** 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 of positive elements in the closed unit ball of such that for all .

## Example

Let be a Hilbert space with infinite orthonormal basis . The C*-algebra is now non-unital. If is the projection onto , then the increasing sequence is an approximate unit for . It will suffice to show that if , since is dense in . Now if , there exist , such that:

.

Hence,

.

Since for all , therefore for each $k$:

.

Hence, .

Let be a C*-algebra and denote by . This set is a poset under the partial order of . In fact, is also upwards directed; that is, if , then there exists such that . Suppose (Murphy actually takes but we can do without this — in particular I want to take to give me the required invertibility: but when .), then ( I’m going to assume it is and take after all because then .). We hence have that . We claim

and (*)

Indeed if , then which implies that , by Theorem 2.2.5, and therefore ; thus proving the claim.

Note that if then . Suppose then that . Let , and . Then , and since , we have that by (*). Similarly and is upwards-directed as claimed.

# Theorem 3.1.1

*Every C*-algebra admits an approximate unit. Indeed, if is the upwards-directed set and for all , then is an approximate unit for (called the *canonical *approximate unit).*

## Proof

We need to show that for each . Since linearly spans (The positive elements span the hermitian elements and the hermitian elements span the C*-algebra.), we can reduce to the case when .

Let and . Let be the Gelfand transformation. If , then is compact, and therefore by Urysohn’s Lemma there is a continuous function of compact support such that for all . Choose such that and . Then (I don’t see this: suppose, roughly, that the supremum is found ‘at’ . Then

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

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 , then and . Now suppose that and . Then , so (Th.2.2.5 (2): we can pre- and post-multiply inequalities by an element and it’s adjoint.). Hence,

(the first inequality here uses submultiplicativity and the C*-equation along with the the result . To see that this is true, not that so that . Therefore . The next equality is just the C*-equation:)

.

(The first and second inequalities here use Th. 2.2.5 (3) () and submultiplicativity.) This shows that

## Remark

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

such that is dense in . Let be any approximate unit for . If , and say, then there exist such that if . Choose such that . Then for all and all . Hence, if and , then there exists such that for all . Also, we can choose the such that for all . Consequently, , for all , and since is dense in , this also holds for all . Therefore is an approximate unit for .

# Theorem 3.1.2

*If is a closed left ideal in a C*-algebra , then there is an increasing net of positive elements in the closed unit ball of such that for all .*

## Proof

Set . Since is a C*-algebra, it admits an approximate unit, say, by the previous theorem. If , then , so

Hence,

,

and therefore

(Where the last inequality uses submultiplicatively — and again requires that . Murphy refers to this afterwards — he states that in this proof we worked in the unitisation of .)

# Theorem 3.1.3

*If is a closed ideal in a C*-algebra , then is self-adjoint and therefore a C^*-subalgebra of . If is an approximate unit for , then for each :*

## Proof

By the previous theorem there is an increasing net of positive elements in the closed unit ball of such that for all . Hence , so , because all of the elements belong to . Therefore is self-adjoint.

Suppose that is an arbitrary approximate unit of , that and that . There is an element such that:

.

Since , there exists such that for all , and therefore (triangle inequality after a trick and the dreaded are the first two inequalities):

.

It follows that , and therefore also

## Remark

Let be a closed ideal in a C*-algebra , and a closed ideal in . Then is also an ideal in . To show this we need only show that and whenever and (since is a C*-algebra, linearly spans ). If is an approximate unit for , then because . Hence, , so because , , and is an ideal in . Therefore also, so , since is self-adjoint.

# Theorem 3.1.4

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

## Proof

Let be an approximate unit for . If and , then

.

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 with a bit of submultiplicativity). Therefore, ; so by Lemma 2.1.3 (if a Banach algebra with an involution satisfies then it is a C*-algebra), is a C*-algebra

# Theorem 3.1.5

*If is an injective *-homomorphism between C*-algebras and , then is necessarily isometric.*

## Proof

It suffices to show that , that is . Thus we may suppose that is abelian (restrict to if necessary), and that is abelian (replace by if necessary). Moreover, by extending to if necessary, we may further assume the , and are unital.

