I’m getting the impression that the bra-ket notation is more useful for linear ON THE LEFT!

An approximate unit for a C*-algebra is an increasing net \{u_\lambda\}_{\lambda\in\Lambda} of positive elements in the closed unit ball of A such that a= \lim_{\lambda }au_\lambda=\lim_\lambda u_\lambda a for all a\in A.

Example

Let H be a Hilbert space with infinite orthonormal basis \{e_n\}. The C*-algebra K(H) is now non-unital. If P_n is the projection onto \langle e_1,\dots,e_n\rangle, then the increasing sequence \{P_n\}\subset K(H) is an approximate unit for K(H). It will suffice to show that T=\lim_np_nT if T\in F(H), since F(H) is dense in K(H). Now if  T\in F(H), there exist x_1,\dots,x_my_1,\dots,y_m\in H such that:

T=\sum_{k=1}^m|x_k\rangle\langle y_k|.

Hence,

P_nT=\sum_{k=1}^m|P_nx_k\rangle\langle y_k|.

Since \lim_n P_nx=x for all x\in H, therefore for each $k$:

\lim_{n\rightarrow \infty}\||P_nx_k\rangle\langle y_k-|x_k\rangle\langle y_k|\|=\lim_{n\rightarrow \infty} \|P_nx_k-x_k\|\|y_k\|=0.

Hence, \lim_{n}P_nT=T.

Let A be a C*-algebra and denote by \Lambda=A^+\cap B_1(\mathbf{0}). This set is a poset under the partial order of A_{\text{SA}}. In fact, \Lambda is also upwards directed; that is, if a,b\in\Lambda, then there exists c\in\Lambda such that a,b\leq c. Suppose a\in \Lambda (Murphy actually takes a\in A^+ but we can do without this — in particular I want to take \|a\|<1 to give me the required invertibility: but 1+a\in G(\tilde{A}) when a\in A^+.), then 1+a\in G(\tilde{A}) ( I’m going to assume it is and take a\in A^+ after all because then 1+a\in G(\tilde{A}).). We hence have that a(1+a)^{-1}=1-(1+a)^{-1}. We claim

a,b\in A^+ and a\leq b\Rightarrow a(1+a)^{-1}\leq b(1+b)^{-1}   (*)

Indeed if 0\leq a\leq b, then 1+a\leq 1+b which implies that (1+a)^{-1}\geq (1+b)^{-1}, by Theorem 2.2.5, and therefore 1-(1+a)^{-1}\leq 1-(1+b)^{-1}; thus proving the claim.

Note that if a\in A^+ then a(1+a)^{-1}\in \Lambda. Suppose then that a,b\in\Lambda. Let a'=a(1-a)^{-1}b'=b(1-b)^{-1} and c=(a'+b')(1+a'+b')^{-1}. Then c\in \Lambda, and since a'\leq a'+b', we have that a=a'(1+a')^{-1}\leq c by (*). Similarly b\leq c and \Lambda is upwards-directed as claimed.

Theorem 3.1.1

Every C*-algebra admits an approximate unit. Indeed, if \Lambda is the upwards-directed set A^+\cap B_1(\mathbf{0}) and u_\lambda=\lambda for all \lambda\in \Lambda, then \{u_\lambda\}_{\lambda\in\Lambda} is an approximate unit for A (called the canonical approximate unit).

Proof

We  need to show that a=\lim_\lambda u_\lambda a for each a\in A. Since \Lambda linearly spans A (The positive elements span the hermitian elements and the hermitian elements span the C*-algebra.), we can reduce to the case when a\in\Lambda.

Let a\in \Lambda and \varepsilon>0. Let \varphi:C^*(a)\rightarrow C_0(\Phi(A)) be the Gelfand transformation. If f=\varphi(a), then K=\{\omega\in \Phi(A):|f(\omega)|\geq \varepsilon\} is compact, and therefore by Urysohn’s Lemma there is a continuous function g:\Phi(A)\rightarrow [0,1] of compact support such that g(\omega)=1 for all \omega\in K. Choose \delta>0 such that \delta<1 and 1-\delta<\varepsilon.  Then \|f-\delta gf\|\leq \varepsilon (I don’t see this: suppose, roughly, that the supremum is found ‘at’ \omega\in K. Then

\underbrace{\|f(\omega)\|}_{\geq \varepsilon}\underbrace{\|1-\delta g(\omega)\|}_{=1-\delta}

that is I can’t make any conclusion. Now if the supremum is found at \omega ‘outside’ K then:

