Short connections inside polynomial lemniscates

A quadratic model

Consider $p(z)=z^2-r^2$ with $0<r<1$. Its critical point is $0$, its critical value is $-r^2$, and the outward slit in the value disc is $J=(-1,-r^2]$. We can compute its preimage without choosing a branch:

This vertical interval separates the two inverse sheets containing $r$ and $-r$. It shrinks to the critical point as $r\uparrow 1$. Yet the horizontal segment $[-r,r]$ stays in $\{|p|<1\}$ and has length $2r<2$. A narrow neck alone therefore does not give the desired counterexample. For the radial-distance estimate below to exceed $2$, the two root distances from the critical point must themselves have sum greater than $2$.

A length estimate for a double cover

The degree on a component counts preimages in that component, with multiplicity; it equals the number of zeros there, not necessarily the degree of the polynomial [8]. Near a simple critical point $c$, write $p(c+z)-p(c)=z^{2}A(z)$. The quadratic model suggests a spatial scale $\sqrt{\delta /M}$: the value slit has length $\delta =1-|p(c)|$, while $M=|A(0)|=|p^{″}(c)|/2$ converts squared distance into value distance. The next lemma controls this comparison on a disc of radius $h$. Degree two serves a different purpose. It ensures that the two local preimages near $c$ exhaust every fibre in the chosen component, so a connection cannot pass through an unexamined part of the slit preimage.

This figure is drawn in the PDF.

The polynomial

Starting from $z^7-1$, we put the roots on $|z|=\rho <1$ and split the multiple critical point at scale $\epsilon$. The linear, quadratic and cubic perturbations below all have order ${\epsilon }^7$ at $z=\epsilon w$, so they alter the critical geometry without comparable root displacement. We must identify the only component containing a pair and make its distance surplus exceed the loss in Lemma 2.1. The fixed coefficients are those of ani’s construction. Put $s={10}^{-6}$, $\epsilon =s^2={10}^{-12}$ and $\rho =1-s^{16}$, and let

Define

The analytic argument

Assume that no pair has a connection of length less than $2$ in ${\mathrm{\Omega }}_f$. At a regular level $t\in (\mu ,1)$, let $C_t$ contain a chosen first merger and let $k\ge 2$ be its root count. Put $a=\mathrm{Area}(C_t)/\pi$ and $x=\log (t/\mu )$; Pólya’s inequality gives $a\le 1$ [7]. Uniformise $C_t$ by the disc, with $d(0,s)=2\mathrm{artanh}s$ for $0\le s<1$. For roots at hyperbolic distance $d$, choose $\phi :\mathbb{D}\to C_t$ sending $\pm s$ to them. Then $d=4\mathrm{artanh}s$ and ${\int }_{\mathbb{D}}|{\phi }^{\prime }{|}^{2}dA=\pi a$. Cauchy–Schwarz with the Bergman kernel gives

Thus pairwise root distances are at least $D$, where $\cosh (D/2)=e^{2/a}$.

To control the distances from the centre as well, choose a point $h$ on a connected set through the first pair in $K_{\mu }(f)$ whose intrinsic distance from every root is at least $1$. Here intrinsic distance means the infimum of Euclidean curve lengths in $C_t$. Such an $h$ exists: the intrinsic open unit balls about distinct roots are disjoint, since an intersection would give a path shorter than $2$, and disjoint open sets cannot cover a connected set containing two roots. Choose the uniformisation to send $0$ to $h$. If $d_j$ are the resulting hyperbolic root distances from $0$, then

The one-root Bergman estimate gives the first bound because every root has intrinsic distance at least $1$ from $h$. For the second, the component-wise Blaschke representation [8] expresses $f/t$ in disc coordinates. Since $h$ is not a root and lies in $K_{\mu }(f)$, evaluation at $0$ gives $\sum _j\lambda (d_j)=\log (t/|f(h)|)\ge x$. Dividing the required sum $x$ by the maximum contribution $\delta (a)/2$ gives $k\ge 2x/\delta (a)$. This uses only distances from the centre. To exploit the pairwise separation too, we intersect the disjoint hyperbolic balls about the roots with a common circle. The passage through a first merger, including simultaneous mergers, is part of the analytic input still to be formalised.

