Y ∩ Uj so that |f (x) − f (z)| < 3/j
and hence f is continuous.]
Example [p-adic integers]: Fix a prime p, let fn : Z/(pn+1 Z) → Z/(pn Z) be fn (a+pn+1 Z) = a+pn Z,
∞
Y
and let Zp = limZ/(pn Z) ⊂
Z/(pn Z). The totally disconnected compact metric group Zp is
←
n=1
S
n
called the group of p-adic integers. Let G = ∞
n=1 Gn , where Gn = {p th roots of unity}. Then G
cp = G and
is a countable subgroup of S 1 . We show that if G is given the discrete topology, then Z
b = Zp (remark: G is called the Pr¨
G
ufer group and is sometimes denoted as Z(p∞ )).
Let a = (1 + pn Z) ∈ Zp . If x = (xn + pn Z) ∈ Zp and k ∈ N, then we see (as in the proof of
surjectivity of g in [164]) that x = (xk + pZ, xk + p2 Z, . . . , xk + pk Z, xk+1 + pk+1 Z, xk+1 + pk+2 Z, . . .)
so that xk a and x agree in the first k coordinates. Therefore Zp has a dense cyclic subgroup
cp is uniquely determined by its value at a. Since (pn a) → e ∈ Zp ,
H = {na : n ∈ Z}. So any g ∈ Z
n
cp , and hence g(a) ∈ G by Exercise-45. The map a 7→ g(a) from Z
cp
we have (g p (a)) → 1 for g ∈ Z
k
to G is a one-one homomorphism. To establish surjectivity of this map, consider b = e2πim/p ∈ G.
Define g : H → S 1 as g(na) = bn for n ∈ Z. If (nj a) → e ∈ Zp , then pk divides nj for all large
j and hence g(nj a) = 1 for all large j. So g : H → S 1 is a homomorphism continuous at e and
cp by
hence continuous on H. Since the subgroup H is dense in Zp , g extends to an element g ∈ Z
cp = G.
Exercise-46. And g(a) = b by construction. Thus Z
TOPOLOGICAL GROUPS - PART 3/3
7
b Then h(Gn ) ⊂ Gn . Since e2πi/pn generates Gn , h is determined by the values
Consider h ∈ G.
n
n
n
h(e2πi/p ), n ∈ N. Let xn ∈ {0, 1, . . . , pn − 1} be such that h(e2πi/p ) = e2πixn /p . Then h
corresponds to the element (xn + pn Z) ∈ Zp . This can be verified to be a topological isomorphism
b to Zp .
from G
b is torsion-free.
[165] (i) Let X be a compact connected metric abelian group. Then X
b is a torsion group.
(ii) Let X be a compact totally disconnected metric abelian group. Then X
b is totally disconnected.
(iii) Let A be a countable discrete abelian torsion group. Then A
b then f (X) is a compact connected subgroup of S 1 and hence f (X) = {1} or
Proof. (i) If f ∈ X,
S 1 . If f n = 1X , then f (X) ⊂ {nth roots of unity} and hence f = 1X .
T
(ii) Let Yn ⊂ X be compact open normal subgroups with Yn+1 ⊂ Yn and ∞
n=1 Yn = {e}. Let
1
b there is n such that the subgroup g(Yn ) of S 1 is contained
V = {z ∈ S : Re(Z) > 1/2}. If g ∈ X,
in V and hence g(Yn ) = {1}. If k is the index of Yn in X and if x ∈ X, there is 0 ≤ i < j ≤ k such
that xi and xj are in the same coset of Yn and then xj−i ∈ Yn . Hence 1 = g(xj−i ) = g(x)j−i . So
g k! = 1X .
(iii) Let A = {a1 , a2 , . . .} and let Fn be the subgroup generated by {a1 , . . . , an }. Then Fn is finite
since A is an abelian torsion group. If V = {z ∈ S 1 : Re(z) > 1/2}, then Γn := S(Fn , V ) is open
b Γn+1 ⊂ Γn and T∞ Γn = {1X }. If g ∈ Γn , then the subgroup g(Fn ) of S 1 is contained in V
in A,
n=1
b : g(Fn ) = {1}}, and hence Γn is also closed since
and hence g(Fn ) = {1}. Therefore, Γn = {g ∈ A
b is that of pointwise convergence. Thus we see that {Γn : n ∈ N} is a clopen base
the topology on A
b is compact.
at {1X }, since A
¤
Remark : Implications in [165] can be replaced with ‘⇔’ if Pontryagin duality is established. Note
that [165] illustrates the principle that properties of compact metric abelian groups translate into
purely algebraic properties of the dual.
2. Sample results about the structure of LCSC abelian groups
First we show that any closed subgroup of (Rn , +) is of the form (a vector subspace)+(a discrete
subgroup), and this will lead to Kronecker’s Theorem.
[166] [Dichotomy] Let Y ⊂ Rn be a closed subgroup.
(i) If Y is discrete, ∃ a linearly independent set {b1 , . . . , bk } ⊂ Rn such that Y ⊂ spanZ {b1 , . . . , bk }.
(ii) If Y is not discrete, there is y ∈ Y \ {0} such that Ry ⊂ Y .
Proof. (i) Choose a linearly independent set {a1 , . . . , ak } ⊂ Y such that Y ⊂ spanR {a1 , . . . , ak }
P
and let K = { ki=1 λi ai : |λi | ≤ 1}. Then K is a compact set containing {a1 , . . . , ak } and Y ∩ K
Pk
is finite since Y is discrete and closed. Fix x =
i=1 ti ai ∈ Y , and for m ∈ N, let z(m) =
Pk
Pk
mx − i=1 [mti ]ai = i=1 (mti − [mti ])ai , where [·] denotes the integer part. Then, z(m) ∈ Y ∩ K
P
and x = z(1) + ki=1 [ti ]a ∈ spanZ (Y ∩ K). Thus Y = spanZ (Y ∩ K). Since Y ∩ K is finite,
there are r < m such that z(r) = z(m) (for a fixed x ∈ Y ) and then mti − [mti ] = rti − [rti ] or
8
T.K.SUBRAHMONIAN MOOTHATHU
[mti] − [rti ]
∈ Q for every i and this means Y ⊂ spanQ {a1 , . . . , ak }. If d ∈ N is the l.c.m.
m−r
of the denominators of the rational coefficients for the finitely many elements in Y ∩ K, and if
ti =
bi = d−1 ai , then Y ∩ K ⊂ spanZ {b1 , . . . , bk }. Hence Y ⊂ spanZ (Y ∩ K) ⊂ spanZ {b1 , . . . , bk }.
(ii) Let D be the closed unit ball in Rn . Since Y is not discrete, there is a sequence (ym ) in
Y ∩ (D \ {0}) converging to 0. For each m let km ∈ N be such that km ym ∈ D and (km + 1)ym ∈
/ D.
Since D is compact and Y is closed, we may suppose that (km ym ) → y ∈ D ∩ Y , after passing onto
a subsequence. Then ((km + 1)ym ) → y also since (ym ) → 0, and therefore y ∈ ∂D. In particular
y 6= 0. If t ∈ R, then [tkm ]ym ∈ Y and |ty − [tkm ]ym | ≤ |t||y − km ym | + |tkm − [tkm ]||ym | → 0 since
(km ym ) → y, (ym ) → 0 and |tkm − [tkm ]| ≤ 1. Hence ty ∈ Y = Y .
¤
Remark : If Y ⊂ Rn is a discrete subgroup, it can be shown (with some more effort) that there is a
∼ Zm for
linearly independent set {c1 , . . . , cm } ⊂ Y such that Y = spanZ {c1 , . . . , cm } and thus Y =
some m ≤ n [see the book by Sidney A. Morris - Pontryagin duality and the structure of locally
compact abelian groups].
[167] Let Y ⊂ Rn be a closed subgroup and let W = {x ∈ Rn : Rx ⊂ Y }. Then W is a vector
subspace of Rn , W ⊂ Y , W ⊥ ∩ Y is discrete and Y = W + (W ⊥ ∩ Y ).
