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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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hyper-resolutions

Published September 30, 2004 by lievenlb

[Last time][1] we saw that for $A$ a smooth order with center $R$ the
Brauer-Severi variety $X_A$ is a smooth variety and we have a projective
morphism $X_A \rightarrow \mathbf{max}~R$ This situation is
very similar to that of a desingularization $~X \rightarrow
\mathbf{max}~R$ of the (possibly singular) variety $~\mathbf{max}~R$.
The top variety $~X$ is a smooth variety and there is a Zariski open
subset of $~\mathbf{max}~R$ where the fibers of this map consist of just
one point, or in more bombastic language a $~\mathbb{P}^0$. The only
difference in the case of the Brauer-Severi fibration is that we have a
Zariski open subset of $~\mathbf{max}~R$ (the Azumaya locus of A) where
the fibers of the fibration are isomorphic to $~\mathbb{P}^{n-1}$. In
this way one might view the Brauer-Severi fibration of a smooth order as
a non-commutative or hyper-desingularization of the central variety.
This might provide a way to attack the old problem of construction
desingularizations of quiver-quotients. If $~Q$ is a quiver and $\alpha$
is an indivisible dimension vector (that is, the component dimensions
are coprime) then it is well known (a result due to [Alastair King][2])
that for a generic stability structure $\theta$ the moduli space
$~M^{\theta}(Q,\alpha)$ classifying $\theta$-semistable
$\alpha$-dimensional representations will be a smooth variety (as all
$\theta$-semistables are actually $\theta$-stable) and the fibration
$~M^{\theta}(Q,\alpha) \rightarrow \mathbf{iss}_{\alpha}~Q$ is a
desingularization of the quotient-variety $~\mathbf{iss}_{\alpha}~Q$
classifying isomorphism classes of $\alpha$-dimensional semi-simple
representations. However, if $\alpha$ is not indivisible nobody has
the faintest clue as to how to construct a natural desingularization of
$~\mathbf{iss}_{\alpha}~Q$. Still, we have a perfectly reasonable
hyper-desingularization $~X_{A(Q,\alpha)} \rightarrow
\mathbf{iss}_{\alpha}~Q$ where $~A(Q,\alpha)$ is the corresponding
quiver order, the generic fibers of which are all projective spaces in
case $\alpha$ is the dimension vector of a simple representation of
$~Q$. I conjecture (meaning : I hope) that this Brauer-Severi fibration
contains already a lot of information on a genuine desingularization of
$~\mathbf{iss}_{\alpha}~Q$. One obvious test for this seemingly
crazy conjecture is to study the flat locus of the Brauer-Severi
fibration. If it would contain info about desingularizations one would
expect that the fibration can never be flat in a central singularity! In
other words, we would like that the flat locus of the fibration is
contained in the smooth central locus. This is indeed the case and is a
more or less straightforward application of the proof (due to [Geert Van
de Weyer][3]) of the Popov-conjecture for quiver-quotients (see for
example his Ph.D. thesis [Nullcones of quiver representations][4]).
However, it is in general not true that the flat-locus and central
smooth locus coincide. Sometimes this is because the Brauer-Severi
scheme is a blow-up of the Brauer-Severi of a nicer order. The following
example was worked out together with [Colin Ingalls][5] : Consider the
order $~A = \begin{bmatrix} C[x,y] & C[x,y] \\ (x,y) & C[x,y]
\end{bmatrix}$ which is the quiver order of the quiver setting
$~(Q,\alpha)$ $\xymatrix{\vtx{1} \ar@/^2ex/[rr] \ar@/^1ex/[rr]
& & \vtx{1} \ar@/^2ex/[ll]} $ then the Brauer-Severi fibration
$~X_A \rightarrow \mathbf{iss}_{\alpha}~Q$ is flat everywhere except
over the zero representation where the fiber is $~\mathbb{P}^1 \times
\mathbb{P}^2$. On the other hand, for the order $~B =
\begin{bmatrix} C[x,y] & C[x,y] \\ C[x,y] & C[x,y] \end{bmatrix}$
the Brauer-Severi fibration is flat and $~X_B \simeq \mathbb{A}^2 \times
\mathbb{P}^1$. It turns out that $~X_A$ is a blow-up of $~X_B$ at a
point in the fiber over the zero-representation.

[1]: http://www.neverendingbooks.org/index.php?p=342
[2]: http://www.maths.bath.ac.uk/~masadk/
[3]: http://www.win.ua.ac.be/~gvdwey/
[4]: http://www.win.ua.ac.be/~gvdwey/papers/thesis.pdf
[5]: http://kappa.math.unb.ca/~colin/

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