Last week, the great Naiomi Cameron visited me for a few days to discuss some new directions about Shabat polynomials. I’ve been horrible about posting on this blog, so now that I’ve been motivated to work on Shabat polynomials again, I figured it’s time for me to write!
is a Shabat polynomial which happens to be the composition of two other Shabat polynomials. The first has monodromy group as a cyclic group, while the second has monodromy group as an alternating group. The monodromy group of the composition has order . Do we have as the wreath product of by ?
Belyi Maps, Degree Sequences, and Passports
Let me begin by establishing some notation and making some remarks. A lot of this to begin with will be background information, so I apologize to those who already know this stuff pretty well.
A Belyi map is a morphism which is branched only above three points . When is a Riemann sphere, a Belyi map must be the ratio of two polynomials having complex coefficients. Rather implicitly:
Once we let be the number of times (, or , respectively) appears in the exponents (, or , respectively), we define the degree sequence of a Belyi map as a multiset of multisets:
and the passport of a Belyi map as shorthand notation for the degree sequence:
These quantities give combinatorial information about the critical points of , namely, the three sets , , and . For example, they correspond to partitions of the integer into , , and parts, respectively.
we may as well assume that a Shabat polynomial is a Belyi map branched only above such that and . In particular, we can factor a Shabat polynomial as
We often omit the information about infinity and write the degree sequence/passport of a Shabat polynomial as
where as before
There is a simple way to generate lots of examples of Shabat polynomials. Say that is one example of a Shabat polynomial. It is well-known that we can generate more Shabat polynomials as the composition of two maps and as long as is a Shabat polynomial which satisfies . One obvious such map is , although of course there are many others. I will return to this construction later.
Dessins d’Enfants and Monodromy Groups
- The “black” vertices are ;
- The “white” vertices are ;
- The “midpoints of the faces” are ; and
- The “edges” are as the inverse image of the line segment .
The Dessin d’Enfant will have vertices, edges, and faces, so the Euler characteristic forces . When is a Shabat polynomial, there is only face, so the corresponding Dessin d’Enfant will be a tree. There is a nice post on MathOverflow which discusses this in a bit more detail.
In addition to Belyi maps/Shabat polynomials and degree sequences /passports , there is a third object of interest. The monodromy group is a transitive subgroup of $S_N$ which is constructed as follows.
- Label the edges of from 1 through .
- For each “black” vertex , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- For each “white” vertex , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- For each “midpoint of edge” , read off the labels of the adjacent edges going counterclockwise, say as a -cycle in the symmetric group.
- Form that subgroup of which is generated by the products of disjoint cycles , and .
If is a different set of permutations arising from this construction but from a different labeling of the edges, then there is a permutation so that these are simultaneously conjugate to , that is, , , and . This permutation must send the edge labels , , and .
It is not too difficult to recover the Dessin d’enfant from a triple . Indeed, the number (, respectively) of disjoint cycles in the decomposition of (, respectively) gives the number of “black” (“white”, respectively) vertices, and one can read off the cycle notation to find the labels of the edges which are adjacent to each “black” (“white”, respectively) vertex. One can connect the edges then slide around the vertices as necessary so that the faces match the disjoint cycle decomposition of .
Relating Belyi Maps and Monodromy Groups
Fix a positive integer , and let denote the symmetric group of degree , and let be a multiset of multisets such that .
Proposition. The following are equivalent:
- is the degree sequence of a Belyi map of degree on the Riemann sphere .
- There is a triple of elements in which can be expressed as a disjoint product of cycles , , and such that and is a transitive subgroup of .
The result was established by Adolf Hurwitz in his 1891 paper Ueber Riemann’sche Flachen mit gegebenen Verzweigungspunkten. I’ve explained how to construct a degree sequence from a Belyi map, but of course the opposite direction is an active area of research.
Proposition. All permutations of cycle type are conjugate, that is, for some . In particular, all triples of “cycle type ” are conjugate, that is, , , and for some .
This is an elementary fact about permutations.
Proposition. There is a one-to-one correspondence between:
- Belyi maps of degree on some compact connected Riemann surface .
- Triples of elements in , modulo simultaneous conjugation, such that and is a transitive subgroup of .
Proposition. Say that is a triple of elements in such that , and denote as the subgroup of generated by them. Denote as the centralizer of in . There is a one-to-one correspondence between:
- The set of all triples , modulo simultaneous conjugation, such that and , for some
- Double cosets in .
These correspondences between Belyi maps, triples , and double cosets were established by Michael Klug, Michael Musty, Sam Schiavone, and John Voight, in their 2013 paper Numerical calculation of three-point branched covers of the projective line. You can find the paper online at http://arxiv.org/abs/1311.2081v3. For the third correspondence, I’ve explained how to construct a triple of permutations from a Belyi map. Here’s a quick sketch of the fourth correspondence: say that and are two triples such that
for some , so denote and . Assume that they generate the same the double coset , that is, there exist and such that . We have the identity
in terms of so that the two triples are simultaneously conjugate. Conversely, assume that the triples are simultaneously conjugate. Then the elements and , so that and generate the same the double coset .
Algorithm: Counting Belyi Maps from Degree Sequences
The monodromy group helps to count the number of Belyi maps of degree . Here’s an algorithm:
- Fix a positive integer .
- Compute all multisets of multisets
such that .
- For each , find a triple of elements in which can be expressed as a disjoint product of cycles , , and such that and is a transitive subgroup of . If no such triple exists, there is no Belyi map for which is a degree sequence.
- Compute and as the centralizers of and in . Given a double-coset representative in , the triple corresponding to and gives a unique Belyi map.
Let me give a relatively simple example. The rational is a Shabat polynomial of degree whose passport is , Dessin d’Enfant is the Star Graph with spokes, and whose monodromy group is as the cyclic group of order . To see why the latter is true, observe that
- the “black” vertex is ,
- the “white” vertices are for ,
- the “midpoint of the face” is , and
- the edges are the segments for .
If we label edge as the positive integer “”, then is an -cycle while is trivial. Hence the group generated by , , and is .
In fact, I can do even better: I claim this is the only Shabat polynomial associated with this passport! Indeed, if we have a triple associated with the passport , then must be the trivial permutation, and so the centralizer . This means there are no nontrivial double cosets in , so there can only be one Shabat polynomial.
is a Shabat polynomial of degree whose passport is . Observe that is the composition of the two rational functions and which are both Shabat polynomials, one of degree with passport , and the other of degree with passport . Since , we see that is the Star Graph and is the cyclic group of order 2. It is easy to see that the monodromy group of has the generators
so that is the alternating group. Here’s the strange part. The monodromy group of the composition has the generators
so that is a group of order .
Question. is a composition of Shabat polynomials, where and . Since , do we have as the wreath product of by ?
I’m really confused by this example because it seems to be part of a general phenomenon… I will post more on this later as I come up with more examples — and possibly a proof!