11
$\begingroup$

Does the algebra of continuous functions from a compact manifold to $\mathbb{C}$ satisfy any specific algebraic property?

I'm not sure what kind of algebraic property I expect, but I feel that because of the Gel'fand transform, it may not be unreasonable to expect something. We can drop the compactness condition if we switch to continuous functions to $\mathbb{C}$ that vanish at infinity.

I'm really hoping for some necessary and sufficient condition, but if anybody knows of any sort of condition, that would be appreciated.

Improve this question
$\endgroup$
15
  • $\begingroup$ Check out: mathoverflow.net/questions/5344/… $\endgroup$
    – Steven Gubkin
    Commented Apr 21, 2010 at 15:27
  • $\begingroup$ @Steven: so far, on MO only $C^{\infty}(M)$ has been discussed, not $C(M)$, as in my question mathoverflow.net/questions/21168/… $\endgroup$
    – Martin Brandenburg
    Commented Apr 21, 2010 at 15:41
  • 1
    $\begingroup$ @Eric: so you're asking for a condition of a commutative $C^*$-algebra (in their language) whose spectrum is a manifold? $\endgroup$
    – Martin Brandenburg
    Commented Apr 21, 2010 at 15:42
  • $\begingroup$ @Martin: yes, that's right. $\endgroup$
    – Eric
    Commented Apr 21, 2010 at 16:02
  • $\begingroup$ @Steven and Martin: thanks for the references. I thought I remembered similar questions, but I couldn't find them for the life of me. I feel like it may be a bit simpler to get some condition on the C^*-algebra of continuous functions to C rather than the R-algebra of smooth functions to R simply because of the Gel'fand transform, but this may be a misplaced suspicion. $\endgroup$
    – Eric
    Commented Apr 21, 2010 at 16:13

1 Answer 1

Reset to default
4
$\begingroup$

I found a reference for a necessary property that might be called algebraic.

Browder proved a theorem relating the number of generators of a complex commutative Banach algebra to the Čech cohomology with complex coefficients of the maximal ideal space, and as a corollary concluded that if $M$ is a compact orientable $n$-dimensional manifold, then $C(M)$ cannot be generated as a Banach algebra by fewer than $n+1$ elements. The paper is very short, but for an even shorter summary here's the MR review.

Just some comments, added later:

One obtains the compact Hausdorff space $X$ (up to homeomorphism) from $C(X)$ by considering the maximal ideal space of $C(X)$ with Gelfand topology, but clearly you want something less tautological than "the maximal ideal space is a manifold." A small step in this direction would be to try to formulate the topological properties of $X$ in terms of the closed ideals of $C(X)$. As alluded to in Qiaochu's comment, there is an analogue of Nullstellensatz: each closed ideal in $C(X)$ consists of all functions vanishing on a (uniquely determined) closed subset of $X$. So for example, the locally Euclidean property could be reformulated for a commutative C*-algebra $A$ as follows: There is an $n$ such that for every maximal ideal $M$ of $A$ there is a closed ideal $I$ of $A$ such that $I$ is not contained in $M$ and $I$ is $*$-isomorphic to $C_0(\mathbb{R}^n)$. Second countability of the maximal ideal space is equivalent to $A$ being separable in the norm topology; that's not algebraic, but might be considered more intrinsic to the C*-algebra.

But this only leads to another, more specific question: Is there a useful or interesting (C*-)algebraic characterization of $C_0(\mathbb{R}^n)$?`

Improve this answer
$\endgroup$
9
  • $\begingroup$ If I remember rightly from having skimmed over that paper en passant a few years ago, an important point is that one gets $n$ rather than $2n$? Or have I misremembered? $\endgroup$
    – Yemon Choi
    Commented Apr 22, 2010 at 18:02
  • $\begingroup$ Browder emphasizes that if we were talking about real-valued generators then the corresponding result would be trivial, but for some compact Hausdorff spaces fewer generators are required in the complex case. So the point seems to be that the lower bound of $n+1$ had previously not been known, even in the case of $S^2$ as Browder (and Rudin in the review) point out. I'm not sure exactly what you mean about $2n$, but I might just be slow. $\endgroup$
    – Jonas Meyer
    Commented Apr 22, 2010 at 18:44
  • $\begingroup$ Oh, I'm sure you're not being slow and I'm just going prematurely senile;) What you describe in your comment sounds familiar -- I just remembered that there was something to do with posssibly, a priori, having different answers in the real and complex cases. $\endgroup$
    – Yemon Choi
    Commented Apr 22, 2010 at 21:50
  • $\begingroup$ @Jonas: Thanks for the paper, it was interesting. While it fits my requirements of an algebraic property (well, C^*-algebraic, but we have no other choice in that), I think I'm going to wait and see what other interesting things crop up :). $\endgroup$
    – Eric
    Commented Apr 23, 2010 at 1:41
  • $\begingroup$ @Eric: For what it's worth, it's better than C*-algebraic; it's Banach algebraic. E.g., $C(S^1)$ is singly generated as a C*-algebra, but requires 2 generators as a Banach algebra. I was tentative about posting, hence the "might be called algebraic" and the community wiki, but I'm glad you find it interesting. $\endgroup$
    – Jonas Meyer
    Commented Apr 23, 2010 at 4:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .