Realize a homomorphism $\mathcal{C}(X) \to \mathbb{R}$ as an evaluation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:55:28Z http://mathoverflow.net/feeds/question/86296 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86296/realize-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation Realize a homomorphism $\mathcal{C}(X) \to \mathbb{R}$ as an evaluation Martin Brandenburg 2012-01-21T11:00:18Z 2012-03-04T23:15:59Z <p>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 <a href="http://math.stackexchange.com/questions/97603/realizing-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation" rel="nofollow">crosspost</a> on math.SE. Feel free to replace $\mathbb{R}$ by $\mathbb{C}$.</p> <p><strong>Question.</strong> 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 $x$?</p> <p>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.</p> http://mathoverflow.net/questions/86296/realize-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation/86343#86343 Answer by Dmitri Pavlov for Realize a homomorphism $\mathcal{C}(X) \to \mathbb{R}$ as an evaluation Dmitri Pavlov 2012-01-22T01:00:49Z 2012-01-22T01:00:49Z <p>Consider a locally compact abelian group G and some character χ: G→<strong>T</strong>. The functional M<sub>χ</sub>: f∈L<sub>1</sub>(G)↦∫fχ∈<strong>C</strong> is multiplicative: M<sub>χ</sub>(f*g)=M<sub>χ</sub>(f)M<sub>χ</sub>(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,<strong>T</strong>)↦M<sub>χ</sub>(f)∈<strong>C</strong>) (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.</p> <p>The easiest concrete example is G=<strong>T</strong>, whose Pontryagin dual is <strong>Z</strong>, and χ=id<sub>G</sub>, or perhaps even χ=1.</p> http://mathoverflow.net/questions/86296/realize-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation/90097#90097 Answer by Ralph for Realize a homomorphism $\mathcal{C}(X) \to \mathbb{R}$ as an evaluation Ralph 2012-03-03T01:17:28Z 2012-03-04T23:15:59Z <p>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. </p> <p>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]</p> <p><strong>Added:</strong> 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). </p> <p>Since $\mu$ is zero on finite sets, it's obviously no Dirac measure. </p> <p><strong>Added 2:</strong> The example also shows that the integral operator for a $\lbrace 0,1 \rbrace$-valued Borel measure is in general no evaluation: </p> <p>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<br> $$F_0: C(X_0) \to \mathbb{R},\; f \mapsto \int_{X_0} f\; d\mu_0$$ is a ring homomorphism that is no evaluation. </p> <p>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 $F_0 = e_x$. </p> <hr> <p><strong>Also note</strong> 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. </p>