Unitary Representations of R

The following theorem shows how spectral measures lead to unitary representa-tions of the group (R,+) and vice versa. It may be considered as a classification of unitary one-parameter groups in terms of spectral measures onRwhich pro-vides important structural information.

Theorem 6.4.1. Let P: S→ P_H be a spectral measure and f: X →R be a measurable function. Then

π(t) :=P(e^itf)

defines a continuous unitary representation π:R→U(H).

Conversely, every continuous unitary representation ofR has this form for X =Randf(x) =x.

Proof. Since R→U(L^∞(X,C)), t7→e^itf is a homomorphism into the unitary group of theC^∗-algebraL^∞(X,C), it follows from Proposition 6.2.8 thatπ(t) :=

P(e^itf) defines a unitary representation ofR. Now letv∈ H. Then Remark 6.2.9 implies that

π^v(t) =hπ(t)v, vi=hP(e^itf)v, vi= Z

X

e^itf(x)dP^v(x),

and the continuity of this function follows from Lebesgue’s Theorem on Domi-nated Convergence.

If, conversely, (π,H) is a continuous unitary representation ofR, then The-orem 6.3.4 implies the existence of a spectral measureP onRb withπ(t) =P(bt) for t∈ R. Identifying the locally compact space Rb with Rin such a way that bt(x) =e^itx(Example 6.3.2), the assertion follows.

Remark 6.4.2. If π is defined as above, then all operators P(E), E ∈ S, commute with π(R).

Example 6.4.3. Note that Theorem 6.4.1 applies in particular forX =Rand S=B(R) (the σ-algebra of Borel sets) andf(x) =x.

If, f.i.,µis a Borel measure onRandH=L²(R, µ) as in Lemma 6.2.5, then we obtain a continuous unitary representation byπ(t)f =e^itxf (Exercise).

Remark 6.4.4. We have seen above that every unitary one-parameter group is of the form π(t) = P(e^itidR) for some spectral measure P on R. If the spectral measure P is supported by a bounded subset of R, i.e., there exist a, b ∈ R with P([a, b]) = 1, then the function id_R is essentially bounded and A:=P(id_R)∈B(H) is a bounded hermitian operator satisfying

π(t) =P(e^itidR) =

∞

X

k=0

1

k!P(itid_R)^k =

∞

X

k=0

1

k!(itA)^k =e^itA. Taking derivatives int, we obtain

A= lim

t→0

1

it π(t)−1 in the operator norm.

For a general unitary one-parameter group the corresponding spectral mea-sureP onRmay have unbounded supported, which is reflected in the fact that the limit

Av:= lim

t→0

1

it π(t)v−v

may not exist for every v ∈ H. Write D(A) for the linear subspace of H consisting of all elements for which this limit exists. Then

A:D(A)→ H

is called anunbounded operator. A closer analysis of this situation leads to the theory of unbounded selfadjoint operators on Hilbert spaces.

Here we only note that for every bounded subsetB⊆Rthe closed subspace HB:=P(B)His invariant, and on this subspace we have

π(t)|_HB =P(e^itidR)|_HB =P(e^itidRχB)|_HB =e^itP(idRχB)|_HB,

and since the operatorP(id_RχB) is bounded, we obtain in particular thatHB⊆ D(A) with

A|HB=P(id_RχB)|HB.

In view of P([−n, n])→P(R) =1 in the strong operator topology, the union S

n∈NH_[−n,n] is dense inH, which implies thatD(A) is a dense subspace ofH.

We say thatAisdensely defined.

This suggests that A should be something likeP(id_R), and to make sense out of that, one has to extend the spectral integral to unbounded measurable functions.

Exercises for Section 6.4

Exercise 6.4.1. (One-parameter groups of U(H))

(1) LetA=A^∗∈B(H) be a bounded hermitian operator. Thenγ_A(t) :=e^itA defines a norm-continuous unitary representation of (R,+).

(2) LetP: (X,S)→B(H) be a spectral measure andf:X →Ra measurable function. Thenγ_f(t) :=P(e^itf) is a continuous unitary representation of (R,+). Show that γ_f is norm-continuous if and only of f is essentially bounded.