\underbrace{\|f(\omega)\|}_{<\varepsilon}\underbrace{\|1-\delta g(\omega)\|}_{\in[1-\delta,1]}

This is unresolved by my supervisor and I but we can prove this result using different techniques so no panic — approximate units still exist!). If \lambda_0=\varphi^{-1}(\delta g), then \lambda_0\in\Lambda and \|a-u_{\lambda_0} a\|\leq \varepsilon. Now suppose that \lambda\in\Lambda and \lambda\geq \lambda_0. Then 1-u_\lambda\leq 1-u_{\lambda_0}, so a(1-u_\lambda)a\leq a(1-u_{\lambda_0})a (Th.2.2.5 (2): we can pre- and post-multiply inequalities by an element and it’s adjoint.). Hence,

\|a-u_\lambda a\|^2=\|(1-u_\lambda)^{1/2}(1-u_\lambda)^{1/2}a\|^2\leq \|(1-u_\lambda)^{1/2}a\|^2

(the first inequality here uses submultiplicativity and the C*-equation along with the the result \|1-u_\lambda\|\leq1. To see that this is true, not that \|u_\lambda\|<1 so that \sigma(u_\lambda)\subset [0,1]. Therefore \sigma(1-u_\lambda)\subset (0,1]\Rightarrow \|1-u_\lambda\|\leq1 The next equality is just the C*-equation:)

\|(1-u_\lambda)^{1/2}a\|^2=\|a(1-u_\lambda)a\|\leq \|a(1-u_{\lambda_0})a\|\leq \|(1-u_{\lambda_0})a\|\leq \varepsilon.

(The first and second inequalities here use Th. 2.2.5 (3) (a\leq b\Rightarrow \|a\|\leq\|b\|) and  submultiplicativity.) This shows that a=\lim_\lambda u_\lambda a \bullet

Remark

If a C*-algebra A is separable, then it admits an approximate unit which is a sequence. For in this case, there exist finite sets

F_1\subset F_2\subset \cdots \subset F_n\subset\cdots

such that F=\cup_n F_n is dense in A. Let \{u_\lambda\}_{\lambda\in\Lambda} be any approximate unit for A. If \varepsilon>0, and F_n=\{a_1,\dots,a_m\} say, then there exist \lambda_1\dots\lambda_m\in\Lambda such that \|a_j-a_j u_\lambda\|<\varepsilon if \lambda\geq \lambda_j. Choose \lambda_\varepsilon\in\Lambda such that \lambda_\varepsilon\geq\lambda_1,\dots,\lambda_m. Then \|a-au_\lambda\|<\varepsilon for all a\in F_n and all  \lambda\geq\lambda_\varepsilon. Hence, if n\in\mathbb{N} and \varepsilon=1/n, then there exists \lambda_n=\lambda_\varepsilon\in\Lambda such that \|a-au_{\lambda_n}\|<1/n for all a\in F_n. Also, we can choose the \lambda_n such that \lambda_n\leq\lambda_{n+1} for all n. Consequently, \lim_n\|a-au_{\lambda_n}\|=0, for all a\in F, and since F is dense in A, this also holds for all a\in A. Therefore \{u_{\lambda_n}\}_{n=1}^\infty is an approximate unit for A.

Theorem 3.1.2

If L is a closed left ideal in a C*-algebra A, then there is an increasing net \{u_\lambda\}_{\lambda\in\Lambda} of positive elements in the closed unit ball of L such that a=\lim_\lambda au_\lambda for all a\in L.

Proof

Set B=L\cap L^*. Since B is a C*-algebra, it admits an approximate unit, \{u_\lambda\}_{\lambda\in\Lambda} say, by the previous theorem. If a\in L, then a^*a\in B, so

\lim_\lambda a^*a(1-u_\lambda)=0.

Hence,

\lim_\lambda \|a-au_\lambda\|^2=\lim_\lambda\|(1-u_\lambda)a^*a(1-u_\lambda)\|\leq \lim_\lambda \|a^*a(1-u_\lambda)\|=0,

and therefore \|a-au_\lambda\|=0 \bullet

(Where the last inequality uses submultiplicatively — and again requires that \|1-u_\lambda\|<1. Murphy refers to this afterwards — he states that in this proof we worked in the unitisation \tilde{A} of A.)

Theorem 3.1.3

If I is a closed ideal in a C*-algebra A, then I is self-adjoint and therefore a C^*-subalgebra of A. If \{u_\lambda\}_{\lambda\in\Lambda} is an approximate unit for I, then for each a\in A:

\|a+I\|=\lim_\lambda\|a-u_\lambda a\|=\lim_\lambda\|a-au_\lambda\|.

