I'm right now learning about Monodromy from self-studying Rick Miranda's fantastic book "Algebraic Curves and Riemann surfaces". Today, I read about monodromy, and the monodromy representation of a holomorphic map between compact Riemann surfaces. I understand that we start by having a holomorphic map $F:X \rightarrow Y$, of degree d, where X and Y are Riemann surfaces, and then we remove the branch points from Y, and all the corresponding points in X mapping to them. Let $B=\{b_1,..,b_n\}$ be the branch points, $A=\{a_1,...,a_m\}$ the ramification points. So, fix a point $q \in V = Y-B$. We have that there are d preimages of q in $U=X-A$.

So for a specific branch point b, we choose some small open neighbourhood W of b so that $F^{-1}(W)$ gives a disjoint union $W_i$ of open neighbourhoods of the points mapping to b. Take some path from our basepoint $q$ to $q_0 \in W$, call this path $\alpha$. Choosing some small loop $\beta$ with basepoint q, with winding number 1 around b, and then considering $\alpha^{-1}\circ \beta \alpha$, gives a loop on V, based at q around b. We can now see that this loop only depends on $\beta$, in some sense.

Say that the points that maps to b has multiplicity $n_i,...,n_j$. Then we have that, according to local normal form, there are local coordinates $z_j$ on the open neighbourhoods from above, so that the map takes the form $z=z_j^{n_j}$. Now, we have that the loop around b, when we lift it up here, will simply yield a cyclic permutation of the preimages in the neighbourhood.

Now, my question is mostly: How do I apply this concretely? Let us take an example (from Miranda's book) : "Let $f(z) = 4z^2(z-1)^2/(2z-1)^2$ define a holomorphic map of degree 4 from $P^1$ to itself. Show that there are three branch points, and that the three permutations in $S_4$ are $\rho_1=(12)(34)$, $\rho_2(13)(24)$ and $\rho_3=(14)(23)$ up to conjugacy." I can find the branch points, and I see that the multiplicity of the two points mapping to it has multiplicity 2, but I don't get how to rigorously show that the above are the associated permutations.

Hope I was clear, and sorry if I wasn't.

UPDATE Now, rereading the question properly, maybe he doesn't want me to find the specific permutations, but just simply showing that they have that conjugacy class. I think that is the case. But I would still be curious of how to find the specific permutation that the monodromy induces.


I'm sure there are better ways to do it, but as nobody has answered I hope this helps ...

I would proceed following the description you give in the following way, first the branch points are $0,-1$ and $\infty$, so if we shade the points that are sent to the upper plane under the map $$ z \to \frac{4z^2(z-1)^2}{(2z-1)^2} $$ we obtain a triangulation of the sphere were the ramification points will be placed in the vertex and the edges are sent to one of the three segments of the real line $(-\infty,-1),(-1,0),(0,\infty)$. Now give a number to each white triangle. A small loop around one of the branch points, say $\beta$ will come from disjoint loops say $\gamma_1, \gamma_2, \dots$ around some of the vertex, now if $\gamma_1$ cuts the triangles numbered $i_1,j_1,k_1, \dots$ (in counter clock wise order), $\gamma_2$ cuts the triangles numbered $i_2,j_2,k_2,\dots$ and so on then associate to the branch point $\beta$ the permutation $(i_1 j_1 k_1\dots)(i_2j_2k_2\dots)\dots$. In your example we have the following figure:


in this figure the two white dots are sent to $0$, the two filled ones to $-1$ and the cross (and $\infty$) to $\infty$ the preimage of a loop around $0$ will then be associated to the permutation $(14)(23)$ as small loops around the left white point cross the triangles 2 and 3 and a loop around the right white point crosses the triangles 1 and 4. Identically a loop around $-1$ with the permutation $(12)(34)$ and one around $\infty$ with $(13)(24)$. Renumbering the triangles in any way lead to conjugate triples.

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    $\begingroup$ Great answer, thank you! $\endgroup$ – Dedalus Jun 22 '11 at 9:19

If you have a branch of order two over a special point, the induced permutation is a two-cycle that switches the sheets connected to the ramified point, and if you have two order-two points over your branch point, you get a product of two disjoint 2-cycles. In $S_4$, there are only three elements of this form, which you listed in the statement of the question.

The remaining problem is to see how the permutations over two branch points relate to each other. The rule to remember is that the product of all three monodromies is trivial, since the composition of three loops yields a contractible loop. If two permutations are equal, then they multiply to identity. This is not allowed, because multiplying with the monodromy around the third point needs to be trivial, and that forces the third element to be trivial. The remaining option is that the three monodromies are matched with the three possible products of two transpositions in some way.

This answer is unique in the sense that the action of $S_4$ by renaming the numbers is transitive on the set of admissible assignments. Indeed, it factors through the permutation action of $S_3$ on the three group elements in question.


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