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Tag: symmetry

way too ambitious

Published January 30, 2007 by lievenlb

Student-evaluation sneak preview : I am friendly and
extremely helpful but have a somewhat chaotic teaching style and am way
too ambitious as regards content… I was about to deny vehemently
all assertions (except for the chaotic bit) but may have to change my
mind after reading this report on
Mark Rowan’s book ‘Symmetry and the monster’ (see also
my post
)

Oxford University Press considers this book
“a must-read for all fans of popular science”. In his blog,
Lieven le Bruyn, professor of algebra and geometry at the University of
Antwerp, suggests that “Mark Ronan has written a beautiful book
intended for the general public”. However, he goes on to say:
“this year I’ve tried to explain to an exceptionally
good second year of undergraduates, but failed miserably Perhaps
I’ll give it another (downkeyed) try using Symmetry and the
Monster as reading material”.

As an erstwhile
mathematician, I found the book more suited to exceptional maths
undergraduates than to the general public and would strongly encourage
authors and/or publishers to pass such works before a few fans of
popular science before going to press.

Peggie Rimmer,
Satigny.

Well, this ‘exceptionally good
year’ has moved on and I had to teach a course ‘Elementary
Algebraic Geometry’ to them last semester. I had the crazy idea to
approach this in a historical perspective : first I did the
Hilbert-Noether period (translating geometry to ideal theory of
polynomial rings), then the Krull-Weil-Zariski period (defining
everything in terms of coordinate rings) to finish off with the
Serre-Grothendieck period (introducing scheme theory)… Not
surprisingly, I lost everyone after 1920. Once again there were
complaints that I was expecting way too much from them etc. etc. and I
was about to apologize and promise I’ll stick to a doable course
next year (something along the lines of Miles Reid’s
‘Undergraduate Algebraic Geometry’) when one of the students
(admittedly, probably the best of this ‘exceptional year’)
decided to do all exercises of the first two chapters of Fulton’s
‘Algebraic Curves’ to become more accustomed to the subject.
Afterwards he told me “You know, I wouldn’t change the
course too much, now that I did all these exercises I realize that your
course notes are not that bad after all…”. Yeah, thanks!

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football representation theory

Published June 9, 2006 by lievenlb

Unless
you never touched a football in your life (that’s a _soccer-ball_
for those of you with an edu account) you will know that the world
championship in Germany starts tonight. In the wake of it, the field of
‘football-science’ is booming. The BBC runs its The
Science of Football-site and did you know the following?

Research indicates that watching such a phenomenon is not
only exciting, it can be good for our health too. The Scottish
researchers found that there were 14% fewer psychiatric admissions in
the weeks after one World Cup than before it started.

But, would you believe that some of the best people in the field
(Kostant and Sternberg to name a few) have written papers on the
representation theory of a football? Perhaps this becomes more plausible
when you realize that a football has the same shape as the buckyball aka Carbon60.
Because the football (or buckyball) is a truncated icosahedron, its
symmetry group is $A_5$, the smallest of all simple groups and its
representations explain some physical properties of the buckyball. Some
of these papers are freely available and are an excellent read. In fact,
I’m thinking of using them in my course on representations of finite
groups, nxt year. Mathematics and the Buckyball by Fan
Chung and Schlomo Sternberg is a marvelous introduction to
representation theory. Among other things they explain how Schur’s
lemma, Frobenius reciprocity and Maschke’s theorem are used to count the
number of lines in the infra red buckyball spectrum! The Graph of the
Truncated Icosahedron and the Last Letter of Galois by Bertram
Kostant explains the observation, first made by Galois in his last
letter to Chevalier, that $A_{5} = PSL_2(\mathbb{F}_5)$ embeds into
$PSL_{2}(\mathbb{F}_{11})$ and applies this to the buckyball.

In effect, the model we are proposing for C60is such that
each carbon atom can be labeled by an element of order 11 in PSl(2,11)
in such a fashion that the carbon bonds can be expressed in terms of the
group structure of PSl(2,11). It will be seen that the twelve pentagons
are exactly the intersections of M with the twelve Borel sub- groups of
PSl(2,11). (A Borel subgroup is any subgroup which is conjugate to the
group PSl(2,11) defined in (2).) In particular the pentagons are the
maximal sets of commuting elements in M. The most subtle point is the
natural existence of the hexagonal bonds. This will arise from a group
theoretic linkage of any element of order 11 in one Borel subgroup with
a uniquely defined element of order 11 in another Borel subgroup.