Proof. Easy to see that W is a vector subspace and W ⊂ Y . If y ∈ Y , then y = y1 + y2 ∈ W + W ⊥
and hence y2 = y − y1 ∈ Y since W ⊂ Y . So Y = W + (W ⊥ ∩ Y ). Now W ⊥ ∩ Y does not contain
Rx for any x ∈ Y \ {0} since W ∩ W ⊥ = {0}, and therefore W ⊥ ∩ Y is discrete by [166](ii).
¤
Remark : In the above, W ⊥ can be replaced with any vector subspace W 0 ⊂ Rn such that Rn =
L 0
W
W . With a little more work, we can show that any closed subgroup Y ⊂ Rn is ∼
= to Ri × Zj
for some integers i, j ≥ 0 with i + j ≤ n.
cn , then y acts as the character x 7→ e2πihx,yi of Rn . Hence Y ⊂ Rn
Note: Recall that if y ∈ Rn = R
separates points of Rn iff for every x ∈ Rn \ {0}, there is y ∈ Y such that hx, yi ∈
/ Z.
[168] Let Y ⊂ Rn be a closed subgroup separating points of Rn . Then Y = Rn .
Proof. First assume that Y is discrete. Then by [166], there is a linearly independent set {b1 , . . . , bk } ⊂
Rn such that Y ⊂ spanZ {b1 , . . . , bk }. By extending the linearly independent set we may assume
k = n. Let L : Rn → Rn be the linear map such that L(ej ) = bj and let [αij ] be the matrix of L.
e : Rn → Rn be the linear map whose matrix is the transpose
Then αij = hei , L(ej )i = hei , bj i. Let L
Pn
e
e
of [αij ]. Now there is x = i=1 λi ei such that L(x)
∈ Zn (true for any linear map). Then [L(x)]
j =
Pn
Pn
Pn
i=1 αij λi =
i=1 hei , bj iλi ∈ Z for every j. Hence for z =
j=1 mj bj ∈ spanZ {b1 , . . . , bn }, we
Pn Pn
Pn e
have hx, zi =
λi mj hei , bj i =
[L(x)]j mj ∈ Z, and therefore hx, zi ∈ Z for every
i=1
j=1
j=1
z ∈ Y , a contradiction.
In the general case, write Y = W + (W ⊥ ∩ Y ) using [167], observe that W ⊥ ∩ Y separates points
of W ⊥ (∵ if x ∈ W ⊥ \ {0}, there is y = y1 + y2 ∈ W + (W ⊥ ∩ Y ) such that hx, yi ∈
/ Z; but
TOPOLOGICAL GROUPS - PART 3/3
9
hx, yi = hx, y2 i since hx, y1 i = 0), and conclude using the first part of the proof that W ⊥ = {0} (∵
W ⊥ ∩ Y is discrete). Thus Y = W = Rn .
¤
Remark : Note the following points: (i) If θ = (θ1 , . . . , θn ) ∈ Rn , then {1, θ1 , . . . , θn } is rationally
independent (i.e., linearly independent over the field Q) iff hz, θi ∈
/ Z for every z ∈ Z \ {0}. (ii)
If θ ∈ R, then {1, θ} is rationally independent iff θ is irrational. (iii) For every n ∈ N there is
(θ1 , . . . , θn ) ∈ Rn such that {1, θ1 , . . . , θn } is rationally independent since any Hamel basis of R
over Q must be uncountable.
[169] [Kronecker’s Theorem] Let θ = (θ1 , . . . , θn ) ∈ Rn . Then the following are equivalent:
(i) {1, θ1 , . . . , θn } is rationally independent.
(ii) Zn + Zθ = Rn , i.e, for every (x1 , . . . , xn ) ∈ Rn and ² > 0, there exist k ∈ Z and (m1 , . . . , mn ) ∈
Zn such that |xi − kθi − mi | < ² for 1 ≤ i ≤ n.
Proof. Let Y ⊂ Rn be the closed subgroup Y = Zn + Zθ. In view of [168] and the fact that Rn has
enough characters, it suffices to show that {1, θ1 , . . . , θn } is rationally independent iff Y separates
points of Rn .
⇒: Let x = (x1 , . . . , xn ) ∈ Rn be such that hx, yi for every y ∈ Y . Since ei ∈ Y , we get
xi = hx, ei i ∈ Z and thus x ∈ Zn . Since θ ∈ Y , we have hx, θi ∈ Z and this implies z = 0 since
{1, θ1 , . . . , θn } is rationally independent. Thus Y separates points of Rn .
⇐: If {1, θ1 , . . . , θn } is not rationally independent, there is z ∈ Zn \ {0} such that hz, thetai ∈ Z.
Then for every k ∈ Z and every a ∈ Zn , hz, a + kθi = hz, ai + khz, θi ∈ Z. That is, hz, yi ∈ Z for
every y ∈ Zn + Zθ. Since y 7→ hz, yi is continuous and Z is closed, we have hz, yi ∈ Z for every
y ∈ Y and this means Y cannot separate z and 0.
¤
Remark : For n = 1, Kronecker’s Theorem is the statement that for θ ∈ R, Z + θZ = R iff θ is
irrational, and this we have seen in the proof of [143](iv).
cn be a closed subgroup separating points of Zn . Then Y = Tn .
Exercise-47 : Let Y ⊂ Tn = Z
[Hint: Let g : Rn → Tn be g(x1 , . . . , xn ) = (e2πix1 , . . . , e2πixn ). If y = g(x) ∈ Tn , then y acts as the
character z 7→ e2πihx,zi on Zn , and hence the image is different from 1 iff hx, zi ∈
/ Z. Observe that
g −1 (Y ) is a closed subgroup of Rn with Zn ⊂ g −1 (Y ); in particular ei ∈ g −1 (Y ). If z ∈ Rn and if
hx, zi ∈ Z for every x ∈ g −1 (Y ), then (by taking x = ei ) z ∈ Zn . Therefore z = 0 since Y separates
points of Zn . Thus g −1 (Y ) separates points of Rn . Now use [168].]
b be a closed
Exercise-48 :Let A be a torsion-free, countable discrete abelian group and let Γ ⊂ A
b [Hint: Let F ⊂ A be a finite set and H ⊂ A be
subgroup separating points of A. Then Γ = A.
b→H
b be φ(f ) = f |H . Then φ(Γ) is a compact (hence
the subgroup generated by F . Let φ : A
b separating points of H. But H ∼
closed) subgroup of H
= Zn for some n ∈ N since A is torsion-free,
b by Exercise-47. In particular, for every g ∈ A,
b there is f ∈ Γ such that
and therefore φ(Γ) = H
b is that of pointwise convergence to deduce
f |F = g|F . Now use the fact that the topology on A
b
that Γ is dense in A.]
10
T.K.SUBRAHMONIAN MOOTHATHU
Exercise-49 : Let X, Y be topological spaces, g : X → Y be an open continuous surjection and let
H ⊂ Y . Then H is dense in Y iff g −1 (H) is dense in X.
[1690 ] [Kronecker’s Theorem for the n-torus] Let θ = (θ1 , . . . , θn ) ∈ Rn and let Hθ ⊂ Tn be the
cyclic subgroup Hθ = {(e2πikθ1 , . . . , e2πikθn ) : k ∈ Z}. Then, Hθ = Tn iff {1, θ1 , . . . , θn } is rationally
independent.
Proof. Let g : Rn → Tn be g(x1 , . . . , xn ) = (e2πix1 , . . . , e2πixn ). Then g is an open continuous
surjective homomorphism, and g −1 (Hθ ) = Zn + Zθ. Now use Exercise-49 and [169].
¤
A topological group is said to be monothetic if it has a dense cyclic subgroup. Examples are finite
cyclic groups, Z, S 1 , Tn (by [1690 ]), and Zp . On the other hand, Z2 , Rn , {0, }N are not monothetic.
[170] Let X be a LCSC abelian group. If X is monothetic, then either X ∼
= Z or X is compact.