Proof

By the previous theorem there is an increasing net \{u_\lambda\}_{\lambda\in\Lambda} of positive elements in the closed unit ball of I such that a=\lim_\lambda au_\lambda for all a\in I. Hence a^*=\lim_\lambda u_\lambda a^*, so a^*\in I, because all of the elements u_\lambda belong to I. Therefore I is self-adjoint.

Suppose that \{u_\lambda\}_{\lambda\in\Lambda} is an arbitrary approximate unit of I, that a\in A and that \varepsilon>0. There is an element b\in I such that:

\|a+b\|<\|a+I\|+\frac{\varepsilon}{2}.

Since b=\lim_\lambda u_\lambda b, there exists \lambda_o\in\Lambda such that\|b-u_\lambda b\|<\varepsilon/2 for all \lambda\geq\lambda_0, and therefore (triangle inequality after a trick and the dreaded \|1-u_\lambda\|<1 are the first two inequalities):

\|a-au_\lambda\|\leq \|(1-u_\lambda)(a+b)\|+\|b-u_\lambda b\|

\leq \|a+b\|+\|b-u_\lambda b\|

<\|a+I\|+\varepsilon/2+\varepsilon/2.

It follows that \|a+I\|=\lim_\lambda\|a-a_\lambda a\|, and therefore also

\|a+I\|=\|a^*+I\|=\lim_\lambda\|a^*-u_\lambda a^*\|=\lim_\lambda \|a-au_\lambda\| \bullet

Remark

Let I be a closed ideal in a C*-algebra A, and J a closed ideal in I. Then J is also an ideal in A. To show this we need only show that ab\in J and ba\in J whenever a\in A and b\in J^+ (since J is a C*-algebra, J^+ linearly spans J). If \{u_\lambda\} is an approximate unit for I, then b^{1/2}=\lim_\lambda u_\lambda b^{1/2} because b^{1/2}\in I. Hence, ab=\lim_\lambda au_\lambda b^{1/2}b^{1/2}, so ab\in J because b^{1/2}\in Jau_\lambda b^{1/2}\in I, and J is an ideal in I. Therefore a^*b\in J also, so ba\in J, since J is self-adjoint.

Theorem 3.1.4

If I is a closed ideal of a C*-algebra A, then the quotient A/I is a C*-algebra under its usual operations and the quotient norm.

Proof

Let \{u_\lambda\} be an approximate unit for I. If a\in A and b\in I, then

\|a+I\|^2=\lim_\lambda \|a-au_\lambda\|^2

=\lim_\lambda \|(1-u_\lambda)a^*a(1-u_\lambda)\|

\leq \sup_\lambda\|(1-u_\lambda)(a^*a+b)(1-u_\lambda)\|+\lim_\lambda \|(1-u_\lambda)b(1-u_\lambda)\|

\leq \|a^*a+b\|+\lim_\lambda \|b-u_\lambda b\|=\|a^*a+b\|.

The first equality is by the previous theorem (the second is the C*-equation. The first inequality is a trick and a triangle inequality. The second inequality again uses \|1-u_\lambda\|\leq 1 with a bit of submultiplicativity). Therefore, \|a+I\|^2\leq \|a^*a+I\|; so by Lemma 2.1.3 (if a Banach algebra with an involution satisfies \|a\|^2\leq \|a^*a\| then it is a C*-algebra), A/I is a C*-algebra \bullet

Theorem 3.1.5

If \varphi:A\rightarrow B is an injective *-homomorphism between C*-algebras A and B, then \varphi is necessarily isometric.

Proof

It suffices to show that \|\varphi(a)\|^2=\|a\|^2, that is \|\varphi(a^*a)\|=\|a^*a\|. Thus we may suppose that A is abelian (restrict to C(a^*a) if necessary), and that B is abelian (replace B by \overline{\varphi(A)} if necessary). Moreover, by extending \varphi:A\rightarrow B to \tilde{\varphi}:\tilde{A}\rightarrow\tilde{B} if necessary, we may further assume the AB and \varphi are unital.