These authors consequently joined forces to write Groups and the
Buckyball in which they give further applications of the Galois
embeddings to the electronic spectrum of the buckyball. Another
account can be found in the Master Thesis by Joris Mooij called The
vibrational spectrum of Buckminsterfullerene – An application of
symmetry reduction and computer algebra. Plenty to read should
tonight’s match Germany-Costa Rica turn out to be boring…

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symmetry and the monster

Published June 2, 2006 by lievenlb

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.

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Alain Connes on everything

Published January 6, 2006 by lievenlb

A few
days ago, Ars Mathematica wrote :

Alain Connes and Mathilde Marcolli have posted a
new survey paper on Arxiv A walk in the
noncommutative garden. There are many contenders for the title of
noncommutative geometry, but Connes’ flavor is the most
successful.

Be that as it may, do
not print this 106 page long paper! Browse through it
if you have to, be dazzled by it if you are so inclined, but I doubt it
is the eye-opener you were looking for if you gave up on reading
Connes’ book Noncommutative
Geometry…. Besides, there is much better
_Tehran-material_ on Connes to be found on the web : An interview
with Alain Connes, still 45 pages long but by all means : print it
out, read it in full and enjoy! Perhaps it may contain a lesson or two
for you. To wet your appetite a few quotes

It is
important that different approaches be developed and that one
doesn’t try to merge them too fast. For instance in noncommutative
geometry my approach is not the only one, there are other approaches
and it’s quite important that for these approaches there is no
social pressure to be the same so that they can develop
independently. It’s too early to judge the situation for instance
in quantum gravity. The only thing I resent in string theory is that
they put in the mind of people that it is the only theory that can
give the answer or they are very close to the answer. That I resent.
For people who have enough background it is fine since they know all
the problems that block the road like the cosmological constant, the
supersymmetry breaking, etc etc…but if you take people who are
beginners in physics programs and brainwash them from the very start
it is really not fair. Young physicists should be completely free,
but it is very hard with the actual system.

And here for some (moderate) Michael Douglas bashing :

Physicists tend to shift often and work on the
last fad. I cannot complain because at some point around 98 that fad was
NCG after my paper with Douglas and Schwarz. But after a while when
I saw Michael Douglas and asked him if he had thought more about
these problems the answer was no because it was no longer the last
fad and he wanted to work on something else. In mathematics one
sometimes works for several years on a problem but these young
physicists have a very different type of working habit. The unit of
time in mathematics is about 10 years. A paper in mathematics which is
10 years old is still a recent paper. In physics it is 3 months. So
I find it very difficult to cope with constant
zapping.

To the suggestion that he is the
prophet (remember, it is a Tehran-interview) of noncommutative geometry
he replies

It is flattering but I don’t think
it is a good thing. In fact we are all human beings and it is a
wrong idea to put a blind trust in a single person and believe in
that person whatever happens. To give you an example I can tell you
a story that happened to me. I went to Chicago in 1996, and gave a
talk in the physics department. A well known physicist was there and
he left the room before the talk was over. I didn’t meet this
physicist for two years and then, two years later, I gave the same
talk in the Dirac Forum in Rutherford laboratory near Oxford. This
time the same physicist was attending, looking very open and convinced
and when he gave his talk later he mentioned my talk quite
positively. This was quite amazing because it was the same talk and
I had not forgotten his previous reaction. So on the way back to
Oxford, I was sitting next to him in the bus, and asked him openly
how can it be that you attended the same talk in Chicago and you
left before the end and now you really liked it. The guy was not a
beginner and was in his forties, his answer was “Witten was seen
reading your book in the library in Princeton”! So I don’t want
to play that role of a prophet preventing people from thinking on
their own and ruling the sub ject, ranking people and all that. I
care a lot for ideas and about NCG because I love it as a branch of
mathematics but I don’t want my name to be associated with it as a
prophet.

and as if that was not convincing
enough, he continues

Well, the point is that what
matters are the ideas and they belong to nobody. To declare that
some persons are on top of the ladder and can judge and rank the
others is just nonsense mostly produced by the sociology (in fact by the
system of recommendation letters). I don’t want that to be true in
NCG. I want freedom, I welcome heretics.