Chapter 7

Closed Subgroups of Banach Algebras

In this chapter we study one of the central tools in Lie theory: the exponential function of a Banach algebra. This function has various applications in the structure theory of Lie groups. First of all, it is naturally linked to the one-parameter subgroups, and it turns out that the local group structure ofA^× for a unital Banach algebra Ain a neighborhood of the identity is determined by its one-parameter subgroups.

In Section 7.1, we discuss some basic properties of the exponential function of a unital Banach algebraA. In Section 7.2, we then use the exponential function to associate to each closed subgroupG⊆ A^×a Banach–Lie algebraL(G), called theLie algebra ofG. We then show that the elements ofL(G) are in one-to-one correspondence with the one-parameter groups of Gand study some functorial properties of the assignment L:G7→L(G). The last section of this chapter is devoted to some tools to calculate the Lie algebras of closed subgroups ofA.

7.1 Elementary Properties of the Exponential Function

LetAbe a unital Banach algebra. Forx∈ Awe define e^x: =

∞

X

k=0

1

k!x^k. (7.1)

The absolute convergence of the series on the right follows directly from the estimate

∞

X

k=0

1

k!kx^kk ≤

∞

X

k=0

1

k!kxk^k =e^kxk 169

and the Comparison Test for absolute convergence of a series in a Banach space.

We define theexponential function ofAby

exp :A → A, exp(x) :=e^x.

Proof. (i) Using the general form of the Cauchy Product Formula (Exercise 7.1.3), we obtain

(ii) From (i) we derive in particular expxexp(−x) = exp 0 =1, which implies (ii).

(iii) is a consequence of gxⁿg⁻¹ = (gxg⁻¹)ⁿ and the continuity of the con-jugation mapcg(x) :=gxg⁻¹onA.

(iv) For the exponential series we have the estimate ke^x−1−xk=kX

Remark 7.1.2. (a) Forn= 1, the exponential function exp :R∼=Mn(R)→R^×∼= GLn(R), x7→e^x is injective, but this is not the case forn >1. In fact,

exp This example is the real picture of the relatione^2πi= 1.

Definition 7.1.3. A one-parameter (sub)group of a group G is a group ho-momorphism γ: (R,+) → G. The following result describes the differentiable one-parameter subgroups ofA^×.

Theorem 7.1.4. (One-parameter Group Theorem) For eachx∈ A, the map γx: (R,+)→ A, t7→exp(tx)

is a smooth group homomorphism solving the initial value problem γx(0) =1 and γ⁰_x(t) =γx(t)x fort∈R.

Conversely, every continuous one-parameter groupγ:R→ A^× is of this form.

Proof. In view of Lemma 7.1.1(i) and the differentiability of exp in 0, we have lim

We first show that each one-parameter groupγ:R→ A^× which is differen-tiable in 0 has the required form. For x:=γ⁰(0), the calculation

γ⁰(t) = lim implies thatγ is continuously differentiable. Therefore

d

dt(e^−txγ(t)) =−e^−txxγ(t) +e^−txγ⁰(t) = 0 implies thate^−txγ(t) =γ(0) =1for eacht∈R, so thatγ(t) =e^tx.

Eventually we consider the general case, whereγ:R→ A^× is only assumed to be continuous. The idea is to construct a differentiable functioneγby applying a smoothing procedure toγand to show that the smoothness ofγeimplies that of γ. So let f: R → R+ be a twice continuously differentiable function with f(t) = 0 for|t|> εandR

Here we use the existence of Riemann integrals of continuous curves with values in Banach spaces, which follows from Theorem 6.1.9. Change of Variables leads to

eγ(t) = Z

R

f(t−s)γ(s)ds,

which is differentiable because is a consequence of the Mean Value Theorem). We further have

because of the inequalitykR

h(s)dsk ≤R

kh(s)kds(cf. Subsection 6.1.3).

Letg:=Rε

−εf(s)γ(−s)ds. In view ofkg−1k ≤ ¹₂ we have g∈ A^× (see the proof of Proposition 1.1.10) and thereforeγ(t) =eγ(t)g⁻¹. Now the differentia-bility ofeγ implies thatγis differentiable, and one can argue as above.

