Quillen – Page 4 – neverendingbooks

noncommutative topology (1)

Published January 29, 2006 by lieven

A couple of days ago Ars Mathematica had a post Cuntz on noncommutative topology pointing to a (new, for me) paper by Joachim Cuntz

A couple of years ago, the Notices of the AMS featured an article on noncommutative geometry a la Connes: Quantum Spaces and Their Noncommutative Topology by Joachim Cuntz. The hallmark of this approach is the heavy reliance on K theory. The first few pages of the article are fairly elementary (and full of intriguing pictures), before the K theory takes over.

A few comments are in order. To begin, the paper is **not** really about noncommutative geometry a la Connes, but rather about noncommutative geometry a la Cuntz&Quillen (based on quasi-free algebras) or, equivalently, a la Kontsevich (formally smooth algebras) or if I may be so bold a la moi (qurves).

About the **intruiging pictures** : it seems to be a recent trend in noncommutative geometry research papers to include meaningless pictures to lure the attention of the reader. But, unlike aberrations such as the recent pastiche by Alain Connes and Mathilde Marcolli A Walk in the Noncommutative Garden, Cuntz is honest about their true meaning

I am indebted to my sons, Nicolas and Michael,
for the illustrations to the examples above. Since
these pictures have no technical meaning, they
are only meant to provide a kind of suggestive
visualization of the corresponding quantum spaces.

As one of these pictures made it to the cover of the **Notices** an explanation was included by the cover-editor

About the Cover :

The image on this month’s cover arose from
Joachim Cuntz’s effort to render into visible art
his own internal vision of a noncommutative
torus, an object otherwise quite abstract. His
original idea was then implemented by his son
Michael in a program written in Pascal. More
explicitly, he says that the construction started
out with a triangle in a square, then translated
the triangle by integers times a unit along a line
with irrational slope; plotted the images thus
obtained in a periodic manner; and stopped
just before the figure started to seem cluttered.
Many mathematicians carry around inside
their heads mental images of the abstractions
they work with, and manipulate these objects
somehow in conformity with their mental imagery. They probably also make aesthetic judgements of the value of their work according to
the visual qualities of the images. These presumably common phenomena remain a rarely
explored domain in either art or psychology.

—Bill Casselman(covers@ams.org)

There can be no technical meaning to the pictures as in the Connes and Cuntz&Quillen approach there is only a noncommutative algebra and _not_ an underlying geometric space, so there is no topology, let alone a noncommutative topology. Of course, I do understand why Cuntz&others name it as such. They view the noncommutative algebra as the ring of functions on some virtual noncommutative space and they compute topological invariants (such as K-groups) of the algebras and interprete them as information about the noncommutative topology of these virtual and unspecified spaces.

Still, it is perfectly possible to associate to a qurve (aka quasi-free algebra or formally smooth algebra) a genuine noncommutative topological space. In this series of posts I’ll explain the little I know of the history of this topic, the thing I posted about it a couple of years ago, why I abandoned the project and the changes I made to it since and the applications I have in mind, both to new problems (such as the birational_classification of qurves) as well as classical problems (such as rationality problems for $PGL_n $ quotient spaces).

Although others have tried to define noncommutative topologies before, I learned about them from Fred Van Oystaeyen. Fred spend the better part of his career constructing structure sheaves associated to noncommutative algebras, mainly to prime Noetherian algebras (the algebras of preference for the majority of non-commutative algebraists). So, suppose you have an ordinary (meaning, the usual commutative definition) topological space X associated to this algebra R, he wants to define an algebra of sections on every open subset $X(\sigma) $ by taking a suitable localization of the algebra $Q_{\sigma}(R) $. This localization is taken with respect to a suitable filter of left ideals $\mathcal{L}(\sigma) $ of R and is defined to be the subalgebra of the classiocal quotient ring $Q(R) $ (which exists because $R$ is prime Noetherian in which case it is a simple Artinian algebra)

$Q_{\sigma}(R) = { q \in Q(R)~|~\exists L \in \mathcal{L}(\sigma)~:~L q \subset R } $

(so these localizations are generalizations of the usual Ore-type rings of fractions). But now we come to an essential point : if we want to glue this rings of sections together on an intersection $X(\sigma) \cap X(\tau) $ we want to do this by ‘localizing further’. However, there are two ways to do this, either considering $~Q_{\sigma}(Q_{\tau}(R)) $ or considering $Q_{\tau}(Q_{\sigma}(R)) $ and these two algebras are only the same if we impose fairly heavy restrictions on the filters (or on the algebra) such as being compatible.

As this gluing property is essential to get a sheaf of noncommutative algebras we seem to get stuck in the general (non compatible) case. Fred’s way out was to make a distinction between the intersection $X_{\sigma} \cap X_{\tau} $ (on which he put the former ring as its ring of sections) and the intersection $X_{\tau} \cap X_{\sigma} $ (on which he puts the latter one). So, the crucial new ingredient in a noncommutative topology is that the order of intersections of opens matter !!!