If \tau is a character on B, then \tau\circ\varphi is one on A. Clearly the map

\varphi':\Phi(B)\rightarrow \Phi(A)\tau\mapsto \tau\circ\varphi

is continuous (as the composition of continuous maps — homomorphisms are always continuous). Hence, \varphi'(\Phi(B)) is compact, because \Phi(B) is compact (Th. 1.3.5: If A is unital and abelian, \Phi(A) is compact and Hausdorff), and therefore \varphi'(\Phi(B)) is closed in \Phi(A). If \varphi'(\Phi(B))\neq \Phi(A), then by Urshohn’s Lemma there is a non-zero continuous function f:\Phi(A)\rightarrow \mathbb{C} that vanishes on \varphi'(\Phi(B)). By the Gelfand representation, f=\hat{a} for some a\in A. Hence, for each \tau\in\Phi(B)\tau(\varphi(a))=\hat{a}(\tau\circ\varphi)=0. Therefore \varphi(a)=0, so a=0 (O.K. — with injectivity of \varphi). But this implies that f is zero, a contradiction. The only way to avoid this is to have \varphi'(\Phi(B))=\Phi(A). Hence, for each a\in A,

\|a\|=\|\hat{a}\|_\infty=\sup_{\tau\in \Phi(A)}|\tau(a)|=\sup_{\tau\in\Phi(B)}|\tau(\varphi(a))|=\|\varphi(a)\|.

Thus \varphi is isometric \bullet

Theorem 3.1.6

If \varphi:A\rightarrow B is a *-homomorphism between C*-algebras, then \varphi(A) is a C*-subalgebra of B.

Proof

The map

A/\text{ker }\varphi\rightarrow Ba+\text{ker }\varphi\mapsto \varphi(a),

is an injective *-homomorphism between C*-algebras and is therefore isometric. It’s image is \Phi(A), so this space is necessarily complete and therefore closed in B \bullet

Theorem 3.1.7

Let B and I respectively be a C*-subalgebra and a closed ideal of in a C*-algebra A. Then B+I is a C*-subalgebra of A.

Proof

We show only that B+I is complete, because the rest is trivial. Since I is complete we need only prove that the quotient (B+I)/I is complete (A/I, I complete implies A complete — a quick calculation). The intersection B\cap I is a closed ideal in B and the map \varphi from B/(B\cap I) to A/I defined by setting \varphi(b+B\cap I)=b+I is a *-homomorphism with range (B+I)/I. By the previous theorem, (B+I)/I is a C*-algebra, and hence complete \bullet

Remark

The map

\varphi: B/(B\cap I)\rightarrow (B+I)/Ib+B\cap I\mapsto b+I,

in the preceding proof is in fact clearly a *-isomorphism.

We return to the topic of multiplier algebras, because we can now say a little more about them using the results of this section.

Suppose that I is a closed ideal in a C*-algebra A. If a\in A, define L_a and R_a\in B(I) by setting L_a(b)=ab and R_a(b)=ba. (L_a,R_a) is a double centraliser on I and the map:

\varphi:A\rightarrow M(A)a\mapsto (L_a,R_a),

is a *-homomorphism. Recall that we identified I as a closed ideal in M(I) by identifying a with (L_a,R_a) if a\in I (We have that \varphi is a homomorphism as

\varphi(a)\varphi(b)=(L_a,R_a)(L_bR_b)

=(L_aL_b,R_bR_a)=(L_{ab},R_{ab})=\varphi(ab).

Also

). Hence, \varphi is an extension of the inclusion map I\mapsto M(I).

If I_1,I_2,\dots,I_n are sets in A, we define I_1I_2\cdots I_n to be the closed linear span of all products a_1a_2\cdots a_n, where a_j\in I. If IJ are closed ideals in A, then I\cap J=IJ. The inclusion IJ\subset I\cap J is obvious. To show the reverse inclusion we need only show that if a is a positive element of I\cap J, then a\in IJ. Suppose then that a\in (I\cap J)^+. Hence a^{1/2}\in I\cap J. If \{u_\lambda\}_{\lambda\in\Lambda} is an approximate unit for I, then a=\lim_\lambda (u_\lambda a^{1/2})a^{1/2}, and since u_\lambda a^{1/2}\in I for all \lambda\in\Lambda, we get a\in IJ, as required.

Let I be a closed ideal in A. We say I is essential if aI=0\Rightarrow a=0. From the previous observations, it is easy to check that I is essential in A iff I\cap J\neq \{\mathbf{0}\} for all non-zero closed ideals J in A (is every element contained in some ideal — how about AaA — but we need a unit for this (and it’s probably not always proper).).

Every C*-algebra I is an essential ideal in it’s multiplier algebra M(I) (a quick calculation).

