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Dear fellows,

I've stuck on a step of proposition 1.2 of Rieffel's article (continuous field of C*-algebras coming from group cocycles and actions, 1989). I think it basically proves that a C(X)-algebras coming from a locally convex Hausdorff space X is upper semicontinuous, but I can only see that happening for the compact case. The question is simple, but let me elaborate on the background.

As definition, of a $C(X)$-algebra, for $X$ compact Hausdorff, it is just a C*-algebra $A$ with a unital $*$ -homomorphism $C(X)$ to $ZM(A)$ (center of its multiplier algebra), so it can be though as the Gelfand transform of a C*-subalgebra of $ZM(A)$ containing its identity. Since we can take non-unital C*-subalgebras of $ZM(A)$, it is also interesting to take locally compact but non compact $X$'es. (I think you can check more on that by googling "C(X)-algebras", apparently Kasparov uses this structure in his KK-theory, but I know nothing about that).

So let $C_0(X)$ be $*$ -isomorphic to a C*-subalgebra of $ZM(A)$. For $x\in X$, take $C_0(X,x)$ as the ideal in $C(X)$ corresponding to functions vanishing at $x$. Then $J_x$ defined as the closure of the linear span generated by $C(X,x)A$ is a closed ideal in $A$. Let $q_x$ be the quotient map from $A \to A_x = A/J_x$. This gives rise to a family of C$^*$-algebras $\{A_x\}_{x\in X}$ over $X$.

Prop. 1.2 of the reffered paper shows that $A$ is upper semicontinuous. This means that, for all $a\in A$ the map $x \mapsto \| q_x(a) \|_{A_x}$ is upper semicontinuous. This follows from the characterization of the quotient norm.

There is a vector of the form $ b= \sum_{i=1}^n f_i b_i \in J_x$ (with $f_i(x)=0$ for all $i$) such that $\| a + b\|_A < \| q_x(a) \|_{A(x)}$. Now, since

Now, every $f_i$ zeroes at $x$, so we can pick a function $g\in C_0(X)$ such that $g$ is one on a small neighborhood $U$ of $x$ and zero outside a small neighborhood of $U$, and such that $\|g b\|<\epsilon$. Then, for all $y\in U$, we have that $(1-g) \in C(X,y)$ [!] and since $(1-g)f \in J_x$, we get $\| g a \|_A = \| a - (1-g)a \|_A > \|q_y(a)\|_{A_y}$. Thus

$ \|q_x(a)|_{A_x} > \|g\|\,\|a+b\| - \epsilon > \|ga\| + \|gb\| + \epsilon > \| q_y(a) \|_{A_y} + 2\epsilon.$

This proves the upper semicontinuity.

NOW HERE IS THE PROBLEM, as we have seen, this is fine for $X$ compact, but if not, I can't see why the constructed $(1-g)a$ is in the ideal $J_y$, since $1$ is an element of the multiplier algebra $M(A)$ and $(1-g)$ would not be an element of $C_0(X)$. Any thoughts on that? Is it true that ANY $C(X)$-algebra is upper semicontinuous then?

Also, why should we take, in usual definitions, the $C_0(X)$ is embedded in ZM(A) instead of $C_b(X)$? Otherwise, the constant fields in $C_0(X,A)$ would not be continuous fields... does that any make any sense?

flag	asked Jun 17 at 20:44 Yul Otani 122●1●5

	The spaces $C_0(X,x)$, $C_0(X)$, and $C(X,x)$ above are all the same, aren't they? – Pietro Majer Jun 18 at 4:39
	Hello Mr. Majer, thank you for your comment. Seems like I wasn't clear enough in the definitions. For any point $x$ of $X$, I call $C_0(X,x)$ the (closed ideal) set of (complex valued) function $f\in C_0(X)$ such that $f(x)=0$ (at the particularly defined $x$). This is, of course, not equal to $C_0(X)$. If we take $C(X,x)$ in the same fashion, as the set of functions $f\in C(X)$ such that $f(x)=0$, then $C_0(X,x)$ and $C(X,x)$ are equal iff $X$ is compact. Thank you. – Yul Otani Jun 18 at 7:17

Nik Weaver · Accepted Answer · 2012-06-17 22:26:18Z UTC

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You can reduce to the compact case by considering the unitization of $C_0(X)$ . Since $ZM(A)$ is unital, if $C_0(X)$ embeds in $ZM(A)$ then so does its unitization $C(X^*)$ where $X^*$ is the one-point compactification of $X$ . If the map $x \mapsto \|q_x(a)\|$ is semicontinuous on $X^*$ then it is semicontinuous on $X$ .

answered Jun 17 at 22:26

Nik Weaver
4,426●1●7●20

Dear Mr. Weaver, thank you for your help! I'll dig in the proof of what you said and look for basic references, since I am illiterate on the subject :( Anyways, if you have other comments on to why take C_0 and not C_b, I should pay attention! Cheers and thanks again! – Yul Otani Jun 18 at 7:12

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I guess people usually stay away from $C_b(X)$ because it's so big (e.g., nonseparable). Remember that $C_b(X)$ is the same as $C(\beta X)$ , the continuous functions on the Stone-Cech compactification of $X$. – Nik Weaver Jun 18 at 14:33

Hello! Thank you again for the comment. I have finally given it some look! I have some considerations. I think that if we evaluate the $x\mapsto \|q_x(a)\|$ defined on $X^*$, the problem is that the ideals and quotients are different to those considering $C_0(X)$ (also, the extension of the injection may not be unique if $C_0(X)$ is degenerate in $ZM(A)$). However, we can simply pick $g$ in $C(X^*)$ instead of in $C_0(X)$, such that all the wanted properties hold, since we only need that $1-g$ to be in $C_0(X)$ (we take the unital extension of the injection $C(X^*)$ in $ZM(A)$). – Yul Otani Jul 22 at 20:56

upper semicontinuity in C(X)-algebras

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