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_m$$y_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 J$$au_\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 $A$$B$ 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 B$$a+\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)/I$$b+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 $I$$J$ 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 A$$b\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 H$$x\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 $g=\varphi(h)$.