Theorem 3.1.8

Let I be a closed ideal in a C*-algebra A. Then there is a unique *-homomorphism \varphi:A\rightarrow M(I) extending the inclusion I\rightarrow M(I). Moreover, \varphi is injective if I is essential in A.

Proof

We have already seen above that the inclusion map I\rightarrow M(I) admits a *-homomorphic extension \varphi: A\rightarrow M(I). Suppose that \psi:A\rightarrow M(I) is another such extension. If  a\in Ab\in I, then \varphi(a)b=\varphi(ab)=ab=\psi(ab)=\psi(a)b. Hence (\varphi(a)-\psi(a))I=0, so \varphi(a)=\psi(a), since I is essential in M(I). Thus \psi=\varphi.

Suppose now that I is essential in A and let a\in\text{ker }\varphi. Then aI=L_a(I)=0, so a=0. Thus, \varphi is injective \bullet

Theorem 3.1.8 tells us that the multiplier algebra M(I) of I is the largest unital C*-algebra containing I as an essential closed ideal.

Example

If H is a Hilbert space, then K(H) is an essential ideal in B(H). For if T\in B(H)  such that T \,K(H)=0, then for all x\in H we have that |Tx\rangle\langle x|=T(|x\rangle\langle x|)=0, so Tx=0. By Theorem 3.1.8, the inclusion map K(H)\rightarrow M(K(H)) extends uniquely to an injective *-homomorphism \varphi:B(H)\rightarrow M(K(H)). We show that \varphi is surjective, that is, a *-isomorphism. Suppose that (L,R)\in M(K(H)), and fix a unit vector e\in H^1. The linear map:

T:H\rightarrow Hx\mapsto (L(|e\rangle\langle e|))(e),

is bounded, since

\|Tx\|\leq \|L(|x\rangle\langle e|)\|\leq \|L\|\||x\rangle\langle e|\|=\|L\|\|x\|.

If x,y,z\in H, then

(L_T(|x\rangle\langle y|))(z)=(|Tx\rangle\langle y|)(z)=\langle z,y\rangle (L(|x\rangle\langle e|))(e)

=(L(|x\rangle\langle e|))(\langle z,y\rangle e)=(L(| x\rangle\langle e|))(|e\rangle\langle y|)(z).

Hence, L_T(|x\rangle\langle y|)=L(|x\rangle\langle e|)(|e\rangle\langle y|)=L(|x\rangle\langle y|) for all x,y\in H. Therefore, (\varphi(T)-(L,R))K(H)=0, so \varphi=(L,R).

Thus we may regard B(H) as the multiplier algebra of K(H).

Example

If X is a locally compact Hausdorff space, then it is easy to check that C_0(X) is an essential ideal in the C*-algebra C_b(X). Therefore, by Theorem 3.1.8 there is a unique injective *-homomorphism \varphi:C_b(X)\rightarrow M(C_0(X)) extending the inclusion C_0(X)\rightarrow C_0(X). We show that \varphi is surjective, that is, a *-isomorphism. To see this, it suffices to show that if g\in M(C_0(X)) is positive, then it is in the range of \varphi. If \{u_\lambda\}_{\lambda\in \Lambda} is an approximate unit for C_0(X), then for each x\in X the net of real numbers \{g(x)u_\lambda(x)\} is increasing and bounded above by \|g\|, and therefore it converges to a number, h(x) say. The function:

f:X\rightarrow \mathbb{C}x\mapsto h(x).

is bounded. Moreover, if f\in C_0(X), then hf=gf, since f=\lim_\lambda fu_\lambda. To see that f is continuous, let \{x_\mu\}_{\mu\in M} be a net in X converging to a point x. Let K be a compact neighbourhood of x in X. To show that h(x)=\lim_\mu h(x_\mu), we may suppose x_\mu\in K for all indices \mu ,(there exists \mu_0 such that x_\mu\in K for all indices \mu\geq \mu_0, so, if necessary, replace the net \{x_\mu\}_{\mu \in M} by the net \{x_\mu\}). Use Urysohn’s Lemma to choose a function f\in C_0(X) such that f=1 on K. Since fh\in C_0(X),

h(x)=fh(x)=\lim_\mu f(x_\mu)h(x_\mu)=\lim_\mu h(x_\mu).

Therefore, h is continuous, so h\in C_b(X). For f an arbitrary function in C_0(X) we have

\varphi(h)f=\varphi(hf)=hf=gf,

so (\varphi(h)-g)C_0(X)=0. Consequently g=\varphi(h).

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