But please, read it all for yourself and draw your own conclusions.

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a Da Vinci chess problem

Published December 30, 2005 by lievenlb

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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micro-sudoku

Published October 9, 2005 by lievenlb

One
cannot fight fashion… Following ones own research interest is a
pretty frustrating activity. Not only does it take forever to get a
paper refereed but then you have to motivate why you do these things
and what their relevance is to other subjects. On the other hand,
following fashion seems to be motivation enough for most…
Sadly, the same begins to apply to teaching. In my Geometry 101 course I
have to give an introduction to graphs&groups&geometry. So,
rather than giving a standard intro to graph-theory I thought it would
be more fun to solve all sorts of classical graph-problems (Konigsberger
bridges, Instant
Insanity, Gas-
water-electricity, and so on…) Sure, these first year
students are (still) very polite, but I get the distinct feeling that
they think “Why on earth should we be interested in these old
problems when there are much more exciting subjects such as fractals,
cryptography or string theory?” Besides, already on the first day
they made it pretty clear that the only puzzle they are interested in is
Sudoku.
Next week I’ll have to introduce groups and I was planning to do
this via the Rubik
cube but I’ve learned my lesson. Instead, I’ll introduce
symmetry by considering micro-
sudoku that is the baby 4×4 version of the regular 9×9
Sudoku. The first thing I’ll do is work out the number of
different solutions to micro-Sudoku. Remember that in regular Sudoku
this number is 6,670,903,752,021,072,936,960 (by a computer search
performed by Bertram
Felgenhauer ). For micro-Sudoku there is an interesting
(but ratther confused) thread on the
Sudoku forum and after a lot of guess-work the consensus seems to be
that there are precisely 288 distinct solutions to micro-Sudoku. In
fact, this is easy to see and uses symmetry. The symmetric group $S_4$
acts on the set of all solutions by permuting the four numbers, so one
may assume that a solution is in the form where the upper-left 2×2
block is 12 and 34 and the lower right 2×2 block consists of the
rows ab and cd. One quickly sees that either this leeds to a
unique solution or so does the situation with the roles of b and c
changed. So in all there are $4! \\times \\frac{1}{2} 4!=24 \\times 12 =
288$ distinct solutions. Next, one can ask for the number of
_essentially_ different solutions. That is, consider the action
of the _Sudoku-symmetry group_ (including things such as
permuting rows and columns, reflections and rotations of the grid). In
normal 9×9 Sudoku this number was computed by Ed Russell
and Frazer Jarvis to be 5,472,730,538 (again,heavily using the
computer). For micro-Sudoku the answer is that there are just 2
essentially different solutions and there is a short nice argument,
given by ‘Nick70′ at the end of the above mentioned thread. Looking a bit closer one verifies easily that the
two Sudoku-group orbits have different sizes. One contains 96 solutions,
the other 192 solutions. It will be interesting to find out how these
calculations will be received in class next week…

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quintominal dodecahedra

Published December 15, 2004 by lievenlb


A _quintomino_ is a regular pentagon having all its sides
colored by five different colours. Two quintominoes are the same if they
can be transformed into each other by a symmetry of the pentagon (that
is, a cyclic rotation or a flip of the two faces). It is easy to see
that there are exactly 12 different quintominoes. On the other hand,
there are also exactly 12 pentagonal faces of a dodecahedron
whence the puzzling question whether the 12 quintominoes can be joined
together (colours mathching) into a dodecahedron.
According to
the Dictionnaire de
mathematiques recreatives this can be done and John Conway found 3
basic solutions in 1959. These 3 solutions can be found from the
diagrams below, taken from page 921 of the reprinted Winning Ways for your Mathematical
Plays (volume 4) where they are called _the_ three
quintominal dodecahedra giving the impression that there are just 3
solutions to the puzzle (up to symmetries of the dodecahedron). Here are
the 3 Conway solutions

One projects the dodecahedron down from the top face which is
supposed to be the quintomino where the five colours red (1), blue (2),
yellow (3), green (4) and black(5) are ordered to form the quintomino of
type A=12345. Using the other quintomino-codes it is then easy to work
out how the quintominoes fit together to form a coloured dodecahedron.

