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connected component coalgebra


Never thought that I would ever consider Galois descent of semigroup
coalgebras
but in preparing for my talks for the master-class it
came about naturally. Let A be a formally smooth algebra
(sometimes called a quasi-free algebra, I prefer the terminology
noncommutative curve) over an arbitrary base-field k. What, if
anything, can be said about the connected components of the affine
k-schemes rep(n,A) of n-dimensional representations
of A? If k is algebraically closed, then one can put a
commutative semigroup structure on the connected components induced by
the sum map

rep(n,A) x rep(m,A) -> rep(n + m,A)   :  (M,N)
-> M + N

as introduced and studied by Kent
Morrison
a long while ago. So what would be a natural substitute for
this if k is arbitrary? Well, define pi(n) to be the
maximal unramified sub k-algebra of k(rep(n,A)),
the coordinate ring of rep(n,A), then corresponding to the
sum-map above is a map

pi(n + m) -> pi(n) \\otimes
pi(m)

and these maps define on the graded
space

Pi(A) = pi(0) + pi(1) + pi(2) + ...

the
structure of a graded commutative k-coalgebra with
comultiplication

pi(n) -> sum(a + b=n) pi(a) \\otimes
pi(b)

The relevance of Pi(A) is that if we consider it
over the algebraic closure K of k we get the semigroup
coalgebra

K G  with  g -> sum(h.h\' = g) h \\otimes
h\'

where G is Morrison\’s connected component
semigroup. That is, Pi(A) is a k-form of this semigroup
coalgebra. Perhaps it is a good project for one of the students to work
this out in detail (and correct possible mistakes I made) and give some
concrete examples for formally smooth algebras A. If you know of
a reference on this, please let me know.

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Brauer’s forgotten group

Non-commutative geometry seems pretty trivial compared
to commutative geometry : there are just two types of manifolds,
points and curves. However, nobody knows how to start classifying
these non-commutative curves. I do have a conjecture that any
non-commutative curve can (up to non-commutative birationality) be
constructed from hereditary orders over commutative curves
by universal methods but I’ll try to explain that another
time.

On the other hand, non-commutative points
have been classified (at least in principle) for at least 50
years over an arbitrary basefield $l$. non-commutative
$l$-points $P$ is an $l$-algebra such that its double
$d(P) = P \\otimes P^o$ ( where $P^o$ is the opposite algebra,
that is with the reverse multiplication) has an element$c=\\sum_i
a_i \\otimes b_i with \\sum_i a_ib_i = 1 (in $P$)$ and such that for
all p in $P$ we have that $(1 \\otimes a).c = (a \\otimes 1).c$ For
people of my generation, c is called a separability idempotent
and $P$ itself is called a separable $l$-algebra.
Examples of $l$-points include direct sums of full matrixrings
(of varying sizes) over $l$ or group-algebras $lG$ for $G$ a
finite group of n elements where n is invertible in $l$. Hence, in
particular, the group-algebra $lG$ of a p-group $G$ over a field $l$
of characteristic p is a non-commutative singular point and
modular representation theory (a theory build almost single
handed by
Richard Brauer) can be viewed as
the methods needed to resolve this singularity. Brauer’s name is
still mentioned a lot in modular representation theory, but another
of his inventions, the Brauer group, seems to be hardly known
among youngsters.

Still, it is a crucial tool
in classifying all non-commutative $l$-points. The algebraic
structure of an $l$-point $P$ is as follows : $$P = S_1 + S_2 + …
+ S_k$$ where each S_i is a simple algebra (that is, it
contains no proper twosided ideals), finite dimensional over
its center $l_i$ which is in its turn a finite dimensional
separable field extension of $l$. So we need to know all
simple algebras $S$, finite dimensional over their center $L$ which
is any finite dimensional separable field extension of $l$. The
algebraic structure of such an $S$ is of the form$$S = M(a,D)$$ that
is, full axa matrices with entries in $D$ where $D$ is a
skew-field (or some say, a division algebra) with
center $L$. The $L$-dimension of such a $D$ is always a square,
say b^2, so that the $L$-dimension of $S$ itself is also a square
a^2b^2. There are usually plenty such division algebras, the simplest
examples being quaternion algebras. Let p and q be two
non-zero elements of $L$ such that the conic $C : X^2-pY^2-bZ^2 =
0$ has no $L$-points in the projective $L$-plane, then the
algebra$D=(p,q)_2 = L.1 + L.i + L.j + L.ij where i^2=p, j^2=q and
ji=-ij$ is a four-dimensional skew-field over $L$. Brauer’s idea to
classify all simple $L$-algebras was to associate a group to them,
the Brauer group, $Br(L)$. Its elements are equivalence
classes
of simple algebras where two simple algebras $S$ and
$S’$ are equivalent if and only if$M(m,S) = M(n,S’)$ for some sizes
m and n. Multiplication on these classes in induced by
the tensor-product (over $L$) as $S_1 \\otimes S_2$ is again a simple
$L$-algebra if $S_1$ and $S_2$ are. The Brauer group $Br(L)$ is an
Abelian torsion group and if we know its structure we know all
$L$-simple algebras so if we know $Br(L)$ for all finite dimensional
separable extensions $L$ of $l$ we have a full classification of
all non-commutative $l$-points.

