Jacobian – neverendingbooks

Let us
take a hopeless problem, motivate why something like non-commutative
algebraic geometry might help to solve it, and verify whether this
promise is kept.

Suppose we want to know all solutions in invertible
matrices to the braid relation (or Yang-Baxter equation)

All such solutions (for varying size of matrices)
form an additive Abelian category , so a big step forward would be to know all its
simple solutions (that is, those whose matrices cannot be brought in
upper triangular block form). A literature check shows that even this
task is far too ambitious. The best result to date is the classification
due to Imre Tuba and
Hans Wenzl of simple solutions of which the matrix size is at most
5.

For fixed matrix size n, finding solutions in is the same as solving a system of cubic
polynomial relations in
unknowns, which quickly becomes a daunting task. Algebraic geometry
tells us that all solutions, say form an affine closed subvariety of -dimensional affine space. If we assume that is a smooth variety (that is, a manifold) and
if we know one solution explicitly, then we can use the tangent space in
this point to linearize the problem and to get at all solutions in a
neighborhood.

So, here is an idea : assume that itself would be a non-commutative manifold, then
we might linearize our problem by considering tangent spaces and obtain
new solutions out of already known ones. But, what is a non-commutative
manifold? Well, by the above we at least require that for all integers n
the commutative variety is a commutative manifold.

But, there
is still some redundancy in our problem : if is a
solution, then so is any conjugated pair where is a basechange matrix. In categorical terms, we are only
interested in isomorphism classes of solutions. Again, if we fix the
size n of matrix-solutions, we consider the affine variety as a variety with a -action
and we like to classify the orbits of simple solutions. If is a manifold then the theory of Luna slices
provides a method, both to linearize the problem as well as to reduce
its complexity. Instead of the tangent space we consider the normal
space N to the -orbit
(in a suitable solution). On this affine space, the stabilizer subgroup
acts and there is a natural one-to-one
correspondence between -orbits
in and -orbits in the normal space N (at least in a
neighborhood of the solution).

So, here is a refinement of the
idea : we would like to view as a non-commutative manifold with a group action
given by the notion of isomorphism. Then, in order to get new isoclasses
of solutions from a constructed one we want to reduce the size of our
problem by considering a linearization (the normal space to the orbit)
and on it an easier isomorphism problem.

However, we immediately
encounter a problem : calculating ranks of Jacobians we discover that
already is not a smooth variety so there is not a
chance in the world that might be a useful non-commutative manifold.
Still, if is a
solution to the braid relation, then the matrix
commutes with both X and Y.

If is a
simple solution, this means that after performing a basechange, becomes a scalar matrix, say . But then, is a solution to

and all such solutions form a
non-commutative closed subvariety, say of and if we know all (isomorphism classes of)
simple solutions in we have solved our problem as we just have to
bring in the additional scalar .

Here we strike gold : is indeed a non-commutative manifold. This can
be seen by identifying
with one of the most famous discrete infinite groups in mathematics :
the modular group . The modular group acts by Mobius
transformations on the upper half plane and this action can be used to
write as the free group product . Finally, using
classical representation theory of finite groups it follows that indeed
all are commutative manifolds (possibly having
many connected components)! So, let us try to linearize this problem by
looking at its non-commutative tangent space, if we can figure out what
this might be.

Here is another idea (or rather a dogma) : in the
world of non-commutative manifolds, the role of affine spaces is played
by the representations of finite quivers Q. A quiver
is just on oriented graph and a representation of it assigns to each
vertex a finite dimensional vector space and to each arrow a linear map
between the vertex-vector spaces. The notion of isomorphism in is of course induced by base change actions in all
of these vertex-vector spaces. (to be continued)

Jacobian conjecture remains open

lieven — Sat, 24 Dec 2005 17:42:14 +0000

Lately some
papers were posted on the arXiv
claiming to solve the plane Jacobian conjecture. Fortunately, T.T. Moh took
the time to crack these attempts and posted the mistakes they made also
on the arXiv : Comment on a Paper by
Yucai Su On Jacobian Conjecture and Comment on a Paper by
Kuo, Parusinski and Paunescu On Jacobian Conjecture. Both papers are
only 2 pages long but are fun reading.

This note
was written on Oct 10, 2005 and was sent to the authors. At once
they replied to insist that they are correct, which was natural.
After a month we checked the website of Parusinski,
and found that a new sentence ‚ÄùThe proof contains some gaps in
section 7‚Äù by the authors without mentioning any objection by
us.

So, the plane Jacobian conjecture remains
open, at least for now..