In preparing to explain this puzzle for my geometry-101 course I
spend a couple of hours working out a possible method to prove that
these are indeed the only three solutions. The method is simple : take
one or two of the bottom pentagons and fill then with mathching
quintominoes, then these more or less fix all the other sides and
usually one quickly runs into a contradiction.
However, along the
way I found one case (see top picture) which seems to be a _new_
quintominal dodecahedron. It can't be one of the three Conway-types
as the central quintomino is of type F. Possibly I did something wrong
(but what?) or there are just more solutions and Conway stopped after
finding the first three of them…
Update (with help from
Michel Van den Bergh) Here is an elegant way to construct
'new' solutions from existing ones, take a permutation $\\sigma
\\in S_5$ permuting the five colours and look on the resulting colored
dodecahedron (which again is a solution) for the (new) face of type A
and project from it to get a new diagram. Probably the correct statement
of the quintominal-dodecahedron-problem is : find all solutions up to
symmetries of the dodecahedron _and_ permutations of the colours.
Likely, the 3 Conway solutions represent the different orbits under this
larger group action. Remains the problem : to which orbit belongs the
top picture??

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icosahedral group

Published October 21, 2004 by lievenlb

In my geometry 101 course I'm doing the rotation-symmetry groups
of the Platonic solids right now. This goes slightly slower than
expected as it turned out that some secondary schools no longer give a
formal definition of what a group is. So, a lot of time is taken up
explaining permutations and their properties as I want to view the
Platonic groups as subgroups of the permutation groups on the vertices.
To prove that the _tetrahedral group_ is isomorphic to $A_4$ was pretty
straigthforward and I'm half way through proving that the
_octahedral group_ is just $S_4$ (using the duality of the octahedron
with the cube and using the $4$ body diagonals of the cube).
Next
week I have to show that the _icosahedral group_ is isomorphic to $A_5$
which is a lot harder. The usual proof (that is, using the duality
between the icosahedron and the dodecahedron and using the $5$ cubes
contained in the dodecahedron, one for each of the diagonals of a face)
involves too much calculations to do in one hour. An alternative road is
to view the icosahedral group as a subgroup of $S_6$ (using the main
diagonals on the $12$ vertices of the icosahedron) and identifying this
subgroup as $A_5$. A neat exposition of this approach is given by John Baez in his
post Some thoughts on
the number $6$. (He also has another post on the icosahedral group
in his Week 79's
finds in mathematical physics).

But
probably I'll go for an “In Gap we
thrust”-argument. Using the numbers on the left, the rotation by
$72^o$ counter-clockwise in the top face we get the permutation in
$S_{20}$
$(1,2,3,4,5)(6,8,10,12,14)(7,9,11,13,15)(16,17,18,19,20)$
and the
rotation by $72^o$ counterclockwise along the face $(1,2,8,7,8)$ gives
the permutation
$(1,6,7,8,2)(3,5,15,16,9)(4,14,20,17,10)(12,13,19,18,11)$
GAP
calculates that the subgroup $dode$ of $S_{20}$ generated by these two
elements is $60$ (the correct number) and with $IsSimplegroup(dode);$ we
find that this group must be simple. Finally using
$IsomorphismTypeInfoFiniteSimplegroup(dode);$
we get the required
result that the group is indeed isomorphic to $A_5$. The time saved I
can then use to tell something about the classification project of
finite simple groups which might be more inspiring than tedious
calculations…

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a cosmic Galois group

Published October 6, 2004 by lievenlb

Are
there hidden relations between mathematical and physical constants such
as

$\frac{e^2}{4 \pi \epsilon_0 h c} \sim \frac{1}{137} $

or are these numerical relations mere accidents? A couple of years
ago, Pierre Cartier proposed in his paper A mad day’s work : from Grothendieck to Connes and
Kontsevich : the evolution of concepts of space and symmetry that
there are many reasons to believe in a cosmic Galois group acting on the
fundamental constants of physical theories and responsible for relations
such as the one above.

