Last time we saw that the algebra $(\OmegaV~C Q,Circ)$ of relative differential forms and equipped with the Fedosov product is again the path algebra of a quiver $\tilde{Q}$ obtained by doubling up the arrows of $Q$. In our basic example the algebra map $C \tilde{Q} \rightarrow \OmegaV~C Q$ is clarified by the following picture of $\tilde{Q}$
(which generalizes in the obvious way to arbitrary quivers). But what about the other direction $\OmegaV~C Q \rightarrow C \tilde{Q}$ ? There are two embeddings $i,j : C Q \rightarrow C \tilde{Q}$ defined by $i : (u,v) \rightarrow (a,x)$ and $j : (u,v) \rightarrow (b,y)$ giving maps
$\forall a \in C Q~:~p(a) = \frac{1}{2}(i(a)+j(a))~\quad~q(a) = \frac{1}{2}(i(a)-j(a))$ Using these maps, the isomorphism $\OmegaV~C Q \rightarrow C \tilde{Q}$ is determined by $ a0 da1 \hdots dan \rightarrow p(a0)q(a1) \hdots q(an)$ In particular, $p$ gives the natural embedding (with the ordinary multiplication on differential forms) $C Q \rightarrow \OmegaV~C Q$ of functions as degree zero differential forms. However, $p$ is no longer an algebra map for the Fedosov product on $\OmegaV~C Q$ as $p(ab) = p(a)Circ p(b) + q(a) Circ q(b)$. In Cuntz-Quillen terminology, $\omega(a,b) = q(a) Circ q(b)$ is the _curvature of the based linear map $p$. I\’d better define this a bit more formal for any algebra $A$ and then say what is special for formally smooth algebras (non-commutative manifolds). If $A,B$ are $V = C \times \hdots \times C$-algebras, then a $V$-linear map $A \rightarrow^l B$ is said to be a based linear map if $ l | V = idV$. The _curvature of $l$ measures the obstruction to $l$ being an algebra map, that is $\forall a,b \in A~:~\omega(a,b) = l(ab)-l(a)l(b)$ and the curvature is said to be nilpotent if there is an integer $n$ such that all possible products $\omega(a1,b1)\omega(a2,b2) \hdots \omega(an,bn) = 0$ For any algebra $A$ there is a universal algebra $R(A)$ turning based linear maps into algebra maps. That is, there is a fixed based linear map $A \rightarrow^p R(A)$ such that for every based linear map $A \rightarrow^l B$ there is an algebra map $R(A) \rightarrow B$ making the diagram commute In fact, Cuntz and Quillen show that $R(A) \simeq (\OmegaV^{ev}~A,Circ)$ the algebra of even differential forms equipped with the Fedosov product and that $p$ is the natural inclusion of $A$ as degree zero forms (as above). Recall that $A$ is said to be _formally smooth if every $V$-algebra map $A \rightarrow^f B/I$ where $I$ is a nilpotent ideal, can be lifted to an algebra morphism $A \rightarrow B$. We can always lift $f$ as a based linear map, say $\tilde{f}$ and because $I$ is nilpotent, the curvature of $\tilde{f}$ is also nilpotent. To get a uniform way to construct algebra lifts modulo nilpotent ideals it would therefore suffice for a formally smooth algebra to have an algebra map $A \rightarrow \hat{R}(A)$ where $\hat{R}(A)$ is the $\mathfrak{m}$-adic completion of $R(A)$ for the ideal $\mathfrak{m}$ which is the kernel of the algebra map $R(A) \rightarrow A$ corresponding to the based linear map $id_A : A \rightarrow A$. Indeed, there is an algebra map $R(A) \rightarrow B$ determined by $\tilde{f}$ and hence also an algebra map $\hat{R}(A) \rightarrow B$ and composing this with the (yet to be constructed) algebra map $A \rightarrow \hat{R}(A)$ this would give the required lift $A \rightarrow B$. In order to construct the algebra map $A \rightarrow \hat{R}(A)$ (say in the case of path algebras of quivers) we will need the Yang-Mills derivation and its associated flow.

• I Love Social Bookmarking
  
  • Subscribe
  • Digg
  • del.icio.us
  • Ma.gnolia
  • StumbleUpon
  • Technorati

No comments
Posted in geometry
Written on Thu, 16 September 2004 at 11:49 am
Tags: art, Cuntz, differential, non-commutative, Quillen, quivers
If you liked this post, then consider subscribing to our full RSS feed.