# Hexagonal circular 3-webs with polar curves of degree three

Sergey I. Agafonov

Department of Mathematics,
São Paulo State University-UNESP,
São José do Rio Preto, Brazil
e-mail<EMAIL_ADDRESS>
###### Abstract

The paper reports the progress with the classical problem, posed by Blaschke
and Bol in 1938. We present new examples and new classifications of natural
classes of hexagonal circular 3-webs. The main results is the classification
of hexagonal circular 3-webs with polar curves of degree 3.

MSC: 53A60.

Keywords: circular hexagonal 3-webs

##  1 Introduction

The problem to describe hexagonal 3-webs formed by circles in the plane
appeared in the first monograph on the web theory published by Blaschke and
Bol in 1938 (see [BB-38] p.31). The authors presented an example with 3
elliptic pencils of circles, each pair of pencils sharing a common vertex, and
observed that one can construct hexagonal circular 3-webs from hexagonal
linear 3-webs, completely described by Graf and Sauer [GS-24] as being formed
by tangent to a fixed curves of third class. The construction involves a
central projection from a plane to a unit sphere followed by sterographic
projection to a plane. The corresponding circular 3-webs were described
earlier by Volk [V-29] and Strubecker [S-32].

Stereographic projection puts the problem into a natural framework of the Lie
sphere geometry: instead of planar circular webs we study circular webs on the
unit sphere, thus treating cirles and straight lines on equal footing. Lie
sphere geometry assigns points outside the unit sphere to circles on this
sphere: the assigned point is the polar point of the plane that cuts the
circle on the sphere. This sphere is called also Darboux quadric. Thus any
circular 3-web on the unit sphere determines locally 3 curve arcs outside the
Darboux quadric, one arc per web foliation. Globally these arcs may glue
together into one curve. In what follows we call this set of polar points a
polar curve of the web. For example, the polar curve of the hexagonal circular
3-web obtained from a linear 3-web is a planar cubic, possibly reducible. The
polar curve of the cited example from [BB-38] splits into 3 non-coplanar
lines.

In the same year as the book [BB-38] appeared, Wunderlich published a new
remarkable example of hexagonal circular 3-web. Its polar curve splits into 3
conics lying in 3 different planes. Since through a point $p$ on the unit
sphere pass the circles whose polar points are intersection of the web polar
curve with the plane tangent to the unit sphere at $p$, the Wunderlich web is
actually 6-web, containing 8 hexagonal 3-subwebs.

Wunderlich gave also a construction of hexagonal 3-webs whose polar curve
splits into 3 non-coplanar lines, two being dual with respect to the Darboux
qudric and the third joining them. These webs were later rediscovered by other
authors.

Further, he presented the following way to construct hexagonal 3-webs: for any
one-parametric group acting in the plane, choose 2 transversally intersecting
curve arcs that are also transversal to the group orbits; acting on the arcs
by the group one gets 2 foliations; the third is composed by the group orbits.
These 3 foliations compose a (local) hexagonal 3-web. Choosing a one-parameter
group either of translation, or of dilatation, or of rotations and taking two
intersecting circles (a straight line counts as a circle), we get circular
hexagonal 3-webs.

Blaschke was well aware of the difficulty of the posed problem and, in his
last book on the web geometry [B-55], discussed the simpler problem of
classifying hexagonal circular 3-webs whose polar curve splits into 3 non-
coplanar lines. Note that to a line corresponds a pencil of circles that is
hyperbolic if the line spears the Darboux quadric, elliptic if the line
completely misses the Darboux quadric, or parabolic if the line touches the
Darboux quadric.

By the year 1977, the list of 6 types (one from [BB-38] and five indicated in
[W-38]) of circular hexagonal 3-webs whose polar curve splits into 3 non-
coplanar lines was completed by Erdoǧan [E-74] and Lazareva [L-77]. The first
attempt to prove that the list is actually complete was published in 1989 by
Erdoǧan [E-89]. Based on direct computational approach, it did not provide the
crucial computation: in fact, a modern computer systems for symbolic
computations shows that there must be a mistake in the proof presented in
[E-89] (see Concluding remarks for further detail).

The Erdoǧan’s claim was proved only in 2005 by Shelekhov [S-05]. His insight
was to look into the singular set of the webs: defined globally, the webs
under study inevitably have singularities. Shelekhov considered the simplest
possible singularities where two of the three circular foliations are tangent.
It turns out that hexagonality imposes a strong restriction: locally, such
singular set is either a circle arc of the 3d foliation or the common circle
arc of the first two. The restriction was rigid enough to obtain all the types
on the list.

