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Tag: Elkies

a Da Vinci chess problem

Published December 30, 2005 by lieven

2005
was the year that the DaVinci code craze hit Belgium. (I started reading Dan Brown’s
Digital Fortress and Angels and Demons a year
before on the way back from a Warwick conference and when I read DVC a
few months later it was an anti-climax…). Anyway, what better way
to end 2005 than with a fitting chess problem, composed by Noam Elkies

The problem is to give an infinite sequence
of numbers, the n-th term of the sequence being the number of ways White
can force checkmate in exactly n moves. With the DVC-hint given, clearly
only one series can be the solution… To prove it, note that
White’s only non-checkmating moves are with the Bishop traveling
along the path (g1,h2,g3,h4) and use symmetry to prove that the number
of paths of length exactly k starting from h2 is the same as those
starting from g3…

If that one was too easy for you,
consider the same problem for the position

Here the solution are the 2-powers of those
of the first problem. The proof essentially is that White has now two
ways to deliver checkmate : Na6 and Nd7… For the solutions and
more interesting chess-problems consult Noam Elkies’ excellent
paper New directions in
enumerative chess problems. Remains the problem which sequences can
arise on an $N \\times N$ board with an infinite supply of chess
pieces!

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a 2006 chess puzzle anyone?

Published December 20, 2005 by lieven

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?

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the Klein stack

Published June 15, 2005 by lieven

Klein’s
quartic $X$is the smooth plane projective curve defined by
$x^3y+y^3z+z^3x=0$ and is one of the most remarkable mathematical
objects around. For example, it is a Hurwitz curve meaning that the
finite group of symmetries (when the genus is at least two this group
can have at most $84(g-1)$ elements) is as large as possible, which in
the case of the quartic is $168$ and the group itself is the unique
simple group of that order, $G = PSL_2(\mathbb{F}_7)$ also known as
Klein\’s group. John Baez has written a [beautiful page](http://math.ucr.edu/home/baez/klein.html) on the Klein quartic and
its symmetries. Another useful source of information is a paper by Noam
Elkies [The Klein quartic in number theory](www.msri.org/publications/books/Book35/files/elkies.pd).
The quotient map $X \rightarrow X/G \simeq \mathbb{P}^1$ has three
branch points of orders $2,3,7$ in the points on $\mathbb{P}^1$ with
coordinates $1728,0,\infty$. These points correspond to the three
non-free $G$-orbits consisting resp. of $84,56$ and $24$ points.
Now, remove from $X$ a couple of $G$-orbits to obtain an affine open
subset $Y$ such that $G$ acts on its cordinate ring $\mathbb{C}[Y]$ and
form the Klein stack (or hereditary order) $\mathbb{C}[Y] \bigstar G$,
the skew group algebra. In case the open subset $Y$ contains all
non-free orbits, the [one quiver](lievenlb.local/master/coursenotes/onequiver.pdf) of this
qurve has the following shape $\xymatrix{\vtx{} \ar@/^/[dd] \\
\\ \vtx{} \ar@/^/[uu]} $ $\xymatrix{& \vtx{} \ar[ddl] & \\
& & \\ \vtx{} \ar[rr] & & \vtx{} \ar[uul]} $ $\xymatrix{& &
\vtx{} \ar[dll] & & \\ \vtx{} \ar[d] & & & & \vtx{} \ar[ull] \\ \vtx{}
\ar[dr] & & & & \vtx{} \ar[u] \\ & \vtx{} \ar[rr] & & \vtx{} \ar[ur]
&} $ Here, the three components correspond to the three
non-free orbits and the vertices correspond to the isoclasses of simple
$\mathbb{C}[Y] \bigstar G$ of dimension smaller than $168$. There are
two such of dimension $84$, three of dimension $56$ and seven of
dimension $24$ which I gave the non-imaginative names \’twins\’,
\’trinity\’ and \’the dwarfs\’. As we want to spice up later this
Klein stack to a larger group, we need to know the structure of these
exceptional simples as $G$-representations. Surely, someone must have
written a paper on the general problem of finding the $G$-structure of
simples of skew-group algebras $A \bigstar G$, so if you know a
reference please let me know. I used an old paper by Idun Reiten and
Christine Riedtmann to do this case (which is easier as the stabilizer
subgroups are cyclic and hence the induced representations of their
one-dimensionals correspond to the exceptional simples).

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