Can
it be that one forgets an entire proof because the result doesn’t seem
important or relevant at the time? It seems the only logical explanation
for what happened last week. Raf Bocklandt asked me whether a
classification was known of all group algebras l G which are
noncommutative manifolds (that is, which are formally smooth a la Kontsevich-Rosenberg or, equivalently, quasi-free
a la Cuntz-Quillen). I said I didn’t know the answer and that it looked
like a difficult problem but at the same time it was entirely clear to
me how to attack this problem, even which book I needed to have a look
at to get started. And, indeed, after a visit to the library borrowing
Warren Dicks
lecture notes in mathematics 790 “Groups, trees and projective
modules” and browsing through it for a few minutes I had the rough
outline of the classification. As the proof is basicly a two-liner I
might as well sketch it here.
If l G is quasi-free it
must be hereditary so the augmentation ideal must be a projective
module. But Martin Dunwoody proved that this is equivalent to
G being a group acting on a (usually infinite) tree with finite
group vertex-stabilizers all of its orders being invertible in the
basefield l. Hence, by Bass-Serre theory G is the
fundamental group of a graph of finite groups (all orders being units in
l) and using this structural result it is then not difficult to
show that the group algebra l G does indeed have the lifting
property for morphisms modulo nilpotent ideals and hence is
quasi-free.
If l has characteristic zero (hence the
extra order conditions are void) one can invoke a result of Karrass
saying that quasi-freeness of l G is equivalent to G being
virtually free (that is, G has a free subgroup of finite
index). There are many interesting examples of virtually free groups.
One source are the discrete subgroups commensurable with SL(2,Z)
(among which all groups appearing in monstrous moonshine), another
source comes from the classification of rank two vectorbundles over
projective smooth curves over finite fields (see the later chapters of
Serre’s Trees). So
one can use non-commutative geometry to study the finite dimensional
representations of virtually free groups generalizing the approach with
Jan Adriaenssens in Non-commutative covers and the modular group (btw.
Jan claims that a revision of this paper will be available soon).
In order to avoid that I forget all of this once again, I’ve
written over the last couple of days a short note explaining what I know
of representations of virtually free groups (or more generally of
fundamental algebras of finite graphs of separable
l-algebras). I may (or may not) post this note on the arXiv in
the coming weeks. But, if you have a reason to be interested in this,
send me an email and I’ll send you a sneak preview.
neverendingbooks Posts
After yesterday’s post I had to explain today what
point-modules and line-modules are and that one can really
describe them as points in a (commutative) variety. Seemingly, the
present focus on categorical methods scares possibly interested students
away and none of them seems to know that this non-commutative projective
algebraic geometry once dealt with very concrete examples.
Let
us fix the setting : A will be a quadratic algebra, that is, A is
a positively graded algebra, part of degree zero the basefield k,
generated by its homogeneous part A_1 of degree one (which we take to be
of k-dimension n 1) and with all defining relations quadratic in these
generators. Take m k-independent linear terms (that is, elements of A_1)
: l1,…,lm and consider the graded left A-module
L = A/(Al1 + ... + Alm)
Clearly, the Hilbert series of this
module (that is, the formal power series in t with coefficient of t^a
the k-dimension of the homogeneous part of L of degree a) starts off
with
Hilb(L,t) = 1 + (n+1-m) t + ...
and
we call L a linear d-dimensional module if the Hilbert series is
the power series expansion of
1/(1-t)^{d +1} = 1 + (d+1)t +(d
+1)(d +2)/2 t^2 ...
In particular, if d=0 (that is, m=n) then L
is said to be a point-module and if d=1 (that is, m=n-1) then L
is said to be a line-module. To a d-dimensional linear module L
one can associate a d-dimensional linear subspace of ordinary (that is,
commutative) projective n-space P^n. To do this, identify
P^n = P(A 1^*)
the projective space of the n 1 dimensional space of
linear functions on the homogeneous part of degree one. Then each of the
linear elements li determines a hyperplane V(li) in P^n and the
intersection of the m hyperplanes V(l1),…,V(lm) is the wanted
subspace. In particular, to a point-module corresponds a point in
P^n and to a line-module a line in P^n. So, where
is the non-commutativity of A hidden? Well, if P is a point-module
P = P0 + P1 + P2 +...
(with all components P_a one dimensional)
then the twisted module
P' = P1 + P2 + P3 + ...
is
again a point-module and the map P–>P’ defines an automorphism on the
point variety. In low dimensions, it is often possible to
reconstruct A from the point-variety and automorphism. In higher
dimensions, one has to consider also the higher dimensional linear
modules.
When I explained all this (far clumsier as it was a
long time since I worked with this) I was asked for an elementary text
on all this. ‘Why hasn’t anybody written a book on all this?’ Well,
Paul Smith wrote such a book so have a look at his
homepage. But then, it turned out that the version one can download from
one of his course pages is a more recent and a lot more
categorical version. There is no more sight of a useful book on
non-commutative projective spaces and their linear modules which might
give starting students an interesting way to learn some non-commutative
algebra and the beginnings of algebraic geometry (commutative and
non-commutative). So, hopefully Paul still has the old version around
and will make it available… The only webpage on this I could find in
short time are the slides of a talk by Michaela Vancliff.
One week to go before the regional and European elections and tension is rising. For me there are two crucial questions : will a racist party get more than 20% of the votes? and will the green party get over the electoral threshold of 5%? If you are not Flemish both probably require some explanation.
A month ago, the extreme right party ‘Vlaams Blok’ was convicted in court for racism and discrimination. They can still participate in the elections because they appealed and Belgian courts are extremely slow. Many people think that this conviction only applies to the party and not to people voting for it. To me, anyone still voting for a party convicted for racism says “I don’t care about society, values and the law!”
Sadly, I wouldn’t be surprised if more than 20% of the electorate will broadcast that message next week. But let us remain optimistic and look at the Vlaams Blog weblogs and their posters ridiculing the Vlaams Blok propaganda.
Then there is the Flemish green party groen!.
Usually they got between 5% and 8% of the votes with one exception in 1999 when they obtained 11%. In 1999 they went into government and among major environmental accomplishments they also voted silly laws such as introducing an electoral threshold of 5%. In last year’s elections they were the first party to be hurt by this when they dived under 4% and had not a single member of parliament left.
I have voted green at every election with one exception : early 80ties the Americans wanted to install cruise missiles in Belgium and with my twenties-naivety I thought to be able to avoid this by casting a strategic (socialist) vote. A traumatic experience because soon afterwards the missiles were flown in…
To me this partly explains the reluctance of groen! to form an alliance with the socialist party as (sadly enough) groen! is run by people of my generation (or older).
Still, in the long run there is no alternative but to form one progressive green-red party. So, I hope that, whatever happens, after the elections competent youngsters such as Tinne Van der Straeten and Els Keytsman will take control of the green party and find equally driven people in the socialist party (not entirely trivial as they seem to specialize in babes whose major accomplishment is the introduction of the sleeveless shirt ministerial look).
In case you wonder : I will vote Tinne Van der Straeten for Europe and Lieve Stallaert for the Flemish
parliament.