As for Kuo and his
collaborators, we believe that they have a good taste of
mathematics, and wish that they will push the analytic method deeper
to solve the Jacobian Conjecture.

fall again

lieven — Tue, 20 Sep 2005 15:27:19 +0000

When the leaves start falling, so does the plane Jacobian conjecture,
or so it seems. The comparison is a bit weak in this case as two of the
authors of the preprint posted today at the arXiv A Proof of the Plane
Jacobian Conjecture are based in Sydney, Australia… A
first glance through the paper shows that it uses Newton-polygons and
the 1975 Abyankar-Moh result on
embeddings of the line in the plane. Techniques that have been tried
before by numerous people in their attempts to tackle the plane Jacobian
conjecture (the reference to Dean in this wikipedia entry is
outdated, as mentioned in an old blog
entry). Still, the paper just might be correct. As there
are several editors chasing me for overdue referee reports I have no
time to go through the proof in detail, but if you hear more on this
paper or have the energy to go through it, please leave a comment.

Jacobian update 2

lieven — Tue, 07 Dec 2004 12:39:09 +0000

Yesterday
a comment was made to the Jacobian update post saying :

The
newest thing I heard was that the proof unfortunately was incorrect at
some point – The jacobian conjecture strikes again..?? Comment by Stefan
12/6/2004 @ 4:16 pm

Clearly I was intrigued and I
asked for more information but (so far) got no reply. Some people
approach me for the latest on this issue (I don’t know a thing about the
‘proof’ but if you do a Google on Carolyn Dean Jacobian this weblog turns up third on
the list and therefore people assume I have to know something…)
so I did try to find out what was going on. I emailed Harm Derksen who is
in Ann Arbor _and_ an expert on polynomial automorphisms, so if
someone knew something about the status of the proof, he definitely
would be the right person. Harm replied instantly, unfortunately with
sad news : it seems that the announced seminar on Carolyn’s proof is
canceled because an error has been found… For the moment at
least, the Jacobian conjecture seems to be entirely open again in two
variables (of course most people expect it to be false in three or more
variables).

Jacobian update

lieven — Sat, 13 Nov 2004 18:52:18 +0000

One way to increase the blogshare-value of this site might be to
give readers more of what they want. In fact, there is an excellent
guide for those who really want to increase traffic on their site
called 26
Steps to 15k a Day. A somewhat sobering suggestion is rule S :

“Think about what people want. They
aren't coming to your site to view “your content”,
they are coming to your site looking for “their
content”.”

But how do we know what
people want? Well, by paying attention to Google-referrals according
to rule U :

“The search engines will
tell you exactly what they want to be fed – listen closely, there is
gold in referral logs, it's just a matter of panning for
it.”

And what do these Google-referrals
show over the last couple of days? Well, here are the top recent
key-words given to Google to get here :

13 :
carolyn dean jacobian conjecture
11 : carolyn dean jacobian

9 : brauer severi varieties
7 : latexrender
7 : brauer severi
7 : spinor bundles
7 : ingalls
azumaya
6 : [Unparseable or potentially dangerous latex
formula Error 6 ]
6 : jacobian conjecture carolyn dean

See a pattern? People love to hear right now about
the solution of the Jacobian conjecture in the plane by Carolyn Dean.
Fortunately, there are a couple of things more I can say about this
and it may take a while before you know why there is a photo of Tracy
Chapman next to this post…

First, it seems I only got
part of the Melvin Hochster
email. Here is the final part I was unaware of (thanks to not even wrong)

Earlier papers established the following: if
there is
a counterexample, the leading forms of $f$ and $g$
may
be assumed to have the form $(x^a y^b)^J$ and $(x^a
y^b)^K$,
where $a$ and $b$ are relatively prime and neither
$J$
nor $K$ divides the other (Abhyankar, 1977). It is known
that
$a$ and $b$ cannot both be $1$ (Lang, 1991) and that one
may
assume that $C[f,g]$ does not contain a degree one
polynomial
in $x, y$ (Formanek, 1994).

Let $D_x$ and $D_y$ indicate partial differentiation with respect

to $x$ and $y$, respectively. A difficult result of Bass (1989)

asserts that if $D$ is a non-zero operator that is a polynomial

over $C$ in $x D_x$ and $y D_y$, $G$ is in $C[x,y]$ and $D(G)$

is in $C[f,g]$, then $G$ is in $C[f,g]$.

The proof
proceeds by starting with $f$ and $g$ that give
a
counterexample, and recursively constructing sequences of
elements and derivations with remarkable, intricate and
surprising relationships. Ultimately, a contradiction is
obtained by studying a sequence of positive integers associated
with the degrees of the elements constructed. One delicate
argument shows that the sequence is bounded. Another delicate
argument shows that it is not. Assuming the results described
above, the proof, while complicated, is remarkably self-contained
and can be understood with minimal background in algebra.

Mel Hochster

Speaking about the Jacobian
conjecture-post at not even wrong and
the discussion in the comments to it : there were a few instances I
really wanted to join in but I'll do it here. To begin, I was a
bit surprised of the implicit attack in the post

Dean hasn't published any papers in almost 15 years and is
nominally a lecturer in mathematics education at Michigan.

