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Let $X$ be a compact Hausdorff space. It is well-known that every homomorphism $F : \mathcal{C}(X) \to \mathbb{R}$ is the evaluation $f \mapsto f(x)$ at some point $x \in X$. The usual proof is not really constructive, but for $X=[0,1]$ there is a constructive one. For details see my crosspost on math.SE. Feel free to replace $\mathbb{R}$ by $\mathbb{C}$.

Question. Is there an explicit example of $X$ and $F : \mathcal{C}(X) \to \mathbb{R}$ as above such that 1) syntactically $F$ is not defined as an evaluation, 2) one does not see directly that $F$ is an evaluation, 3) some computation has to be done to find the point $x \in X$ such that $F$ is the evaluation at $x$?

Background: Gelfand duality states that the adjunction between $\mathrm{Spec}$ and $\mathcal{C}$ is actually an equivalence, which means that A) for every compact Hausdorff space $X$ the unit $X \to \mathrm{Spec}(\mathcal{C}(X))$ is an isomorphism and B) for every commutative unital $C^*$-algebra $A$ the counit $A \to \mathcal{C}(\mathrm{Spec}(A))$ is an isomorphism. There are many important applications for B), for example the existence of the functional calculus, but I don't know of any specific application for A) (as a result independent from this duality). It would be nice to have some computational example which shows the relevance of A). Actually I'm after another duality, where A) is already proven but its significance is unclear.

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    $\begingroup$ The classical statement of Gelfand-Neumark duality can be split into two parts. The first part is Gelfand-Neumark duality proper and asserts a contravariant equivalence between the categories of commutative C*-algebras and compact regular locales. It can be proved in any elementary topos, in particular the proof is constructive: ncatlab.org/nlab/show/constructive+Gelfand+duality+theorem. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 21, 2012 at 12:05
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    $\begingroup$ The second part asserts that compact regular locales are spatial, in particular the category of compact regular locales is equivalent to the category of compact Hausdorff topological spaces. This statement is equivalent to a weak form of the axiom of choice. Your question belongs to the second part, but the connection to Gelfand-Neumark duality seems to be rather weak. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 21, 2012 at 12:14
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    $\begingroup$ However, perhaps this example will do. Consider a locally compact abelian group G and some character χ. Integrating with respect to χ gives you a multiplicative functional (with respect to the convolution of functions, not the usual multiplication). On the first glance it is unclear why this functional is evaluation at some point. This is only revealed by the Fourier transform. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 21, 2012 at 12:20
  • $\begingroup$ (@DP, Why not to post as an answer something wich is an answer?) $\endgroup$
    – Qfwfq
    Commented Jan 21, 2012 at 13:35
  • $\begingroup$ @Qfwfq: I have doubts whether my comment can be seen as an answer. In particular, conditions (2) and (3) are rather vague and I am not sure how to interpret them. Let's see what Martin has to say. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 21, 2012 at 13:52

2 Answers 2

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Let $\mu$ be a Borel measure on $X$ that satisfies a zero-one law (i.e. $\mu$ takes only values $0,1$) and has $\mu(X) = 1$. Then $$F: C(X) \to \mathbb{R},\; f \mapsto \int_X f\; d \mu$$ defines a ring homomorphism.

I'm not sure, if there are not even examples for $(X, \mu)$ such that $F$ isn't an evalution (of course in this case $X$ can't be a Q-space). At least, there are such examples for Baire measures. [Edit: See "Added 2" for an affirmative answer]

Added: To give an example, let $\omega_1$ be the first uncountable ordinal and let $X := [0,\omega_1]$ be the set of all ordinals $0 \le \alpha\le\omega_1$ considered as topological space with the order topology. Then $X$ is a compact Hausdorff space. Futhermore, it can be shown that if $B$ is a Borel set, then $$\mu(B) := \begin{cases} 1, \text{ if } B \textstyle \text{ has an unbounded, closed subset of } X \setminus \lbrace \omega_1 \rbrace \newline 0, \textstyle\text{ otherwise}\end{cases}$$ defines a Borel measure on $X$ (cf. Halmos, Measure Theory, Exercise 52.10).

Since $\mu$ is zero on finite sets, it's obviously no Dirac measure.

Added 2: The example also shows that the integral operator for a $\lbrace 0,1 \rbrace$-valued Borel measure is in general no evaluation:

Let $X_0 := [0,\omega_1) \subseteq X$. Then $\mu_0 := \mu|X_0$ is a $\lbrace 0,1 \rbrace$-valued Borel measure on $X_0$ with $\mu_0(X_0) = 1$ and
$$F_0: C(X_0) \to \mathbb{R},\; f \mapsto \int_{X_0} f\; d\mu_0$$ is a ring homomorphism that is no evaluation.

