# For which polynomials $p\in\mathbb{C}[w]$ are the branches of the inverse $p^{-1}$ expressible using algebraic operations?

The collection of all degree-$n$ polynomials in the variable $w$ (call this set $\mathbb{C}[w]_n$) can be identifies with $\mathbb{C}^{n+1}$ by the bijection $F:\mathbb{C}^{n+1}\to\mathbb{C}[w]_n$ defined by $$F:(a_0,a_1,\ldots,a_n)\mapsto w=p(z)=a_0+a_1z+\cdots+a_nz^n.$$ Let $\mathcal{A}_n\subset\mathbb{C}^{n+1}$ denote the set of polynomials (using the above identification) defined by saying that $p(z)\in\mathcal{A}_n$ if and only if each branch of the inverse $z=p^{-1}(w)$ is expressible as an explicit formula using finitely many algebraic operations (ie addition/subtraction, multiplication/division, and roots). That is, something like $$z=p^{-1}(w)=\sqrt{w+\sqrt{2/w}}.$$ Note that for any polynomial $p(z)\in\mathbb{C}^{n+1}$, either each branch of $p^{-1}$ is so expressible, or none is (since every branch can be reached as an analytic continuation of any other).

My Question: What is the structure of $\mathcal{A}_n$ in $\mathbb{C}^{n+1}$? Is it an algebraic variety? What is its dimension? What can we say about its topology?

• are you basically asking about the polynomials that are solvable in radicals ? Commented Aug 4, 2018 at 20:08
• @mercio, yes, that is right. Commented Aug 5, 2018 at 22:59

So you are asking if we can tell when an extension $$\Bbb C(w = p(z)) \subset \Bbb C(z)$$ is solvable by radicals, so you want to study the Galois group of that extension and see if it is a solvable group. The answer to your question is that yes, this can be given by algebraic equations.

This is a pretty geometric question, because there is a correspondance between extensions of $$\Bbb C(w)$$ of dimension $$n$$ and $$n$$-sheeted branched coverings of the Riemann sphere, and the Galois group in the algebraic side corresponds to the group of deck transformations in the geometric side.

If you choose a finite subset $$\{w_1 ; \ldots ; w_k\}$$ of the Riemann sphere and fix some non-intersecting branch cuts between $$w_0$$ and each $$w_i$$, then you can make a new Riemann surface by building an $$n$$-sheeted covering branched at those points, by appropriately choosing permutations of the branches $$\alpha_i$$ when looping around each of the $$w_i$$. Given a $$k$$-uple of elements of $$S_n$$ $$\alpha = (\alpha_1, \ldots, \alpha_k)$$, the corresponding covering is connected if and only if the subgroup $$G_\alpha$$ of $$S_n$$ generated by the $$\alpha_i$$ is transitive. If so the covering defines a Riemann surface, and the meromorphic functions on it are a field extension of $$\Bbb C(w)$$ with Galois group $$G_\alpha$$.
Finally, two $$k$$-uples $$\alpha^1,\alpha^2$$ give isomorphic coverings if they are the same after reordering the sheets, that is, if there is a $$\beta \in S_n$$ such that $$\alpha_i^1 = \beta \alpha_i^2 \beta^{-1}$$ for $$i \in \{1 \cdots k\}$$, so we will only be interested in equivalence classes with regard to this. Let us denote this relation with $$\sim$$.

So to understand the Galois group of polynomial maps we simply have to look at their possible branching data.

Because polynomials have an $$n$$-uple pole at infinity (around which a loop makes an $$n$$-cycle permutation of the sheets), there is an $$n-1$$-uple branch point above $$w_n = \infty$$, and there are another $$n-1$$ branch points (counted with multiplicity) above the critical values of $$p$$, that is the $$w_i = p(z_i)$$ where the $$z_i$$ are the critical points of $$p$$, the points where $$p'(z)$$ vanishes.
Generically, those $$n-1$$ critical values are all distinct and the loops induce simple transpositions there, so the choice of the branching data amounts to the choice of $$n-1$$ transpositions the composition of which give an $$n$$-cycle.
Let $$T_n \subset S_n$$ be the subset of transpositions, and let $$G_n = \{ (\alpha_1,\ldots,\alpha_{n-1}) \in T_n^{n-1} \mid \alpha_1\cdots \alpha_{n-1} = (123\cdots n)\} / \sim$$, the set of possible polynomial branching data on $$n-1$$ distinct branch points.

In the non-generic case, some critical values have higher multiplicity, and you can obtain their data by starting with a generic one and composing some of the transpositions together. This is also the only way that the group can possibly be smaller than $$S_n$$.

