Archive for the ‘Azumaya’ tag
M-geometry (3)
For any finite dimensional A-representation S we defined before a character $\chi(S) $ which is an linear functional on the noncommutative functions $\mathfrak{g}_A = A/[A,A]_{vect} $ and defined via
$\chi_a(S) = Tr(a | S) $ for all $a \in A $
We would like to have enough such characters to separate simples, that is we would like to have an embedding
$\mathbf{simp}~A \hookrightarrow \mathfrak{g}_A^* $
from the set of all finite dimensional simple A-representations $\mathbf{simp}~A $ into the linear dual of $\mathfrak{g}_A^* $. This is a consequence of the celebrated Artin-Procesi theorem.
Michael Artin was the first person to approach representation theory via algebraic geometry and geometric invariant theory. In his 1969 classical paper “On Azumaya algebras and finite dimensional representations of rings” he introduced the affine scheme $\mathbf{rep}_n~A $ of all n-dimensional representations of A on which the group $GL_n $ acts via basechange, the orbits of which are exactly the isomorphism classes of representations. He went on to use the Hilbert criterium in invariant theory to prove that the closed orbits for this action are exactly the isomorphism classes of semi-simple -dimensional representations. Invariant theory tells us that there are enough invariant polynomials to separate closed orbits, so we would be done if the caracters would generate the ring of invariant polynmials, a statement first conjectured in this paper.
Claudio Procesi was able to prove this conjecture in his 1976 paper “The invariant theory of $n \times n $ matrices” in which he reformulated the fundamental theorems on $GL_n $-invariants to show that the ring of invariant polynomials of m $n \times n $ matrices under simultaneous conjugation is generated by traces of words in the matrices (and even managed to limit the number of letters in the words required to $n^2+1 $). Using the properties of the Reynolds operator in invariant theory it then follows that the same applies to the $GL_n $-action on the representation schemes $\mathbf{rep}_n~A $.
So, let us reformulate their result a bit. Assume the affine $\mathbb{C} $-algebra A is generated by the elements $a_1,\ldots,a_m $ then we define a necklace to be an equivalence class of words in the $a_i $, where two words are equivalent iff they are the same upto cyclic permutation of letters. For example $a_1a_2^2a_1a_3 $ and $a_2a_1a_3a_1a_2 $ determine the same necklace. Remark that traces of different words corresponding to the same necklace have the same value and that the noncommutative functions $\mathfrak{g}_A $ are spanned by necklaces.
The Artin-Procesi theorem then asserts that if S and T are non-isomorphic simple A-representations, then $\chi(S) \not= \chi(T) $ as elements of $\mathfrak{g}_A^* $ and even that they differ on a necklace in the generators of A of length at most $n^2+1 $. Phrased differently, the array of characters of simples evaluated at necklaces is a substitute for the clasical character-table in finite group theory.
neverendingbooks-geometry (2)
Here pdf-files of older NeverEndingBooks-posts on geometry. For more recent posts go here.
noncommutative curves and their maniflds
Last time we have
seen that the noncommutative manifold of a Riemann surface can be viewed
as that Riemann surface together with a loop in each point. The extra
loop-structure tells us that all finite dimensional representations of
the coordinate ring can be found by separating over points and those
living at just one point are classified by the isoclasses of nilpotent
matrices, that is are parametrized by the partitions (corresponding
to the sizes of the Jordan blocks). In addition, these loops tell us
that the Riemann surface locally looks like a Riemann sphere, so an
equivalent mental picture of the local structure of this
noncommutative manifold is given by the picture on teh left, where the surface is part of the Riemann surface
and a sphere is placed at every point. Today we will consider
genuine noncommutative curves and describe their corresponding
noncommutative manifolds.
Here, a mental picture of such a
_noncommutative sphere_ to keep in mind would be something
like the picture on the right. That is, in most points of the sphere we place as before again
a Riemann sphere but in a finite number of points a different phenomen
occurs : we get a cluster of infinitesimally nearby points. We
will explain this picture with an easy example. Consider the
complex plane $\mathbb{C} $, the points of which are just the
one-dimensional representations of the polynomial algebra in one
variable $\mathbb{C}[z] $ (any algebra map $\mathbb{C}[z] \rightarrow \mathbb{C} $ is fully determined by the image of z). On this plane we
have an automorphism of order two sending a complex number z to its
negative -z (so this automorphism can be seen as a point-reflexion
with center the zero element 0). This automorphism extends to
the polynomial algebra, again induced by sending z to -z. That
is, the image of a polynomial $f(z) \in \mathbb{C}[z] $ under this
automorphism is f(-z).