Here are some examples
of Brauer groups : if $L$ is algebraically closed (or separable
closed), then $Br(L)=0$ so in particular, if $l$ is algebraically
closed, then the only non-commutative points are sums of matrix rings.
If $R$ is the field of real numbers, then $Br(R) = Z/2Z$ generated by
the Hamilton quaternion algebra (-1,-1)_2. If $L$ is a complete
valued number field, then $Br(L)=Q/Z$ which allows to describe also
the Brauer group of a number field in terms of its places. Brauer groups
of function fields of (commutative) varieties over an algebraically
closed basefield is usually huge but there is one noteworthy
exception $Tsen’s theorem$ which states that $Br(L)=0$ if $L$ is the
function field of a curve C over an algebraically closed field. In 1982
Merkurjev and Suslin proved a marvelous result about generators of
$Br(L)$ whenever $L$ is large enough to contain all primitive roots
of unity. They showed, in present day lingo, that $Br(L)$
is generated by non-commutative points of the quantum-planes
over $L$ at roots of unity. That is, it is generated by cyclic
algebras
of the form$(p,q)_n = L
\\< X,Y>/(X^n=p,Y^n=q,YX=zXY)$where z is an n-th primitive root of
unity. Next time we will recall some basic results on the relation
between the Brauer group and Galois cohomology.

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now I see you, now I don’t

For
someone as clumsy as me, it is no real surprise to loose one in three
hard disks, but what happened yesterday was a bit puzzling at first. I
tried to replace the original 4 Gb hard disk of an original iMac (a tray
loading iMac) following the instructions of the MacWorld : how to upgrade an iMac-page I used last
time together with Jan to replace two hard disks in slot-loading iMacs.
The whole process is a bit scary : unplug 4 connections, remove the
motherboard, remove the CD-driver in order to get at the hard disk, but
to my own surprise I managed to do all this fairly quickly and replaced
the hard disk by a 120 Gb Seagate Barracuda hard disk. I
put the iMac back together and started up from the OS 9 CD (last time I
forgot this and it is becoming fairly impossible to get a working
System9 defacto on 10.3). I opened DiskUtility and to my surprise the
utility found the new disk, so I managed to install everything properly.
I could even initialize and partition the disk (to run OS X on a first
generation iMac one needs to install it on a partition which is no
larger than 8 Gb) in two partitions (one 8Gb, the other the rest) and
installed System9 on the first partition. So far, so good but when I
restarted the iMac, a blinking question-mark appeared on the screen
indicating that it could not find the installed System9! Then I tried to
start-up from the 10.3-installation disk, started up the DiskUtility and
this time it found no hard disk at all. So I started up again from the
System9 CD and the two partitions appeared on my Desktop, seemingly in
perfect order. What was going on? There was an hard disk, I put System9
on one of its partitions but somehow it refused to find it, and starting
from the 10.3 CD it looked as if there was no hard disk whatsoever. If
you are knowledgeable, you know already where the problem was situated
but as I am more a software than a hardware guy I looked for similar
problems on the net and found an entry in which the solution was
obtained by installing System9 on the larger partition. So I tried this,
but again met the same problems.

So it must be a
hardware problem and I downloaded the product manual and began browsing through it until
I found one of these marvelous computer-terms : the master-slave
jumper settings
. Who invents this kind of terminology? The
master-slave jumper… Anyway, here are the possibilities for a
Barracuda
I
admit I didn’t look at the jumper-setting when I inserted the hard
disk. The previous two times it was not necessary and I assumed that the
default position would be the master-setting but wasn’t certain. Hence,
there was only one way to find out and that was redoing the whole
replacement-process… So, this morning I did this and found out that
the jumper-settings were set at Cable select which according to
Geert is the best setting for Windows-computers as
they then automatically decide whether to use the disk as master or
slave, so perhaps for Seagate there is some marketing logic in choosing
this as their default setting. Hence, I changed the setting to
master, quickly put back the iMac and in the end discovered that
I was left with two screws… As they must have been the screws
connecting the hard-disk cage to the motherboard I had little choice but
to redo the whole process a third time. Surprisingly, I began to like
the whole procedure, one should be forced as a computer-user to take
your computer apart a couple of time before working on it. Finally, I
tried to install OS X again, the DiskUtility recognized the two
partitions without any problem and the installation went smoothly.
Probably System9 can find a Cable-select connected hard disk, whereas OS
X cannot…

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