On a fixed hyperbolic circle of radius $r>0$ centred at $0$, the balls of radius $D/2$ cut out disjoint arcs. Their angular half-widths satisfy

Here ${\mathrm{clamp}}_{[-1,1]}$ truncates to $[-1,1]$. This circle-slice inequality is formalised separately. To turn it into a root count, majorise the contribution of one root by weighted slice widths: choose ${\sigma }_i\ge 0$, $r_i>0$ and $U>0$ such that $\lambda (d)\le U+\sum _i{\sigma }_{i}w(d,r_i)$ whenever $d>0$ and $\lambda (d)\le \delta (a)/2$. Summing at the roots and using the angular bound at each $r_i$ gives

The circle must be fixed before taking these intersections: angular projections of disjoint balls at different distances can overlap.

We now convert the root count into area growth. Lift one common value radius from all $k$ roots to $\partial C_t$ and join successive endpoints by boundary arcs. All pieces stay in ${\mathrm{\Omega }}_f$ because $t<1$. Splitting the lift integral at level $\mu$ bounds the mean total lift length by $\sqrt{ka(x+2)/2}$. Each of the $k$ resulting root-to-root connections has length at least $2$; their sum counts each lift twice and the boundary once. Hence $2k\le \sqrt{2ka(x+2)}+{\mathcal{H}}^1(\partial C_t)$. Combining this with ${\mathcal{H}}^1(\partial C_t{)}^2\le 2\pi kt(d/dt)\mathrm{Area}(C_t)$ yields the ordinary differential inequality

Here the derivative is with respect to $x$, between merger levels. The companion proves the lift estimate and accounts for the nonnegative area jumps at mergers; those jumps do not weaken the lower bound.

The recorded rational comparison

At each level use the largest integer lower bound for the current root count $k$: (5), $k\ge 2x/\delta (a)$, or the companion’s two further packing estimates. The starting bound $a(3/{10}^5)>{10}^{-6}$ follows by integrating (6) with $k\ge 2$ and $a\le 1$; it is not an extra assumption on the polynomial. The recorded rational calculation then reaches $a>1$ by

It used $18$ area levels, $126$ certified weighted inequalities and a maximum step $1/400$. On each interval $[u,v]$ of possible root distances, the calculation bounds $\lambda (d)-\sum _i{\sigma }_{i}w(d,r_i)$ by $\lambda (u)-\sum _i{\sigma }_{i}min(w(u,r_i),w(v,r_i))$; on the tail, $\lambda$ itself suffices. Thus numerical optimisation only proposes the weights. Rational inequalities certify a valid $U$ on the whole admissible half-line, as required by (5). The endpoint minimum is valid because each slice width, as a function of $d$, increases and then decreases, or is monotone; it has no interior dip. This is an interval bound, not sampling between grid points.

Since $t=\mu e^x$, the contradiction $a>1$ occurs before level $1$ provided that $\mu e^X<1$. The recorded comparison gives $(13/25)e^X<0.982000386<1$. It also gives $(529/1000)e^X<0.998996547$, so the ordinary argument supports that sharper threshold as well. Neither threshold is claimed optimal, and this arithmetic does not formalise the analytic reduction.

A related ordinary averaging argument, also unformalised in its analytic construction, gives the following bounds. For a monic polynomial of degree $n\ge 2$, two zero occurrences can be joined in $K_{2\mu }(f)$ with length at most $(71/10){\mu }^{1/n}$; they are distinct in the squarefree case, and a repeated root gives a constant path. When $\mu \le 1/2$, the argument gives a path of length at most $5.7$ in ${\mathrm{\Omega }}_f$. The proof, together with refinements that retain the number of zeros in the component or its logarithmic capacity, is given in the inverse-ray averaging argument.

Square roots and component area

The ordinary argument first removes the scale of $v$, then constructs a curve, and finally bounds its length. Choose ${\alpha }^n=v$ and set $P(z)=v^{-1}f(c+\alpha z)$. This polynomial is monic, with its chosen critical point at $0$ and critical value $1$; Euclidean lengths are multiplied by $|\alpha |=|v{|}^{1/n}$ on returning to the original plane. The component $U$ of $P^{-1}(D(w_0,S))$ containing $0$ has degree two. To apply Riemann–Hurwitz, exhaust it by components above regular inner discs: the outer boundary itself may contain other critical values. Since $U$ is simply connected and $1-P$ has only a double zero at $0$ there, it has a single-valued analytic square root $\xi$. This square root removes the simple branching. It is a proper local biholomorphism onto $V=\{\xi :|{\xi }^2-(1-w_0)|<S\}$, a star-shaped and hence simply connected domain. Thus it is biholomorphic. Its inverse maps $-1$ and $1$ to the two roots and $[-1,1]$ to their connection. Along this interval $P=1-{\xi }^2\in [0,1]$, which gives the required containment. Let $L$ be the Euclidean length of this normalised curve.