With this data we can form a noncommutative
algebra, the _skew-group algebra_ $\mathbb{C}[z] \ast C_2 $ the
elements of which are either of the form $f(z) \ast e $ or $g(z) \ast g $ where
$C_2 = \langle g : g^2=e \rangle $ is the cyclic group of order two
generated by the automorphism g and f(z),g(z) are arbitrary
polynomials in z.
The multiplication on this algebra is determined by
the following rules
$(g(z) \ast g)(f(z) \ast e) = g(z)f(-z) \astg $ whereas $(f(z) \ast e)(g(z) \ast g) = f(z)g(z) \ast g $
$(f(z) \ast e)(g(z) \ast e) = f(z)g(z) \ast e $ whereas $(f(z) \ast g)(g(z)\ast g) = f(z)g(-z) \ast e $
That is, multiplication in the
$\mathbb{C}[z] $ factor is the usual multiplication, multiplication in
the $C_2 $ factor is the usual group-multiplication but when we want
to get a polynomial from right to left over a group-element we have to
apply the corresponding automorphism to the polynomial (thats why we
call it a _skew_ group-algebra).
Alternatively, remark that as
a $\mathbb{C} $-algebra the skew-group algebra $\mathbb{C}[z] \ast C_2 $ is
an algebra with unit element 1 = 1\aste and is generated by
the elements $X = z \ast e $ and $Y = 1 \ast g $ and that the defining
relations of the multiplication are
$Y^2 = 1 $ and $Y.X =-X.Y $
hence another description would
be
$\mathbb{C}[z] \ast C_2 = \frac{\mathbb{C} \langle X,Y \rangle}{ (Y^2-1,XY+YX) } $
It can be shown that skew-group
algebras over the coordinate ring of smooth curves are _noncommutative
smooth algebras_ whence there is a noncommutative manifold associated
to them. Recall from last time the noncommutative manifold of a
smooth algebra A is a device to classify all finite dimensional
representations of A upto isomorphism Let us therefore try to
determine some of these representations, starting with the
one-dimensional ones, that is, algebra maps from
$\mathbb{C}[z] \ast C_2 = \frac{\mathbb{C} \langle X,Y \rangle}{ (Y^2-1,XY+YX) } \rightarrow \mathbb{C} $
Such a map is determined by the image of X and that of
Y. Now, as $Y^2=1 $ we have just two choices for the image of Y
namely +1 or -1. But then, as the image is a commutative algebra
and as XY+YX=0 we must have that the image of 2XY is zero whence the
image of X must be zero. That is, we have only
two one-dimensional representations, namely $S_+ : X \rightarrow 0, Y \rightarrow 1 $
and $S_- : X \rightarrow 0, Y \rightarrow -1 $
This is odd! Can
it be that our noncommutative manifold has just 2 points? Of course not.
In fact, these two points are the exceptional ones giving us a cluster
of nearby points (see below) whereas most points of our
noncommutative manifold will correspond to 2-dimensional
representations!
So, let’s hunt them down. The
center of $\mathbb{C}[z]\ast C_2 $ (that is, the elements commuting with
all others) consists of all elements of the form $f(z)\ast e $ with f an
_even_ polynomial, that is, f(z)=f(-z) (because it has to commute
with 1\ast g), so is equal to the subalgebra $\mathbb{C}[z^2]\ast e $.
The
manifold corresponding to this subring is again the complex plane
$\mathbb{C} $ of which the points correspond to all one-dimensional
representations of $\mathbb{C}[z^2]\ast e $ (determined by the image of
$z^2\ast e $).
We will now show that to each point of $\mathbb{C} – { 0 } $
corresponds a simple 2-dimensional representation of
$\mathbb{C}[z]\ast C_2 $.
If a is not zero, we will consider the
quotient of the skew-group algebra modulo the twosided ideal generated
by $z^2\ast e-a $. It turns out
that
$\frac{\mathbb{C}[z]\ast C_2}{(z^2\aste-a)} =
\frac{\mathbb{C}[z]}{(z^2-a)} \ast C_2 = (\frac{\mathbb{C}[z]}{(z-\sqrt{a})}
\oplus \frac{\mathbb{C}[z]}{(z+\sqrt{a})}) \ast C_2 = (\mathbb{C}
\oplus \mathbb{C}) \ast C_2 $
where the skew-group algebra on the
right is given by the automorphism g on $\mathbb{C} \oplus \mathbb{C} $ interchanging the two factors. If you want to
become more familiar with working in skew-group algebras work out the
details of the fact that there is an algebra-isomorphism between
$(\mathbb{C} \oplus \mathbb{C}) \ast C_2 $ and the algebra of $2 \times 2 $ matrices $M_2(\mathbb{C}) $. Here is the
identification
$~(1,0)\aste \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $
$~(0,1)\aste \rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $
$~(1,0)\astg \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $
$~(0,1)\astg \rightarrow \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $
so you have to verify that multiplication
on the left hand side (that is in $(\mathbb{C} \oplus \mathbb{C}) \ast
C_2 $) coincides with matrix-multiplication of the associated
matrices.