Five new types of hexagonal circular 3-webs were presented by Nilov [N-14] in
2014. Polar curves for four of them split into a line and a conic. The fifth
example may be viewed as a 5-web whose polar curve is a union of a line and
two conics. Taking the line and two arcs on different conics as the polar
curve, one gets a hexagonal 3-subwebs.

One can not help to observe that the polar curves of all the known examples
are algebraic. Motivated also by the dual reformulation of the Graf and Sauer
Theorem, we consider the following natural class of 3-webs: hexagonal circular
3-webs with polar curve of degree three. The main result of the paper is the
complete classification of such webs.

The case of planar polar curve follows immediately from the Graf and Sauer
Theorem: 3 points on the polar curve corresponding to 3 circles through a
point $p$ on the sphere are the ones where the plane, tangent to the sphere at
$p$ meets the polar curve. This plane cuts the polar curve plane along the
line. On the polar curve plane we get the configuration dual to the Graf and
Sauer Theorem.

The case of non-planar set of 3 lines was finally settled by Shelekhov [S-05].

We prove that there is no hexagonal circular 3-web whose polar curve is a
rational normal curve and obtain a classification of 3-webs whose planar polar
curve splits into a line and a smooth conic. Up to Möbius transformation,
there are 15 types, most of them depending on one parameter. Four types of
five in Nilov’s paper [N-14] are webs of this list, namely, of the types 6,
10, 11 and 15, presented in Section 5. (In fact, Nilov has found only one
Möbius orbit from one-parametric family of orbits of our type 6.)

Another natural class that we study in this paper is the set of hexagonal
circular 3-webs symmetric by action of one-parameter subgroup of the Möbius
group. We also give a complete classification of such webs.

To select candidates for hexagonal webs we exploit further the above mentioned
observation of Shelekhov on simplest singularities of hexagonal 3-webs. The
proof of the observed property in [S-05], based on considering the normal form
of the web function is not complete: this normal form often does not exists at
singular points (see Concluding remarks for more detail). We make precise the
ideas about the type of singularities and then prove the key singularity
property.

For completeness, we also present the classification of hexagonal webs with 3
non-coplanar polar lines. The proof mainly follows the line taken by Shelekhov
in [S-05].

##  2 Hexagonal 3-webs, Blaschke curvature, singularities.

A planar 3-web ${\cal W}_{3}$ in a planar domain is a superposition of 3
foliations $\mathcal{F}_{i}$, which may be given by integral curves of three
ODEs

$\sigma_{1}=0,\ \ \ \ \sigma_{2}=0,\ \ \ \ \sigma_{3}=0,\ \ \ \ $

where $\sigma_{i}$ are differential one-forms. At non-singular points, where
the kernels of these forms are pairwise transverse, we normalize the forms so
that $\sigma_{1}+\sigma_{2}+\sigma_{3}=0.$ The connection form of the web
${\cal W}_{3}$ is a one-form $\gamma$ determined by the conditions

$d\sigma_{i}+\gamma\wedge\sigma_{i}=0,\ \ \ \ i=1,2,3.$

The connection form depends on the normalization of the forms $\sigma_{i}$,
the Blaschke curvature $d\gamma$ does not.

###### Definition 1

A 3-web is hexagonal if for any non-singular point there are a neighbourhood
and a local diffeomorphism sending the web leaves of this neighbourhood in 3
families of parallel line segments.

Topologically, hexagonality means the following incidence property that has
given its name to the notion: for any point $m$, each sufficiently small
curvilinear triangle with the vertex $m$ and sides formed by the web leaves,
may be completed to the curvilinear hexagon, whose sides are web leaves and
whose ”large” diagonals are the web leaves meeting at $m$ (see the gallery of
pictures illustrating hexagonal webs in the next section). Computationally,
hexagonality amounts to vanishing of the Blaschke curvature [BB-38].