Proof. If X is discrete, then X ∼
= Z or X is a finite cyclic group, and we are done. So assume X is
not discrete and let H = {an : n ∈ Z} be a dense cyclic subgroup, which is obviously infinite. Let U
S
n
be a symmetric neighborhood of e with compact closure. Claim-1 : X = ∞
n=1 a U . If x ∈ X there
is m ∈ Z such that am ∈ xU . Let W be a symmetric neighborhood of e with am W ⊂ xU . Since
W is symmetric and X is not discrete, there is k < m such that ak ∈ W . Then n = m − k ∈ N and
an ∈ xU or x ∈ an U −1 = an U , proving Claim-1.
let n ∈ N be the smallest with x ∈
such that
x−1 an
∈
ak U .
Then x ∈
an U .
Then
an−k U −1
=
Sp
n
n=1 a U .
x−1 an ∈ U −1 =
Since U is compact, there is p ∈ N such that U ⊂
an−k U .
Claim-2 : X =
Sp
n=1 U .
If x ∈ X,
U ⊂ U , so there is k ∈ {1, . . . , p}
By the choice of n, we get n − k ≤ 0 or
n ≤ k ≤ p.
¤
Remark : [167], [168] and [170] are the initial steps in the investigation to understand the structure
of LCSC abelian groups. The final theorem is stated as [177] later.
3. Some Major Theorems (without proof) and their consequences
Some of the results are stated not in full generality; the hypothesis of second countability may
be redundant.
[171] [Major Theorem] Every LCSC abelian group has enough characters to separate points.
Remark : This Theorem is a consequence of Gelfand-Raikov Theorem from the theory of unitary
representations of locally compact groups (p.343 of Hewitt and Ross). For compact (metric) abelian
groups, the conclusion of [171] follows from Peter-Weyl Theorem about group representations.
[1710 ] Let X be a LCSC abelian group, Y ⊂ X be a closed subgroup and let x ∈ X \ Y . Then
b such that f |Y ≡ 1 and f (x) 6= 1.
there is f ∈ X
[ such that δ(xY ) 6= 1. Let f = δ ◦ q, where q : X →
Proof. Applying [171] to X/Y , find δ ∈ X/Y
X/Y is the quotient map.
¤
TOPOLOGICAL GROUPS - PART 3/3
11
b is a closed subgroup
Exercise-50 : Let H be a finitely generated discrete abelian group. If Γ ∈ H
b [Hint: If Γ 6= H,
b then by [1710 ], there is a non-trivial character
separating points of H, then Γ = H.
b
∼ Zn × G for some finite abelian group G so that H
b such that β|Γ ≡ 1. By hypothesis H =
β∈H
satisfies Pontryagin duality. So there is x ∈ H \{e} with αx = β. But then f (x) = αx (f ) = β(f ) = 1
for every f ∈ Γ, a contradiction.]
b be a closed subgroup separating
[172] (i) Let A be a countable discrete abelian group and let Γ ⊂ A
b
points of A. Then Γ = A.
(ii) Every compact metric abelian group satisfies Pontryagin duality.
(iii) Every countable discrete abelian group satisfies Pontryagin duality.
b in the argument.
Proof. (i) Imitate the solution of Exercise-47. Use Exercises-50 to say φ(Γ) = H
bb
(ii) Let X be a compact metric abelian group. We know α : X → X is a continuous homomorphism.
bb
Since X has enough characters by [171], α is one-one. Now, α(X) is a compact subgroup of X
bb
b Applying (i) with A = X,
b we get α(X) = X.
separating points of X.
Then by Open Mapping
Theorem α−1 is continuous.
bb
(iii) Let A be a countable discrete abelian group and let B = A.
Enough to show that α : A → B
0
b such that δ|α(A) ≡ 1. Since A
b
is onto. If α(A) 6= B, by [171 ] there is a non-trivial character δ ∈ B
b \ {1A } such that αf = δ. But f (a) = αa (f ) = αf (αa ) =
satisfies Pontryagin duality, there is f ∈ A
δ(αa ) = 1 for every a ∈ A so that f ≡ 1, contradiction.
¤
[173] Let X be a compact metric abelian group. Then:
b is a torsion group.
(i) X is totally disconnected ⇔ X
b is torsion-free.
(ii) X is connected ⇔ X is divisible ⇔ X
Proof. (i) Combine [165] and [172].
b torsion-free’. Now suppose that X
b is torsion free.
(ii) We know from [165] that ‘X connected ⇒ X
b \ {1X }, then f n 6= 1X and so there is
Fix n ∈ N and let φ : X → X be φ(x) = xn . If f ∈ X
x ∈ X such that 1 6= f n (x) = f (xn ) = f (φ(x)) = αφ(x) (f ). Thus the closed subgroup α(φ(X))
bb
bb
b Hence by [172,], α(φ(X)) = X,
of X
separates points of X.
and hence φ(X) = X since α is a
topological isomorphism. Next assume X is divisible. If Y ⊂ X is the connected component of e,
[ is torsion by (i).
then Y is a closed subgroup and X/Y is totally disconnected by [148]. Then X/Y
[ and let n ∈ N be such that δ n ≡ 1. If x ∈ X there is z ∈ X with z n = x and therefore
Let δ ∈ X/Y
δ(xY ) = δ(z n Y ) = δ n (zY ) = 1. That is, δ ≡ 1. This implies by [1720 ] that Y = X.
¤
cp = Z(p∞ ) satisfy Pontryagin duality by [172]. If A = Q
Example: If p is prime, then Zp and Z
b is a compact connected
with discrete topology, then A satisfies Pontryagin duality by [172]; also A
divisible metric abelian group by [173].
Exercise-51 : If Y ⊂ Tn is a closed subgroup, then Y ∼
= Tk × F for some k ≤ n and some finite
cn = Zn : f |Y ≡ 1}. Then Γ is a closed subgroup
discrete abelian group F . [Hint: Let Γ = {f ∈ T
12
T.K.SUBRAHMONIAN MOOTHATHU
of Zn and β : Zn /Γ → Yb defined as β(f Γ)(y) = f (y) is a one-one (continuous) homomorphism.
Since β(Zn /Γ) separates points of Y , β(Zn /Γ) = Yb by [172]. Thus Yb ∼
= Zn /Γ. But Zn /Γ, being a
k×F =
∼ Zk × F . Hence Y =
∼ Yb
∼ Z\
∼ Tk × F .]
b =
finitely generated discrete abelian group, is =
[174] Let X be a compact metric abelian group. Then:
(i) X is topologically isomorphic to a closed subgroup of
Q∞
n=1 S
1.
(ii) If e ∈ X has a neighborhood U such that {e} is the only subgroup of X contained in U , then
X∼
= Tk × F for some integer k ≥ 0 and some finite discrete abelian group F .
(iii) If X is connected, and if e ∈ X has a neighborhood U such that {e} is the only subgroup of
X contained in U , then X ∼
= Tk for some integer k ≥ 0.
b = {f1 , f2 , . . .} and define φ : X → Q∞ S 1 as φ(x) = (fn (x)) (we may assume X
b
Proof. Let X
n=1
is infinite). Then φ is a continuous homomorphism and φ(X) is compact (hence closed). Since
b separates points of X, φ is one-one. Also φ : X → φ(X) is open by Open Mapping
by [171] X
∼ φ(X).
Theorem. Thus X =
(ii) Since φ(U ) is a relatively open subset of φ(X) with (1, 1, . . .) ∈ φ(U ), there are open sets
V1 , . . . , Vn ⊂ S 1 containing 1 such that for V = V1 ×· · ·×Vn ×S 1 ×S 1 ×· · · we have V ∩φ(X) ⊂ φ(U ).
T
Let H = ni=1 ker(fi ). If x ∈ H, then φ(x) ∈ V ∩ φ(X) ⊂ U , and hence H = {e} by the property
of U . Use this to check that ψ : X → Tn defined as ψ(x) = (f1 (x), . . . , fn (x)) is a topological
isomorphism from X to the closed subgroup ψ(X) of Tn . Now use Exercise-51.
(iii) follows from (ii) since F must be {0}.
