Author: lievenlb

[Last
time][1] we saw that the algebra $(\Omega_V~C Q,Circ)$ 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 $C \tilde{Q} \rightarrow
\Omega_V~C Q$ is clarified by the following picture of $\tilde{Q}$
$\xymatrix{\vtx{} \ar@/^/[rr]^{a=u+du} \ar@/_/[rr]_{b=u-du} & &
\vtx{} \ar@(u,ur)^{x=v+dv} \ar@(d,dr)_{y=v-dv}} $ (which
generalizes in the obvious way to arbitrary quivers). But what about the
other direction $\Omega_V~C Q \rightarrow C \tilde{Q}$ ? 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 $\Omega_V~C
Q \rightarrow C \tilde{Q}$ is determined by $ a_0 da_1 \ldots da_n
\rightarrow p(a_0)q(a_1) \ldots q(a_n)$ In particular, $p$ gives the
natural embedding (with the ordinary multiplication on differential
forms) $C Q \rightarrow \Omega_V~C Q$ of functions as degree zero
differential forms. However, $p$ is no longer an algebra map for the
Fedosov product on $\Omega_V~C Q$ 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
$V = C \times \ldots \times C$-algebras, then a $V$-linear map $A
\rightarrow^l B$ is said to be a _based linear map_ if $ l | V = id_V$.
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 $\omega(a_1,b_1)\omega(a_2,b_2) \ldots
\omega(a_n,b_n) = 0$ 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 $\xymatrix{A \ar[r]^l \ar[d]^p & B
\\\ R(A) \ar[ru] &} $ In fact, Cuntz and Quillen show that $R(A)
\simeq (\Omega_V^{ev}~A,Circ)$ 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.

[1]: http://www.matrix.ua.ac.be/index.php?p=354