But this was immediately addressed and retracted in
the comments :

Just curious. What exactly did
you mean by “nominally a lecturer”?
Posted by mm
at November 10, 2004 10:54 PM

I don't know
anything about Carolyn Dean personally, just that one place on the
Michigan web-site refers to her as a “lecturer”, another
as a “visiting lecturer”. As I'm quite well aware from
personal experience, these kinds of titles can refer to all sorts of
different kinds of actual positions. So the title doesn't tell you
much, which is what I was awkwardly expressing.
Posted by Peter
at November 10, 2004 11:05 PM

Well, I know a few things
about Carolyn Dean personally, the most relevant being that she is a
very careful mathematician. I met her a while back (fall of 1985) at
UCSD where she was finishing (or had finished) her Ph.D. If Lance
Small's description of me would have been more reassuring, we
might even have ended up sharing an apartment (quod non). Instead I
ended up with Claudio
Procesi… Anyway, it was a very enjoyable month with a group
of young starting mathematicians and I fondly remember some
dinner-parties we organized. The last news I heard about Carolyn was
10 to 15 years ago in Oberwolfach when it was rumoured that she had
solved the Jacobian conjecture in the plane… As far as I recall,
the method sketched by Hochster in his email was also the one back
then. Unfortunately, at the time she still didn't have all pieces
in place and a gap was found (was it by Toby Stafford? or was it
Hochster?, I forgot). Anyway, she promptly acknowledged that there was
a gap.
At the time I was dubious about the approach (mostly
because I was secretly trying to solve it myself) but today my gut
feeling is that she really did solve it. In recent years there have
been significant advances in polynomial automorphisms (in particular
the tame-wild problem) and in the study of the Hilbert scheme of
points in the plane (which I always thought might lead to a proof) so
perhaps some of these recent results did give Carolyn clues to finish
off her old approach? I haven't seen one letter of the proof so
I'm merely speculating here. Anyway, Hochster's assurance that
the proof is correct is good enough for me right now.
Another
discussion in the NotEvenWrong-comments was on the issue that several
old problems were recently solved by people who devoted themselves for
several years solely to that problem and didn't join the parade of
dedicated follower of fashion-mathematicians.

It is remarkable that the last decade has seen great progress in
math (Wiles proving Fermat's Last Theorem, Perelman proving the
Poincare Conjecture, now Dean the Jacobian Conjecture), all achieved
by people willing to spend 7 years or more focusing on a single
problem. That's not the way academic research is generally
structured, if you want grants, etc. you should be working on much
shorter term projects. It's also remarkable that two out of three
of these people didn't have a regular tenured position.

I think particle theory should learn from this. If
some of the smarter people in the field would actually spend 7 years
concentrating on one problem, the field might actually go somewhere
instead of being dead in the water
Posted by Peter at November
13, 2004 08:56 AM

Here we come close to a major problem of
today's mathematics. I have the feeling that far too few
mathematicians dedicate themselves to problems in which they have a
personal interest, independent of what the rest of the world might
think about these problems. Far too many resort to doing trendy,
technical mathematics merely because it is approved by so called
'better' mathematicians. Mind you, I admit that I did fall in
that trap myself several times but lately I feel quite relieved to be
doing just the things I like to do no matter what the rest may think
about it. Here is a little bit of advice to some colleagues : get
yourself an iPod and take
some time to listen to songs like this one :

Don't be tempted by the shiny apple
Don't you eat
of a bitter fruit
Hunger only for a taste of justice
Hunger only for a world of truth
'Cause all that you have
is your soul

from Tracy Chapman's All
that you have is your soul

congrats carolyn

lieven — Wed, 10 Nov 2004 19:20:46 +0000

Rumour has it (see for example here or here)
that Carolyn Dean proved the Jacobian
conjecture in two variables!!!
Melvin Hochster seems to have
checked the proof and is convinced it is ok. Here is what he mailed to
seminar participants

The Jacobian conjecture
in the plane has been an open problem since 1939 (Keller). The simple
statement is this: given a ring map $F$ of $C[x,y]$ (the polynomial ring
in two variables over the complex numbers $C$) to itself that fixes $C $
and sends $x, y$ to $f, $g, respectively, $F$ is an automorphism if and
only if the Jacobian determinant $f_x g_y – f_y g_x$ is a nonzero
element of $C$. The condition is easliy seen to be necessary.
Sufficiency is the challenge.

Carolyn Dean has
proved the conjecture and will give a series of talks on it beginning
Thursday, December 2, 3-4 pm, continuing on December 9 and December 16.
Because there have been at least five published incorrect proofs and
innumerable incorrect attempts, any announcement of a proof tends to be
received with skepticism. I have spent approximately one hundred hours
(beginning in mid-August) checking every detail of the argument. It is
correct.

Many congratulations Carolyn and I hope to see you once
again somewhere, sometime.