Dessins d'enfants: a long explanation and some fun
Although I am not the most competent person for this and may make a few
mistakes, in this long post I will try to explain what are "Dessins d'Enfants",
what is there use, and how you can have a lot of fun playing with them, even
if you don't know much mathematics. I will NOT try to be rigorous, just to
make you get a feel, hence a few of the things I will say may be slightly
wrong.
As mentioned by a previous poster, a dessin is a graph drawn on a Riemann
surface (see below the precise definition). Behind this definition,
let us look at a very important example, i.e.
a graph drawn on the Riemann sphere, or in other words on the complex
projective line P_1(C). This is the set of complex numbers together with a
single point at infinity. The natural operations on the projective line are the
linear fractional transformations (LFT for short) x goes to (ax+b)/(cx+d)
where adbc is non zero. For instance, under this transformation the point
x=d/c goes to infinity and infinity goes to a/c (which can be infinity if
c=0). The important elementary fact about this is that an LFT is defined
uniquely by the image of three distinct points. In other words, given
3 distinct points x,y,z in the projective line, and three other distinct points
x',y',z', there exists a unique LFT sending x to x', y to y', z to z'.
The basic triple we will take (which is not better than any other but which
is simply more convenient) is 0, 1, infinity. Hence given x,y,z distinct, there
exists a unique LFT sending them to 0,1,infinity respectively.
Let us now come to the heart of the subject. We are interested in ramified
coverings of P_1(C). These are maps from a Riemann surface X onto P_1(C).
We will for simplicity only consider the case where X=P_1(C), but to avoid
confusion which can occur very easily, we will continue to denote it by X,
although we remember that X=P_1(C). In this case, such a map is simply
a rational function f, i.e. a quotient of two polynomials. We say that
a point a in X is ramified if f'(a)=0. More correctly, one should take
a local parameter g(x) around a and a local parameter h(y) around f(a),
and ask that w'(0)=0, where w(x)=h^1(f(g(x)). When a is not
infinity and not a pole of f, we can take g(x)=x+a, h(y)=y+f(a), hence
w(x)=f(x+a)f(a) and the condition is f'(a)=0 as stated. In other words,
a must be a zero of order at least 2 of the function f(x)f(a).
When a is not infinity but is a pole of f, we can take g(x)=x+a, h(y)=1/y,
hence w(x)=1/f(x+a) and the condition is that the limit of (f'/f^2)(x)
as x tends to a must vanish. This is easily seen to be equivalent to the
fact that f has a pole of order at least 2 at a.
Finally if a is infinity and is a pole of f (i.e. if the degree of the
numerator of f is strictly greater than that of the denominator),
then we can take g(x)=1/x, h(y)=1/y, hence w(x)=1/f(1/x), so the condition
is that the limit of x^2f'(x)/f^2(x) as x tends to infinity must vanish.
This in fact means that the degree of the numerator of f must be at least
2 more than the degree of the denominator.
The ramified a above are called the ramification POINTS of the function f.
The values of f(a) at the ramification points are called the ramification
VALUES.
Now there are two remarkable theorems on the subject, which give a very
surprising description of functions f which have at most 3 ramified VALUES.
The first theorem is due to Andre Weil, and states that if f is such a
function, then we can take the coefficients of f to be in a specific number
field, considered as a subfield of C (please do not link this with the
eternal thread about this subject; here all number fields are going to
be subfields of C, but this does not mean that it is the best way to look
at them). This is already surprising in many respects. First, we started
from a purely geometric situation, and we find an unexpected arithmetic
aspect. In some sense, this is a little similar to the famous theorem of
Riemann stating that all Riemann surfaces are algebraic: you start with
an object defined by purely analytical means, and you obtain an algebraic
object.
The second surprising thing is that one can give a more precise definition
of the number field associated to f than that given above, and then it is
completely defined and not only up to conjugacy. In other words, it is a
