# Do the bounded isophase lines of a complex polynomial $f$ through the zeroes of $f’$ define a spanning tree?

Let $$f: \mathbb{C} \to \mathbb{C}$$ be a polynomial and let $$\arg(f(z))$$ be the phase of $$f(z) = | f(z)| \exp(\mathrm{i} \arg(f(z)))$$. The zeroes of $$f'(z)$$ are saddle points of $$\arg(f(z))$$, i.e. intersections of isochromatic lines (if we color-encode the phase, cf. http://www.ams.org/notices/201106/rtx110600768p.pdf, page 772ff). In the following we assume $$f$$ to have only simple zeroes.

Questions:

a1) Is it correct that through every (generic) saddle point of $$\arg(f)$$ (= zero of $$f’$$) there are two isochromatic lines, one of which is unbounded and one of which connects two zeroes of $$f$$?

a2) If a1) is correct: Is there a simple (geometric) criterion determining which zeroes are directly connected via an isochromatic line?

b1) Is it correct that the set of isochromatic lines which connect two zeroes define a spanning tree (of all zeroes)?

b2) If b1) is correct: Can the (graph theoretical, i.e. topolocial) structure of the spanning tree be solely defined by the position of the zeroes of $$f$$ (e.g. minimal weigth spanning tree where weight is square of Euclidean distance of zeroes)?

The figure shows the color-encoded phase plot of an example of a polynomial of degree four (the thin black lines are lines of equal absolute value, they are orthogonal to the isochromatic lines). The four zeroes of $$f$$ can be seen as the minima of the absolute value. The isochromatic lines through the zeroes of $$f’$$ are the thick black lines. The three chromatic saddle points (=zeroes of $$f’$$) are clearly visible. In this example the spanning tree has one central vertex (the zero of $$f$$ below the middle) from which there are edges to the other three zeroes of $$f$$.

An electrostatic interpretation could possibly be helpful. If we consider the zeroes of $$f$$ to be equal charges in 2d, the question amounts to how can one know which charges are connected via a field line of the electrostatic field.

Motivation:

Every complex polynomial with $$n$$ distinct zeroes has $$n-1$$ critial points (zeroes of $$f’$$). For the location of the zeroes there is huge literature (for a starting point see e.g. Q.I. Rahman und G. Schmeisser, Analytic Theory of Polynomials, Oxford University Press, 2002).

From numerical studies it seems to be interesting to ask which spanning tree of the zeroes could be helpful in describing the location of the critical points (note that the spanning tree through $$n$$ vertices has $$n-1$$ edges).

The question was inspired by Alth\“ofer/Voigt, Spiele, R\“atsel, Zahlen, Springer Spektrum 2014 and Kaffka, Zuordnungen zwischen Nullstellen und kritischen Punkten von Polynomen, Diploma Thesis supervised by Prof. Dr. I. Alth\"ofer, Jena, 2012, page 42, http://www.althofer.de/kaffka-diplom.pdf.

In this context I have proven a slight strengthening of the theorem of Gauss-Lucas for polynomials of degree four (http://arxiv.org/abs/1405.0689).

• You should include in your question the precise definition of "isochromatic lines". And of the $\arg z$ too. Argument is not a function. – Alexandre Eremenko Jul 13 '15 at 7:13

Edited

As @Alexandre Eremenko noted, in the generic case the answer is yes, and in fact it can be shown that any connected finite tree can appear as the isochromatic tree graph of some polynomial (see Theorem 6.1 here, this essentially has the parallel result for level curves, an induction argument will give you the desired application to isochromatic trees).

In the non-generic case the picture is even more interesting, and the isochromatic tree graphs must allow critical points as vertices as well, as I describe below.

The sets you are considering I will call "gradient lines" of $f$, namely components of the sets $\{z:\operatorname{Arg}(z)=\theta\}$ for various values of $\theta\in[0,2\pi)$. First I need a bit of info about the orthogonal objects, namely the level curves of $f$ (sets where $|f|$ is constant). (I call them "orthogonal" because at any point $z$ not a critical point of $f$, the level curve and gradient line of $f$ containing $z$ are orthogonal to each other at $z$.)

EDIT: I just realized that I don't think I need the second bullet point in the theorem below for this answer, but it is none the less true and I think interesting.

THEOREM Let $p$ be a polynomial, and let $\Lambda$ be a level curve of $p$ (ie. a component of the set $\{z:|p(z)|=\epsilon\}$ for some $\epsilon>0$).

• $\Lambda$ consists of a connected finite graph whose vertices are critical point of $f$, and satisfying the following properties.

1.) There are evenly many (and more than two) edges of $\Lambda$ incident to each vertex of $\Lambda$.

2.) Each edge of $\Lambda$ is incident to both the unbounded face of $\Lambda$ and a bounded face of $\Lambda$.

That is, $\Lambda$ is a sort of complicated "figure eight" graph.

• Let $D$ be a bounded face of $\Lambda$. Then either $f$ has a single distinct zero of $f$ in $D$, or there is some unique level curve $\Lambda_D$ in $D$ which is maximal in the sense that every critical point of $f$ in $D$ is either in $\Lambda_D$ or contained in a bounded face of $D$, and every zero of $f$ in $D$ is contained in a bounded face of $D$.

