Skip to content →

Category: featured

noncommutative topology (1)

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…

Leave a Comment

dvonn (2) overload

In the
[previous post](http://www.neverendingbooks.org/index.php?p=309) we have
seen that it is important to have lots of mobile pieces around in the
endgame and that it is hard for a computer-program to evaluate a
position correctly. In fact, we illustrated this with a position which
‘clearly’ looks much better for Black (the computer) whereas it is
already lost! In fact, the computer lost this particular game already 7
plies earlier. Consider the position

$\xymatrix@=.3cm @!C
@R=.7cm{.& & & & & & & & & & & & & \\ & & & \SBlack \connS & &
\bull{d}{5} \conn & & \bull{e}{5} \conn & & \bull{f}{5} \conn & &
\bull{g}{5} \conn & & \bull{h}{5} \conn & & \SWhite \connS & & \SWhite
\connS & & \SWhite \conneS & & & \\ & & \SBlack \connS & & \SBlack
\connS & & \Black{6} \connS & & \bull{e}{4} \conn& & \bull{f}{4} \conn &
& \bull{g}{4} \conn & & \bull{h}{4} \conn & & \SWhite \connS & &
\SWhite \connS & & \SWhite \conneS & & \\ & \SBlack \connbeginS & &
\SBlack \connS & & \BDvonn{2} \connS & & \bull{d}{3} \conn & & \SBlack
\connS & & \BDvonn{3} \connS & & \White{4} \connS & & \SWhite \connS &
& \Dvonn \connS & & \SWhite \connS & & \SWhite \connendS & . \\ & &
\Black{5} \connbeginS & & \SBlack \connS & & \SBlack \connS & &
\bull{d}{2} \conn & & \SBlack \connS & & \bull{f}{2} \conn & &
\bull{g}{2} \conn & & \SWhite \connS & & \SWhite \connS & & \SWhite
\connendS & & \\ & & & \bull{a}{1} \con & & \bull{b}{1} \con & &
\Black{5} \conS & & \bull{d}{1} \con & & \bull{e}{1} \con & &
\bull{f}{1} \con & & \bull{g}{1} \con & & \bull{h}{1} \con & & \White{2}
& & & \\ .& & & & & & & & & & & & & } $

Probably, Black lost the
game with its last move d1-f3 thereby disconnecting its pieces into two
clusters. White (the human player) must already have realized at this
moment he had a good chance of winning (as indicated in the previous
post) by letting Black run out of moves by building large stacks on the
third row, White building a stack of the appropriate size which then
jumps on the largest Black stack on the final move. Btw. this technique
is called *sharpshooting* in Dvonn-parlance

The concept
of manipulating the height of a stack so that it can land precisely on a
critical space. It’s a matter of counting and one-digit addition. Notice
that this doesn’t necessarily mean putting your own stacks atop one
another – the best sharpshooting moves are moves which also neutralize.
To counter a sharpshooting move is called “spoiling”.

But
for this strategy to have a chance, White must keep the Black stacks
containing the Dvonn pieces on the third row. At the moment the stack on
c3 can move to c1 or to c5 and with his next move White counters this
by *overloading* the stack, that is

To spoil a move or
prevent a lifting move by moving atop the enemy stack. Even if the
opponent has enough control to retake the stack, he cannot move it
because it has become taller.

So, White sacrifies his
height 4 stack on g3 with the move g3-c3. Black must take back
immediately (if not, White moves c3-i3 and all Black’s material in the
farmost right cluster is lost) but now the previously mobile Black
height 2 stack at c3 has become an immobile (or *old stack*) height 7
stack which has no option but to stay on c3 (clearly Black will never
move it to j3…). Next, White performs a similar startegy to
neutralize the *young* height 3 Black stack on f3 by overloading it by 2
and hence after the forced recapture it becomes a height 6 Black stack
which must remain on f3 forever. Here are the actual moves 1) g3-c3
b2-c3 2) h2-h3 b4-c5 3) h3-f3 e2-f3 and we end up with the
situation we analyzed last time, that is