Okay, this begins to look like what we are after. To
every point of the complex plane minus zero (or to every point of the
Riemann sphere minus the two points ${ 0,\infty } $) we have
associated a two-dimensional simple representation of the skew-group
algebra (btw. simple means that the matrices determined by the images
of X and Y generate the whole matrix-algebra).
In fact, we
now have already classified ‘most’ of the finite dimensional
representations of $\mathbb{C}[z]\ast C_2 $, namely those n-dimensional
representations
$\mathbb{C}[z]\ast C_2 =
\frac{\mathbb{C} \langle X,Y \rangle}{(Y^2-1,XY+YX)} \rightarrow M_n(\mathbb{C}) $
for which the image of X is an invertible $n \times n $ matrix. We can show that such representations only exist when
n is an even number, say n=2m and that any such representation is
again determined by the geometric/combinatorial data we found last time
for a Riemann surface.
That is, It is determined by a finite
number ${ P_1,\dots,P_k } $ of points from $\mathbb{C} – 0 $ where
k is at most m. For each index i we have a positive
number $a_i $ such that $a_1+\dots+a_k=m $ and finally for each i we
also have a partition of $a_i $.
That is our noncommutative
manifold looks like all points of $\mathbb{C}-0 $ with one loop in each
point. However, we have to remember that each point now determines a
simple 2-dimensional representation and that in order to get all
finite dimensional representations with det(X) non-zero we have to
scale up representations of $\mathbb{C}[z^2] $ by a factor two.
The technical term here is that of a Morita equivalence (or that the
noncommutative algebra is an Azumaya algebra over
$\mathbb{C}-0 $).
What about the remaining representations, that
is, those for which Det(X)=0? We have already seen that there are two
1-dimensional representations $S_+ $ and $S_- $ lying over 0, so how
do they fit in our noncommutative manifold? Should we consider them as
two points and draw also a loop in each of them or do we have to do
something different? Rememer that drawing a loop means in our
geometry -> representation dictionary that the representations
living at that point are classified in the same way as nilpotent
matrices.
Hence, drawing a loop in $S_+ $ would mean that we have a
2-dimensional representation of $\mathbb{C}[z]\ast C_2 $ (different from
$S_+ \oplus S_+ $) and any such representation must correspond to
matrices
$X \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $Y \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $
But this is not possible as these matrices do
_not_ satisfy the relation XY+YX=0. Hence, there is no loop in $S_+ $
and similarly also no loop in $S_- $.
However, there are non
semi-simple two dimensional representations build out of the simples
$S_+ $ and $S_- $. For, consider the matrices
$X \rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $ and $Y \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $
then these
matrices _do_ satisfy XY+YX=0! (and there is another matrix-pair
interchanging $\pm 1 $ in the Y-matrix). In erudite terminology this
says that there is a _nontrivial extension_ between $S_+ $ and $S_- $
and one between $S_- $ and $S_+ $.
In our dictionary we will encode this
information by the picture
$\xymatrix{\vtx{}
\ar@/^2ex/[rr] & & \vtx{} \ar@/^2ex/[ll]} $
where the two
vertices correspond to the points $S_+ $ and $S_- $ and the arrows
represent the observed extensions. In fact, this data suffices to finish
our classification project of finite dimensional representations of
the noncommutative curve $\mathbb{C}[z] \ast C_2 $.
Those with Det(X)=0
are of the form : $R \oplus T $ where R is a representation with
invertible X-matrix (which we classified before) and T is a direct
sum of representations involving only the simple factors $S_+ $ and
$S_- $ and obtained by iterating the 2-dimensional idea. That is, for
each factor the Y-matrix has alternating $\pm 1 $ along the diagonal
and the X-matrix is the full nilpotent Jordan-matrix.
So
here is our picture of the noncommutative manifold of the
noncommutative curve $\mathbb{C}[z]\ast C_2 $ : the points are all points
of $\mathbb{C}-0 $ together with one loop in each of them together
with two points lying over 0 where we draw the above picture of arrows
between them. One should view these two points as lying
infinetesimally close to each other and the gluing
data
$\xymatrix{\vtx{} \ar@/^2ex/[rr] & & \vtx{}
\ar@/^2ex/[ll]} $
contains enough information to determine
that all other points of the noncommutative manifold in the vicinity of
this cluster should be two dimensional simples! The methods used
in this simple minded example are strong enough to determine the
structure of the noncommutative manifold of _any_ noncommutative curve.