¤
[175] Let X be a LCSC abelian group. Then X is compactly generated iff there is a neighborhood
b of 1X such that {1X } is the only subgroup of X
b contained in Γ.
Γ⊂X
Proof. Suppose X is generated by compact K ⊂ X. We may assume e ∈ int[K] and K is symmetric.
S
n
1
Then X = ∞
n=1 K . Let V = {z ∈ S : Re(z) > 1/2} and put Γ = S(K, V ). If Π ⊂ Γ is a subgroup
b and f ∈ Π, then f n ∈ Π ⊂ Γ and therefore f n (K) ⊂ V for every n ∈ N. This implies f |K ≡ 1
of X
S
n
for every f ∈ Π. Since X = ∞
n=1 K , we have Π = {1X }.
Conversely assume Γ exists as stated. Let V ⊂ X be a symmetric open neighborhood of e with
b : |1−f (x)| < ² for every x ∈ V } ⊂ Γ.
compact closure and ² > 0 be such that B(1X , V , ²) = {f ∈ X
S∞
b : f |Y ≡ 1}, then
Then Y = n=1 V n is an open, hence closed, subgroup of X. If Π = {f ∈ X
Π is a subgroup and f |V ≡ 1 for every f ∈ Π since V ⊂ V 2 . Hence Π ⊂ B(1X , V , ²) ⊂ Γ and so
Π = {1X }. Therefore by [1710 ], Y = X and thus X is generated by compact V .
¤
See p.376 and p.90 of Hewitt and Ross for more information about the results [176] and [177]
stated below.
[176] [Major Theorem - Pontryagin duality] Let X be a LCSC abelian group. Then X satisfies
bb
αx (f ) = f (x), is a topological isomorphism.
Pontryagin duality, i.e., α : X → X,
b be a closed subgroup separating points
Exercise-52 : Let X be a LCSC abelian group and let Γ ⊂ X
b [Hint: [172] and [176].]
of X. Then Γ = X.
TOPOLOGICAL GROUPS - PART 3/3
13
[177] [Major Theorem - Structure of LCSC abelian groups] Let X be a LCSC abelian gruop. Then:
(i) X has a clopen subgroup ∼
= Ri × K for some integer i ≥ 0 and some compact metric abelian
group K.
(ii) If X is compactly generated, then X ∼
= Ri × Zj × K for some integers i, j ≥ 0 and some compact
metric abelian group K.
(iii) If X is compactly generated, then every neighborhood U of e contains a compact subgroup H
∼ Ri × Zj × Tk × F for some integers i, j, k ≥ 0 and some finite discrete abelian
such that X/H =
group F .
Note: (i) Q with discrete topology is a LCSC abelian group, but it is not compactly(finitely)
generated. (ii) X/H ∼
=Y ;X∼
= H × Y ; for example R/Z ∼
= S 1 but R Z × S 1 .
Remark : There are some implications among the three statements of [177]. Note that (i) follows
S
S∞
n
n
from (ii) since X has a compactly generated clopen subgroup Y = ∞
n=1 V =
n=1 V , where V
is any symmetric neighborhood of e with compact closure. It is non-trivial to show (iii) ⇒ (ii) but
∼ Ri × Zj × Tk × F and if q : X → X/H is the quotient map, let
the idea is as follows: if X/H =
K = q −1 (Tk × F ) and one tries to show X ∼
= Ri × Zj × K. Also observe that (ii) generalizes the
result that any finitely generated abelian group is isomorphic to Zj × F for some integer j ≥ 0 and
some finite abelian group F .
Remark : Pontryagin duality for compactly generated LCSC abelian groups follows from [177](ii)
and [172](ii). Now, [176] can be proved along the following lines: X has a clopen subgroup Y that
is compactly generated, which satisfies Pontryagin duality by the first sentence of the Remark. And
X/Y , being discrete and countable, satisfies Pontryagin duality by [172](iii) (see p.84, Pontryagin
duality and the structure of locally compact abelian groups, Sidney A. Morris).
[178] [Corolary] Let X be a LCSC abelian group. Then:
(i) If X is connected, then X ∼
= Ri × K for some integer i ≥ 0 and some compact connected metric
abelian group K.
b are connected, then X ∼
(ii) If X and X
= Ri for some i ≥ 0.
Proof. Since X is connected, X is compactly generated by [147]. Now apply [177](ii) to get X ∼
=
i
b is connected then K = {0} since K
b is discrete.
R × K. If X
¤
Recall that a topological group X is monotheic if it has a dense cyclic subgroup; or equivalently,
if there is a homomorphism f : Z → X such that f (Z) = X. A topological group X is called
solenoidal if there is a continuous homomorphism f : R → X such that f (R) = X (this condition
is sometimes phrased as: ‘X has a dense one-parameter subgroup’). Note that solenoidal groups
must be connected. Compare with [170] the following.
Exercise-53 : Let X be a LCSC abelian group. If X is solenoidal, then either X = R or X is
compact. [Hint: X = Ri × K for some compact K by [178]. If i > 0, write X = R × Ri−1 × K
and let π : X → R be the projection to the first coordinate. If f : R → X is a continuous
14
T.K.SUBRAHMONIAN MOOTHATHU
homomorphism with f (R) = X, then π ◦ f : R → R is a continuous homomorphism and hence a
linear map. Since π(f (R)) is dense in R, π ◦ f is onto and hence one-one. Thus π is one-one and
therefore Ri−1 × K = {0}.]
b are closed subgroups, then their annihilators
Definition: If X is a LCSC group, and Y ⊂ X, Γ ⊂ X
b Y ) = {f ∈ X
b : f |Y ≡ 1} and A(X, Γ) = {x ∈ X : f (x) =
are defined respectively as A(X,
b Y)⊂X
b and A(X, Γ) ⊂ X are closed subgroups.
1 for every f ∈ Γ}. Then A(X,
bb
bb
∼ Y (∵ since X satisfies Pontryagin duality, A(X,
b Y )) =
b Y )) = {αx :
Observation: A(X,
A(X,
A(X,
b Y )} ∼
b Y )} = Y by [1710 ]).
αx ≡ 1 on A(X,
= {x ∈ X : f (x) = 1 for every f ∈ A(X,
b and if Y ⊂ X is a closed subgroup, then clearly f |Y ∈ Yb . But it can happen
If f ∈ X
b and this suggests that we should consider a quotient space
that f |Y = g|Y for distinct f, g ∈ X,
b Similar to what we
construction by identifying such f, g if we wish to express Yb in relation to X.
have in Functional Analysis, we show that dual of a subgroup is quotient of the dual, and dual of
a quotient is subgroup of the dual.
b
b Y)
[179] Let X be a LCSC abelian group and Y ⊂ X be a closed subgroup. Then, Yb ∼
X,
= X/A(
[ ∼
b Y ).
and X/Y
= A(X,
b Y ). Define φ : X/Y → Γ as φ(δ) = δ ◦ q, where q : X → X/Y is the
Proof. Write Γ = A(X,
quotient map. Then φ is a one-one continuous homomorphism. If f ∈ Γ, define δ : X/Y → S 1 as
δ(xY ) = f (x); then δ ∈ Γ and φ(δ) = f , so φ is onto. By Open Mapping Theorem, we conclude
[ ∼
X/Y
= Γ.
bb
[
∼
b
By the first part of the proof and the Observation above, we have X/Γ
Γ) ∼
= A(X,
= Y . Then
[
[
b ∼
b
by duality, Yb ∼
¤
= X/Γ
= X/Γ.
An extension Theorem similar to Hahn-Banach, Tietze, etc. is the following.
[180] Let X be a LCSC abelian group and let Y ⊂ X be a closed subgroup. If g ∈ Yb , there is
b such that f |Y = g.
f ∈X
bb
[
∼
b Y ) and Λ = A(X,
b
Proof. Let Γ = A(X,
Γ). The idea of the proof is simple. We have X/Γ
=Λ=
[
[
∼
b = X/Γ
b
b
b
{αy : y ∈ Y } ∼
So there is f ∈ X
= Y . So g may be considered as an element of Λ
= X/Γ.
such that g = f Γ.
