IJMMS 28:4 (2001) 243–246 PII. S0161171201011589 http://ijmms.hindawi.com
© Hindawi Publishing Corp.
ON THE ZEROS AND CRITICAL POINTS OF A RATIONAL MAP
XAVIER BUFF (Received 23 January 2001)
Abstract.Letf:P1→P1be a rational map of degreed. It is well known thatf hasd zeros and 2d−2 critical points counted with multiplicities. In this note, we explain how those zeros and those critical points are related.
2000 Mathematics Subject Classification. 30C15.
In this note,f:P1→P1is a rational map. We denote by {αi}i∈I the set of zeros off, and by{ωj}j∈Jthe set of critical points off which are not zeros off(the sets I and J are finite). Moreover, we denote byni the multiplicity ofαi as a zero off and bymjthe multiplicity ofωjas a critical point off. The local degree off atαi
isniand the local degree off atωj isdj=mj+1. In particular, whenωj≠∞and f (ωj)≠∞, the pointωjis a zero offof ordermj.
Our goal is to understand the relations that exist between the pointsαi and the pointsωj.
Proposition1. Given a finite collection of distinct pointsαi∈P1with multiplicities niandωj∈P1with multiplicitiesmj, there exists a rational mapfvanishing exactly at the pointsαi with multiplicitiesni and having extra critical points exactly at the pointsωjwith multiplicitiesmjif and only if
(i)
(ni+1)−
mj=2,and (ii) for anyksuch thatαk∈C,
res
ωj∈C z−ωj
mj
αi∈C z−αi
ni+1dz, αk
=0. (1)
We will give a geometric interpretation of (ii) in the case whereαkis a simple zero off: working in a coordinate whereαk= ∞, the barycentre of the remaining zeros weighted with their multiplicities is equal to the barycentre of the critical points off weighted with their multiplicities (seeProposition 3below).
Proof. The proof is elementary. It is based on the observation that the 1-forms d(1/f )and
φ=
ωj∈C z−ωj
mj
αi∈C z−αi
ni+1dz (2)
are proportional. The differential equationd(1/f )=φhas a rational solution if and only ifφis exact, if and only if the residues ofφat all finite poles are equal to zero.
244 XAVIER BUFF
Lemma2. Letfbe a rational map. Denote byαiits zeros and bynitheir multiplici- ties. Denote byωjthe critical points offwhich are not multiple zeros offand bymj
their multiplicities. The zeros of the1-formd(1/f )are exactly the pointsωjwith order mjand its poles are exactly the pointsαiwith orderni+1.
Proof. A singularity of the 1-formd(1/f )= −df /f2is necessarily a zero or a pole off, a zero off, or∞(whereφis defined by analytic continuation). Considering the Laurent series off at each of those points, one immediately gets the result.
Now assume that there exists a rational map f with the required properties.
Lemma 2shows that the 1-formsφand d(1/f ) have the same poles and the same zeros inC, with the same multiplicities. Hence, their ratio is a rational function which does not vanish inC. Thus,φ and d(1/f )are proportional. In particular, φhas a singularity at∞if and only ifd(1/f )has a singularity at∞and the singularity is of the same kind for both 1-forms. Since the number of poles minus the number of zeros of any nonzero 1-form onP1is equal to 2 (the Euler characteristic ofP1), we see that (ni+1)−
mj=2 which is precisely condition (i) inProposition 1. Besides, since φis exact, it follows that the residues at all the polesαkvanish and condition (ii) is satisfied.
Conversely, the 1-formφhas poles of orderni+1 at the pointsαi∈Cand zeros of ordermjat the pointsωj∈C. Condition (ii) implies thatφis exact, that is, there exists a rational mapg:P1→P1such thatφ=dg. Since the number of poles ofφin P1minus the number of zeros ofφinP1is equal to 2, condition (i) implies that when
∞is neither a pointαinor a pointωj, it is a regular point ofφ, when∞ =αi0, it is a pole ofφof orderni0, and when∞ =ωj0, it is a zero ofφof ordermj0. Finally, φ=d(1/f ), withf=1/g, andLemma 2shows that the rational mapf=1/gvanishes exactly at the pointsαiwith multiplicitiesniand has extra critical points exactly at the pointsωjwith multiplicitiesmj.
We will now give a geometric interpretation of (ii) whenαkis a simple zero off.