For every such branching data, the covering will be a Riemann surface of genus $$0$$, so another Riemann sphere, and by deciding that the preimage of $$w_n = \infty$$ should be $$z_n = \infty$$ the covering becomes a polynomial map. Two polynomial maps give the same branching data if and only if they differ by pre-composition with an automorphism of the Riemann sphere, and since polynomials fix $$\infty$$, they have to differ by a map of the form $$z_1 = az_2+b$$ from some nonzero $$a \in \Bbb C$$. Then, if we restrict ourselves to polynomials with $$a_n=1$$ and $$a_{n-1}=0$$, then for $$n \ge 3$$ there are exactly $$n$$ possible polynomials $$p(z), p(\zeta_n z), p(\zeta^2 z)$$ etc (for $$n=2$$ there is only one because if $$p(x) = x^2+a_0$$, $$p(-x) = p(x)$$).

Let $$g_n$$ be the cardinal of $$G_n$$. Geometrically, the space of polynomials (up to precomposition) is a $$g_n$$-sheeted branched covering of $$Sym^{n-1}(\Bbb C)$$, and we again have an isomorphism of groups between the (very large !!) Galois group and the deck transformations of that covering. This group, as a subgroup of permutations of $$G_n$$, is generated by the $$n-2$$ operations you get by replacing $$(\alpha_i, \alpha_{i+1})$$ with $$(\alpha_i\alpha_{i+1}\alpha_i^{-1},\alpha_i)$$ (think of this as moving two critical values in order to swap them). For $$G_n$$, it is transitive, but that won't be the case when you start merging elements of $$\alpha$$, then the number of orbits tells you the number of components you should expect.

Now let us go to the algebraic side.

Given a degree $$n$$ polynomial $$p(z) = \sum_{i=0}^n a_i z^i$$, if you denote $$z_1 \ldots z_{n-1}$$ the critical points, and $$w_1,\ldots,w_{n-1}$$ the critical values, then the polynomial expression $$q(w) = \prod_{i=1}^{n-1} (w - w_i) = \prod_{i=1}^{n-1} (w - p(z_i))$$ is symmetric in the $$z_i$$.
Call $$b_i$$ the coefficients of $$q$$ ($$q(w) = w^{n-1} + \sum_{i=0}^{n-2} b_i w^i$$).
Given that $$p'(z) = na_n \prod (z - z_i)$$, it is possible to express $$b_i$$ as polynomial expressions in terms of $$p'(z)/na_n$$, which are themselves algebraic in the $$a_i$$.

When you assume $$a_{n-1}=0$$ and $$a_n=1$$, they become polynomial in the $$a_i$$, and the field extension $$K_n = \Bbb C(q_0,\ldots,q_{n-2}) \subset L_n = \Bbb C(a_0,a_1, \ldots, a_{n-2})$$ has finite degree $$d_n$$. From what we know of the geometric side, we should have $$d_n = n g_n$$ for $$n \ge 3$$ and for $$n = 2$$, $$d_n = g_n$$.
(it is possible to compute $$d_n = n^{n-2}$$, so $$g_n = n^{n-3}$$ for $$n \ge 3$$)

The places where this cover can branch is when two critical values coincide, which means when the discriminant $$\Delta$$ of $$q$$ vanishes. Meanwhile, the only chance for your polynomial to not have Galois group $$S_n$$ is when you merge some critical values together in order to push all of the generators of the deck transformation into a subgroup of your choice. And this gives a polynomial equation $$\Delta_n(a_0, \ldots, a_{n-2}) = 0$$ that has to vanish for your polynomial to be non-generic.

There is a lot of information encoded in the discriminant $$\Delta_n(a_0, \ldots,a_{n-2})$$. It doesn't actually depend on $$a_0$$. The number of components and how they intersect themselves or each other into a number of singularity loci and so on all can all be described combinatorially on the geometric side.

The discriminant of $$p'$$ has to divide it three times (because if two critical points coincide, then two critical values will also coincide, and do so by making a $$3$$-cycle holonomy).
Its other irreducible factors have exponent $$2$$ (there is one for each geometrically different way to merge two commuting critical values).

At this point I should illustrate with what happens for small $$n$$

I don't know how to easily get nice systems of equations for the lower dimensional pieces, nor do I know how to interpret the various multiplicites you see in the geometric/combinatoric side into algebraic statements (at least in the case $$n=5$$, I did see lots of factorisations happening)

degree $$2$$ : $$p(z) = z^2 + a_0$$

$$G_2$$ only has one element, $$[(12)]$$, so $$g_2=1,d_2=1$$ and so $$K_2 = L_2$$.
There is only one possible geometry here. Given a critical value $$w_1$$, there is only one $$2$$-sheeted covering branched at $$w_1$$ and $$\infty$$, and there is only one possible polynomial, $$p(z) = z^2 + w_1$$.

degree $$3$$ : $$p(z) = z^3 + a_1z + a_0$$

Again $$G_3$$ is the singleton $$[(12),(23)]$$, so we have $$g_3=1$$ and $$d_3=3$$.