[
b →Λ
The details are as follows. We will make use of Pontryagin duality and [179]. Let ψ : X/Γ
b → X/Γ
b
be the topological isomorphism given by ψ(γ) = γ ◦ q where q : X
is the quotient map.
[
[
[
b
b
Define δ : X/Γ
→ S 1 as δ(γ) = g(y) if ψ(γ) = αy . Then δ ∈ X/Γ
(and δ corresponds to g). Let
[
[
b be such that δ = αf Γ , i.e., δ(γ) = γ(f Γ) for every γ ∈ X/Γ.
b
b
f ∈X
If y ∈ Y and γ ∈ X/Γ
is such
that ψ(γ) = αy , then g(y) = δ(γ) = γ(f Γ) = γ(q(f )) = ψ(γ)(f ) = αy (f ) = f (y).
¤
[1800 ] Let X be a LCSC abelian group, Y ⊂ X be a closed subgroup, g ∈ Yb and let x ∈ X \ Y .
b such that f |Y = g and f (x) 6= 1.
Then there is f ∈ X
TOPOLOGICAL GROUPS - PART 3/3
15
b be such that h|Y = g, by [180]. If h(x) 6= 1, take f = h. If h(x) = 1, choose
Proof. Let h ∈ X
b by [1710 ] such that h1 |Y ≡ 1 and h1 (x) 6= 1. Then put f = hh1 .
h1 ∈ X
¤
We may ask whether we could have used R in the place of S 1 in the definition of a character. In
view of the important result [171], we may ask the following. If X is a LCSC abelian group and if
x ∈ X \ {e}, does there exist a continuous homomorphism f : X → R such that f (x) 6= 0? The
result below says why R is not a suitable substitute for S 1 .
[181] Let X be a LCSC abelian group. Then the following are equivalent:
(i) For every x ∈ X \ {e} there is a continuous homomorphism f : X → R such that f (x) 6= 0.
(ii) {e} is the only compact subgroup of X.
Proof. (i) ⇒ (ii) is clear since {0} is the only compact subgroup of R. We now prove (ii) ⇒ (i).
Given x ∈ X \ {e}, let V ⊂ X be a symmetric neighborhood of e with x ∈ V and V compact. Then
S
S∞
n
n
i
j
Y = ∞
n=1 V =
n=1 V is a compactly generated clopen subgroup of X. By [177](ii), Y = R ×Z
for some integers i, j ≥ 0 (∵ Y does not have non-trivial compact subgroups). Clearly there is a
continuous homomorphism (consider the restriction of a linear map) f : Y → R such that f (x) 6= 0.
Then there is a homomorphism fe : X → R such that fe|Y = Y (∵ if X is an abelian group, Y ⊂ X
is a subgroup, D is a divisible abelian group and if f : Y → D is a homomorphism, then f extends
to a homomorphism fe : X → D; argue as in the proof of [161](ii)). Since fe is continuous on the
open set Y , fe is continuous on X.
¤
Definition: Let X be a LCSC abelian group. We say x ∈ X is a compact element if x is contained
in a compact subgroup of X (note that this is a generalization of an element with finite order).
Exercise-54 : Let X be a LCSC abelian group and let Y ⊂ X be the collection of all compact
elements of X. Then Y is a closed subgroup, but Y may not be compact. [Hint: If a, b ∈ Y
and if A, B ⊂ X are the corresponding compact subgroups, then AB is a compact subgroup and
ab−1 ∈ AB. If x ∈ Y , let H be a compactly generated clopen subgroup of X with x ∈ H. Now,
H = Ri × Zj × K, K compact, by [177](ii). The only compact elements of H are of the form (0, 0, z)
with z ∈ K; and {0} × {0} × K is closed. Hence x ∈ {0} × {0} × K ⊂ Y , and so Y is closed. To
L
see Y need not be compact, consider ∞
n=1 {0, 1} with discrete topology.]
[182] Let X be a LCSC abelian group, let Y ⊂ X be the closed subgroup of all compact elements
b be the connected component of 1X . Then A(X,
b Y ) = Γ and A(X, Γ) = Y .
of X and let Γ ⊂ X
Proof. Enough to show A(X, Γ) = Y , the other one will follow by duality.
b is connected. Then A(X, Γ) = A(X, X)
b = {e} by [171]. If H ⊂ X is a
Step-1 : Suppose X
b = X/A(
b
b H) is discrete and connected, and hence H = {e}. Thus
compact subgroup, then H
X,
Y = {e} = A(X, Γ).
b is totally disconnected. Then A(X, Γ) = A(X, {1X }) = X. If x ∈ X, let K ⊂ X
Step-2 : Suppose X
be a compact set with x, e ∈ int[K] and consider S(K, V ), where V = {z ∈ S 1 : Re(z) > 1/2}.
b by [150]. Since Π is open, X/Π
b
Choose a compact open subgroup Π ⊂ S(K, V ) of X
is discrete (and
16
T.K.SUBRAHMONIAN MOOTHATHU
[
b
countable) and therefore A(X, Π) = X/Π
is a compact subgroup of X. Now Π(x) = {f (x) : f ∈ Π}
is a subgroup of S 1 contained in V and hence Π(x) = {1}. Thus x ∈ A(X, Π) and therefore x ∈ Y .
Thus Y = X.
b
Step-3 : Consider the general case. Then X/Γ
is totally disconnected by [148], and hence every
[
b
element of A(X, Γ) = X/Γ is compact by Step-2. Thus A(X, Γ) ⊂ Y . Next, since Γ is connected,
b has no non-trivial compact subgroup by Step-1. If y ∈ Y \A(X, Γ) and if H ⊂ X is a
X/A(X, Γ) = Γ
compact subgroup containing y, then q(H) would be a non-trivial compact subgroup of X/A(X, Γ)
(where q is the quotient map), which is not possible. Therefore Y = A(X, Γ).
¤
Generalizing [173], we have the following corollary.
[183] Let X be a LCSC abelian group. Then:
b is totally disconnected.
(i) Every element of X is compact iff X
b is connected (iff for every x ∈ X \ {e}, there is a
(ii) {e} is the only compact subgroup of X iff X
continuous homomorphism f : X → R with f (x) 6= 0, by [181]).
[184] Let X be a compact metric abelian group. Then:
b∼
(i) X is monothetic iff X
= to a subgroup of S 1 with discrete topology.
b∼
(ii) X is solenoidal iff X
= to a subgroup of R with discrete topology.
b → S 1 , δ(f ) = f (a), is a one-one homomorphism.
Proof. (i) Suppose X = {an : n ∈ Z}. Then δ : X
bb
b → S 1 is a one-one homomorphism, then δ ∈ X
b is discrete, and therefore
Conversely if δ : X
since X
b and 1 = f (a) = αa (f ) = δ(f ), then f = 1X
δ = αa for some a ∈ X by Pontryagin duality. If f ∈ X
since δ is one-one. Hence {an : n ∈ Z} = X by [1710 ].
b = R defined as
b →R
(ii) If φ : R → X is a continuous homomorphism with φ(R) = X, then δ : X
b → R is a one-one homomorphism,
δ(f ) = f ◦ φ is a one-one homomorphism. The other way, if δ : X
bb
b → S 1 for each a ∈ R as βa (f ) = e2πiaδ(f ) . Then βa ∈ X
define βa : X
so that there is φ : R → X
such that βa = αφ(a) for every a ∈ R. It may be checked that φ is a continuous homomorphism. If
b and if f ≡ 1 on φ(R), then e2πiaδ(f ) = βa (f ) = αφ(a) (f ) = f (φ(a)) = 1 for every a ∈ R and
f ∈X
therefore δ(f ) = 0. Hence f = 1X since δ is one-one. This shows that φ(R) = X by [1710 ].
¤
Note: By [184] and duality, the dual of (Q, +, discrete topology) is solenoidal.
Remark : With more machinery, we can prove the following for a compact metric abelian group X:
X is solenoidal ⇔ X is connected ⇒ X is monothetic [p.407, p.409, Hewitt and Ross]; the example
of Zp shows that ‘X is monothetic ; X is connected’.