Product and Commutator Formula

We have seen in Lemma 7.1.1 that the exponential image of a sum x+y can be computed easily if x and y commute. In this case we also have for the commutator [x, y] := xy−yx = 0 the formula exp[x, y] = 1. The following proposition gives a formula for exp(x+y) and exp([x, y]) in the general case.

If g, h are elements of a group G, then (g, h) := ghg⁻¹h⁻¹ is called their commutator. On the other hand, we call for two elementa, b∈ Athe expression

[a, b] :=ab−ba theircommutator bracket.

Proposition 7.1.5. Forx, y∈ A, the following assertions hold:

(i) lim_k→∞

e^k1xe^k1yk

=e^x+y (Trotter Product Formula).

(ii) limk→∞ e^k1xe^k1ye^−1kxe^−k1yk²

=e^xy−yx (Commutator Formula).

Proof. We start with a general consideration. We shall have to estimate an expression of the formA^k−B^k. To this end we write

Let us assume that there exists a constantC >0 with

kA^jk,kB^jk ≤C for 1≤j≤k. (7.2)

Then we get the estimate kA^k−B^kk ≤

k−1

X

j=0

kA^jk · kB^k−1−jk · kA−Bk ≤kC²kA−Bk.

(i) We apply the estimate from above to some special situations. First we consider k=n,A=e^1nxe^n1y andB=e^1n(x+y). Then

Aⁿ−Bⁿ= e^n1xe^n1yn

−e^(x+y),

and we have to show that this expression tends to zero. We therefore check the assumptions from above. We have

kAk ≤ ke^n1xk ke^n1yk ≤e^1nkxke^1nkyk=e^n1(kxk+kyk),

and therefore kA^jk ≤e^kxk+kyk for j≤n. We likewise obtainkB^jk ≤e^kxk+kyk forj≤n, and hence

kAⁿ−Bⁿk ≤e^2(kxk+kyk)nkA−Bk.

We further have

n(A−B) =e^n1xe^1ny−e^n1(x+y)

1 n

=e^n1x(e^1ny−1) + (e^1nx−1) +1−e^n1(x+y)

1 n

→e⁰·dexp(0)y+dexp(0)x−dexp(0)(x+y)

=y+x−(x+y) =0.

This impliesAⁿ−Bⁿ→0 and hence the Trotter Formula.

(ii) Now letk=n²,A=e^n1xe^n1ye^−n1xe^−n1y andB=e^n12(xy−yx). Then Aⁿ²−Bⁿ² =

e^n1xe^n1ye^−1nxe^−1nyn2

−e^xy−yx,

and we have to show that this expression tends to zero. Again, we verify (7.2).

In view of

kBk ≤e^n12kxy−yxk≤e^{n12(2kxk·kyk)},

we have kB^jk ≤ e^2kxk·kyk for j ≤ n². To estimate the A-part, let us write O(t^k) for a function for whicht^−kO(t^k) is bounded fort→0. We likewise write O(n^k), k∈ Z, for a function for whichn^−kO(n^k) is bounded forn→ ∞. We consider the smooth curve

γ: R→ A^×, t7→e^txe^tye^−txe^−ty.

Thene^tx=1+tx+^t₂²x²+O(t³) leads to γ(t) = 1+tx+t²

2x²+O(t³)

1+ty+t²

2y²+O(t³)

· 1−tx+t²

2x²+O(t³)

1−ty+t²

2y²+O(t³)

=1+t(x+y−x−y) +t²(x²+y²+xy−x²−xy−yx−y²+xy) +O(t³)

=1+t²(xy−yx) +O(t³).

This implies that γ⁰(0) = 0 andγ⁰⁰(0) = 2(xy−yx). Moreover, for j ≤n² we have for eachε >0

kA^jk=kγ(¹_n)^jk ≤ kγ(_n¹)k^j ≤ 1 +_n¹2kxy−yxk+O(n⁻³)^j

≤ 1 + 1

n²kxy−yxk+O(n⁻³)ⁿ² . For sufficiently largen, we thus obtain for allj≤n²:

kA^jk ≤ 1 + 1

n²(kxy−yxk+ 1)n²

≤e^kxy−yxk+1. This proves the existence of a constantC >0 (independent ofn) with

kAⁿ²−Bⁿ²k ≤Cn²kA−Bk for all n∈N. We further have

n²(A−B) = γ(_n¹)−e^n12(xy−yx)

1 n²

= γ(¹_n)−1+1−e^n12(xy−yx)

1 n²

→ ¹₂γ⁰⁰(0)−(xy−yx) = (xy−yx)−(xy−yx) = 0.

