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A for aggregates

Let us
begin with a simple enough question : what are the points of a
non-commutative variety? Anyone? Probably you\’d say something like :
standard algebra-geometry yoga tells us that we should associate to a
non-commutative algebra $A$ on object, say $X_A$ and an arbitrary
variety is then build from \’gluing\’ such things together. Ok, but what
is $X_A$? Commutative tradition whispers $X_A=\mathbf{spec}~A$ the
[prime spectrum][1] of $A$, that is, the set of all twosided prime
ideals $P$ (that is, if $aAb \subset P$ then either $a \in P$ or $b \in
P$) and \’points\’ of $\mathbf{spec}~A$ would then correspond to
_maximal_ twosided ideals. The good news is that in this set-up, the
point-set comes equipped with a natural topology, the [Zariski
topology][2]. The bad news is that the prime spectrum is rarely
functorial in the noncommutative world. That is, if $\phi~:~A
\rightarrow B$ is an algebra morphism then $\phi^{-1}(P)$ for $P \in
\mathbf{spec}~B$ is not always a prime ideal of $A$. For example, take
$\phi$ the inclusion map $\begin{bmatrix} C[x] & C[x] \\ (x) & C[x]
\end{bmatrix} \subset \begin{bmatrix} C[x] & C[x] \\ C[x] & C[x]
\end{bmatrix}$ and $P$ the prime ideal $\begin{bmatrix} (x) & (x) \\ (x)
& (x) \end{bmatrix}$ then $P Cap \begin{bmatrix} C[x] & C[x] \\ (x) &
C[x] \end{bmatrix} = P$ but the corresponding quotient is
$\begin{bmatrix} C & C \\ 0 & C \end{bmatrix}$ which is not a prime
algebra so $\phi^{-1}(P)$ is not a prime ideal of the smaller algebra.
Failing this, let us take for $X_A$ something which obviously is
functorial and worry about topologies later. Take $X_A = \mathbf{rep}~A$
the set of all finite dimensional representations of $A$, that is
$\mathbf{rep}~A = \bigsqcup_n \mathbf{rep}_n~A$ where $\mathbf{rep}_n~A
= \{ Chi~:~A \rightarrow M_n(C)~\}$ with $Chi$ an algebra morphism. Now,
for any algebra morphism $\phi~:~A \rightarrow B$ there is an obvious
map $\mathbf{rep}~B \rightarrow \mathbf{rep}~A$ sending $Chi \mapsto Chi
Circ \phi$. Alernatively, $\mathbf{rep}_n~A$ is the set of all
$n$-dimensional left $A$-modules $M_{Chi} = C^n_{Chi}$ with $a.m =
Chi(m)m$. As such, $\mathbf{rep}~A$ is not merely a set but a
$C$-_category_, that is, all objects are $C$-vectorspaces and all
morphisms $Hom(M,N)$ are $C$-vectorspaces (the left $A$-module
morphisms). Moreover, it is an _additive_ category, that is if
$Chi,\psi$ are representations then we also have a direct sum
representation $Chi \oplus \psi$ defined by $a \mapsto \begin{bmatrix}
Chi(a) & 0 \\ 0 & \psi(a) \end{bmatrix}$. Returning at the task at
hand let us declare a _non-commutative variety_ $X$ to be (1) _an
additive_ $C$-_category_ which \’locally\’ looks like $\mathbf{rep}~A$
for some non-commutative algebra $A$ (even if we do not know at the
momemt what we mean by locally as we do not have defined a topology,
yet). Let is call objects of teh category $X$ the \’points\’ of our
variety and $X$ being additive allows us to speak of _indecomposable_
points (that is, those objects that cannot be written as a direct sum of
non-zero objects). By the local description of $X$ an indecomposable
point corresponds to an indecomposable representation of a
non-commutative algebra and as such has a local endomorphism algebra
(that is, all non-invertible endomorphisms form a twosided ideal). But
if we have this property for all indecomposable points,our category $X$
will be a Krull-Schmidt category so it is natural to impose also the
condition (2) : every point of $X$ can be decomposed uniquely into a
finite direct sum of indecomposable points. Further, as the space of
left $A$-module morphisms between two finite dimensional modules is
clearly finite dimensional we have also the following strong finiteness
condition (3) : For all points $x,y \in X$ the space of morphisms
$Hom(x,y)$ is a finite dimensional $C$-vectorspace. In their book
[Representations of finite-dimensional algebras][3], Peter Gabriel and
Andrei V. Roiter call an additive category such that all endomorphism
algebras of indecomposable objects are local algebras and such that all
morphism spaces are finite dimensional an _aggregate_. So, we have a
first tentative answer to our question **the points of a
non-commutative variety are the objects of an aggregate** Clearly, as
$\mathbf{rep}~A$ has stronger properties like being an _Abelian
category_ (that is, morphisms allow kernels and cokernels) it might also
be natural to replace \’aggregate\’ by \’Abelian Krull-Schmidt category
with finite dimensional homs\’ but if Mr. Abelian Category himself finds
the generalization to aggregates useful I\’m not going to argue about
this. Are all aggregates of the form $\mathbf{rep}~A$ or are there
other interesting examples? A motivating commutative example is : the
category of all coherent modules $Coh(Y)$ on a _projective_ variety $Y$
form an aggegate giving us a mental picture of what we might expect of a
non-commutative variety. Clearly, the above tentative answer cannot be
the full story as we haven\’t included the topological condition of
being locally of the form $\mathbf{rep}~A$ yet, but we will do that in
the next episode _B for Bricks_. [1]:
http://planetmath.org/encyclopedia/PrimeSpectrum.html [2]:
http://planetmath.org/encyclopedia/ZariskiTopology.html [3]:

