Sharp bounds for the valence of certain harmonic polynomials

D. Khavinson and G. Swiatek proved that harmonic polynomials p(z)+q(z), where p is holomorphic, q is antiholomorphic, and deg p = n>1 = deg q, can have at most 3n-2 complex zeros. We show that this bound is sharp for all n by proving a conjecture of Sarason and Crofoot about the existence of holomorphic polynomials p which map all critical points to their complex conjugates. We also count the number of equivalence classes of real polynomials solving this system of equations. The methods employed rely on Thurston's characterization of post-critically finite rational maps, in particular the results on polynomials by S. Levy and A. Poirier.


Introduction and Results
In [KS] it was proved that harmonic polynomials z − p(z), deg p = n > 1, have at most 3n − 2 complex zeros. An example was provided that this bound is sharp for n = 3 and it was conjectured that it is sharp for every n. The main result of this note is the following theorem, which was conjectured by Crofoot and Sarason [KS,Conj. 1].
Theorem 1. For every n > 1 there exists a complex analytic polynomial p of degree n and mutually distinct points z 1 , z 2 , . . . , z n−1 with p ′ (z j ) = 0 and p(z j ) = z j .
In particular, this immediately implies the sharpness of the bound 3n − 2 for every n > 1.
Corollary 2. For every n > 1 there exists a complex analytic polynomial p of degree n such that p(z) − z has 3n − 2 zeros.
It was proved in [KS] that Theorem 1 implies Corollary 2. However, for the convenience of the reader we will provide an alternative argument to derive this Corollary after the proof of the Theorem. The main ingredient in the proof of Theorem 1 is an explicit construction of a "topological polynomial" and a result of Levy that certain topological polynomials are Thurstonequivalent to polynomials.
In addition we are able to calculate the number of real polynomials p satisfying the conditions in Theorem 1, up to equivalence. Two polynomials p and q are conjugate if there is an affine linear map T with p = T −1 • q • T . They are equivalent if there are affine linear maps S and T with p = S •q •T . Theorem 3. The number E n of equivalence classes of real polynomials p of degree n having n − 1 distinct critical points c 1 , . . . , c n−1 and satisfying p(c j ) = c j for j = 1, . . . , n − 1 is E n = C ⌊(n−1)/2⌋ , where C m = 1 m+1 2m m is the m-th Catalan number. The number of conjugacy classes is Q n = E n +C k if n = 4k + 3, and Q n = E n if n ≡ 3 (mod 4).
Remark. There are non-real polynomials satisfying the same equations. In particular, for any real solution p, and ω on the unit circle, the function q(z) = ωp(ωz) is also a solution. We conjecture that all non-real solutions are equivalent to real solutions.
The proof of Theorem 3 relies on Poirier's results about Hubbard trees for post-critically finite polynomials from [Po1] and [Po2].
If one allows rational functions p instead of polynomials, the number of zeros of p(z) − z is at most 5n − 5 (see [KN], which also outlines the connection of this problem to gravitational lensing), and the proof that this bound is sharp is much easier and very explicit [Rh].
Acknowledgment. I would like to thank Dmitry Khavinson for helpful discussions.

Proof of Theorem 1
We will construct a polynomial with real coefficients. There is a minor difference between the cases where n is even and the one where n is odd. In the case of even n, one of the points will be real, in the odd case all z j will be non-real. Note also that the polynomial p itself will map all z j to their complex conjugates, so the second iterate fixes all z j . These polynomials are special cases of post-critically finite rational maps. The main ingredient in the proof is Thurston's topological characterization of rational maps [DH], and a result of Levy on topological polynomials. For background see also [BFH] and [Pi].
A topological polynomial is an orientation-preserving branched covering f of the sphere to itself with f −1 (∞) = {∞}. Critical points are points where f is not locally injective, the post-critical set P (f ) is the closure of {f n (c) : n > 1, c critical}. An orientation-preserving branched covering of the sphere is post-critically finite if the post-critical set contains finitely many points, or equivalently if all critical points are eventually periodic under iteration. Two post-critically finite maps f and g are Thurston equivalent if there exist orientation-preserving homeomorphisms φ and ψ, homotopic rel P (f ), with Thurston gave a criterion for post-critically finite branched coverings to be equivalent to rational maps, and Levy proved the following about topological polynomials (see [Le], [BFH,Cor. 5.13]).