Haar measure: Recall that the Lebesgue measure on R is a regular Borel measure that is translationinvariant (i.e., µ[x + A] = µ[A]) and locally finite.
Definition: Let X be a LCSC group. A Borel measure µ on X is called a left Haar measure if
µ 6= 0, µ is locally finite and µ is left invariant in the sense that µ[xA] = µ[A] for every x ∈ X
TOPOLOGICAL GROUPS - PART 3/3
17
and Borel A ⊂ X (recall that locally finite Borel measures on LCSC spaces are regular by [122]).
Similarly we define right Haar measure.
[185] [Major Theorem] Let X be a LCSC group. Then there is a left Haar measure (similarly, right
Haar measure) µ on X. If β is another left Haar measure on X, then β = λµ for some λ > 0.
Remarks: (i) The idea of the proof is to produce a non-zero positive linear functional φ : Cc (X, R) →
R such that φ(f ◦ Ly ) = φ(f ) for every f ∈ Cc (X, R) and for every y ∈ X, and obtain a correR
sponding locally finite measure µ on X such that φ(f ) = X f dµ by Riesz Representation Theorem
[12900 ]. This µ will be left invariant (∵ if K ⊂ X is compact, let (gn ) be a decreasing sequence in
R
R
Cc (X, R) converging to χK pointwise. Then µ[yK] = X χyK (x)dµ(x) = limn→∞ X gn (yx)dµ(x) =
R
R
limn→∞ φ(gn ◦ Ly ) = limn→∞ φ(gn ) = limn→∞ X gn (x)dµ(x) = X χK (s)dµ(x) = µ[K], and then
use regularity). (ii) If µ is a left Haar measure on X, then β[A] = µ[A−1 ] is a right Haar measure.
(iii) If X is a compact metric abelian group and µ is a left Haar measure on X, then µ[X] < ∞ so
we may assume µ[X] = 1 after normalizing.
[186] Let X be a LCSC group and µ be a left Haar measure on X. Then:
(i) µ[U ] > 0 for every nonempty open U ⊂ X.
(ii) If µ[X] < ∞, then X is compact.
Proof. If µ[U ] = 0, after a translation assume U is a neighborhood of e. Let {an : n ∈ N}
be a countable dense subset of X and let x ∈ X. Since xU −1 is a neighborhood of X, there is
S
P∞
P∞
an ∈ xU −1 or x ∈ an U . Thus X = ∞
n=1 an U and therefore µ[X] ≤
n=1 µ[an U ] =
n=1 µ[U ] = 0,
a contradiction.
(ii) Suppose that X is not compact. Let U be a neighborhood of e with compact closure and
let V be a symmetric neighborhood of e with V 2 ⊂ U . Inductively choose x1 , x2 , . . . X such
S
that xn+1 ∈ X \ ni=1 xi U . Note that if m < n, then xm V ∩ xn V = ∅ (∵ if xm v1 = xn v2 ,
then xn = xm v1 v2−1 ∈ xm V V −1 ⊂ xm U , contradiction). Since µ[V ] > 0 by (i), we get µ[X] ≥
P∞
P∞
¤
n=1 µ[V ] = ∞, contrary to the hypothesis.
n=1 µ[xn V ] =
Let X be a LCSC group and µ be a left Haar measure on X. For x ∈ X, define βx [A] = µ[Ax].
Then βx is also a left Haar measure on X and therefore there is a function ∆ : X → (0, ∞) such
that βx = ∆(x)µ for every x ∈ X. Then ∆(x) = µ[Ax]/µ[A] for any Borel set A ⊂ X with
0 < µ[A] < ∞ (for instance, let V ⊂ X be a nonempty open set with compact closure and let
µ[Ax] µ[Axy]
µ[Axy]
A = V ). Now, ∆(xy) =
=
= ∆(x)∆(y) so ∆ is a homomorphism. Now we
µ[A]
µ[A] µ[Ax]
show ∆ is continuous at X by checking continuity at e.
Given ² > 0, let A ⊂ X be a compact set with µ[A] > 0 and let B ⊂ X be compact with
R
A ⊂ int[B]. Choose f ∈ Cc (X, [0, 1]) such that f ≡ 1 on A, supp(f ) ⊂ int[B], and X |χA − f |dµ <
²µ[A]/3. Let U be a neighborhood of e such that |f (a) − f (b)| < ²µ[A]/(3µ[B]) for every a, b ∈ X
with a−1 b ∈ U , by Exercise-38. And let V ⊂ U be a symmetric neighborhood of e such that
18
T.K.SUBRAHMONIAN MOOTHATHU
R
R
supp(f )V ⊂ B, by Exercise-26. If x ∈ V , then |µ[A] − µ[Ax]| ≤ X |χA − f |dµ + X |f (y) −
R
f (yx−1 )|dµ(y) + X |f (yx−1 ) − χA (yx−1 )|dµ(y) = I1 + I2 + I3 , say (∵ y ∈ Ax ⇔ yx−1 ∈ A).
Now, I1 and I2 are < ²µ[A]/3 by the choice of f . If yx−1 ∈ supp(f ), then y = y(yx−1 ) ∈
R
R
supp(f )V ⊂ B, and therefore I2 = B |f (y) − f (yx−1 )|dµ(y) ≤ B ²µ[A]/(3µ[B])dµ(y) = ²µ[A]/3
since y −1 (yx−1 ) = x−1 ∈ V ⊂ U . Hence for every x ∈ V , we have |1−∆(x)| = |µ[A]−µ[Ax]|/µ[A] <
². Thus we have:
[187] Let X be a LCSC group and µ be a left Haar measure on X. Then there is a continuous
homomorphism ∆ : X → (0, ∞) (called the modular function) such that µ[Ax] = ∆(x)µ[A] for
every Borel set A ⊂ X and every x ∈ X. We have ∆ ≡ 1 iff µ is also a right Haar measure on X.
Finally, ∆ is independent of µ (∵ any other left Haar measure on X is a positive scalar multiple of
µ).
Definition A LCSC group X is called unimodular if the modular function ∆ is ≡ 1; equivalently if
every left Haar measure on X is a right Haar measure; equivalently if every right Haar measure on
X is a left Haar measure.
[188] Let X be a LCSC group. If X is abelian, discrete or compact, then X is unimodular.
Proof. Abelian case is trivial. If X is discrete, let µ[A] = |A| and then ∆ ≡ 1. If X is compact,
then ∆(X) is a compact subgroup of (0, ∞) and hence equal to {1}.
¤
Definition: If µ is a left Haar measure on X, the corresponding left Haar integral φ : Cc (X, R) → R
R
is φ(f ) = X f dµ. Then the modular function ∆ satisfies φ(f ◦ Rx−1 ) = ∆(x)φ(f ) for every
f ∈ Cc (X, R) and for every x ∈ X (∵ yx−1 ∈ A ⇔ y ∈ Ax). This property can also be taken as
the definition of ∆.
Exercise-55 : Let X be a LCSC group and let µ be a left Haar measure on X. Then X is unimodular
iff µ is inversion-invariant (i.e., µ[A−1 ] = µ[A]). [Hint: If µ is inversion-invariant, then µ is also a
right Haar measure. Conversely if ∆ ≡ 1, then µ and β defined as β[A] = µ[A−1 ] are right Haar
measures. If β = λµ, then we see λ = 1 by considering symmetric A ⊂ X with 0 < µ[A] < ∞.]
Example-1 : On Rn , Lebesgue measure is a Haar measure, and the corresponding Haar integral on
Cc (Rn , R) is just the Riemann integral.
Example-2 : On S 1 , we get the normalized Haar measure by identifying S 1 with [0, 1) and then using
R1
R 2π
1
iθ
the Lebesgue measure on [0, 1). The Haar integral for S 1 is φ(f ) = 0 f (e2πit )dt = 2π
0 f (e )dθ.