Now we can define the non-generic version of the isochromatic graphs. For any point $z\in\mathbb{C}$, let $\Lambda_z$ denote the level curve of $p$ containing $z$, and let $\Gamma_z$ denote the gradient line of $p$ containing $z$.

Let $z_o$ be a critical point of $p$. Let $D_1,D_2,\ldots,D_k$ denote the bounded faces of $\Lambda_{z_o}$ which are incident to $z_o$. For each $i\in\{1,2,\ldots,k\}$ a portion of the gradient line $\Gamma_{z_o}$ extends from $z_o$ into $D_i$, and ends at a zero or critical point of $p$ in $D_k$. Let therefore $A_{z_o}$ denote the portion of $\Gamma_{z_o}$ which is contained in the bounded faces $D_1,D_2,\ldots,D_k$, stopping when a zero or other critical point of $p$ is reached (ie. if a gradient line of $p$ reaches down from $z_o$ to a zero of $p$ which has multiplicity greater than $1$, then other branches of that gradient line will emanate from that zero, and we want to disregard those other branches, similarly for other critical points).

If we now let $\mathcal{T}$ denote the union of the $A_z$ over all critical points $z$ of $p$, then $\mathcal{T}$ has the following nice properties.

• $\mathcal{T}$ is a connected finite tree, the vertices of which are either critical points or zeros of $p$.
• Every zero and critical point of $p$ appears as a vertex of $p$.
• $\operatorname{Arg}(p(z))$ is constant on each edge of $\mathcal{T}$.
• $\mathcal{T}$ is the minimal such tree that satisfies the above criteria (that is, it is the intersection of all such trees).

In my own work I have classified the way in which the critical level curves of the polynomial $p$ (level curves containing critical points) may be arranged. This geometric property of of polynomials is a strong conformal invariant. That is, for any two polynomials $p_1$ and $p_2$, they have (geometrically) the same configuration of critical level curves if and only if there is a degree $1$ polynomial $\phi:\mathbb{C}\to\mathbb{C}$ such that $p_2=p_1\circ\phi$ on $\mathbb{C}$. A similar analysis might be done using the trees described above. See papers 2, 3, and 4 here for this work.

EDITED.

As I understand from the reference you give, "isochromatic lines" are simply the lines where the argument is constant. In other words they are preimages of the rays from the origin.

Assume that we are in general position, that is critical points are simple and no critical values have the same argument.

Then indeed from every critical value there is a segment to 0 which contains no other critical values. Preimage of this segment is an "isochromatic line" which contains two zeros. Through the same critical value also a ray goes to infinity. Thus in the generic case, from each critical point there is an "isochromatic lines" which ends at two zeros, and another, perpendicular one whose both ends go to infinity.

Now, consider those isochromatic lines that connect zeros. Their union is evidently connected. It also cannot separate the plane, because if there is a bounded component of the complement, then what will its image be? Thus this is a tree which has zeros as vertices.

It is not difficult to classify topologically all pictures which will arise from polynomials.

EDIT 2. The moral is that the really useful geometric picture of an analytic function is the Riemann surface of the inverse function spread over the plane/sphere.

• If I understand correctly, the questions are about isochromatic lines connecting a pair of zeros, not a pair of critical points. – Malik Younsi Jul 13 '15 at 12:13
• @Mark Younsi: Yes, thanks, I corrected my answer. – Alexandre Eremenko Jul 13 '15 at 14:14

You are asking about the foliation (with some singular points) provided by the preimages of radial lines. Clearly you can say the following:

a1. Generically, every critical point is simple, and all critical values have different arguments. Hence clearly there are four lines meeting at each critical point, and analytic continuation of two of these will lead to a zero, while the other two will lead to infinity. Hence the answer is positive.

a2. I am not quite sure what you are looking for here. There are a number of ways to describe the mapping behaviour of the polynomial (classical "line complexes", "dessins d'enfants" in the case where there are only two critical values, various constructions connected with studies of Hurwitz numbers), and most of these should allow you to read off the desired information. But your construction is very similar to this already.

b1. To see that the set $A$ consisting of the lines that you are considering indeed forms a spanning tree of the zeros, we must only prove that it is connected with connected complement. Note that $A$ is the preimage of a "star" $S$ connecting zero to each of the critical values. The map from the complement of $A$ to the complement of $S$ is a covering map on each component. Since the complement of $S$ is topologically a punctured disc, and since your map is a polynomial and hence has no poles, it follows easily that there is only one complementary component, which is homeomorphic to a punctured disc. So the answer is again positive.

EDIT. Rereading your question, I now see that by "geometric", you mean, "in terms of the position of the zeros". I think that, in the way you mean the question, the answer is negative. For example, by work of Bishop ("True trees are dense"), your set can approximate any given compact set arbitrarily closely - even for polynomials with just two critical values, say $0$ and $1$, where your tree is the "dessin d'enfant" (and the polynomial is called a "Shabat polynomial"). In particular, you can make a polynomial of this type for which the tree contains some vertices that are very close together, but not connected by any short path - and some vertices relatively far together that are connected by an edge. [Basically, you start Bishop's construction with an arc of a circle that just leaves out some very small interval.] This seems to exclude any of the kind of answers that you are looking for.