Theorem (Levy). If f is a topological polynomial such that every critical point eventually lands in a periodic cycle containing a critical point, then f is Thurston-equivalent to a polynomial. The polynomial is unique up to affine conjugation.
We will first construct a topological polynomial f satisfying f (z) = f (z) for all z, and f (c j ) = c j for all critical points c j . The cases of even and odd degrees are slightly different, so we will start with the case of odd degree n = 2m + 1 for some m ≥ 1. Let g be a polynomial of degree 2m + 1 with real coefficients, m distinct critical points c 1 , . . . , c m in the upper halfplane such that the corresponding critical values v j = g(c j ) are mutually distinct points in the lower halfplane. E.g., one could choose a small constant ǫ > 0, define c 2k−1 = ik, c 2k = i(k + ǫ), and It is easy to check that g satisfies all the conditions if ǫ is sufficiently small. Now choose a homeomorphism h of the closed lower halfplane with h(v j ) = c j . Extend h by reflection to a homeomorphism of the plane and define f = h • g. Then f is a topological polynomial of degree 2m + 1, commuting with reflection in the real line, mapping all its critical points to their complex conjugates. The post-critical set of f consists of all the critical points c 1 , . . . , c m , c 1 , . . . , c m and ∞. By Levy's result f is Thurstonequivalent to a polynomial p. The uniqueness part of Levy's theorem implies that p is conjugate to p, so p is conjugate to a real polynomial, and we may as well assume that p itself is real. It also follows that both φ and ψ are symmetric with respect to the real line. By the definition of Thurston-equivalence we know that z j = φ(c j ) = ψ(c j ), so p has critical points z 1 , . . . , z m , z 1 , . . . , z m , and p(z j ) = z j for all j.
If the degree n is even, we have to modify the initial step of the construction. Write n = 2m + 2 with some m ≥ 0, and find a real polynomial g of degree n with distinct simple critical points c 0 ∈ R, c 1 , . . . , c m in the upper halfplane, such that v j = g(c j ) are all distinct and in the lower halfplane for 1 ≤ j ≤ m.
Here one can take with c k chosen as above for k ≥ 1. Now choose a homeomorphism h of the closed lower halfplane onto itself such that h(g(c 0 )) = c 0 and h(g(c k )) = c k for k ≥ 1. The rest of the proof is identical to the case of odd degree.
Proof of Corollary 2. We will show that z − p(z) has at least 3n − 2 zeros. Together with the result from [KS], this shows that it has exactly 3n − 2 zeros. First we observe that p has no degenerate fixed points, i.e. all fixed points of z → p(z) satisfy |p ′ (z)| = 1. (For holomorphic maps, a fixed point is degenerate if the derivative is exactly 1, for anti-holomorphic maps it is degenerate whenever the derivative has modulus 1.) If this were the case, then p(p(z)) would have a parabolic fixed point, which would have a critical point in its attracting basin. However, all critical points of this maps are mapped to the critical points of p and those are fixed. Having only nondegenerate fixed points, the sum of the Lefschetz indices of the fixed points in the Riemann sphere has to be equal to −n + 1, as the induced maps on 0-and 2-dimensional homology are the identity and multiplication by −n, respectively. It already has super-attracting fixed points at ∞ and z 1 , . . . , z n−1 . The Lefschetz index of a super-attracting fixed point is 1, thus the sum of the indices of all those fixed points is n. We conclude that p must have at least 2n − 1 fixed points with index −1, i.e. repelling fixed points for p. This proves sharpness.

Counting Conjugacy and Equivalence Classes
We will use results about Hubbard trees by Poirier (see [Po1] and [Po2]) in order prove Theorem 3. Hubbard trees were first introduced in [BFH] in order to classify post-critically finite polynomials without periodic critical points. In that case the Julia set is a dendrite, and the Hubbard tree is the embedded subtree of the Julia set spanned by the post-critical set. It has a finite number of vertices and edges. The polynomial induces a map on the tree, and conversely an embedded finite tree with a map satisfying certain conditions gives rise to a post-critically finite polynomial, unique up to affine conjugation. Poirier generalized this to include the case of postcritically finite polynomials with critical periodic points. In this case all Fatou components are simply connected with a distinguished base point, the unique preperiodic point in the component. The Hubbard tree in this case intersects both the Julia and the Fatou set, and the part in a Fatou set consists of preimages of radial lines under the Riemann map mapping the component to the unit disk and the basepoint to 0. This construction determines an angle between any two edges meeting at a point in the Fatou set. The important information needed to reconstruct the map from the (marked) tree is the local degree at every vertex, the angles between any two edges meeting at a Fatou vertex, and the mapping on vertices. There is no well-defined angle between edges meeting at points in the Julia set, but there is a well-defined cyclic order, and we define the "angles" between the edges to be all the same. In that way we end up with an angle function on the tree defined for any pair of incident edges.