Put $a=1-w_0$ and $p=w_0(1-w_0)$. Squaring maps $V$ onto $D(a,S)$. A Möbius map from this disc to the unit disc fixing $0$, followed by taking a square root, gives the conformal map

Choose the square-root factor positive at $0$. The endpoints $\xi =\pm 1$ then map to $\zeta =\pm q$, where $q^2=S/(S^2+p)$. The hypothesis on $S$ gives $S^2-S+p=(S-w_0)(S-(1-w_0))>0$, so $0<q<1$. The Bergman segment estimate used in Section 4 now applies to $[-q,q]$ and gives

The factor $n-1$ comes from the roots outside $U$, not from Pólya’s inequality alone. For monic $F$ of degree $n$ and regular level $t>0$, let $W$ be a component of $\{|F|<t\}$ containing $k<n$ zeros counted with multiplicity. Reflecting the $n-k$ exterior-root factors into the disc gives a Blaschke product $B$ of degree $n-k$ with $|B(0)|=\mathrm{cap}(\bar{W}{)}^n/t$ and $|B^{\prime }|<n$ on the boundary. The boundary-fibre identity gives

Choosing $w$ opposite $B(0)$ gives $(n-k)/n<(1-|B(0)|)/(1+|B(0)|)$, hence $\mathrm{cap}(\bar{W}{)}^n/t<k/(2n-k)$.

Apply this to $F=P-w_0$ on the regular inner components exhausting $U$. Their degree is two, so $k=2$ and $k/(2n-k)=1/(n-1)$. Here $k$ counts preimages of $w_0$ under $P$. These need not be the roots joined by the curve; the component degree gives both counts. Pólya’s area–capacity inequality [15], [7], followed by exhaustion, now gives $\mathrm{Area}(U)\le \pi (S/(n-1){)}^{2/n}$. The original curve has length $\mathrm{length}(\mathrm{\Gamma })=|v{|}^{1/n}L$, which gives (7). Dubinin’s relative area inequality [14] has a different full-covering hypothesis; the absolute component estimate is needed here.

For the conditional threshold, decrease $S$ to $4/3$. Since $p\ge 0$ and $n\ge 3$, the logarithm is at most $\log 7<2$ and $(S/(n-1){)}^{2/n}\le 1$, giving $\mathrm{length}(\mathrm{\Gamma })<2|v{|}^{1/n}<2$. For $n=3,w_0=1,S=6/5$, use $(3/5{)}^{2/3}\log 11<2$. The chosen critical value must be nonzero and have modulus less than one. Root locations by themselves do not isolate it from the others.

Root distances do not ensure containment

Order the root distances at a non-root critical point $c$ as $0<d_1\le \cdots \le d_n$. The balance $\sum _j(c-z_j{)}^{-1}=0$ and this ordering give

The ratio bound $1\le d_2/d_1\le n-1$ permits the scalar comparison

Neither inequality supplies containment.

When all zeros lie in the open unit disc, a different estimate holds: at every non-root critical point the sum of the distances to the two nearest zeros is strictly less than $2$. The companion gives both root-distance estimates. The nearest pair is chosen among all roots of $f$. It need not be the pair in the two-root component of the degree-seven example, where containment restricts which pair can be joined.

Even a short segment from a critical point can leave the lemniscate. For the quintic

all roots lie in the disc and $0$ is a non-root critical point, but its unique nearest-root segment leaves the unit lemniscate. Moreover, every root-pair midpoint of $z^3-(99/100{)}^3$ lies outside it. This cubic still admits the two-segment connections of Theorem 3.1; it is the straight chords that fail. The exact counterexamples include the escape calculations.

Nor do the two critical-value criteria cover every polynomial in the root-disc class. For $z^3+3z/100-3/4$, the least critical-value modulus exceeds $13/25$, and the two normalised critical values are too close for the isolation criterion.

A further cubic has two roots at its first merger but normalised component capacity one at level $2\mu$. By that level all three roots have merged, so the initial two-root count cannot be used at the later level. The component is now the whole filled lemniscate, whose capacity is given by $\mathrm{cap}(K_t(f))=t^{1/n}$ [23].

Finally, $z^8-(3/2)z$ has least critical-value modulus $\mu >1$, yet a one-root unit-level component has perimeter larger than $\mathrm{\Gamma }(1/4{)}^2/(2\sqrt{\pi })$; see the companion’s component examples. No root-location restriction is imposed. The open question asks for a constant $\beta$ bounding each component perimeter of $K_{\sigma }(f)$ by $\beta {\sigma }^{1/n}$, uniformly over squarefree monic $f$ of degree $n\ge 2$ and $0<\sigma <\mu$. This level restriction is stronger than a one-root condition on a single component. Although [21] reports a Runge construction of unbounded one-root perimeters, its proof is unavailable and we do not use it.

