permutation representations of monodromy groups

By lieven

Dessins d'enfants

Monsieur Mathieu
The best rejected proposal ever
The cartographers’ groups
the cartographers’ groups (2)
the noncommutative manifold of a Riemann surface
noncommutative curves and their maniflds
permutation representations of monodromy groups
anabelian geometry

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 (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 points of lying over any other point of $\mathbb{P}^1$ (we call the degree of $\pi$ ) and that the number of points over ~0,1~ and $~\infty$ is smaller than . To such a map we associate a dessin d\’enfant, a drawing on linking the pre-images of and with exactly edges (the preimages of the open unit-interval). Next, we look at the preimages of and associate a permutation $\tau_0$ of letters to it by cycling counter-clockwise around these preimages and recording the edges we meet. We repeat this procedure for the preimages of 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 map $\mathbb{P}^1 \rightarrow \mathbb{P}^1$ and if we let the black (resp. starred) vertices be the preimages of (respectively of ), 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 from the 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 $\sigma1$ 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 . 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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anabelian, Brauer, Galois, Grothendieck, groups, Klein, Mathieu, permutation representation, representations, Riemann

1 comment
Posted in geometry, groups, modular
Written on Tue, 20 March 2007 at 3:22 pm
Tags: anabelian, Brauer, Galois, Grothendieck, groups, Klein, Mathieu, permutation representation, representations, Riemann
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One Response to “permutation representations of monodromy groups”

recycled : dessins at neverendingbooks Says:
December 27th, 2007 at 3:57 pm
[...] Permutation representations of monodromy groups how dessins give such representations. [...]