An abstract Hubbard tree is a finite tree embedded in the plane, together with a mapping F : V → V on the vertices, a local degree function δ : V → {1, 2, . . .} on the vertices, and an angle function α defined for any pair of incident edges l k , l j . It has to satisfy α(l k , l j ) ∈ R \ Q, α(l j , l k ) = −α(l k , l j ), and α(l j , l k ) + α(l k , l m ) = α(l j , l m ) whenever l k , l j , l m meet at a vertex.
The map F satisfies F (v) = F (w) whenever v and w are adjacent vertices. In this way F induces a mapping on the edges, mapping the edge [v, w] homeomorphically to the unique path between F (v) and F (w), i.e. mapping an edge to a union of edges. If two edges l j and l k meet at v, then F (l j ) and F (l k ) meet at F (v), and we require that α(F (l j ), F (l k )) = δ(v)α(l j , l k ), which is just the fact that holomorphic maps of local degree δ multiply angles by δ. A vertex is a Fatou vertex if it eventually lands in a periodic vertex v with δ(v) > 1. All other vertices are Julia vertices. The last and not quite as obvious requirement for the Hubbard tree is that F is expanding, i.e. whenever v and w are adjacent Julia vertices, there exists n such that F n (v) and F n (w) are not adjacent.
The degree of the Hubbard tree is d = 1 + v∈V (δ(v) − 1). Poirier's main result in [Po2] is that any such abstract Hubbard tree corresponds to a polynomial of degree d, unique up to affine conjugation. The uniqueness part easily implies that any symmetries of the Hubbard tree correspond to symmetries of the polynomial.
Proof of Theorem 3. We will first prove the result about the number of conjugacy classes. By Poirier's theorem we only have to count the number of corresponding Hubbard trees up to symmetry. We will first establish some general properties of the corresponding Hubbard trees, after which we will treat odd and even degrees separately. The following claim gives a complete characterization of the Hubbard trees in question, the rest of the argument is counting.
Claim 1. The induced map on the Hubbard tree is z → z. The only vertices in the Hubbard tree are the critical points. All angles are 1/3 or 2/3. Conversely, any finite tree T which is symmetric with respect to complex conjugation, has local degree 2 at every vertex, with the map f : T → T , f (z) = z, arises as a Hubbard tree of a polynomial satisfying the requirements in Theorem 3.
Proof. Let c 1 and c 2 be critical points such that the path γ between them does not contain any other critical point. Then γ will be mapped homeomorphically to the path connecting p(c 1 ) = c 1 and p(c 2 ) = c 2 , which is the complex conjugate of γ. Since the whole post-critical set consists of critical points, this argument covers the whole Hubbard tree. Any noncritical vertices would have to be branch points, and the induced map z → z reverses the cyclic order of edges at branch points, contradicting the fact that p is locally conformal at all non-critical vertices. Finally, let l 1 and l 2 be two edges meeting at a critical vertex c. Then δ(c) = 2, and 2α(l 1 , l 2 ) = α(p(l 1 ), p(l 2 )) = α(l 1 , l 2 ) = −α(l 1 , l 2 ) in R/Z, thus 3α(l 1 , l 2 ) ∈ Z.
Conversely, if all local degrees are 2, and all angles are 1/3 or 2/3, complex conjugation f (z) = z satisfies δ(v)α(l 1 , l 2 ) = α(f (l 1 ), f (l 2 )) for any two edges meeting at a vertex v. The property that f on the tree T be expanding is trivially satisfied as there are no Julia vertices at all. R-symmetric Hubbard trees give rise to R-symmetric polynomials, so this shows the last part of the claim.
