Today we will explain how curves defined over
$\overline{\mathbb{Q}}$ determine permutation representations
of the carthographic groups. We have seen that any smooth projective
curve $C$ (a Riemann surface) defined over the algebraic
closure $\overline{\mathbb{Q}}$ of the rationals, defines a
_Belyi map_ $\xymatrix{C \ar[rr]^{\pi} & & \mathbb{P}^1}$ which is only ramified over the three points
$\\{ 0,1,\infty \\}$. By this we mean that there are
exactly $d$ points of $C$ lying over any other point
of $\mathbb{P}^1$ (we call $d$ the degree of
$\pi$) and that the number of points over $~0,1~$ and
$~\infty$ is smaller than $~d$. To such a map we
associate a _dessin d\’enfant_, a drawing on $C$ linking the
pre-images of $~0$ and $~1$ with exactly $d$
edges (the preimages of the open unit-interval). Next, we look at
the preimages of $~0$ and associate a permutation
$\tau_0$ of $~d$ letters to it by cycling
counter-clockwise around these preimages and recording the edges we
meet. We repeat this procedure for the preimages of $~1$ and
get another permutation $~\tau_1$. That is, we obtain a
subgroup of the symmetric group $\langle \tau_0,\tau_1 \rangle \subset S_d$ which is called the monodromy
group
of the covering $\pi$. For example, the
dessin on the right is
associated to a degree $8$ map $\mathbb{P}^1 \rightarrow \mathbb{P}^1$ and if we let the black (resp. starred) vertices be
the preimages of $~0$ (respectively of $~1$), then the
corresponding partitions are $\tau_0 = (2,3)(1,4,5,6)$
and $\tau_1 = (1,2,3)(5,7,8)$ and the monodromy group
is the alternating group $A_8$ (use
GAP ).

But wait! The map is also
ramified in $\infty$ so why don\’t we record also a
permutation $\tau_{\infty}$ and are able to compute it from
the dessin? (Note that all three partitions are needed if we want to
reconstruct $C$ from the $~d$ sheets as they encode in
which order the sheets fit together around the preimages). Well,
the monodromy group of a $\mathbb{P}^1$ covering ramified only
in three points is an epimorphic image of the fundamental
group
of the sphere
minus three points $\pi_1(\mathbb{P}^1 – { 0,1,\infty })$ That is, the group of all loops beginning and
ending in a basepoint upto homotopy (that is, two such loops are the
same if they can be transformed into each other in a continuous way
while avoiding the three points). This group is generated by loops
$\sigma_i$ running from the basepoint to nearby the i-th
point, doing a counter-clockwise walk around it and going back to be
basepoint $Q_0$ and the epimorphism to the monodromy group is given by sending

$\sigma_1 \mapsto \tau_0~\quad~\sigma_2 \mapsto \tau_1~\quad~\sigma_3 \mapsto \tau_{\infty}$

Now,
these three generators are not independent. In fact, this fundamental
group is

$\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) = \langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2 \sigma_3 = 1 \rangle$ To understand this, let us begin
with an easier case, that of the sphere minus one point. The fundamental group of the plane minus one point is
$~\mathbb{Z}$ as it encodes how many times we walk around the
point. However, on the sphere the situation is different as we can make
our walk around the point longer and longer until the whole walk is done
at the backside of the sphere and then we can just contract our walk to
the basepoint. So, there is just one type of walk on a sphere minus one
point (upto homotopy) whence this fundamental group is trivial. Next,
let us consider the sphere minus two points Repeat the foregoing to the walk $\sigma_2$, that
is, strech the upper part of the circular tour all over the backside of
the sphere and then we see that we can move it to fit with the walk
$\sigma_1$ BUT for the orientation of the walk! That is, if we do this
modified walk $\sigma_1 \sigma_2^{\’}$ we just made the
trivial walk. So, this fundamental group is $\langle \sigma_1,\sigma_2~\mid~\sigma_1 \sigma_2 = 1 \rangle = \mathbb{Z}$ This is also the proof of the above claim. For,
we can modify the third walk $\sigma_3$ continuously so that
it becomes the walk $\sigma_1 \sigma_2$ but
with the reversed orientation ! As $\sigma_3 = (\sigma_1 \sigma_2)^{-1}$ this allows us to compute the
\’missing\’ permutation $\tau_{\infty} = (\tau_0 \tau_1)^{-1}$ In the example above, we obtain
$\tau_{\infty}= (1,2,6,5,8,7,4)(3)$ so it has two cycles
corresponding to the fact that the dessin has two regions (remember we
should draw ths on the sphere) : the head and the outer-region. Hence,
the pre-images of $\infty$ correspond to the different regions of the
dessin on the curve $C$. For another example,
consider the degree 168 map

$K \rightarrow \mathbb{P}^1$

which is the modified orbit map for the action of
$PSL_2(\mathbb{F}_7)$ on the Klein quartic.
The corresponding dessin is the heptagonal construction of the Klein
quartic Here, the pre-images of 1 correspond to the midpoints of the
84 edges of the polytope whereas the pre-images of 0 correspond to the
56 vertices. We can label the 168 half-edges by numbers such that
$\tau_0$ and $\tau_1$ are the standard generators b
resp. a of the 168-dimensional regular representation (see the atlas
page
).
Calculating with GAP the element $\tau_{\infty} = (\tau_0 \tau_1)^{-1} = (ba)^{-1}$ one finds that this permutation
consists of 24 cycles of length 7, so again, the pre-images of
$\infty$ lie one in each of the 24 heptagonal regions of the
Klein quartic. Now, we are in a position to relate curves defined
over $\overline{Q}$ via their Belyi-maps and corresponding
dessins to Grothendiecks carthographic groups $\Gamma(2)$,
$\Gamma_0(2)$ and $SL_2(\mathbb{Z})$. The
dessin gives a permutation representation of the monodromy group and
because the fundamental group of the sphere minus three
points $\pi_1(\mathbb{P}^1 – \\{ 0,1,\infty \\}) = \langle \sigma_1,\sigma_2,\sigma_3~\mid~\sigma_1 \sigma_2 \sigma_3 = 1 \rangle = \langle \sigma_1,\sigma_2 \rangle$ is the free group op two generators, we see that
any dessin determines a permutation representation of the congruence
subgroup $\Gamma(2)$ (see this
post
where we proved that this
group is free). A clean dessin is one for which one type of
vertex has all its valancies (the number of edges in the dessin meeting
the vertex) equal to one or two. (for example, the pre-images of 1 in
the Klein quartic-dessin or the pre-images of 1 in the monsieur Mathieu
example
) The corresponding
permutation $\tau_1$ then consists of 2-cycles and hence the
monodromy group gives a permutation representation of the free
product $C_{\infty} \ast C_2 = \Gamma_0(2)$ Finally, a clean dessin is said to be a
quilt dessin if also the other type of vertex has all its valancies
equal to one or three (as in the Klein quartic or Mathieu examples).
Then, the corresponding permutation has order 3 and for these
quilt-dessins the monodromy group gives a permutation representation of
the free product $C_2 \ast C_3 = PSL_2(\mathbb{Z})$ Next time we will see how this lead
Grothendieck to his anabelian geometric approach to the absolute Galois
group.

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