# Boolean to C* Algebras

This is something I noticed a while ago, and I don’t seem to be going anywhere with it recently, so I thought I’d blog about it. It’s a cool factoid which may be new (At least one person has muttered something along the lines of ‘I vaguely recall having heard about something like that’, but was unable to give any specifics), but hasn’t really got enough content to be publishable.

What motivated the question was the following:

We have full and faithful functors

$$S : Boolean^{op} to KHauss$$

$$C : KHauss^{op} to C^*$$

The categories of Boolean algebras, Compact Hausdorff spaces and C* algebras respectively. S is the Stone space functor, C is the ‘continuous functions on X’ functor.

Now, in a slight abuse of notation, these combine to give a functor

$$CS : Boolean to C^*$$

In particular, this functor is covariant! It’s an equivalence of categories between the category of Boolean algebras and some subcategory of the category of commutative C* algebras.

So, I asked two questions: What’s the image of this functor, and is there a natural way of realising it? (The Stone functor is fairly nonconstructive, so is in some sense not ‘natural’).

The image of the Stone functor is precisely the class of zero-dimensional compact Hausdorff spaces, and the C functor is an isomorphism of categories, so rephrasing the first question we ask how can we tell if $$X$$ is zero-dimensional from looking at $$C(X)$$?

This turns out to have a nice answer. $$X$$ is zero-dimensional iff $$C(X)$$ is generated by its idempotents (i.e. they generate a dense sub-algebra).

The way to see this is an easy application of the following lemma:

Lemma: Let $$X$$ be compact Hausdorff. $$A subseteq C(X)$$ generates a dense subalgebra iff the usual topology on $$X$$ is the topology generated by $$A$$ (i.e. the coarsest topology with respect to which each element of $$A$$ is continuous). Equivalently, if the set $${ f^{-1}(U) : f in A, U subseteq mathbb{C} }$$, where $$U$$ is restricted to range over the open sets, is a basis.

I won’t bother proving this lemma here. It’s just a vaguely clever application of Stone-Weierstrass + the fact that compact Hausdorff spaces are minimally Hausdorff.

Theorem: Let $$X$$ be compact Hausdorff. $$X$$ is zero-dimensional iff $$C(X)$$ is generated by its idempotents.

Proof:

If $$X$$ is zero-dimensional then the set of characteristic functions of clopen sets generates a dense subalgebra, by the above.

Any idempotent is in fact the characteristic function of a clopen set, so again by the above lemma if the idempotents generate a dense subalgebra then the clopen sets form a basis.

QED

Right. So that’s that sorted.

Some definitions. We’ll (confusingly) call a C* algebra a Boolean C* algebra if it is commutative and generated by its idempotents. The above shows that the category of Boolean algebras is equivalent to the category of Boolean C* algebras. Now we’d just like to construct a more direct equivalence.

It turns out to be much easier to go from the Boolean C* algebras to the Boolean algebras, so we’ll do that first.

Let $$A$$ be a commutative C* algebra. Define $$I(A) = { f in A : f^2 = f }$$

Now, if $$f, g in I(A)$$ then we have $$fg in I(A)$$ and $$1 – f in I(A)$$. So, define the following:

$$f wedge g = fg$$

$$neg f = 1 – f$$

$$f vee g = neg( neg f wedge neg g)$$

So, $$I(A)$$ is closed under these operations. Easy check: They make I(A) into a Boolean algebra.

It’s easy to see that if $$f : A to B$$ is a homomorphism then $$f(I(A)) subseteq I(B)$$. Further this is a Boolean algebra homomorphism. Hence $$I$$ is functorial in a natural way.

So, what we want to do is clear. We want to realise an arbitrary Boolean algebra as $$I(A)$$ of some C* algebra.

We’ll study the structure of $$I(A)$$ a little more before we set off to do so.

First of all, note that if $$f perp g$$ (i.e. $$f wedge g = fg = 0$$) then $$f vee g = f + g$$. This means that we can write any $$f in left< I(A) right>$$ as $$f = sum z_k f_k$$ with $$f_k in I(A)$$ non-zero and $$f_k perp f_l$$ for $$k not= l$$.

This is important for a number of reasons. Among them, because it means that the C* condition fixes the norm on $$left< I(A) right>$$ (Of course, the norm on $$A$$ is uniquely determined, but < I(A) right> will almost never be complete).

We have $$||f||^2 = ||f f^*|| = || sum |z_k|^2 f_k ||$$.

So $$||f|| = || sum |z_k|^{2^n} f_k ||^{2^{-n}} to max |z_k|$$

This formula for the norm is as important as the fact that the norm is determined uniquely. The following consequences in particular will be of crucial importance:

Corollary : Let $$f, g in I(A)$$ be distinct. Then $$||f – g|| = 1$$.

Proof: $$f – g = f(neg g) – g(neg f)$$. These are disjoint, hence $$||f – g|| = max{ 1 } = 1$$.
Theorem: Let $$B subseteq I(A)$$ be a Boolean subalgebra which generates a dense subalgebra of $$A$$ (remember, that’s C* subalgebra, not Boolean). Then $$B = I(A)$$.