Claim 2. The number of vertices on the real line is at most 2. There is a bijective correspondence between Hubbard trees with 0 real vertices and those with 2 real vertices.
Remark. The bijective correspondence replaces the vertical edge which intersects the real line by a horizontal edge which is a real interval. See Figure 1 for examples of Hubbard trees up to degree 8.
Proof. Let c 1 and c 2 be real critical points. Since the Hubbard tree is a tree symmetric with respect to the real line, the subtree spanned by c 1 and c 2 has to be the real interval between them. Since no angle in the Hubbard tree is 1/2, there can be no other vertex in that interval, i.e. there can not be other critical points between c 1 and c 2 . This shows the first claim. In order to construct the bijective correspondence, we only need to give the modification of the tree in the upper halfplane. The part in the lower halfplane is determined by symmetry and the map on the tree is always complex conjugation. If there are two real critical points c 1 < c 2 , let T 1 and T 2 be the subtrees in the upper halfplane attached at c 1 and c 2 , resp. The correspondence is given by replacing the edge l = [c 1 , c 2 ] bỹ l = [−i, i], and attaching T 1 and T 2 to the left and right of i, resp., so that α(T 2 , T 1 ) = α(T 1 ,l) = α((l), T 2 ) = 1/3. The inverse of this operation is given by replacing the edge which crosses the real line by a real interval and attaching the left and right subtrees at the endpoints of this interval, again with angles 1/3. Observe also that any symmetry with respect to the imaginary axis will be preserved by this correspondence and its inverse.
It remains to count the Hubbard trees. Let us start with the easier case of even degree n = 2m + 2 with m ≥ 0. Here we have an odd number of critical points, so by Claim 2 there is exactly one real critical point c 1 . The number of edges meeting at this point must be even by symmetry. In the case m = 0, there is only one critical point, so there are no edges. In all other cases the number of edges meeting at c 0 must be 2, since it can be at most 3 by Claim 1. Also the angles between the real axis and the edges must be 1/6 and 1/3, respectively. There are exactly two choices, and they correspond to each other under the symmetry z → −z. Since we count Hubbard trees up to symmetry, we may fix an orientation. It also follows that all those Hubbard trees have no rotational symmetry. We can view the part of the tree in the upper halfplane as a finite part of an infinite rooted binary tree. The number of vertices, excluding the root, is m, and it is well-known (and 2m m of those. In the case of odd degree n = 2m + 1 there are 2m critical points. Let us count the number of trees which have no vertex on the real line. Then the part in the upper halfplane is again a finite part of a rooted binary tree with m vertices, so there are C m of those. However, for every such tree the image under z → −z has the same form. If m = 2k, this implies that there are C m /2 up to symmetry. However, if m = 2k + 1, there are trees which are symmetric under z → −z. The number of those corresponds to the number of subtrees with k vertices attached to the right of the first vertex in the upper halfplane. This means that there are C k symmetric trees, so we have (C m + C k )/2 inequivalent Hubbard trees in this case. By Claim 2, we have the same number of trees with 2 vertices on the real line, so the number of inequivalent Hubbard trees is C 2k for n = 4k + 1 and C 2k+1 + C k for n = 4k + 3.
Let us now count the number of equivalence classes. The equivalence of two solutions p and q implies the existence of affine linear maps S and T with SpS −1 = Sq, i.e. the existence of an affine map R = ST such that R maps the critical points of q to itself. In particular this implies that RqR −1 is again a solution. The only rotations R which map one of our symmetric Hubbard trees to a symmetric Hubbard tree are R(z) = i p z with p ∈ {1, 2, 3}. In the case R(z) = −z, the map p = Rq has the same underlying Hubbard tree as q with the map z → −z on it. This corresponds to the Hubbard tree of q rotated by 90 degrees with complex conjugation as the map. This can only happen if n = 4k + 3, and the number of trees which admit this is C k . The case R(z) = ±i can actually not occur, since the map p = Rq would have the same Hubbard tree and the map on it would be reflection in the line (1 ± i)R. However, none of the Hubbard trees has one of those lines as a line of symmetry.
Combining all this, the number of equivalence classes is the same as the number of conjugacy classes except in the case n = 4k + 3, where it becomes C 2k+1 . Thus the number of equivalence classes is Q n = C ⌊(n−1)/2⌋ for all n.