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curvatures

[Last
time][1] we saw that the algebra (ฮฉV CQ,Circ) of relative
differential forms and equipped with the Fedosov product is again the
path algebra of a quiver Q~ obtained by doubling up the arrows
of Q. In our basic example the algebra map CQ~โ†’ฮฉV CQ is clarified by the following picture of Q~
Misplaced & (which
generalizes in the obvious way to arbitrary quivers). But what about the
other direction ฮฉV CQโ†’CQ~ ? There are two
embeddings i,j:CQโ†’CQ~ defined by i:(u,v)โ†’(a,x) and j:(u,v)โ†’(b,y) giving maps
โˆ€aโˆˆCQ : p(a)=12(i(a)+j(a))  q(a)=12(i(a)โˆ’j(a)) Using these maps, the isomorphism ฮฉV CQโ†’CQ~ is determined by a0da1โ€ฆdanโ†’p(a0)q(a1)โ€ฆq(an) In particular, p gives the
natural embedding (with the ordinary multiplication on differential
forms) CQโ†’ฮฉV CQ of functions as degree zero
differential forms. However, p is no longer an algebra map for the
Fedosov product on ฮฉV CQ as p(ab)=p(a)Circp(b)+q(a)Circq(b). In Cuntz-Quillen terminology, ฯ‰(a,b)=q(a)Circq(b) is
the _curvature_ of the based linear map p. I\โ€™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=Cร—โ€ฆร—C-algebras, then a V-linear map Aโ†’lB 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 โˆ€a,bโˆˆA : ฯ‰(a,b)=l(ab)โˆ’l(a)l(b) and
the curvature is said to be _nilpotent_ if there is an integer n such
that all possible products ฯ‰(a1,b1)ฯ‰(a2,b2)โ€ฆฯ‰(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 Aโ†’pR(A) such that for every based
linear map Aโ†’lB there is an algebra map R(A)โ†’B making the diagram commute Misplaced & In fact, Cuntz and Quillen show that R(A)โ‰ƒ(ฮฉVev 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 Aโ†’fB/I where
I is a nilpotent ideal, can be lifted to an algebra morphism Aโ†’B. We can always lift f as a based linear map, say
f~ and because I is nilpotent, the curvature of 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_ Aโ†’R^(A) where
R^(A) is the m-adic completion of R(A) for the
ideal m which is the kernel of the algebra map R(A)โ†’A corresponding to the based linear map idA:Aโ†’A. Indeed, there is an algebra map R(A)โ†’B
determined by f~ and hence also an algebra map R^(A)โ†’B and composing this with the (yet to be constructed)
algebra map Aโ†’R^(A) this would give the required lift
Aโ†’B. In order to construct the algebra map Aโ†’R^(A) (say in the case of path algebras of quivers) we
will need the Yang-Mills derivation and its associated flow.

[1]: https://lievenlb.local/index.php?p=354

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