Stone Čech Compactification And Multiplier Algebras

26 Aug 2022 -

There is an isomorphism theorem of Gelfand, detailed in the next paragraph. It ends up giving an (anti)-equivalence of categories between $C^\ast$-algebras and the category of locally compact haussdorf spaces. It gives rise to the philosophy that $C^\ast$ algebras are really the study of "non-commutative topology". One of the ideas here is that phenomena in topology should have their counterparts in $C^\ast$-algebras. One of the most important examples of this is the extension of $K$-theory to $C^\ast$-algebras. Today, we do something simpler but still very much striking. We talk about compactifications and their analogues, and how we can use this interplay between topology and $C^\ast$-algebras to learn more about compactifications.

The Gelfand IsomorphismPermalink

Recall a Banach space is a normed $\bb{C}$-vector space complete under the induced metric. A Banach algebra in turn is a Banach space with a multiplication, such that $\norm{ab}\leq \norm{a}\norm{b}$. The simplest such objects to study are the abelian Banach algebras. For each unital abelian Banach algebra $A$, take $\Spec(A)$ to be the space of characters, i.e homomorphisms $A\lra{\gamma}\bb{C}$. We can give $\Spec(A)$ the weak*-topology and its easy to check due after applying the Banach-Alaoglu theorem that this is infact a compact space. Now we get a representation $$\Gamma: A \lra{} C(\Spec(A)),\quad \Gamma(a)(\gamma) = \gamma(a)$$ The norm on $C(X)$ for compact haussdorf $X$, the space of continous functions $X\lra{}\bb{C}$, is $\norm{f}_\infty= \mathrm{sup}_{x\in X} |f(x)|$. We call this the Gelfand representation. Note we could have asked for $A$ to be non-unital, and then the character space would have been locally compact haussdorf. We would have to use $C_0(\Spec(A))$ then, the space of continous functions like before but they go to zero at infinity (think about infinity by thinking about the one point compactification.).

A $C^\ast$ algebra is an algebra with an involution $\ast$ such that $\norm{a^\ast a} = \norm{a}^2$. Note that if pointwise complex conjugation is the involution, both $C(Y)$ and $C_0(X)$ are $C^\ast$-algebras.Now the big theorem due to Gelfand is that for commutative $C^\ast$ algebras, the Gelfand representation is an isomorphism. This tells us each non-unital commutative $C^\ast$-algebra is dual to some LCH space, and that every unital commutative is dual to some compact haussdorf space. A lot of $C^\ast$ theory now is to see how much of this isomorphism theorem we can translate over to the non-commutative case. So in a sense $C^\ast$-algebra is the study of non-commutative topology. With that being said, we expect concepts in topology to carry over to $C^\ast$-algebras and vice versa. We will see that for compactification today.

I should mention that this really is an (anti-)equivalence of categories between commutative $C^\ast$ algebras and the category of LCH spaces with continuous maps. Send a space to its $C_0$ and a commutative algebra to its spectrum.

Unitizations and CompactificationsPermalink

We shall say compact $Y$ is a compactification of LCH $X$ (non-compact) if

The easiest example of this is $\opc{X}$, the one point compactification. This is just adding a single point to $X$, call it $\infty$, to make it compact. We can also write it by its universal property, that any compactification of $X$ maps into $\opc{X}$, i.e $\opc{X}$ is the "smallest" compactificaiton.

As mentioned before, the philosophy is that every topological concept should have an analogue in operator theory. So lets look at $C(\opc{X})$ towards that end. Notice that we have a map $C(\opc{X})\lra{\cong} C_0(X) \oplus \bb{C}$ sending $f\mapsto ((f-f(\infty))\mid_{X},f(\infty))$. It is easy to see this is an isomorphism. So adjoining a unit like this is the analogue of the one-point compactification. Lets try to find the analogues for compactifications in general. There is a unique norm on each $C^\ast$-algebra that satisfies the $C^\ast$ identity, I leave it as an exercise to figure out the norm on $C_0(X)\oplus \bb{C}$ making it a $C^\ast$-algebra.

Let $X\lra{\iota}Y$ be a compactification of $X$. Consider $f\in C_0(X)$, then it extends to a $\opc{f} \in C(\opc{X})$. We can combine the map $Y\lra{}\opc{X}$ given by the universal property of $\opc{X}$ and $\opc{f}$ to get an element of $C(Y)$.

A unitization of a $C^\ast$-algebra $A$ is:

I have already shown compactifications lead to unitizations.I leave it as an exercise to show the converse.

Multiplier Algebras.Permalink

The Stone-Čech compactification of a LCH space $X$, $\beta X$, is the biggest compactification of $X$. As we saw above, we expect this to correspond to the biggest unitization of a non-unital $C^\ast$-algebra $A$. The advantage of thinking in these terms is that for $C^\ast$ algebras this is quite explicit. Indeed, the biggest unitization of $A$, called $M(A)$(its multiplier algebra) is given as follows:

$M(A)$ is the set of double centralizers $(L,R)$ on $A$. This means:

Suppose $B$ contains $A$ as an ideal. Then notice we have a map $\psi:B\lra{}M(A)$, by sending $c\mapsto (L_c,R_c)$ again and using the fact that $A$ is an ideal. This extends the cannonical map $A\lra{} M(A)$. This is infact unique, for say $\phi:B\lra{} M(A)$ also extended it. Then for $b\in B, a\in A$, we would have $$\psi(b) (L_a,R_a) = \psi(ba) = (L_{ba},R_{ba}) =\phi(ba) = \phi(b) (L_a,R_a)$$ So that $(\phi(a)-\psi(a)) A =0$ in $M(A)$, but $A$ is essential in $M(A)$ so $\phi(a)=\psi(a)$. Finally suppose $A$ is essential in $B$, then if $b\in \ker(\psi)$ then $L_b(A) = bA=0$, so that $b=0$. Hence the map is injective.

To summarize, whenever $B$ contains $A$ as an essential ideal, $B$ has a unique embedding into $M(A)$. So $M(A)$ is the biggest unitization as promised. We will get to the commutative case in the next section, but a good example for now is that if $A=K(H)$, compact operators on some hilbert space, then $M(A) = B(H)$, bounded operators on that hilbert space.

Stone-Čech CompactificationPermalink

The Stone-Čech Compactification, $\beta X$,is the biggest compactification of a space, i.e if $Y$ is a compactification than $\beta X$ uniquely maps into $Y$ making the obvious diagram commute. This is literally the dual condition of the multiplier algebra, so we just need to figure out what $M(C_0(X))$ is. Luckily for us, the answer is easy: $C_b(X)$, bounded continuous functions on $X$ with sup norm. One can easily check that $C_0(X)$ is essential in $C_b(X)$, so that there is a injective map $\psi: C_b(X) \lra{} M(C_0(X))$. All we need to do is show surjection, which is easily done but the proof I know is a bit annoying and uses some other tools (like approximate units). I will hence refer the user to Gerard Murphy's book on the topic, page 83.

By the Gelfand Isomorphism, we know that $C_b(X)\cong C(\beta X)$ for some $\beta X$, and by the universal property discussed before, this $\beta X$ is indeed the Stone-Čech compactification of $X$. This proves that such a compactification exists in the first place but also gives a very nice explicit form for what the space should be.