Saturday, 8 May 2010

c star algebras - What is the commutative analogue of a C*-subalgebra?

Using the duality between locally compact Hausdorff spaces and commutative $C^*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions regarding its ring of functions $C_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences:
$$
;;;text{open subset of $X$}quad longleftrightarrowquadtext{ideal in $C_0(X)$}
$$
$$
;;;;;quadtext{dense open subset of $X$}quad longleftrightarrowquadtext{essential ideal in $C_0(X)$}
$$
$$
;;;quadtext{closed subset of $X$}quad longleftrightarrowquadtext{quotient of $C_0(X)$}
$$
$$
text{locally closed subset of $X$}quad longleftrightarrowquadtext{subquotient of $C_0(X)$}
$$
$$
;;;quadqquadqquadqquadqquadtext{???}qquadqquad longleftrightarrowquadtext{$C^*$-subalgebra in $C_0(X)$}
$$
By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.



Well, I can't quite see how to reconvert a $C^*$-subalgebra in $C_0(X)$ into something topological involving only the space $X$ (and some data describing the subalgebra in topological terms). Can you come up with something handy?




Example: A simple example of a subalgebra of a commutative $C^*$-algebra not being an ideal is
$$
mathbb Ccdot(1,1)subset mathbb Coplusmathbb C.
$$




First attempts: Instead of talking about a subalgebra, we should probably talk about the injective $^* $-homomorphism given by the inclusion of this subalgebra. But is this inclusion proper (i.e., does it preserve approximate units) in general? Well, at least when we restrict to compact spaces. Then an injective $^* $-homomorphism $C(Y)to C(X)$ will induce a surjective continuous map $Xto Y$. How to proceed?




Remark: Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.




Disclaimer: I posted this question yesterday on MSE. I also got an interesting answer. However, I'm not yet fully satisfied. If I violate any policy by reposting the question here, please tell me about it.

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