Mark
Ronan has written a beautiful book intended for the general public
on Symmetry and the Monster. The
book’s main theme is the classification of the finite simple groups. It
starts off with the introduction of groups by Galois, gives the
classifivcation of the finite Lie groups, the Feit-Thompson theorem and
the construction of several of the sporadic groups (including the
Mathieu groups, the Fischer and Conway groups and clearly the
(Baby)Monster), explains the Leech lattice and the Monstrous Moonshine
conjectures and ends with Richard Borcherds proof of them using vertex
operator algebras. As in the case of Music of the
Primes it is (too) easy to be critical about notation. For example,
whereas groups are just called symmetry groups, I don’t see the point of
calling simple groups ‘atoms of symmetry’. But, unlike du Sautoy,
Mark Ronan stays close to mathematical notation, lattices are just
lattices, characer-tables are just that, j-function is what it is etc.
And even when he simplifies established teminology, for example
‘cyclic arithmetic’ for modular arithmetic, ‘cross-section’
for involution centralizer, ‘mini j-functions’ for Hauptmoduln
etc. there are footnotes (as well as a glossary) mentioning the genuine
terms. Group theory is a topic with several colourful people
including the three Johns John Leech, John
McKay and John Conway
and several of the historical accounts in the book are a good read. For
example, I’ve never known that the three Conway groups were essentially
discovered in just one afternoon and a few telephone exchanges between
Thompson and Conway. This year I’ve tried to explain some of
monstrous moonshine to an exceptionally good second year of
undergraduates but failed miserably. Whereas I somehow managed to give
the construction and proof of simplicity of Mathieu 24, elliptic and
modular functions were way too difficult for them. Perhaps I’ll give it
another (downkeyed) try using ‘Symmetry and the Monster’ as
reading material. Let’s hope Oxford University Press will soon release a
paperback (and cheaper) version.
Tag: modular
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)
Noam Elkies is one of
those persons I seem to bump into (figuratively speaking) wherever my
interests take me. At the moment I’m reading (long overdue, I
know, I know) the excellent book Notes on
Fermat’s Last Theorem by Alf Van der
Poorten. On page 48, Elkies figures as an innocent bystander in the
1994 April fools joke e-perpetrated by
Henri Darmon in the midst of all confusion about ‘the
gap’ in Wiles’ proof.
There has
been a really amazing development today on Fermat’s Last Theorem.
Noam Elkies has announced a counterexample, so that FLT is not true
after all! He spoke about this at the institute today. The solution to
Fermat that he constructs involves an incredibly large prime exponent
(larger than $10^{20}$), but it is constructive. The main idea seems to
be a kind of Heegner-point construction, combined with a really
ingenious descent for passing from the modular curves to the Fermat
curve. The really difficult part of the argument seems to be to show
that the field of definition of the solution (which, a priori, is some
ring class field of an imaginary quadratic field) actually descends to
$\\mathbb{Q}$. I wasn’t able to get all the details, which were
quite intricate…
Elkies is also an
excellent composer of chess problems. The next two problems he composed
as New Year’s greetings. The problem is : “How many shortest
sequences exists (with only white playing) to reach the given
position?”
$\\begin{position}
\\White(Kb5,Qd1,Rb1,Rh1,Nc3,Ne5,Bc1,Bf1,a2,b2,c4,d2,e2,f3,g3,h2)
\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard
}xc $
Here’s Elkies’ solution
:
There are 2004 sequences of the minimal length 12.
Each consists of the sin- gle move g3, the 3-move sequence
c4,Nc3,Rb1, and one of the three 8-move sequences
Nf3,Ne5,f3,Kf2,Ke3,Kd3(d4),Kc4(c5),Kb5. The move g3 may be played at
any point, and so contributes a factor of 12. If the King goes
through c5 then the 3- and 8-move sequences are independent, and can
be played in $\\binom{11}{3}$ orders. If the King goes through c4 then
the entire 8-move sequence must be played before the 3-move sequence
begins, so there are only two possibilities, depending on the choice
of Kd3 or Kd4. Hence the total count is $12(\\binom{11}{3}+2)=2004$ as
claimed.
A year later he composed the
problem
$\\begin{position}
\\White(Kh3,Qe4,Rc2,Rh1,Na4,Ng1,Bc1,Bf1,a2,b2,c3,d3,e2,f4,g2,h2)
\\end{position}{\\font\\mathbb{C}hess=chess10 \\showboard
}xd $
of which Elkies’ solution is
:
There are 2005 sequences of the minimal length 14.
This uses the happy coincidence $\\binom{14}{4}=1001$. Here White
plays the 4-move sequence f4,Kf2,Kg3,Kh3 and one of the five
sequences Nc3,Na4,c3,Qc2,Qe4,d3,Bd2(e3,f4,g5,h6),Rc1,Rc2,Bc1 of
length 10. If the Bishop goes to d2 or e3, the sequences are
independent, and can be played in $\\binom{14}{4}$ orders. Otherwise
the Bishop must return to c1 before White plays f4, so the entire
10-move sequence must be played before the 4-move sequence begins. Hence
the total count is $2 \\binom{14}{4}+3 =
2005$.
With just a few weeks remaining, anyone in for
a 2006 puzzle?