So, let us look at a real-life example. Once again, take the
Kleinian quartic In a previous
course-post we recalled that
there is an action by automorphisms on the Klein quartic K by the
finite simple group $PSL_2(\mathbb{F}_7) $ of order 168. Hence, we
can form the noncommutative Klein-quartic $K \ast PSL_2(\mathbb{F}_7) $
(take affine pieces consisting of complements of orbits and do the
skew-group algebra construction on them and then glue these pieces
together again).
We have also seen that the orbits are classified
under a Belyi-map $K \rightarrow \mathbb{P}^1_{\mathbb{C}} $ and that this map
had the property that over any point of $\mathbb{P}^1_{\mathbb{C}}
- { 0,1,\infty } $ there is an orbit consisting of 168 points
whereas over 0 (resp. 1 and $\infty $) there is an orbit
consisting of 56 (resp. 84 and 24 points).
So what is
the noncommutative manifold associated to the noncommutative Kleinian?
Well, it looks like the picture we had at the start of this
post
For all but three points of the Riemann sphere
$\mathbb{P}^1 – { 0,1,\infty } $ we have one point and one loop
(corresponding to a simple 168-dimensional representation of $K \ast
PSL_2(\mathbb{F}_7) $) together with clusters of infinitesimally nearby
points lying over 0,1 and $\infty $ (the cluster over 0
is depicted, the two others only indicated).
Over 0 we have
three points connected by the diagram
$\xymatrix{& \vtx{} \ar[ddl] & \\ & & \\ \vtx{} \ar[rr] & & \vtx{} \ar[uul]} $
where each of the vertices corresponds to a
simple 56-dimensional representation. Over 1 we have a cluster of
two points corresponding to 84-dimensional simples and connected by
the picture we had in the $\mathbb{C}[z]\ast C_2 $ example).
Finally,
over $\infty $ we have the most interesting cluster, consisting of the
seven dwarfs (each corresponding to a simple representation of dimension
24) and connected to each other via the
picture
$\xymatrix{& & \vtx{} \ar[dll] & & \\ \vtx{} \ar[d] & & & & \vtx{} \ar[ull] \\ \vtx{} \ar[dr] & & & & \vtx{} \ar[u] \\ & \vtx{} \ar[rr] & & \vtx{} \ar[ur] &} $
Again, this noncommutative manifold gives us
all information needed to give a complete classification of all finite
dimensional $K \ast PSL_2(\mathbb{F}_7) $-representations. One
can prove that all exceptional clusters of points for a noncommutative
curve are connected by a cyclic quiver as the ones above. However, these
examples are still pretty tame (in more than one sense) as these
noncommutative algebras are finite over their centers, are Noetherian
etc. The situation will become a lot wilder when we come to exotic
situations such as the noncommutative manifold of
$SL_2(\mathbb{Z}) $…
TheLibrary (demo)

It is far from finished but you can already visit a demo-version of
TheLibrary which I hope will one day be a useful collection of
online courses and books on non-commutative algebra & geometry. At
the moment it just contains a few of my own things but I do hope that
others will find the format interesting enough to allow me to include
their courses and/or books. So, please try this demo out! But before you
do, make sure that you have a good webbrowser-plugin to view
PDF-documents from within your browser (rather than having to download
the files). If you are using Macintosh 10.3 or better there is a very
nice plugin freely
available whch you only have to drag into your _/Library/Internet
Plug-Ins/_-folder to get it working (after restarting Safari).
/> If you click on the title you will get a page with hyper-links to all
bookmarks of the pdf-file (for example, if you have used the hyperref package to
(La)TeX your file, you get these bookmarks for free). If you only have a
PDF-file you can always include the required bookmarks using Acrobat.
/> No doubt the most useful feature (at this moment) of the set-up is
that all files are fully searchable for keywords.
For example, if
you are at the page of my 3 talks on noncommutative
geometry@n-course and fill out “Azumaya” in the Search
Document-field you will get a screen like the one below
That is, you wlll get all occurrences of 'Azumaya' in
the document together with some of the context as well as page- or
section-links nearby that you can click to get to the paragraph you are
looking for. In the weeks to come I hope to extend the usability of
_TheLibrary_ by offering a one-page view, modular security
enhancements, a commenting feature as well as a popularity count. But,
as always, this may take longer than I want…
If you think
that the present set-up might already be of interest to readers of your
courses or books and if you have a good PDF-file of it available
(including bookmarks) then email and we will try to include your
material!
Jacobian update

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