We first work in a coordinate where∞is neither one of the pointsαinor a pointωj. Define
R(z)=
j
z−ωj
mj
i≠k
z−αi
ni+1. (3)
Then,
res
j
z−ωj
mj
i
z−αini+1dz, αk
=res R(z)
z−αk2dz, αk
=R αk
. (4)
SinceR(αk)≠0, this residue vanishes if and only if R
αk
R
αk =
j
mj
αk−ωj−
i≠k
ni+1
αk−αi=0. (5)
Letdbe the number of zeros counted with multiplicities, that is,d=
ini. The total number of critical points is 2d−2=
jmj+
i(ni−1)(the critical points off are
ON THE ZEROS AND CRITICAL POINTS OF A RATIONAL MAP 245 the pointsωjand the multiple zeros off). Then, (5) can be rewritten as
1 2d−2
j
mj
αk−ωj+
i≠k
ni−1 αk−αi
= 1 d−1
i≠k
ni
αk−αi
. (6)
This last equality can be interpreted in the following way.
Proposition3. Assume thatf is a rational map having a simple zero at∞. Then, the barycentre of the remaining zeros weighted with their multiplicities is equal to the barycentre of the critical points off weighted with their multiplicities.
Remark4. One can prove this proposition directly. We may writef=P /Q, where
P=
d−1
k=0
akzk, Q= d k=0
bkzk, (7)
are co-prime polynomials with deg(Q)=deg(P )+1=d. Without loss of generality, we may assume that the barycentre of the zeros offis equal to 0. In other words, we may assume thatP is a centered polynomial, that is,ad−2=0. A simple calculation shows that
PQ−QP=
2d−2 k=0
ckzk (8)
is a polynomial of degree 2d−2 and thatc2k−1=0. Therefore, the barycentre of the zeros ofPQ−QP, that is, the barycentre of the critical points off, is equal to 0.
Apply this geometric interpretation in order to re-prove two known results. The first corollary is related to the Sendov conjecture (cf. [1] and more particularly Section 4).
This conjecture asserts that if a polynomialPhas all its roots in the closed unit disk, then, for each zeroαithere exists a critical pointω(possibly a multiple zero) such that|αi−ω| ≤1.
Corollary5. LetP:C→Cbe a polynomial. Assume the zeros ofPare all contained in the closed unit disk andα0∈S1is a zero ofP. Then, the closed disk of diameter[0, α0] contains at least one critical point off.
Proof. Denote bydthe degree ofP. Ifα0is a multiple zero ofP, then the result is trivial. Thus, assumeα0 is a simple zero of P. We work in the coordinateZ = α0/(α0−z). The rational mapf:ZP (α0−α0/Z)has a simple zero atZ= ∞and the remaining zeros are contained in the half-plane{Z∈P1| (Z)≥1/2}. Thus the barycentreβof those zeros satisfies(β)≥1/2. Moreover,f has a critical point of multiplicitydatZ=0. Thus, the barycentre of thedremaining critical points is 2β.
Since(2β)≥1, we see thatf has at least one critical pointωcontained in the half plane{Z∈P1| (Z)≥1}. Then,α0−α0/ωis a critical point ofP contained in the closed disk of diameter[0, α0].
The second corollary has been proved by Videnskii [2]. Our result provides an al- ternate proof.
246 XAVIER BUFF
Corollary6. Assume thatf :P1→P1is a rational map and∆⊂P1 is a closed disk or a closed half-plane containing all the zeros off. Then,∆contains at least one critical point off.
Proof. Without loss of generality, we may assume that the zeros are simple and that at least one zero, sayα0, is on the boundary of∆. In a coordinate whereα0= ∞,
∆is a closed half-plane. The barycentre of the remaining zeros is contained in this half-plane. Consequently, the barycentre of the critical points is contained in∆. Thus,
∆contains at least one critical point.
Videnskii also proved that this result is optimal in the sense that there exist rational maps of arbitrary degrees with simple zeros contained in a disk∆but only one critical point in∆.
References
[1] M. Marden,Conjectures on the critical points of a polynomial, Amer. Math. Monthly90 (1983), no. 4, 267–276.MR 84e:30007. Zbl 535.30010.
[2] I. Videnskii,On the zeros of the derivative of a rational function and invariant subspaces for the backward shift operator on the Bergman space, in preparation.
Xavier Buff: Université Paul Sabatier, Laboratoire Emile Picard, 118, Route de Narbonne,31062Toulouse Cedex, France
E-mail address:[email protected]
