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?

  • $\begingroup$ The spaces $C_0(X,x)$, $C_0(X)$, and $C(X,x)$ above are all the same, aren't they? $\endgroup$ Jun 18, 2012 at 4:39
  • $\begingroup$ 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. $\endgroup$
    – Yul Otani
    Jun 18, 2012 at 7:17

1 Answer 1


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 $X$.

  • $\begingroup$ 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! $\endgroup$
    – Yul Otani
    Jun 18, 2012 at 7:12
  • 1
    $\begingroup$ 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 $X$. $\endgroup$
    – Nik Weaver
    Jun 18, 2012 at 14:33
  • $\begingroup$ 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 $ZM(A)$). $\endgroup$
    – Yul Otani
    Jul 22, 2012 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.