Up to a suitable affine transformation, the forms $\sigma_{i}$ may be
normalized as follows:

$\sigma_{1}=(Q-R)(dy-Pdx),\ \ \ \sigma_{2}=(R-P)(dy-Qdx),\ \ \
\sigma_{3}=(P-Q)(dy-Rdx),$

where $P(x,y),Q(x,y),R(x,y)$ are the slopes of the tangent lines to the web
leaves at $(x,y)$. Vanishing of the curvature writes as

$(R-Q)[P_{xx}+(Q+R)P_{xy}+QRP_{yy}]+(P-R)[Q_{xx}+(P+R)Q_{xy}+PRQ_{yy}]+$
---
$(Q-P)[R_{xx}+(P+Q)R_{xy}+PQR_{yy}]+$
$\frac{(Q-R)(2P-Q-R)[P^{2}_{x}+(Q+R)P_{x}P_{y}+QRP^{2}_{y}]}{(P-Q)(P-R)}+\frac{(R-P)(2Q-P-R)[Q^{2}_{x}+(P+R)Q_{x}Q_{y}+PRQ_{y}^{2}]}{(Q-R)(Q-P)}+$
$\frac{(P-Q)(2R-P-Q)[R_{x}^{2}+(P+Q)R_{x}R_{y}+PQR_{y}^{2}]}{(R-Q)(R-P)}+$
$\frac{(2R-P-Q)P_{x}Q_{x}}{P-Q}+\frac{(2P-Q-R)Q_{x}R_{x}}{Q-R}+\frac{(2Q-R-P)R_{x}P_{x}}{R-P}+$
$\frac{(R^{2}-PQ)[P_{x}Q_{y}+P_{y}Q_{x}]}{(P-Q)}+\frac{(P^{2}-QR)[Q_{x}R_{y}+Q_{y}R_{x}]}{Q-R}+\frac{(Q^{2}-PR)[R_{x}P_{y}+R_{y}P_{x}]}{R-P}+$
$\frac{(2PQR-(P+Q)R^{2})P_{y}Q_{y}}{Q-P}+\frac{(2PQR-(Q+R)P^{2})Q_{y}R_{y}}{R-Q}+\frac{(2PQR-(P+R)Q^{2})R_{y}P_{y}}{P-R}=0.$

If only one slope, say $R$, is given as an explicit functions of $x,y$ and
$P,Q$ are roots of a quadratic equation $P^{2}+AP+B=0$ then one finds the
first derivatives of $P,Q$ by differentiating the Vieta relations

$P+Q=-A,\ \ \ PQ=B,$ (1)

as a functions of $P,Q$ and the first derivatives of $A,B$. Differentiating
these expressions, one gets also the second derivatives. Finally, excluding
$P$ and $Q$ with the help of (1), one can rewrite (LABEL:curvature) in terms
of $A,B,R$ and their derivatives. The result is presented in the Appendix.

The webs considered later will inevitably have singularities: some kernels of
the forms $\sigma_{i}$ can be not transverse or the forms can vanish at some
points. We call a singular 3-web hexagonal if its Blaschke curvature vanishes
identically at regular points. The simplest type of singularities of hexagonal
3-webs have the following remarkable property, first observed by Shelekhov
[S-05].

###### Lemma 1

Suppose that a hexagonal 3-web, defined by three smooth (possibly singular)
direction fields $\xi_{1},\xi_{2},\xi_{3}$, has a singular point $p_{0}$ such
that

  1. 1.

all $\xi_{i}$ are well defined at $p_{0}$,

  2. 2.

$\xi_{1}$ and $\xi_{2}$ are transverse at $p_{0}$,

  3. 3.

$\xi_{1}=\xi_{3}$ at $p_{0}$,

then either the leaves of $\mathcal{F}_{1}$ and $\mathcal{F}_{3}$ trough
$p_{0}$ coincide or $\xi_{1}=\xi_{3}$ along the leaf of $\mathcal{F}_{2}$
trough $p_{0}$.

Proof: The property is a consequence of separation of variables for hexagonal
webs. The second condition implies that we can rectify $\xi_{1}$ and
$\xi_{2}$, i.e. choose some local coordinates $u,v$ so that
$\xi_{1}=\partial_{v}$ and $\xi_{2}=\partial_{u}$. Then
$\xi_{3}=-f(u,v)\partial_{u}+\partial_{v}$ with $f(u_{0},v_{0})=0$, where
$p_{0}=(u_{0},v_{0}).$ Now the hexagonality amounts to
$\partial_{v}\left(\frac{\partial_{u}f}{f}\right)=0$ hence $f(u,v)=a(u)b(v)$.
If $a(u_{0})=0$ then the integral curves of $\xi_{1}$ and $\xi_{3}$ passing
through $p_{0}$ coincide. If $b(v_{0})=0$ then the the leaves of
$\mathcal{F}_{1}$ and $\mathcal{F}_{3}$ are tangent along the line $v=v_{0}$,
which is the leaf of $\mathcal{F}_{2}$. $\Box$