Of course, this is just the germ of an idea. He then went on to properly define what a noncommutative topology (and even more generally a noncommutative Grothendieck topology) should be by using this localization-example as guidance. I will not state the precise definition here (as I will have to change it slightly later on) but early version of it can be found in the Antwerp Ph.D. thesis by Luc Willaert (1995) and in Fred’s book Algebraic geometry for associative algebras.

Although _qurves_ are decidedly non-Noetherian (apart from trivial cases), one can use Fred’s idea to associate a noncommutative topological space to a qurve as I will explain next time. The quick and impatient may already sneak at my old note a non-commutative topology on rep A but please bear in mind that I changed my mind since on several issues…

From Galois to NOG

Published December 22, 2004 by lieven

Evariste Galois (1811-1832) must rank pretty high on the all-time
list of moving last words. Galois was mortally wounded in a duel he
fought with Perscheux d\’Herbinville on May 30th 1832, the reason for
the duel not being clear but certainly linked to a girl called
Stephanie, whose name appears several times as a marginal note in
Galois\’ manuscripts (see illustration). When he died in the arms of his
younger brother Alfred he reportedly said “Ne pleure pas, j\’ai besoin
de tout mon courage pour mourir ‚àö‚Ä† 20 ans”. In this series I\’ll
start with a pretty concrete problem in Galois theory and explain its
elegant solution by Aidan Schofield and Michel Van den Bergh.
Next, I\’ll rephrase the problem in non-commutative geometry lingo,
generalise it to absurd levels and finally I\’ll introduce a coalgebra
(yes, a co-algebra…) that explains it all. But, it will take some time
to get there. Start with your favourite basefield $k$ of
characteristic zero (take $k = \mathbb{Q}$ if you have no strong
preference of your own). Take three elements $a,b,c$ none of which
squares, then what conditions (if any) must be imposed on $a,b,c$ and $n
\in \mathbb{N}$ to construct a central simple algebra $\Sigma$ of
dimension $n^2$ over the function field of an algebraic $k$-variety such
that the three quadratic fieldextensions $k\sqrt{a}, k\sqrt{b}$ and
$k\sqrt{c}$ embed into $\Sigma$? Aidan and Michel show in \’Division
algebra coproducts of index $n$\’ (Trans. Amer. Math. Soc. 341 (1994),
505-517) that the only condition needed is that $n$ is an even number.
In fact, they work a lot harder to prove that one can even take $\Sigma$
to be a division algebra. They start with the algebra free
product $A = k\sqrt{a} \ast k\sqrt{b} \ast k\sqrt{c}$ which is a pretty
monstrous algebra. Take three letters $x,y,z$ and consider all
non-commutative words in $x,y$ and $z$ without repetition (that is, no
two consecutive $x,y$ or $z$\’s). These words form a $k$-basis for $A$
and the multiplication is induced by concatenation of words subject to
the simplifying relations $x.x=a,y.y=b$ and $z.z=c$.

Next, they look
at the affine $k$-varieties $\mathbf{rep}(n) A$ of $n$-dimensional
$k$-representations of $A$ and their irreducible components. In the
parlance of $\mathbf{geometry@n}$, these irreducible components correspond
to the minimal primes of the level $n$-approximation algebra $\int(n) A$.
Aidan and Michel worry a bit about reducedness of these components but
nowadays we know that $A$ is an example of a non-commutative manifold (a
la Cuntz-Quillen or Kontsevich-Rosenberg) and hence all representation
varieties $\mathbf{rep}n A$ are smooth varieties (whence reduced) though
they may have several connected components. To determine the number of
irreducible (which in this case, is the same as connected) components
they use _Galois descent, that is, they consider the algebra $A
\otimes_k \overline{k}$ where $\overline{k}$ is the algebraic closure of
$k$. The algebra $A \otimes_k \overline{k}$ is the group-algebra of the
group free product $\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}
\ast \mathbb{Z}/2\mathbb{Z}$. (to be continued…) A digression : I
cannot resist the temptation to mention the tetrahedral snake problem
in relation to such groups. If one would have started with $4$ quadratic
fieldextensions one would get the free product $G =
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z}$. Take a supply of
tetrahedra and glue them together along common faces so that any
tertrahedron is glued to maximum two others. In this way one forms a
tetrahedral-snake and the problem asks whether it is possible to make
such a snake having the property that the orientation of the
\’tail-tetrahedron\’ in $\mathbb{R}^3$ is exactly the same as the
orientation of the \’head-tetrahedron\’. This is not possible and the
proof of it uses the fact that there are no non-trivial relations
between the four generators $x,y,z,u$ of $\mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/2\mathbb{Z} \ast
\mathbb{Z}/2\mathbb{Z}$ which correspond to reflections wrt. a face of
the tetrahedron (in fact, there are no relations between these
reflections other than each has order two, so the subgroup generated by
these four reflections is the group $G$). More details can be found in
Stan Wagon\’s excellent book The Banach-tarski paradox, p.68-71.

curvatures

Published September 16, 2004 by lieven

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

Tag: Quillen

noncommutative topology (1)

From Galois to NOG

curvatures