Some quick computation shows that
$$b_1 = -2a_0, b_0 = a_0^2 + 4a_1^3/27 : K_3 = \Bbb C(a_0, a_1^3)$$.

$$\Delta_3(a_0,a_1) = b_1^2 - 4b_0 = 4a_1^3/27$$, so it only contains the cube of the discriminant of $$p'$$, as expected.

The Galois group of $$p$$ is $$S_3$$ unless $$a_1=0$$ and then it is a cyclic group of order $$3$$.

degree $$4$$ : $$p(z) = z^4 + a_2z^2 + a_1z + a_0$$

Now things start getting a bit more interesting : $$G_4 = \{[(12),(23),(34)],[(12),(23),(24)],[(12),(23),(14)],[(12),(34),(13)]\}$$.

Merging two critical values gives two orbits, that of $$[(123),(34)]$$ and that of $$[(12)(34),(13)]$$, which reflects the factorisation $$\Delta_4 = (-27a_1^2-8a_2^3)^3 a_1^2/2^{16}$$.

If the first factor vanishes the Galois group is still $$S_4$$, but if $$a_1$$ vanishes then the Galois group of $$p$$ is the diedral group of order $$8$$. If both vanish ($$a_1=a_2=0$$) then you get a cyclic group of order $$4$$

Also note that the curve $$(-27a_1^2-8a_2^3)=0$$ is pinched at $$a_1=a_2=0$$.

degree $$5$$ : $$p(z) = z^5 + a_3z^3 + a_2z^2 + a_1z + a_0$$.

The number of branching data is increasing quickly, since now $$G_5$$ has $$25$$ elements (in a single orbit) So $$L_5$$ should be an extension of degree $$125$$ of $$K_5$$ (I didn't check that).

Moving down a dimension, all the possible ways to merge two critical values fall (again, and forever) in two orbits, which reflects the factorisation (up to a multiplicative constant)

$$\Delta_5 = (-16a_3^6+224a_3^4a_1-88a_3^3a_2^2-1040a_3^2a_1^2+360a_3a_2^2a_1+135a_2^4+1600a_1^3)^2 \times \\ (-81a_3^4a_1+27a_3^3a_2^2+360a_3^2a_1^2-540a_3a_2^2a_1+135a_2^4-400a_1^3)^3$$

The surface $$S_1$$ given by the first factor corresponds to the orbit of $$[(12),(24),(23)(45)]$$, while the surface $$S_2$$ corresponds to the orbit of $$[(12),(23),(345)]$$.

Polynomials falling on any smooth point of the $$\Delta_5=0$$ surfaces still have Galois Group $$S_5$$

Going down a dimension, we now get five orbits, so five curves. (by the way, the curves will always have genus $$0$$ because they are branched coverings of the Riemann sphere with only two branch points).
Hopefully, I paired the curves with the orbits properly :

$$[(12), (2345)]$$ : where $$S_1$$ and $$S_2$$ intersect and $$S_2$$ is singular.
It is parametrized with $$p(z) = z^5-10t^2z^3+20t^3z^2-15t^4z+a_0$$.
The Galois group is $$S_5$$.

$$[(13), (345)(12)]$$ : where $$S_1$$ and $$S_2$$ intersect and $$S_1$$ is singular.
It is parametrized with $$p(z) = z^5-15t^2z^3+10t^3z^2+60t^4z+a_0$$.
The Galois group is $$S_5$$.

$$[(123), (345)]$$ : where $$S_2$$ intersects itself.
It is parametrized with $$p(z) = z^5+10tz^3+45t^2z+a_0$$.
The Galois group is $$A_5$$.

$$[(124), (45)(23)]$$ : where $$S_1$$ and $$S_2$$ intersect transversally.
It is parametrized with $$p(z) = z^5+30t^2z^3+100t^3z^2+105t^4z+a_0$$.
The Galois group is $$A_5$$.

$$[(23)(14), (45)(13)]$$, where $$S_1$$ intersects itself.
It is parametrized with $$p(z) = z^5+5tz^3+5t^2z+a_0$$.
The Galois group is the diedral group of order $$10$$, so there you have your solvable degree $$5$$ polynomials.

Down another dimension, you end up at the point $$(a_1=a_2=a_3=0)$$, at the intersection of every curve, the Galois group there is cyclic of order $$5$$.