specific subfield of C, not just the image by some embedding. I will describe
an example later to make this clear.
The second theorem was discovered only much later by the russian mathematician
Belyi, and is even more surprising: it says that the converse of Weil's
theorem is also true. More precisely, if you have a map f, which may have
more than 3 ramification values, and whose coefficients lie in a number field,
then you can find a birational map g such that g composed with f will
have at most 3 ramification values. The proof of Belyi's theorem is in
fact remarkably simple.
These two theorems were one of the main themes behind Grothendieck's
"Esquisse", although this type of functions was considered since Riemann.
Thanks to these theorems and also other properties, dessins have become a
central part of investigations in certain areas of mathematics and of
physics. For example it gives a new angle of attack on the inverse Galois
problem: given a finite group G, does there exist a Galois extension K of
Q with Galois group isomorphic to G?
Now finally let us come to the dessins themselves. Let f be a map from
X to P_1(C) (recall that X is also P_1(C)), with at most three ramification
values Thanks to the theorem on LFT mentioned at the beginning, we may
assume that the ramification values of f are 0, 1 and infinity. Now consider
the inverse image of the closed real interval [0,1] by the function f.
It can easily be shown that this will be a connected graph D in X, with edges
joining the inverse images of 0 to the inverse images of 1. The inverse
images of infinity will lie ouside of the graph, and there will be exactly one
per connected component of XD. The graph D will be called the
"dessin d'enfant" corresponding to f. It can be viewed either as a bicolored
graph with vertices marked with O for points above 0 and X for points
above 1, an edge joining only a O to a X. Or it can also be viewed as a
triangulation of X, the vertex of each triangles being labeled O, X and
A (above infinity) in a consistent way.
But what is much more fun is to go the other way. An easy consequence of
the above theorems is that ANY connected bicolored graph (with a finite
number of vertices and edges) is topologically equivalent to a dessin
d'enfant. In other words, we should be able to find the rational function f
and draw the corresponding dessin. In addition, we find the number field
corresponding to the dessin.
Let us start with a simple example. Consider the following graph in X=P_1(C):
OXOXO
We would like to compute a function f whose points (ramified or not) above 0
are the points marked O, whose points above 1 (ramified or not) are the points
marked X, and whose point above infinity (ramified or not) is infinity itself.
It is easily seen from the definition of ramification, that the recipe is as
follows.
There are no poles other than infinity, hence we must take f to be a
polynomial. The middle O is ramified of order 2. By a LFT (which must
preserve infinity which has already been chosen), we may assume that it is
at 0 (zero!). The two other O give two other zeros, and they are nonramified
since there is only one edge from them. Hence the function f is of the form
f(x)=x^2P(x) for some second degree polynomial. Now the two points marked X
are ramified of order 2 above 1. This simply means that the function f(x)1
is the square of a polynomial. Hence we must solve
x^2P(x)1=(ax^2+bx+c)^2
A priori this could have no solution or an infinite number. But it is a
remarkable consequence of the above theorems that in fact it always has a
finite nonzero number of solutions, up to a LFT (in our case we already
fixed the image of two points, hence we should have only one degree of
freedom left).
In any case we immediately obtain c=+i and b=0, hence
f(x)=x^2P(x)=a^2x^4+2aix^2. To obtain the simplest possible number field
(in this case Q), we will choose a=i, hence we finally obtain
f(x)=2x^2x^4
Now although all this is simple, there is some amount of magic involved
here: if you take the inverse image in X=P_1(C) by f(x) of the real interval
[0,1], you will get a true dessin, whose shape will be very similar to
the graph we started with, and you will immediately notice some properties
of the ramification points.
Let us now look at a more complicated but more instuctive example.
Consider the following graph
X
/ \
 