##  3 Projective model of Möbius geometry

Following Blaschke [B-29], we call the subgroup $PSO(3,1)$ of projective
pransformations of $\mathbb{RP}^{3}$, leaving invariant the quadric

$X^{2}+Y^{2}+Z^{2}-U^{2}=0,$

the Möbius group. For the reference, we present here infinitesimal generators
of Möbius group in homogeneous coordinates $[X:Y:Z:U]$ in $\mathbb{P}^{3}$,
affine coordinates $x=\frac{X}{U},\ y=\frac{Y}{U},\ z=\frac{Z}{U}$ in
$\mathbb{R}^{3}$ and cartesian coordinate $(\bar{x},\bar{y})$ in
$\mathbb{R}^{2}$ related to points $(x,y,z)$ on the unit sphere via
stereographic projection:

$x=\frac{2\bar{x}}{1+\bar{x}^{2}+\bar{y}^{2}},\ \ \
y=\frac{2\bar{y}}{1+\bar{x}^{2}+\bar{y}^{2}},\ \ \
z=\frac{1-\bar{x}^{2}-\bar{y}^{2}}{1+\bar{x}^{2}+\bar{y}^{2}}.$

There are 3 rotations around the affine axes:

$\begin{array}[]{l}R_{z}=Y\partial_{X}-X\partial_{Y}=y\partial_{x}-x\partial_{y}=\bar{y}\partial_{\bar{x}}-\bar{x}\partial_{\bar{y}},\\\
\\\
R_{x}=Z\partial_{Y}-Y\partial_{Z}=z\partial_{y}-y\partial_{z}=\bar{x}\bar{y}\partial_{\bar{x}}+\frac{1}{2}(1-\bar{x}^{2}+\bar{y}^{2})\partial_{\bar{y}},\\\
\\\
R_{y}=X\partial_{Z}-Z\partial_{X}=x\partial_{z}-z\partial_{x}=-\frac{1}{2}(1+\bar{x}^{2}-\bar{y}^{2})\partial_{\bar{x}}-\bar{x}\bar{y}\partial_{\bar{y}},\\\
\end{array}$

and 3 boosts (or ”hyperbolic rotations”):

$\begin{array}[]{l}B_{x}=U\partial_{X}+X\partial_{U}=\partial_{x}-x(x\partial_{x}+y\partial_{y}+z\partial_{z})=\frac{1}{2}(1-\bar{x}^{2}+\bar{y}^{2})\partial_{\bar{x}}-\bar{x}\bar{y}\partial_{\bar{y}},\\\
\\\
B_{y}=U\partial_{Y}+Y\partial_{U}=\partial_{y}-y(x\partial_{x}+y\partial_{y}+z\partial_{z})=-\bar{x}\bar{y}\partial_{\bar{x}}+\frac{1}{2}(1+\bar{x}^{2}-\bar{y}^{2})\partial_{\bar{y}},\\\
\\\
B_{z}=U\partial_{Z}+Z\partial_{U}=\partial_{z}-z(x\partial_{x}+y\partial_{y}+z\partial_{z})=-\bar{x}\partial_{\bar{x}}-\bar{y}\partial_{\bar{y}}.\\\
\end{array}$

The identity component of $PSO(3,1)$ is well known to be isomorphic to the
group $PSL_{2}(\mathbb{C})$, the isomorphism being given by the action
$A(V)=AVA^{*}$ of $A\in SL_{2}(\mathbb{C})$ on the vector space of matrices

$V=\left(\begin{array}[]{lll}X+U&Y+iZ\\\ Y-iZ&U-X\end{array}\right)$

with real $X,Y,Z,U$. This action preserves determinant of $V$, which is
$U^{2}-X^{2}-Y^{2}-Z^{2}$. By this isomorphism, the generators are represented
by the following matrices:

$R_{x}=\frac{1}{2}\left(\begin{array}[]{lll}-i&0\\\ 0&i\end{array}\right),\ \
R_{y}=\frac{1}{2}\left(\begin{array}[]{lll}0&-i\\\ -i&0\end{array}\right),\ \
R_{z}=\frac{1}{2}\left(\begin{array}[]{lll}0&1\\\ -1&0\end{array}\right),$
$B_{x}=iR_{x},\ \ B_{y}=iR_{y},\ \ \ B_{z}=iR_{z}.$