1/ref=sr_1_8_1/026-3923724-4530018

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ISBN prefix

How to sell
modular arithmetic to students only interested in literature? Well, try
to explain the structure of ISBN-numbers
“The ISBN (International Standard Book Number) is a unique
machine-readable identification number, which marks any book
unmistakably. This number is defined in ISO Standard 2108. The number
has been in use now for 30 years and has revolutionised the
international book-trade. 166 countries and territories are officially
ISBN members. The ISBN accompanies a publication from its production
onwards.” The ISBN-number of a book is a ten-digit number divided into
four parts, separated by hyphens telling you a lot about the ambitions
and location of the book’ publisher. I’ll explain some of it by
telling how I obtained the barcode for the first book to be published by
neverendingbooks.org (see
picture). The first part is the group
identifier
and identifies a country, area or language area
participating in the ISBN system. For the Netherlands and the Flemish
speaking part of Belgium this identifier is 90. Hence,
depending on your location you have to approach different agencies in
order to obtain an ISBN-number. If you are living in the US all you have
to do is to invent a name for your PublishingHouse, get your Visa-card
out and visit isbn.org
. For smaller groups the process is more personal. The first time I
tried to apply for an ISBN-number with De Boekenbank I
messed up and got immediately an email telling me what I did wrong. I
replied explaining what NeverEndingBooks had in mind and asked advice on
how to set it up properly. I’m sure I’ll need this personal contact
again in the near future. The second part is the _Publisher
Identifier_ or _prefix_. I didn’t know this before but the
very definition of a _publisher_ is the person or company
registering a book’s ISBN. Hence, if you intend to publish a series of
books your local ISBN-agency has to reserve a certain amount of
ISBN-numbers for you, all having the same start-block (the prefix). The
shorter the prefix the more ambitious the PublishingHouse. The
registered prefix of NeverEndingBooks is 90809390 which
tells the experienced ISBN-watcher that we intend in the next years to
publish “only” ten books. If you have more energy you can also apply
for a series of 100, 1000 or even 10000 ISBN-numbers but the amount of
money needed to register these series increases quickly… The third
part of the ISBN-number is the _title identifier_ so for our
first book it is just 1. However, in order to register
it you have to provide the agency (minimally) with a title and
publication date (fortunately, author, price, number of pages etc. are
optional at this stage). Anyway, the first real deadline for
NeverEndingBooks will be may 15th 2005! And now it is time to return
to modular arithmetic, the fourth part is a _check digit_. The
check digit is the last digit of an ISBN. It is calculated on a modulus
11 with weights 10-2, using X in lieu of 10 where ten would occur as a
check digit. This means that each of the first nine digits of the ISBN
excluding the check digit itself is multiplied by a number
ranging from 10 to 2 and that the resulting sum of the products, plus
the check digit, must be divisible by 11 without a remainder. In our
case, we have the following numbers(weights) 9(10) 0(9) 8(8)
0(7) 9(6) 3(5) 9(4) 0(3) 1(2)
We have to multiply the numbers
with their weight and add them all up,
90+0+64+0+54+15+36+0+2=261=8(mod 11) whence the check
digit should be 3. For example, I know already that the
ISBN-number of the second book to be published by NeverEndingBooks will
be 90-809390-2-1 but, due to lacking information, it will take a while
before it can be registered.

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From Galois to NOG


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‚Ć 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.

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