This proves (ii).

Exercises for Section 7.1

Exercise 7.1.1. LetX1, . . . , Xn be Banach spaces andβ:X1×. . .×Xn→Y a continuousn-linear map.

(a) Show that there exists a constantC≥0 with

kβ(x1, . . . , xn)k ≤Ckx1k · · · kxnk for xi∈Xi. (b) Show thatβ is differentiable with

dβ(x₁, . . . , x_n)(h₁, . . . , h_n) =

n

X

j=1

β(x₁, . . . , x_j−1, h_j, x_j+1, . . . , x_n).

Exercise 7.1.2. LetY be a Banach space anda_n,m, n, m∈N, elements in Y

and that both iterated sums exist.

(b) Show that for each sequence (S_n)_n∈Nof finite subsetsS_n ⊆N×N,n∈N,

Exercise 7.1.3. (Cauchy Product Formula) LetX, Y, Zbe Banach spaces and β: X×Y → Z a continuous bilinear map. Suppose that if x := P∞

n=0xn is absolutely convergent inX and ify:=P∞

n=0y_n is absolutely convergent inY, then where P is a polynomial.

(b) For λ >0 the function Ψ(t) := Φ(t)Φ(λ−t) is a non-negative smooth

Exercise 7.1.6. Show that for X =−X^∗ ∈M_n(C) the matrix e^X is unitary and that the exponential function

exp : Ahermn(C) :={X∈Mn(C) :X^∗=−X} →Un(C), X7→e^X is surjective.

Exercise 7.1.7. Show that forX^>=−X ∈M_n(R) the matrixe^Xis orthogonal and that the exponential function

exp : Skewn(R) :={X∈Mn(R) :X^>=−X} →On(R)

is not surjective. Can you determine which orthogonal matrices are contained in the image? Can you interprete the result geometrically in terns of the geometry of the flowR×Rⁿ→Rⁿ,(t, v)7→e^tXv.

Exercise 7.1.8. Show that forA:=C(S¹) the exponential function exp : Aherm(A) :={a∈ A:a^∗=−a} →U(A) =C(S¹,T), a7→e^a is not surjective. It requires some covering theory to determine which elements f ∈C(S¹,T) lie in its image. Hint: Use the winding number with respect to 0.

Exercise 7.1.9. Show that for any measure space (X,S) and the C^∗-algebra A:=L^∞(X,S), the exponential function

exp : Aherm(A)→U(A), a7→e^a is surjective.

Exercise 7.1.10. Show that for every von Neumann algebraA, the exponential function

exp : Aherm(A)→U(A), a7→e^a

is surjective. This applies in particular toA=B(H), so that for every complex Hilbert spaceH, the exponential function exp : Aherm(H)→U(H), a7→e^a is surjective.

Exercise 7.1.11. (a) Calculatee^tN fort∈Kand the matrix

N =













0 1 0 . . . 0

· 0 1 0 ·

· · · ·

· · 1

0 . . . 0













∈M_n(K).

(b) If A is a block diagonal matrix diag(A1, . . . , Ak), then e^A is the block diagonal matrix diag(e^A1, . . . , e^Ak).

(c) Calculate e^tA for a matrix A ∈ Mn(C) given in Jordan normal form.

Hint: Use (a) and (b).

Exercise 7.1.12. Leta, b∈M_n(C) be commuting elements.

(a) Ifaandb are nilpotent, thena+bis nilpotent.

(b) Ifaandbare diagonalizable, then a+bandab are diagonalizable.

(c) Ifaandb are unipotent, thenabis unipotent.

Exercise 7.1.13. ForA∈M_n(C) we havee^A=1if and only ifAis diagonal-izable with all eigenvalues contained in 2πiZ. Hint: Exercise 2.2.10.

Im Dokument An Introduction to Unitary Representations of Lie Groups (Seite 170-182)