The Euler-Zagier numbers are infinite
sums over $n_1 > n_2 > ! > n_r \geq 1 $ of the form

$\zeta(k_1,\dots,k_r) = \sum n_1^{-k_1} \dots n_r^{-k_r} $

and there are polynomial relations with rational coefficients between
these such as the product relation

$\zeta(a)\zeta(b)=\zeta(a+b)+\zeta(a,b)+\zeta(b,a) $

It is
conjectured that all polynomial relations among Euler-Zagier numbers are
consequences of these product relations and similar explicitly known
formulas. A consequence of this conjecture would be that
$\zeta(3),\zeta(5),\dots $ are all trancendental!

Drinfeld
introduced the Grothendieck-Teichmuller group-scheme over $\mathbb{Q} $
whose Lie algebra $\mathfrak{grt}_1 $ is conjectured to be the free Lie
algebra on infinitely many generators which correspond in a natural way
to the numbers $\zeta(3),\zeta(5),\dots $. The Grothendieck-Teichmuller
group itself plays the role of the Galois group for the Euler-Zagier
numbers as it is conjectured to act by automorphisms on the graded
$\mathbb{Q} $-algebra whose degree $d $-term are the linear combinations
of the numbers $\zeta(k_1,\dots,k_r) $ with rational coefficients and
such that $k_1+\dots+k_r=d $.

The Grothendieck-Teichmuller
group also appears mysteriously in non-commutative geometry. For
example, the set of all Kontsevich deformation quantizations has a
symmetry group which Kontsevich conjectures to be isomorphic to the
Grothendieck-Teichmuller group. See section 4 of his paper Operads and motives in
deformation quantzation for more details.

It also appears
in the renormalization results of Alain Connes and Dirk Kreimer. A very
readable introduction to this is given by Alain Connes himself in Symmetries Galoisiennes
et renormalisation. Perhaps the latest news on Cartier’s dream of a
cosmic Galois group is the paper by Alain Connes and Matilde Marcolli posted
last month on the arXiv : Renormalization and
motivic Galois theory. A good web-page on all of this, including
references, can be found here.

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explaining symmetry

Published September 10, 2004 by lievenlb


PseudonomousDaughterTwo learned vector-addition at school and
important formulas such as the _Chasles-Moebius_ equation

$\forall A,B,C \in \mathbb{R}^2~:~\vec{AB}+\vec{BC} = \vec{AC} $

Last evening I helped her a bit with her homework and there was one
problem she could not do immediately (but it was a starred exercise so
you didn't have to do it, but…) :

consider a regular pentagon
with center $\vec{0} $. Prove that

$\vec{0A} + \vec{0B} +
\vec{0C} + \vec{0D} + \vec{0E} = \vec{0} $

PD2 : How would
_you_ do this? (with a tone like : I bet even you can't do
it)
Me : Symmetry!
PD2 : Huh?
Me : Rotate the plane
1/5 turn, then $A \mapsto B $, $B \mapsto C $ and so on. So the vector
giving the sum of all five terms must be mapped to itself under this
rotation and the only vector doing this is the zero vector.
PD2 :
That cannot be the solution, you didn't take sums of vectors and all
other exercises did that.
Me : I don't care, it is an elegant
solution, you don't have to compute a thing!

But clearly
she was not convinced and I had to admit there was nothing in her
textbook preparing her for such an argument. I was about to explain that
there was even more symmetry : reflecting along a line through a vertex
giving dihedral symmetry when I saw what the _intended solution_
of the exercise was :

Me : Okay, if you _have_ to do
sums let us try this. Fix a vertex, say A. Then the sum
$\vec{0E}+\vec{0B} $ must lie on the line 0A by the parallellogram-rule
(always good to drop in a word from the textbook to gain some
trust…), similarly the sum $\vec{0C}+\vec{0D} $ must lie on the
line 0A. So you now have to do a sum of three vectors lying on the
line 0A so the result must lie on 0A
PD2 : Yes, and???
Me : But there was nothing special about $A$. I could have started with
B and do the whole argument all over again and then I would get that
the sum is a vector on the line 0B
PD2 : And the only vector
lying on both 0A and 0B is $\vec{0} $
Me : Right! But
all we did now was just redoing the symmetry argument because the line
0A is mapped to 0B
PD2 : Don't you get started on
_that symmetry_ again!

I wonder which of the two
solutions she will sell today as her own. I would love to see the face
of a teacher when a 15yr old says “Clearly that is trivial because
the zero vector is the only one left invariant under
pentagon-symmetry!”

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