6
$\begingroup$

Dictionary between algebra and geometry is somewhat one of the main concepts in modern mathematics. So commutative $C^*$ algebras are one-to-one with locally compact Hausdorff spaces. So it is natural to be curious how one can see properties of the manifold from algebra of functions. In the realm of algebraic functions we can easily characterize algebraic functions on $C^n$ (i.e. $C[x_1,...,x_n]$ ) as free commutative algebra with $n$-generators.

Question: So I wonder is there any (hopefully simple) characterization of $R^n$ in $C^*$ world ? I.e. can one somehow characterize the $C^*$-algebra of continuous functions on $R^n$ in an alternative way ?

Or may be the question for $S^n$ would be simpler, since it is compact.

The answers given below clarify the question a lot for me. Let me stress what I found somehow surprising for myself.

The natural description for $C^*$-algebra of $R^n$ would be just the same as in algebraic geometry setup - commutative $C^*$-algebra freely generated by $x_1,...x_n$... But this does not work because $x_i$ are unbounded functions on R^n and hence they do not fit to $C^*$ setup. So it is probably natural to extend class of $C^*$-algebras to include some unbounded operators, rather than ask about "simple" $C^*$-description of R^n.

I had a misunderstanding that there is NO description of $C^*$-algebras by generators and relations. Indeed, usually we take free algebra and factorize it over relations, but in $C^*$-world free algebra is NOT $C^*$ in any reasonable sense. Other way to describe $C^*$-algebra by generators and relations would be to define appropriate completion of the algebra of polynomials in generators - but this also does not seem to work since: how to describe norms of |a+b| in terms of a,b ? how to describe what kind of power series in generators belong to completion - this seems cannot be done explicitly, it is like one to describe continuous functions in terms theirs Fourier series, which " there is no way to characterize the Fourier series of continuous functions by means of a naive "sequence space" condition on the sequence. "

So it is somehow surprising for me that there is such a simple way to define $C^*$ by generators and relations - just say "it is universal $C^*$-algebra with such relations"... That seems to be an example where abstract thinking does seem to have a way round... Well not all the relations will work, e.g. ab-ba=1 will not - see MO151809, but that is another story.

Okay, thanks for answers.

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ Note that in the world of C*-algebras there is no such thing as e.g. the free C*-algebra on a generator. This is because the "universal element" of such a C*-algebra has some fixed norm, and so cannot map to elements with larger norm. There is such a thing as the free C*-algebra on a generator of norm $\le r$, which is the C*-algebra of continuous functions on the closed disk of radius $r$ in $\mathbb{C}$. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 29, 2016 at 20:29
  • 2
    $\begingroup$ @QiaochuYuan: I think you mean "on a normal generator of norm $\leq r$". Without this qualification you get the universal contraction algebra. $\endgroup$
    – Nik Weaver
    Commented Jan 29, 2016 at 22:54
  • $\begingroup$ Right. I had in mind the commutative case. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 29, 2016 at 23:47

2 Answers 2

Reset to default
10
$\begingroup$

Yes, it can be defined as the univeral commutative $C^*$-algebra with unit, generated by $n+1$ self adjoint elements $x_1,...,x_{n+1}$ subject to the relation $x_1^2+...+x_{n+1}^2=1$. Here universal mean the following: $A$ with generators $(a_j)_j$ and relations $(r_k)_k$ is called universal if whenever there is $C^*$-algebra $B$ generated by $(b_j)_j$ (the same indexing set) and satysfying relations $(r_k)_k$ then there is an epimorphim $\varphi:A \to B$ such that $\varphi(a_j)=b_j$. The problem with existence of such universal algebras is because some relations don't impose any restriction on norms of elements. For example there is no universal unital $C^*$-algebra generated by a single self adjoint element but there is universal unital $C^*$-algebra generated by a single unitary element (this algebra is in fact $C(S^1)$). However you can prove that if there is any $C^*$-algebra $B$ with generating set $S=(b_s)_s$ such that $\|b_s\| \leq 1$ and those elements satisfy some relations $(r_j)_j$ then there is a universal $C^*$-algebra $A$ with generating set $(a_s)_s$ (the same cardinality as $S$) where $(a_s)_s$ satisfy relations $r_j$. So this difficulty concerning the norms is somehow the only obstruction

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much very helpful answer ! May be it is worth to link Universal C*-algebra en.wikipedia.org/wiki/Universal_C*-algebra if you consider it well enough written. $\endgroup$
    – Alexander Chervov
    Commented Jan 31, 2016 at 9:59
8
$\begingroup$

Podles defined quantum 2-spheres in such a universal way. In a special case they restrict to $C(S^2)$. The description in this case is: the universal unital C*-algebra generated by operators $A$ and $B$ satisfying

  • $A^* = A$
  • $AB = BA$
  • $BB^* = B^*B = I - A^2$.

The generalization to higher dimensions can be found in Section 2 of this paper.

In the case of $\mathbb{R}^n$, remember that $\mathbb{R}^n$ is homeomorphic to $S^n$ minus a point. So $C_0(\mathbb{R}^n)$ is realized as (any) maximal ideal of $C(S^n)$.

Improve this answer
$\endgroup$

You must log in to answer this question.