Example-3 : On a finite discrete group X with |X| = n, the normalized Haar measure is µ[A] = |A|/n
P
P
and the Haar integral is φ(f ) = n−1 x∈X f (x) since f = x∈X f (x)χ{x} . On a countable discrete
P
group X, the counting measure µ[A] = |A| is a Haar measure and φ(f ) =
x∈X f (x) is the
corresponding Haar integral.
Example-4 : (Q, +, usual topology) is not locally compact and has no ‘Haar measure’. This is
argued as follows. If µ is a translation invariant non-zero Borel measure on Q, then µ[{x}] =
TOPOLOGICAL GROUPS - PART 3/3
19
µ[{0}] > 0 for every x ∈ Q since µ 6= 0. Then µ[U ] = ∞ for any nonempty open set U ⊂ Q (since
U cannot be finite) and µ[K] = ∞ for the compact set K = {0} ∪ {1/n : n ∈ N}, so µ is not locally
finite.
Example-5 : Haar measure and Haar integral can be transferred using a topological isomorphism.
We know that y → ey is a topological isomorphism from (R, +) to ((0, ∞), ·) with inverse x 7→
log (x). Hence a Haar measure on (0, ∞) is given by µ[A] = Lebesgue measure of log (A), and the
R∞
R∞
Haar integral for (0, ∞) is φ(f ) = −∞ f (ey )dy = 0 x−1 f (x)dx (putting x = ey ). This can be
seen in another way with the help of [189] below.
Let us mention that if X is a LCSC group, a non-zero positive linear φ : Cc (X, R) → R is a left
Haar integral if φ(f ◦ Ly ) = φ(f ) for every f ∈ Cc (X, R) and every y ∈ X (that is, we need not
invoke a left Haar measure to define a left Haar integral; however, such a left Haar integral comes
from a left Haar measure by Riesz Theorem).
[189]: Let X be a LCSC group, let ψ : Cc (X, R) → R be a non-zero positive linear functional and
R
let β be the locally finite Borel measure on X such that ψ(f ) = X f dβ. If there is a continuous
function δ : X → (0, ∞) such that ψ(f ◦ Ly ) = δ(y)ψ(f ) for every f ∈ Cc (X, R) and every y ∈ X,
then:
(i) δ is a homomorphism.
(ii) φ : Cc (X, R) → R defined as φ(f ) = ψ(δf ) is a left Haar integral for X.
Proof. (i) If f is such that ψ(f ) 6= 0, then δ(ab)ψ(f ) = ψ(f ◦Lab ) = ψ(f ◦Lb ◦La ) = δ(a)ψ(f ◦Lb ) =
δ(a)δ(b)ψ(f ) so that δ(ab) = δ(a)δ(b).
(ii) Clearly φ is a non-zero positive linear functional. Now, φ(f ◦ Ly ) = ψ(δ(f ◦ Ly )) =
R
R
R
−1 yx)f (yx)dβ(x) = δ(y)−1
−1
δ(x)f
(yx)dβ(x)
=
δ(y
X
X
X δ(yx)f (yx)dβ(x) = δ(y) ψ((δf )◦Ly ) =
δ(y)−1 δ(y)ψ(δf ) = φ(f ).
¤
Example-5 continued: If β is the Lebesgue measure on (0, ∞) (note that β is not a Haar measure
R∞
R∞
R∞
on (0, ∞)), then ψ(f ) = 0 f (x)dx and therefore ψ(f ◦ Ly ) = 0 f (yx)dx = 0 y −1 f (t)dt so that
R∞
δ(y) = y −1 . Hence by [189] the Haar integral φ is φ(f ) = ψ(δf ) = 0 x−1 f (x)dx. And Haar
R∞
R
R
measure is µ[A] = X δ(x)χA (x)dβ(x) = 0 x−1 χA (x)dx = A x−1 dx = β[log (A)].
Example-6 : Note that C \ {0} ∼
= (0, ∞) × S 1 . Hence a Haar integral φ for (C \ {0}, ·) is given
R ∞ R 2π
by φ(f ) = r=0 θ=0 f (reiθ )r−1 dθdr (ignoring a factor of 1/2π). Now |det(Jacobian)| of the map
(r, θ) 7→ (r cos θ, r sin θ) is r and hence rdθdr = dxdy by Change of Variable Formula. Hence
RR
R
φ(f ) =
f (x + iy)|x + iy|−2 dxdy = C\{0} f (z)|z|−2 dz.
R
Example-7 : Let G = GL(2, R) ⊂ R4 , ψ : Cc (R4 , R) → R be ψ(f ) = R4 f dβ where β is the Lebesgue
measure on R4 , and note that every f ∈ Cc (G, R) and be extended
to!f ∈ CC (R4 , R) by putting
Ã
a b
f ≡ 0 on R4 \ G. To apply [189], let us find ψ(f ◦ LY ) for Y =
∈ G. Let M be the 4 × 4
c d
matrix representing the linear map X 7→ Y X from R4 to R4 . We have
20
T.K.SUBRAHMONIAN MOOTHATHU
Ã
Y e1 =
Ã
Y e2 =
Ã
a b
c d
a b
c d
a b
!Ã
!Ã
!Ã
1 0
0 0
0 0
1 0
0 1
Ã
!
=
!
Ã
=
!
a 0
c 0
0 b
0 d
!
∼ (a, c, 0, 0) ∈ R4 ,
!
∼ (0, 0, b, d) ∈ R4 ,
= (0, 0, a, c) ∈ R4 , and Y e4 = (0, 0, b, d) ∈ R4 .
c d
0 0
Ã
!
Y 0
Thus M =
, and therefore, |det(Jacobian)| of X 7→ Y X is |det(M )| = |det(Y )|2 . By
0 Y
R
R
change of variable, ψ(f ◦ LY ) = R4 f (Y X)dβ(X) = R4 f (Z)|det(Y )|−2 dβ(Z) = |det(Y )|−2 ψ(f )
Y e3 =
and similarly ψ(f ◦ RY ) = |det(Y )|−2 ψ(f ) so that δ(Y ) = |det(Y )|−2 in both the cases. Hence by
R
[189] the (left as well as right) Haar integral φ for GL(2, R) is φ(f ) = R4 f (X)|det(X)|−2 dβ(X),
and GL(2, R) is thus unimodular. More generally, GL(n, R) is unimodular and a Haar integral for
R
GL(n, R) is φ(f ) = Rn2 f (X)|det(X)|−n dβ(X).
Ã
!
x y
Example-8 [Not unimodular]: Let H = {
: x > 0, y ∈ R} ⊂ GL(2, R). Check that H
0 1
Ã
!
a
b
is a closed subgroup of GL(2, R). Identify H with (0, ∞) × R ⊂ R2 . Let A =
∈ H
0 1
!
Ã
!
Ã
ax ay + b
x y
, then AX =
and consider the maps X 7→ AX and X 7→ XA. If X =
0 1
0
1
Ã
!
ax bx + y
and XA =
so that X 7→ AX corresponds to (x, y) 7→ (ax, bx + y) and X 7→ XA
0
1
corresponds to (x, y) 7→ (ax, bx + y). Therefore |det(Jacobian)| of X 7→ AX and X 7→ XA
R∞ R∞
are respectively a2 and a. If ψ(f ) = x=0 y=−∞ f (x, y)dxdy, we get by change of variable that
R∞ R∞
R∞ R∞
ψ(f ◦ LA ) = x=0 y=−∞ f (x, y)a−2 dxdy and ψ(f ◦ RA ) = x=0 y=−∞ f (x, y)a−1 dxdy. Hence δ
function is A 7→ a−2 and A 7→ a−1 respectively and by [189], the left Haar integral is φ(f ) =
R∞ R∞
R∞ R∞
−2
−1 dxdy. We
e
x=0 y=−∞ f (x, y)x dxdy and the right Haar integral is φ(f ) = x=0 y=−∞ f (x, y)x Ã
!
a b
have almost proved that H is not unimodular. To be precise, we find ∆. Let A =
0 1
Ã
!