\ /
\ /
\ /
O

X

OXOXO

X

O
(sorry for the bad quality of ASCII graphics!)
First consider the points above 0, marked O. We have 1 of order 4, 1 of
order 3, and 3 of order 1. Using a LFT, we can send the point of order 4
to 0 and the point of order 3 to 1. Hence the NUMERATOR of the
rational function f(x) is of the form x^4(x1)^3P(x) where P is a polynomial
of degree 3.
Now let us look at the points above infinity. They correspond to the
connected components of XD, so here there are 2: 1 inside the loop, and
one completely ouside. I will decide that the one outside is the point
at infinity, which I still can do with a LFT since I still had one wish
left. The situation is now completely rigid. The one inside the loop
is some complex number a.
To compute the multiplicity of the poles is not difficult. The recipe is
as follows: each X must be considered as the concatenation of two arrows
> and <. You then count the number of arrows inside each component.
We see in this way that the point a has multiplicity 1, since half of the
upper X is outside the loop, while the point at infinity has multiplicity
9 (4 full X, plus a half one). This means finally that the function f
is of the form
x^4(x1)^3P(x)
f(x) =  with P a polynomial of degree 3.
xa
Finally, since all the points X have multiplicity 2, the ramification
condition above 1 is that f(x)1 should have 5 double zeros, or in
other words that f(x)1=Q(x)^2/(xa) for a polynomial Q of degree 5.
Hence we must solve
x^4(x1)^3P(x) Q(x)^2
  1 = 
xa xa
with P of degree 3 and Q of degree 5. Note that since we have used our
three wishes with the LFT, the situation is completely rigid hence
we must have only a finite nonzero number of solutions.
To solve the above, we could expand by brute force. It is slightly
simpler to compute the derivative of each side. It is clear that (xa)^2
times the derivative of the LHS is equal to x^3(x1)^2 multiplied by some
polynomial of degree 5, while (xa)^2 times the derivative of the RHS is
equal to Q(x)(2(xa)Q'(x)Q(x)). Since Q(x) is coprime to x and x1 by
the formula, we deduce that 2(xa)Q'(x)Q(x) is a constant multiple of
x^3(x1)^2. We can now set Q(x)=sum_{0<=i<=5} a_ix^i and identify
coefficients. We thus can express all the a_i in terms of a and a_5 only.
Plugging back into the initial equation, the extra condition is that
Q(x)^2+(xa) must vanish at 0 and 1. This gives two conditions on a and a5.
In this special case, these conditions can easily be expressed because
a5^2 can be expressed as a rational function of a in both equations. Replacing,
we get an equation for a only (in the general case, we would have to
compute a resultant, and/or Groebner bases). After a short computation on
a CAS, we finally get
(a  1) ( 105 a^3  99 a^2  9 a  1 ) = 0
The value a=1 can be discarded because the denominator xa=x1 would
cancel the numerator (or in other words because a point cannot be
simulataneously a pole and a zero of f). We are left with an irreducible
third degree equation 105 a^3  99 a^2  9 a  1 = 0, and voila!, here are
our three number fields K.
A little calculation shows that a = (13 t^2 + 18 t + 33) / 105, where t
is a root of t^3  6 =0, i.e. a cube root of 6. But now, which cube root
should be taken ? This is really a remarkable phenomenon: only 1 of
the 3 possible values of t will correspond to our dessin, the two others
giving socalled "conjugate" dessins (a notion closely linked, but not
identical to, the notion of conjugate number fields).
There are several ways to lift the indeterminacy. The simplest in this
case is to notice that the dessin has an axis of symmetry, and this is
characteristic of REAL dessins, in other words of dessins whose associated
number field is real (at least when there is no automorphism, which is
the case here). Hence the value of t is the real cube root of 6, and we
can then replace and find the values of a and all the coefficients of
the function f.
Another way which is more fun is to use the definition, and draw the
inverse image in P_1(C) of the real interval [0,1] by the function f.
This is easily done on a computer, and has several merits. First, it shows
the exact shape of the dessin (initially, we simply took a topological
graph). Second, it gives the conjugate dessins. In our case, the other
2 (complex) cube roots of 6 give the dessins
X X
/ \ / \
   
\ / \ /
\ / \ /
\ / \ /
O O
/ \ and / \
X X X X
/ \ / \
O  X  O O O O  X  O
 
X X
 
O O
You can play around with dessins for a very long time with the help of a
CAS, and note the number fields which occur.
Henri Cohen