Two points $p_{1}=[X_{1}:Y_{1}:Z_{1}:U_{1}]$ and
$p_{2}=[X_{2}:Y_{2}:Z_{2}:U_{2}]$ in $\mathbb{RP}^{3}$ determine a line with
the Plücker coordinates

$a:=X_{1}U_{2}-X_{2}U_{1},\ \ \ \ b:=Y_{1}U_{2}-Y_{2}U_{1},\ \ \ \
c:=Z_{1}U_{2}-Z_{2}U_{1}$ $f:=Y_{1}Z_{2}-Y_{2}Z_{1},\ \ \ \
g:=Z_{1}X_{2}-Z_{2}X_{1},\ \ \ \ h:=X_{1}Y_{2}-X_{2}Y_{1}.$

By direct computation one proves the following fact.

###### Lemma 2

All points of a line with Plücker coordinates $[a:b:c:f:g:h]$ are stable with
respect to subgroup with the infinitesimal generator
$aR_{x}+bR_{y}+cR_{z}+fB_{x}+gB_{y}+hB_{z}$.

Observe that the line dual to $[a:b:c:f:g:h]$ is the one with coordinates
$[-f:-g:-h:a:b:c]$, which corresponds to multiplication by $i$ of the
corresponding matrix representation of the generator.

A line in $\mathbb{RP}^{3}$ can be hyperbolic, elliptic or parabolic with the
respect to the Darboux quadric. Considering simple representatives of these
classes one sees that

1) for hyperbolic line, the corresponding operator ”rotates” the dual line
(consider $R_{z}$) thus not having extra stable points in in
$\mathbb{RP}^{3}$,

2) for elliptic line, the corresponding operator hyperbolically ”rotates” the
dual line (consider $B_{z}$) and leaves invariant two extra points $[0:0:\pm
1:1]$ on the dual line.

3) for parabolic line, the corresponding operator moves all points on the dual
line (consider $\partial_{y}=R_{x}+B_{y}$) . Thus the converse to the Lemma is
not true in general.

In the following Proposition we summarize further properties of the above
correspondence.

###### Proposition 3

Let $\xi=aR_{x}+bR_{y}+cR_{z}+fB_{x}+gB_{y}+hB_{z}$ be an infinitesimal
operator of the Möbius group.

  1. 1.

The corresponding action of the one-parameter group is not loxodromic and
therefore Möbius equivalent (i.e. conjugated) either to rotation, or dilation,
or translation of the plane $(\bar{x},\bar{y})$ if and only if

$af+bg+ch=0.$ (2)

  2. 2.

This action is Möbius equivalent to rotation if and only if (2) and
$a^{2}+b^{2}+c^{2}>f^{2}+g^{2}+h^{2}$ are true, the line with Plücker
coordinates $[a:b:c:f:g:h]$ being the set of polar points of the circular
orbits.

  3. 3.

This action is Möbius equivalent to dilatation if and only if (2) and
$a^{2}+b^{2}+c^{2}<f^{2}+g^{2}+h^{2}$ are true, the line with Plücker
coordinates $[a:b:c:f:g:h]$ being the set of polar points of the orbits.

  4. 4.

The action is Möbius equivalent to translation if and only if (2) and

$a^{2}+b^{2}+c^{2}=f^{2}+g^{2}+h^{2}$ (3)

are true, the line with Plücker coordinates $[a:b:c:f:g:h]$ being the set of
polar points of the orbits.

Proof: Consider the characteristic polynomial for $\xi$

$\Psi(\lambda)=\lambda^{2}+\frac{1}{4}(a^{2}+b^{2}+c^{2}-f^{2}-g^{2}-h^{2})+\frac{i}{2}(af+bg+ch)$

and the matrices for non-loxodromic Möbius representatives $R_{z}$, $B_{z}$
and $\partial_{y}=R_{x}+B_{y}$. $\Box$

##  4 Polar curve splits into 3 non-coplanar lines

Recall that limit circles of hyperbolic pencil correspond to intersection
points of the hyperbolic line with the Darboux quadric under the stereographic
projection. Vertexes of elliptic pencil are two points common to all the
circles of the pencil, the vertexes correspond to the intersection of the
Darboux quadric with the line, dual to the polar elliptic line. Considering
the parabolic pencil as a limit case of elliptic, one calls the point
corresponding to the point of tangency of the Darboux quadric with the
parabolic line also the vertex. The vertex of a parabolic pencil is the common
point for all circles of this parabolic pencil.