Proof:

Suppose not, then we can find some idempotent $$f notin left< B right>$$. Let $$epsilon > 0$$. We can find $$delta > 0$$ such that for $$||f – g|| < delta|| we have ||g^2 - g|| < epsilon$$. $$g in left< B right>$$ with $$||f – g|| < delta$$. Then $$g = sum z_k g_k$$ for $$g_k in B$$ perpendicular. So $$g^2 - g = sum (z_k^2 - z_k)$$. In particular $$||g^2 - g|| = max ||z_k^2 - z_k||$$. So, for every $$k$$ we have $$||z_k^2 - z_k|| < epsilon$$. Now, choose $$epsilon$$ small enough so that this requires that $$z_k < frac{1}{4}$$ or $$z_k > frac{3}{4}$$. Let $$h = sumlimits_{z_k > frac{3}{4}} g_k$$. Then $$h$$ is an idempotent and $$||h – g|| < frac{1}{4}[/tex. Hence [tex]||h - f|| < frac{1}{4} + delta$$. We may choose $$delta$$ to be as small as we like, so let $$delta < frac{1}{4}$$. Then $$||f - h|| < frac{1}{2}$$. But $$f$$ and $$h$$ are both idempotents, and by our previous corollary if they were distinct then we'd have $$||f - h|| = 1$$. Hence $$f = h$$, and so $$f in left< B right>$$.

Hence $$B = I(A)$$
QED

You may not believe it, but we’re now essentially done!

Start with some Boolean algebra $$B$$. Generate the free algebra with $$B$$ as its generators. For $$x in B$$ we’ll denote the corresponding element of the algebra by $$x^C$$. Quotient out by the requirement that $$1^C = 1$$ and $$0^C = 0$$, that $$(x wedge y)^C = x^C y^C$$ and that $$(x vee y)^C = x^C + y^C – xy$$.

Now, for $$f = sum z_k x_k^C$$ define $$f^* = sum z_k^* x_k^C$$.

Again, we can write $$f$$ as a disjoint sum of the $$x_k^C$$. Having done so define $$||f|| = max |z_k|$$.

Easy checks show that this gives an algebra norm which satisfies the C* condition. Thus its completion is a C* algebra. Call this algebra $$C(B)$$
You can quickly see that morphisms between Boolean algebras extend to morphisms between the generated algebras, and so extend to the completion. Thus this is functorial.

Our previous theorem shows that $$B = I(C(B))$$, as $$B$$ is a Boolean subalgebra of $$I(C(B))$$ which generates a dense C* subalgebra of $$C(B)$$.

So, we’re basically done. Some trivial checking of details remains, but I hope I’ve now convinced you that these functors give an equivalence of categories.

But do they do what we originally wanted?

Sure. Requiring that $$C$$ is isomorphic to $$CS$$ (sorry for the overloading of labels) is the same as showing there’s a natural isomorphism from the character space of C(B) to the Stone space of B.

But elements of the character space are just homomorphisms $$f : C(B) to Bbb{C}$$. $$Bbb{C}$$ is $$C(mathbb{2})$$ where $$mathbb{2}$$ is the 2-element Boolean algebra $${0, 1}$$. So these homomorphisms naturally biject with the Boolean algebra homomorphisms $$B to mathbb{2}$$. i.e. the elements of the Stone space. It’s then easy to check that this bijection is a homeomorphism.

Phew. So, that’s all over and done with.

Now, why do we actually care about this? I’m afraid it’s not because this gives us tons of exciting new examples of Boolean algebras or C* algebras – intriguingly it seems like natural examples of one correspond to natural examples of others. This is actually quite surprising, and more than a little cool! It gives us multiple ways to look at existing objects. So far I’ve only found one case where this was actually useful (and it could be done without it), but I’m sure there are more. I’ve not been looking very hard.

Here are some examples:

• If $$B$$ is the Boolean algebra of finite and cofinite subsets of $$mathbb{N}$$ then $$C(B)$$ is the C* algebra of convergent complex valued sequences.
• $$C(P(mathbb{N})) = l^{infty}$$
• If Meas is the Boolean algebra of measurable subsets of $$[0, 1]$$ modulo the ideal of null sets, then $$C(Meas) = L^{infty}[0, 1]$$

This last example is the one I found useful – it’s much more obvious what the Stone space of Meas is than what the character space of L^{infty}[0, 1] is (although once you know what it is it isn’t that hard to prove directly). I needed an example of a commutative C* algebra with an inseparable character space but a separable Hilbert space representation, and this was it.

Questions I’d like to go on to answer:

What’s the noncommutative generalisation of this? You can certainly get a poset out of it, and the poset has an order reversing operator analagous to $$neg$$. Unfortunately I can’t seem to prove that $$I(A)$$ is always a lattice here – it certainly seems to be in the cases I’ve seen. The problem is that it isn’t closed under multiplication of noncommuting elements, so the obvious candidate for $$wedge$$ doesn’t work.

For topological spaces we have an additional way of getting a Boolean algebra out – the regular open sets. This then gives us the notion of the absolute of a space (the stone space of the algebra of regular open sets, which is also the maximal irreducible preimage of $$X$$). Does this correspond to anything nice in the C* algebra setting?

Other stuff I haven’t yet thought of. :-)

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