For, let $x \in X_0$ and set $f(\alpha) := 1$ if $\alpha \le x$, $f(\alpha) := 0$ if $\alpha > x$. $f$ is continuous and since $\mu([\alpha_0+1,\omega_1)) = 1$, $f = 0$ almost everythere. Hence $F_0(f) = 0$ but $e_x(f) = f(x) = 1$. Consequently there is no $x \in X_0$ such that $F_0 = e_x$.

Also note that if $X$ is compact, then, by the Riesz representation theorem, the only non-zero Radon measures with values in $\lbrace 0,1 \rbrace$ are the Dirac measures.

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  • $\begingroup$ Can you give some specific example, especially where $\mu$ isn't a dirac measure? $\endgroup$
    – Martin Brandenburg
    Commented Mar 3, 2012 at 8:49
  • $\begingroup$ Space $\mathbb N$, the zero-one finitely additive measure is an ultrafilter. They define homomorphisms, and the "points" corresponding to them are the points of the Stone-Cech compactification $\beta \mathbb N$. So: it is not an evaluation to start with, but then we "invent" the points to make it so. $\endgroup$
    – Gerald Edgar
    Commented Mar 3, 2012 at 12:39
  • $\begingroup$ As Ralph hints, when the space $X$ is not a Q-space (or, in other language, not realcompact) there is a countably-additive example of a zero-one Baire measure that is not fixed. The set of such things constitutes $\upsilon X$, the Hewitt realcompactification of $X$. $\endgroup$
    – Gerald Edgar
    Commented Mar 3, 2012 at 12:42
  • $\begingroup$ Ok, but isn't this just a fancy way to state the old result that prime ideals of a product of fields, indexed by a set $N$, correspond to ultrafilters on $N$? $\endgroup$
    – Martin Brandenburg
    Commented Mar 3, 2012 at 16:00
  • $\begingroup$ @Ralph: Thanks for the update! Do we have $F(f)=f(\omega_1)$ in your example? $\endgroup$
    – Martin Brandenburg
    Commented Mar 3, 2012 at 21:17
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Consider a locally compact abelian group G and some character χ: G→T. The functional Mχ: f∈L1(G)↦∫fχ∈C is multiplicative: Mχ(f*g)=Mχ(f)Mχ(g), where f*g is the convolution of f and g. At the first glance it is unclear why this functional should be given by the evaluation at some point. However, Pontryagin duality tells us that the Fourier tranform f↦(χ∈Hom(G,T)↦Mχ(f)∈C) (with the appropriate domain and codomain) is an isomorphism of algebras, where the first algebra structure is given by the convolution and the second algebra structure is given by the pointwise multiplication.

The easiest concrete example is G=T, whose Pontryagin dual is Z, and χ=idG, or perhaps even χ=1.

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  • $\begingroup$ As someone who learned about Banach algebras before learning about Pontryagin duality, my way of looking at this would be that the Gelfand transform and associated theory gives us the canonical homomorphism $L^1(G) \to C_0(\Omega)$ for suitable $\Omega$, and then some analysis will be required to show that this homomorphism is injective. Is that what you are saying, or have I misunderstood? The power of the Gelfand representation is that it tells us we can view any non-zero multiplicative linear functional on a commutative (Banach) algebra as point evaluation, in a natural (pun intended) way $\endgroup$
    – Yemon Choi
    Commented Jan 22, 2012 at 2:51
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    $\begingroup$ Besides, my question asks about $C(X)$. $\endgroup$
    – Martin Brandenburg
    Commented Jan 22, 2012 at 17:54
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    $\begingroup$ @Martin: Or in other words, to define a functional on C(X) compose the inverse Fourier transform with M_χ. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 22, 2012 at 19:23
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    $\begingroup$ @Dmitri: for a given commutative Banach algebra, one may need to do some work to show that the Gelfand representation is faithful. For a commutative B*-algebra (defined as an abstract algebra, not presuming the GN-theorem) I think this is not too hard, but for something like $L^1(T)$ you in effect need the fact that an integrable function whose Fourier coefficients are all zero is itself zero, and that is usually deduced from something like FEjer's theorem $\endgroup$
    – Yemon Choi
    Commented Jan 22, 2012 at 20:07
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    $\begingroup$ I guess the mention of $L^1(G)$ is superfluous, what you are doing is taking a character of $G$ and saying that it gives you a nonzero multiplicative linear functional on the full group C*-algebra $C^*(G)$, via the universal property of the full group $C^*$-algebra construction. Then GN-duality kicks in, to say that this functional can be viewed naturally as point evaluation. But since you started with an element of $\widehat{G}$, rather than discovering it from the functional $C^*(G)\to C$, this doesn't seem quite what Martin was asking for. $\endgroup$
    – Yemon Choi
    Commented Jan 23, 2012 at 3:16

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