Sparse families and binomial chords

Let $f(z)=P((z-h{)}^q)$, where $q\ge 2$ and $P$ is a monic cubic, and suppose all zeros of $f$ lie in the open unit disc. For any root $r=y^q$ of $P$, averaging $|h+y\zeta {|}^2$ over the $q$th roots of unity cancels the cross term and gives $|h{|}^2+|y{|}^2<1$. Thus the quotient roots satisfy $|r|=|y{|}^q<1$. If $f$ has two distinct zeros, the companion selects a nonzero root $r$ with $[0,r]\subset \{|P|<1\}$. Choose $y^q=r$. Then $f(h+ty\zeta )=P(t^{q}r)$ for $0\le t\le 1$, so the whole fibre gives contained segments through $h$. Two distinct fibre points are joined with length $2|y|<2$, using the same average.

For $f(z)=z^5+a{z}^4+bz+c$ with open-disc roots, Abel summation gives a contained radial segment whenever $|b{z}_j+c|<1$. For $a\ne 0$, the missing coefficients fix the first three power sums. After rotation, the companion’s harmonic function sums to $5-2|a|$ over the roots. It is bounded above by $4-2|a|$ on the disc and by $2/31$ at roots with $|b{z}_j+c|\ge 1$. Four such indices would therefore give

a contradiction. Hence two indices satisfy $|b{z}_j+c|<1$. When $a=0$, the root equation gives $|b{z}_j+c|=|z_j{|}^5<1$ directly. Coincident selected root values give the constant path. The companion gives the cubic-fibre proof and the quintic selection argument, including repeated root occurrences.

For $z^n-r^n$, $0<r<1$, the maximum modulus on an adjacent chord of radius $0<s\le r$ is $r^n+(s\cos (\pi /n){)}^n$. The outer root chord therefore works below $r_{\ast }=(1+{\cos }^n(\pi /n){)}^{-1/n}$. At and above this value, take a slightly contracted inner radius $s>0$ making that maximum less than one. The radial legs remain contained, and for $n\ge 3$ the total length is

The chord calculation proves the maximum on the entire segment, not geodesic optimality. For $n=2$, use the diameter.

Collinear roots

For collinear roots of diameter $D$, a Chebyshev alternation argument gives two adjacent root occurrences whose segment satisfies

The constant is sharp, with equality for the appropriately scaled zeros of the Chebyshev polynomial $T_n$. Qualitative existence of a contained root segment was already proved in [2]. For distinct roots, first reduce to the real monic case. If every gap maximum exceeded the bound, subtracting the endpoint-normalised monic Chebyshev polynomial would give a polynomial of degree at most $n-1$ with $n$ distinct zeros: two endpoints and $n-2$ intervening sign changes. See Eremenko and Yuditskii [10] for the critical-sequence viewpoint. The companion contains the alternation proof and equality case.

Critical-value means

For $c_j\in \bar{\mathbb{D}}$ and positive weights of sum one, set $G(z)=\prod _j|1-{\overline{c}}_{j}z{|}^{w_j}$. Then $\sum _j{w}_{j}G(c_j{)}^2\le 1$, with equality only when all $c_j=0$.

For interior centres, the nonvanishing factors give an analytic $g(z)=1+\sum _{\nu \ge 1}{a}_{\nu }{z}^{\nu }$ with $|g|=G$. On $\zeta =e^{i\theta }$ the weighted Poisson kernel is $P=\sum _j{w}_j(1-|c_j{|}^2)/|\zeta -c_j{|}^2=1-2\mathrm{Re}(\zeta g^{\prime }/g)$. With $dm=d\theta /(2\pi )$, subharmonicity gives the inequality below, and orthogonality gives the identity:

Indeed, $\int |g{|}^{2}dm=1+\sum |a_{\nu }{|}^2$ and $\int \zeta g^{\prime }\bar{g}dm=\sum \nu |a_{\nu }{|}^2$. Each nonconstant coefficient therefore contributes $1-2\nu <0$. For boundary centres, replace $c_j$ by $r{c}_j$ and let $r\uparrow 1$ with a finite coefficient sum first; no boundary logarithm is needed. Equality forces every $a_{\nu }$ to vanish. Logarithmic differentiation of $g=1$, grouping repeated centres, then forces all the positive-weight centres to be zero.