$\xymatrix@=.3cm @!C
@R=.7cm{.& & & & & & & & & & & & & \\ & & & \Black{2} \connS & &
\bull{d}{5} \conn & & \bull{e}{5} \conn & & \bull{f}{5} \conn & &
\bull{g}{5} \conn & & \bull{h}{5} \conn & & \SWhite \connS & & \SWhite
\connS & & \SWhite \conneS & & & \\ & & \bull{b}{4} \conn & & \SBlack
\connS & & \Black{6} \connS & & \bull{e}{4} \conn& & \bull{f}{4} \conn &
& \bull{g}{4} \conn & & \bull{h}{4} \conn & & \SWhite \connS & &
\SWhite \connS & & \SWhite \conneS & & \\ & \SBlack \connbeginS & &
\SBlack \connS & & \BDvonn{7} \connS & & \bull{d}{3} \conn & & \SBlack
\connS & & \BDvonn{6} \connS & & \bull{g}{3} \conn & & \bull{h}{3}
\conn & & \Dvonn \connS & & \SWhite \connS & & \SWhite \connendS & . \\
& & \Black{5} \connbeginS & & \bull{b}{2} \conn & & \SBlack \connS & &
\bull{d}{2} \conn & & \bull{e}{2} \conn & & \bull{f}{2} \conn & &
\bull{g}{2} \conn & & \bull{h}{2} \conn & & \SWhite \connS & & \SWhite
\connendS & & \\ & & & \bull{a}{1} \con & & \bull{b}{1} \con & &
\Black{5} \conS & & \bull{d}{1} \con & & \bull{e}{1} \con & &
\bull{f}{1} \con & & \bull{g}{1} \con & & \bull{h}{1} \con & & \White{2}
& & & \\ . & & & & & & & & & & & & & } $

Leave a Comment

dvonn (1) mobility

[Dvonn](http://www.gipf.com/dvonn $ is
the fourth game in the [Gipf Project](http://www.gipf.com/project_gipf/index.html) and the most
mathematical of all six. It is a very fast (but subtle) game with a
simple [set of rules](http://www.gipf.com/dvonn/rules/rules.html). Here
is a short version

DVONN is a stacking game. It is played
on an elongated hexagonal board, with 23 white, 23 black and 3 red
DVONN-pieces. In the beginning the board is empty. The players first
place the DVONN-pieces on the board and next their own pieces. Then they
start stacking pieces on top of each other. A single piece may be moved
1 space in any direction, a stack of two pieces may be moved two spaces,
etc. A stack must always be moved as a whole and a move must always end
on top of another piece or stack. If pieces or stacks lose contact with
the DVONN pieces, they must be removed from the board. The game ends
when no more moves can be made. The players put the stacks they control
on top of each other and the one with the highest stack is the winner.
That’s all!

All this will become clearer once we fix a
specific end-game, for example

$\xymatrix@=.3cm @!C @R=.7cm{ & &
\Black{2} \connS & & \bull{d}{5} \conn & & \bull{e}{5} \conn & &
\bull{f}{5} \conn & & \bull{g}{5} \conn & & \bull{h}{5} \conn & &
\SWhite \connS & & \SWhite \connS & & \SWhite \conneS & & \\ &
\bull{b}{4} \conn & & \SBlack \connS & & \Black{6} \connS & &
\bull{e}{4} \conn& & \bull{f}{4} \conn & & \bull{g}{4} \conn & &
\bull{h}{4} \conn & & \SWhite \connS & & \SWhite \connS & & \SWhite
\conneS & \\ \SBlack \connbeginS & & \SBlack \connS & & \BDvonn{7}
\connS & & \bull{d}{3} \conn & & \SBlack \connS & & \BDvonn{6} \connS &
& \bull{g}{3} \conn & & \bull{h}{3} \conn & & \Dvonn \connS & & \SWhite
\connS & & \SWhite \connendS \\ & \Black{5} \connbeginS & &
\bull{b}{2} \conn & & \SBlack \connS & & \bull{d}{2} \conn & &
\bull{e}{2} \conn & & \bull{f}{2} \conn & & \bull{g}{2} \conn & &
\bull{h}{2} \conn & & \SWhite \connS & & \SWhite \connendS & \\ & &
\bull{a}{1} \con & & \bull{b}{1} \con & & \Black{5} \conS & &
\bull{d}{1} \con & & \bull{e}{1} \con & & \bull{f}{1} \con & &
\bull{g}{1} \con & & \bull{h}{1} \con & & \White{2} & &} $

