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Hi everyone. My question is about the absolute Galois group action on the set of the Grothendieck dessins. The dessins I am interested in are trees with only one vertex of valency more then 2. (I don't know if there is a generally accepted term in graph theory. Starlike trees?). What exactly is known about them? Is the action transitive, at least for trees with 3 terminal vertices (with the same valency lists)?

EDIT: I see some clarification is necessary. In fact, I consider the alternating dessins, so there are two valency lists for each of them. (It is convenient to assume that the "center" is always, say, black). As Will Savin pointed out, a dessin with rotational symmetry cannot turn into one with less symmetry. I confess I missed this obvious fact, but certainly this is not what I had in mind. The question was actually about nontrivial obstacles to transitivity. (I know there are some for general trees).

Then, "what is known" part of the question. The best answer would be a long list of references.

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  • $\begingroup$ clean dessins, or alternating black and white vertices? $\endgroup$
    – Will Sawin
    Jun 4, 2012 at 17:20
  • $\begingroup$ I'd be very surprised if the transitivity were known, just because there are not very many methods for proving things of that kind. $\endgroup$
    – JSE
    Jul 3, 2012 at 13:17

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No. The automorphism group of a dessin is the automorphism group of that covering, and is thus Galois-invariant. Choose a tree that has 3-fold rotational symmetry, and a tree with the same valency list that does not. Then these two trees are not Galois conjugate.

The smallest example is, for clean dessins, a 7-vertex symmetric tree and any other 7-vertex tree, and for alternating desins, a 10-vertex symmetric tree and any other 10-vertex tree with white terminal vertices.

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    $\begingroup$ daseins = Heidegger's drawings? $\endgroup$
    – user2035
    Jun 4, 2012 at 17:40
  • $\begingroup$ Fixed, thus unfortunately ruining the joke. $\endgroup$
    – Will Sawin
    Jun 4, 2012 at 17:55

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