R∞ R∞
1/a −b/a
and B = A−1 =
. Now, φ(f ◦ RB −1 ) = φ(f ◦ RA ) = x=0 y=−∞ f (ax, bx +
0
1
R∞ R∞
R∞ R∞
−2
y)x dxdy = a x=0 y=−∞ f (x, y)a(ax)−2 dxdy = x=0 y=−∞ f (s, y)s−2 dsdt = aφ(f ) by change of
variable. Since ∆ must satisfy φ(f ◦ RB −1 ) = ∆(B)φ(f ), we conclude that ∆(B) = a, and thus H
is not unimodular.
Observation: Let X be a totally disconnected compact metric group and let µ be the normalized
Haar measure on X. We know that there are compact open normal subgroups Yn ⊂ X such that
{Yn : n ∈ N} is a base at {e}. If kn = [X : Yn ], then µ[Yn ] = µ[xYn ] = 1/kn . Since any open set
TOPOLOGICAL GROUPS - PART 3/3
21
U ⊂ X can be written as a countable disjoint union of cosets of the form xYn (x ∈ X, n ∈ Yn ), we
can find µ[U ], and then µ[A] for any Borel A ⊂ X by regularity. We use this observation below.
Example-9 : If X = Zp and Yn = {(x1 + pZ, x2 + p2 Z, . . .) ∈ Zp : xi = 0 for 1 ≤ i ≤ n}, then
{Yn : n ∈ N} is a countable base at e and [Zp : Yn ] = pn . Hence µ[Yn ] = p−n if µ is the normalized
Haar measure on Zp .
Example-10 : Let X =
Q∞
n=1 Xn ,
where Xn ’s are finite discrete groups, and let Yn = {(x1 , x2 , . . .) ∈
Q
Qn
−1 if µ is the
X : xi = ei for 1 ≤ i ≤ n}. Then [X : Yn ] = ∞
i=1 |Xi | and hence µ[Yn ] = [ i=1 Xi ]
normalized Haar measure. As a special case, if X = {0, 1}N and Yn = {(x1 , x2 , . . .) ∈ X : xi =
0 for 1 ≤ i ≤ n}, then [X : Yn ] = 2n and hence µ[Yn ] = 2−n .
Approximating the Haar integral: Let X be a totally disconnected compact metric group and let
µ be the normalized Haar measure on X. Given f ∈ Cc (X, R) and ² > 0, choose a compact open
normal subgroup Y (which is a neighborhood of e) such that |f (a) − f (b)| < ² for every a, b ∈ X
S
P
with a−1 b ∈ Y , let n = [X : Y ], and suppose X = ni=1 xi Y . Define g = ni=1 f (xi )χxi Y . Now
if x ∈ X, there is a unique i such that x ∈ xi Y and then x−1
i x ∈ Y so that ² > |f (x) − f (xi )| =
R
R
R
R
P
|f (x) − g(x)| and therefore | f dµ − gdµ| ≤ |f − g|dµ ≤ ². Since gdµ = n−1 ni=1 f (xi ), we
P
get |φ(f ) − n−1 ni=1 f (xi )| for the Haar integral φ.
4. Abstract Fourier Transform
We consider the simplest case - that of a compact metric abelian group.
[190] Let X be a compact metric abelian group and let µ be the normalized Haar measure on X.
b is an orthonormal basis for L2 (X, µ).
Then X
C
R
R
b then |g(x)|2 dµ = 1dµ = µ[X] = 1 < ∞ so that X
b ⊂ L2 (X, µ) and kgk = 1
Proof. If g ∈ X,
C
b If g ∈ X
b \ {1X }, let y ∈ X be such that g(y) 6= 1. Then by the left invariance
for g ∈ X.
R
R
of the Haar integral φ, we have hg, 1X i = g(x)dµ(x) = φ(g) = φ(g ◦ Ly ) = g(yx)dµ(x) =
R
b are distinct then
g(y) g(x)dµ(x) = g(y)hg, 1x i so that hg, 1X i = 0 since g(y) 6= 1. Now if f, g ∈ X
R
R
b is an orthonormal set. To
0 = hf g −1 , 1X = f (x)g(x)−1 dµ(x) = f (x)g(x)dµ = hf, gi. Thus X
b is dense in L2 (X, µ). Since
show it is an orthonormal basis, we need to show now that A = spanC X
C
C(X, C) is dense in
L2C (X, µ)
by
on C(X, C) is stronger than the
[1240 ]
and since the topology induced by the supremum metric d
L2 -topology,
it suffices to show that A is dense in (C(X, C), d).
We observe that A is a complex algebra satisfying the following: (i) f ∈ A implies f ∈ A (since
b as f = f −1 ), (ii) x ∈ X implies there is f ∈ A with f (x) 6= 0 (take f = 1X ),
this is true for f ∈ X
b does this already by [171]; note that X being abelian is
(iii) A separates points of X (since X
used here). Therefore, by the complex version of the Stone-Weierstrass Theorem (p.165, Rudin,
Principles of Mathematical Analysis), A is dense in (C(X, C), d).
¤
[191] [Some facts from Functional Analysis] Let H be a complex Hilbert space with orthonormal
basis {bn : n ∈ N}. Then:
22
T.K.SUBRAHMONIAN MOOTHATHU
P∞
(i) If y ∈ H, then kyk2 = n=1 |hy, bn i|2 (Parseval’s identity; p.170 of Kreyszig).
P∞
2 , then
(ii) If (αn ) ∈ lC
n=1 αn bn is absolutely convergent and belongs to H (p.164 of Kreyszig).
2 given by T y = (hy, b i) is a Hilbert space isomorphism, i.e., T is a linear bijection
(iii) T : H → lC
n
with hT x, T yi = hx, yi for every x, y ∈ H.
[Linearity of T is clear. Bijectivity follows from (i) and (ii). And hT x, T yi = hx, yi can be verified
by using the linearity and continuity of the inner product.]
b = {f1 , f2 , . . .} (we may parametrize
Definition: Let X be a compact metric abelian group and let X
b with other countable sets such as Z also). If µ is the normalized Haar measure on X, then the
X
2 (X)
b is defined as F (g) = (hg, fn i). Here hg, fn i’s are called
Fourier transform F : L2C (X, µ) → lC
the Fourier coefficients of g.
2 (X)
b defined above as F (g) = (hg, fn i) is a Hilbert
[192] The Fourier transform F : L2C (X, µ) → lC
P
space isomorphism. In particular F is a linear bijective isometry. And F −1 ((αn )) = ∞
n=1 αn fn ∈
2
2
b
L (X, µ) for (αn ) ∈ l (X).
C
C
b is an orthonormal basis for L2 (X, µ). Now use [191].
Proof. By [190], X
¤
C
Pk
P
Remark : When we say g = ∞
n=1 αn fn k = 0, and
n=1 αn fn , this only means that limk→∞ kg −
P∞
we may not have the pointwise convergence g(x) = n=1 αn fn (x), x ∈ X.
Example-1 : Note that 2π-periodic functions h : R → C correspond to functions g : S 1 → C via
b = {fn : n ∈ Z} where fn (eit ) = eint . If g ∈ L2 (S 1 , µ) and
h(t) = g(eit ). If X = S 1 , then X
R 2π
R ∞C
R
h(t) = g(eit ), then hg, fn i = S 1 g(x)fn (x)dµ = (1/2π) 0 g(eit )e−int dt = (1/2π) 0 h(t)e−int dt,
which is the standard formula for the nth Fourier coefficient.
b = {f1 , . . . , fk }. The norExample-2 : Let X = {x1 , . . . , xk } be a finite abelian group and let X
malized Haar measure µ on X is µ[A] = |A|/k, and L2C (X, µ) = {all functions from X to C} ∼
= Ck
P
via the correspondence g ↔ (g(x1 ), . . . , g(xk )). If g ∈ L2C (X, µ), then g = kn=1 g(xn )χ{xn } so that
P
hg, hi = k −1 kn=1 g(xn )h(xn ) for g, h ∈ L2C (X, µ). Thus if g ∈ L2C (X, µ), then the mth Fourier
P
coefficient is hg, fm i = k −1 kn=1 g(xn )fm (xn ).
Further reading (sample): Harmonic Analysis - H.Helson, Fourier Analysis on Groups - W.Rudin.
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