with
White to move. Some comments about notation : the left-slanted columns
are denoted by letters from a (left) to k (right) and the rows are
labeled 1 to 5 from bottom to top (surprisingly this ‘standard’
webgame-notation differs from the numbering on my Dvonn-board where the
rows are labeled from top to bottom…). So, for example, the three
spots on the upper right are k3,k4 and k5 (there are no k1 or k2 spots).
The three Dvonn pieces are colored red and in the course of the game a
stack may land on a Dvonn piece and so stacks containing a Dvonn piece
are denoted with a red halo. For example, the symbol on spot f3 stands
for for a stack of 6 pieces, one of which is a red Dvonn piece, under
the control of Black (that is, the top-piece is Black). Further note
that a piece or stack can only move if it is not surrounded by 6 other
pieces or stacks (so the White pieces on j3 and j4 cannot (yet) move). A
piece can only move by one step in either line-direction provided there
is another piece or stack on that position. The same applies for stacks
: an height 3 stack for example can move in each lin-direction by
exactly 3 steps provided there is a piece or stack to jump onto. For
example, the height 6 stack on d4 can only move to j4 whereas the height
6 stack on f3 cannot move at all! Similarly, the two black height 5
stacks are immobile. At the moment black has all its stacks defended,
that is, if White should be able to jump onto one of them (which White
cannot at the moment), Black can use one of its neighbouring pieces to
take the stack back under its control. So, any computer program would
‘evaluate’ the position as favourable for Black : Black has stacks of
total height 34 safely under control (there are no immediate threats to
be seen : the [horizon effect](http://www.comp.lancs.ac.uk/computing/research/aai-aied/people/paulb/old243prolog/subsection3_7_5.html) in such programs) whereas White
can only claim potential stacks of total height 13… Still, Black
has already lost the game. White has more pieces which are quite mobile
as opposed to the immobile black stacks, so Black will soon run out of
moves to make and his end position will have some large stacks on the
third row. All white has to do is to let Black run out of moves and then
continue (Dvonn forces each player to make a move if they still can and
to pass the move otherwise, so the most mobile player can still continue
long after the other player was forced to stop) to build a White stack
of the appropriate height on the third row to jump on the highest Black
stack with its last move! Here is how the play continued : 1) j2-k3 ;
a3-b3 2) i1-k3 ; c5-c3 3) i2-i3 ; c2-c3 4) i3-k3 ; d4-j4 5)
j3-j4 ; e3-f3 6) i4-j4 ; c4-b3 to arrive at the position where
Black is no longer able to make any moves at all