If is a character on , then is one on . Clearly the map

,

is continuous (as the composition of continuous maps — homomorphisms are always continuous). Hence, is compact, because is compact (Th. 1.3.5: If is unital and abelian, is compact and Hausdorff), and therefore is closed in . If , then by Urshohn’s Lemma there is a non-zero continuous function that vanishes on . By the Gelfand representation, for some . Hence, for each , . Therefore , so (O.K. — with injectivity of ). But this implies that is zero, a contradiction. The only way to avoid this is to have . Hence, for each ,

.

Thus is isometric

# Theorem 3.1.6

*If is a *-homomorphism between C*-algebras, then is a C*-subalgebra of .*

## Proof

The map

, ,

is an injective *-homomorphism between C*-algebras and is therefore isometric. It’s image is , so this space is necessarily complete and therefore closed in

# Theorem 3.1.7

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

## Proof

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

## Remark

The map

, ,

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 is a closed ideal in a C*-algebra . If , define and by setting and . is a *double centraliser *on and the map:

, ,

is a *-homomorphism. Recall that we identified as a closed ideal in by identifying with if (We have that is a homomorphism as

.

Also

). Hence, is an extension of the inclusion map .

If are sets in , we define to be the closed linear span of all products , where . If , are closed ideals in , then . The inclusion is obvious. To show the reverse inclusion we need only show that if is a positive element of , then . Suppose then that . Hence . If is an approximate unit for , then , and since for all , we get , as required.

Let be a closed ideal in . We say is *essential* if . From the previous observations, it is easy to check that is essential in iff for all non-zero closed ideals in (is every element contained in *some *ideal — how about — but we need a unit for this (and it’s probably not always proper).).

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

# Theorem 3.1.8

*Let be a closed ideal in a C*-algebra . Then there is a unique *-homomorphism extending the inclusion . Moreover, is injective if is essential in .*

## Proof

We have already seen above that the inclusion map * *admits a *-homomorphic extension *. *Suppose that* *is another such extension. If , , then . Hence , so , since is essential in . Thus .

Suppose now that is essential in and let . Then , so . Thus, is injective

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

## Example

If is a Hilbert space, then is an essential ideal in . For if such that , then for all we have that , so . By Theorem 3.1.8, the inclusion map extends uniquely to an injective *-homomorphism . We show that is surjective, that is, a *-isomorphism. Suppose that , and fix a unit vector . The linear map:

, ,

is bounded, since

.

If , then

.

Hence, for all . Therefore, , so .

Thus we may regard as the multiplier algebra of .

## Example

If is a locally compact Hausdorff space, then it is easy to check that is an essential ideal in the C*-algebra . Therefore, by Theorem 3.1.8 there is a unique injective *-homomorphism extending the inclusion . We show that is surjective, that is, a *-isomorphism. To see this, it suffices to show that if is positive, then it is in the range of . If is an approximate unit for , then for each the net of real numbers is increasing and bounded above by , and therefore it converges to a number, say. The function:

, .

is bounded. Moreover, if , then , since . To see that is continuous, let be a net in converging to a point . Let be a compact neighbourhood of in . To show that , we may suppose for all indices ,(there exists such that for all indices , so, if necessary, replace the net by the net ). Use Urysohn’s Lemma to choose a function such that on . Since ,

.

Therefore, is continuous, so . For an arbitrary function in we have

,

so . Consequently .

## 2 comments

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December 4, 2014 at 2:00 am

momumomuif $(U_{n})$ is an approximate unit for a C^{∗}-algebra A. Is (U_{n})^2$ is an approximateunit for a C^{*}−algebra A? Thank in advance.

December 10, 2014 at 5:15 pm

J.P. McCarthyhttp://math.stackexchange.com/questions/1050761/existence-of-a-approximate-unit-u-n2-for-a-c-algebra/1061102#1061102