Edit: Nope, they aren't.
I didn't realize this until I was writing my talk. I wanted to post last night, but I didn't have time. Indeed, I barely have time now!
In any case, let $\mathcal{U}$ be an $\'{E}tale$ groupoid. We define its convolution algebra as the set of all functions $f: \mathcal{U} \to \mathbb{C}$ with the convolution product
$f_1 \ast f_2 (g) = \sum_{g_1 g_2=g} f_1(g_1) f_2(g_2)$
The problem isn't so strikingly obvious if we let $\mathcal{U}$ be the graph $X \cross X$ of an equivalence relation, e.g., the algebra of the Bost-Connes system. In this case, the convolution is:
$f_1 \ast f_2 (x,y) = \sum_{x \sim r \sim y} f_1(x,r) f_2 (r,y)$
The pairs $(x,r)$ and $(r,y)$ are always going to be swapped at some point, so this product is commutative too.
Does the completion into a $C^{\ast}$-algebra somehow not preserve commutativity? That would just be weird.
In any case, I don't have much more time to keep thinking about it, as I have to run to UCL to give my talk on the Bost Connes system! Can anyone help me out here? It wouldn't damage the analysis of the states on the Bost Connes system if it were commutative, but it would certainly weaken the motivation for using KMS states, et cetera. (Not to mention make my talk seem a bit silly. I suppose if Connes can present in a commutative algebra while giving motivation for a noncommutative one, then so can I!)
No, they aren't! Think of the most simple example of groupoid: a group $G$. You can look at it as the groupoid with only one object and arrows indexed by the elements of the group, i.e. $\mathcal{G}_0 = {\ast\}$ and $\mathcal{G}_1 = G$.
ReplyDeleteThe convolution algebra (forget about topology for now, even assume G is finite so no worries about infinite sums) is the set of functions from the arrows of the groupoid to the complex numbers. In this case it would be the set $\{ f:G \to \mathbb{C} \} $; now, in this set we can give two different algebra structures. If we take pointwise sum and multiplication we just obtain the commutative and boring $G^\mathbb{C}$, direct product of $|G|$ copies of the complex numbers. But if we take the convolution product instead, what one gets is the group-ring $\mathbb{C}[G]$. You can easily prove this identifying each function $f$ on $G$ with the element $\sum_{x\in G} f(x)\cdot x$ in $\mathbb{C}[G]$. Now one can prove that the group algebra is commutative if and only if $G$ is commutative.
For general groupoids you have something similar. SInce every going arrow has the corresponding inverse coming back, one ends up having to deal with all the compositions of a given arrow with things in the isotropy group of the source or target. If the isotropy group is noncommutative (as it is for the BC-system) then the convolution algebra will also be noncommutative. Even if the isotropy group is commutative one often has to deal with a crossed-product action that becomes noncommutative.
Dude, if things were just commutative we'd just use prime/primitive spectrums and work happily with classical geometry rather than undertaking all this major mess! ;-)
An easy example that should convince you: take a set $X=\{a,b\}$ with two elements, consider the total equivalence relation $a\sim b$, and construct the convolution algebra for the equivalence groupoid (without bothering about completion). What algebra do you get?
ReplyDeletethanks for the comments!
ReplyDelete"You can easily prove this identifying each function $f$ on $G$ with the element $sum_{x \in G} f(x)$ in $\mathbb{C}[G]$. Now one can prove that the group algebra is commutative if and only if $G$ is commutative."
That's actually the most helpful statement. I was thinking yesterday of the equiv statement for a group algebra, if $G$ is a finite group, then $\mathbb{C}[G]$ is just the $\mathbb{C}$ algebra with $G$ as a basis and the group operation as multiplication. In that case it's obviously commutative if and only if G is commutative. But I wasn't able to see that in the function description of a groupoid. http://en.wikipedia.org/wiki/Convolution#Properties especially confused me. :)
Glad I could help (sorry it didn't arrive on time for your talk). I personally find easier to think of groupoids as certain kinds of directed graphs and work with a few finite examples to get some intuition. Infinite ones, though more geometrical, make the algebraic properties harder to grasp.
ReplyDelete