$\xymatrix@=.3cm
@!C @R=.7cm{ & & \bull{c}{5} \conn & & \bull{d}{5} \conn & & \bull{e}{5}
\conn & & \bull{f}{5} \conn & & \bull{g}{5} \conn & & \bull{h}{5} \conn
& & \SWhite \connS & & \SWhite \connS & & \SWhite \conneS & & \\ &
\bull{b}{4} \conn & & \bull{c}{4} \conn & & \bull{d}{4} \conn & &
\bull{e}{4} \conn& & \bull{f}{4} \conn & & \bull{g}{4} \conn & &
\bull{h}{4} \conn & & \bull{i}{4} \connS & & \White{9} \connS & &
\SWhite \conneS & \\ \bull{a}{3} \connbegin & & \Black{3} \connS & &
\BDvonn{10} \connS & & \bull{d}{3} \conn & & \bull{e}{3} \conn & &
\BDvonn{7} \connS & & \bull{g}{3} \conn & & \bull{h}{3} \conn & &
\bull{i}{3} \conn & & \bull{j}{3} \conn & & \WDvonn{6} \connendS \\ &
\Black{5} \connbeginS & & \bull{b}{2} \conn & & \bull{c}{2} \conn & &
\bull{d}{2} \conn & & \bull{e}{2} \conn & & \bull{f}{2} \conn & &
\bull{g}{2} \conn & & \bull{h}{2} \conn & & \bull{i}{2} \conn & &
\bull{j}{2} \connend & \\ & & \bull{a}{1} \con & & \bull{b}{1} \con & &
\bull{c}{1} \con & & \bull{d}{1} \con & & \bull{e}{1} \con & &
\bull{f}{1} \con & & \bull{g}{1} \con & & \bull{h}{1} \con & &
\bull{i}{1} & &} $

Note that all pieces and stacks no longer
connected to a Dvonn piece must be removed. So, for example, after the
third move by Black, the Black height 5 stacks on c1 was removed. All
white now has to do is to built an height 8 stack on k3 and jump onto
the height 10 Black stack on c3 to win the game. The (only) way to do
this is by 7. j5-k5 and 8. k5-k3 to finish with 9. k3-c3 with final
position (note again that the White right-hand pieces and stacks are no
longer connected to a Dvonn piece and are hence removed)

$\xymatrix@=.3cm @!C @R=.7cm{ & & \bull{c}{5} \conn & & \bull{d}{5}
\conn & & \bull{e}{5} \conn & & \bull{f}{5} \conn & & \bull{g}{5} \conn
& & \bull{h}{5} \conn & & \bull{i}{5} \conn & & \bull{j}{5} \conn & &
\bull{k}{5} \conne & & \\\ & \bull{b}{4} \conn & & \bull{c}{4} \conn &
& \bull{d}{4} \conn & & \bull{e}{4} \conn& & \bull{f}{4} \conn & &
\bull{g}{4} \conn & & \bull{h}{4} \conn & & \bull{i}{4} \conn & &
\bull{j}{4} \conn & & \bull{k}{4} \conne & \\\ \bull{a}{3} \connbegin
& & \Black{3} \connS & & \WDvonn{18} \connS & & \bull{d}{3} \conn & &
\bull{e}{3} \conn & & \BDvonn{7} \connS & & \bull{g}{3} \conn & &
\bull{h}{3} \conn & & \bull{i}{3} \conn & & \bull{j}{3} \conn & &
\bull{k}{3} \connend \\\ & \Black{5} \connbeginS & & \bull{b}{2} \conn
& & \bull{c}{2} \conn & & \bull{d}{2} \conn & & \bull{e}{2} \conn & &
\bull{f}{2} \conn & & \bull{g}{2} \conn & & \bull{h}{2} \conn & &
\bull{i}{2} \conn & & \bull{j}{2} \connend & \\\ & & \bull{a}{1} \con &
& \bull{b}{1} \con & & \bull{c}{1} \con & & \bull{d}{1} \con & &
\bull{e}{1} \con & & \bull{f}{1} \con & & \bull{g}{1} \con & &
\bull{h}{1} \con & & \bull{i}{1} & & } $

So White wins with 18 to
Black’s 15. This shows that it is important to maintain mobility and
also that it is possible to win a Dvonn-game from computers. In fact,
the above end-game was played against a computer-program (Black). The
entire game can be found
[here](http://www.littlegolem.net/jsp/game/game.jsp?gid=426457&nmove=91)
.

One Comment