The reader may visualize the pencil fixed by a line $L$ thinking of its
circles as cut on the Darboux quadric by the pencil of planes containing the
dual line $L^{*}$.

Let us list hexagonal 3-webs with non-planar polar lines. There are 9 types of
such webs up to Möbius transformation.

Figure 1: Hexagonal circular 3-webs. 3 hyperbolic pencils (left), 1 hyperbolic
and 2 elliptic pencils (center), 1 elliptic and 2 hyperbolic pencils (right).

Take 3 polar lines intersecting inside the Darboux quadric so that each of the
line contains the point dual to the plane of the other two polar lines, i.e.
each pencil has a circle orthogonal to all the circles of the other two
pencils. A representative of this web orbit, having one limit circle at
infinity, is shown in Figure 1 on the left. This type was described by
Lazareva [L-77].

Replacing two pencils by their orthogonal, we get the web in the center of
Figure 1. In the projective model, we replace two polar lines by their dual
ones. This web was also described by Lazareva [L-77].

Figure 2: Hexagonal circular 3-webs. 1 hyperbolic and 2 elliptic pencils
(left), 1 parabolic, 1 elliptic, and 1 hyperbolic pencil (center), 1
hyperbolic and 2 parabolic pencils (right).

Wunderlich [W-38] mentioned the following construction, used later also by
Balabanova and Erdoǧan, to produce hexagonal 3-webs: take two dual polar lines
and supplement it by a third intersecting that dual pair. There are four webs
in the list, obtained in this way (see also [B-73]): with two hyperbolic and
one elliptic pencils on the right of Figure 1; with one hyperbolic and two
elliptic pencils on the left of Figure 2; with one hyperbolic, one elliptic,
and one parabolic pencils in the center of Figure 2; and with one hyperbolic
and two parabolic pencils on the right of Figure 2.

Another web with two elliptic and one hyperbolic pencils is depicted on the
left of Figure 3. Its projective model has two elliptic polar lines, lying in
a plane tangent to the Darboux quadric at a point, and a hyperbolic line
passing through the points (different from the above tangency point), where
duals to elliptic lines intersect the Darboux quadric. Its elliptic pencils
share one common vertex at infinity, the other two vertices are also the limit
circles of the hyperbolic pencil. This web was described by Erdoǧan [E-74].

In the center of Figure 3, there is a web with three elliptic pencils, the
vertices of the pencils are two of three fixed points. It is historically the
first hexagonal circular 3-web described in the literature [BB-38]. The chosen
representative has a vertex at infinity.

Figure 3: Hexagonal circular 3-webs. 1 hyperbolic and 2 elliptic pencils
(left),

3 elliptic pencil (center), 1 hyperbolic, 1 elliptic and 1 parabolic pencil
(right).

Erdoǧan [E-74] found a web with one hyperbolic, one elliptic and one parabolic
pencil, arranged so that the vertex $P$ of the parabolic pencil coincides with
one vertex of the elliptic pencil, while the other vertex $E$ of the elliptic
pencil coincides with one of the limiting circles of the hyperbolic, the
common circle of the elliptic and the parabolic pencils being orthogonal to
the circle passing through the second limiting circle of the hyperbolic pencil
and the points $P$, $E$. On the right of Figure 3 is a Möbius representative
of this type web with $P$ at infinity. On the projective model, we have 3
pairwise distinct points on the Darboux quadric: $E,P$ and $H$. The hyperbolic
polar line $L_{h}$ spears the quadric at $E$ and $H$, the parabolic polar
$L_{p}$ touches the quadric at $P$ and intersects $L_{h}$, and the elliptic
polar $L_{e}$ is dual to the line through $P$ and $E$ (and intersects
$L_{p}$).

Finally, we present a family of hexagonal 3-webs formed by 3 pencils, whose
Möbius orbits are parameterized by one parameter. Two pencils are parabolic
with distinct vertexes, and the third is elliptic, whose dual to the polar
line meets the Darboux quadric at these vertexes. A representative of the
family is shown in Figure 4, the vertexes being the origin and the infinite
point. In this normalization one can fix the direction of one parabolic line,
the direction of the other is arbitrary. Any web in this normalization is
symmetric by dilatation $x\partial_{x}+y\partial_{y}$.

Figure 4: Hexagonal circular 3-web. 1 elliptic and 2 parabolic pencils.

Surprising is not only the fact that the ”largest” family was explicitly
described only in 1977 by Lazareva [L-77], even more amazing is that the
family falls within the general construction (see Introduction), which seems
to have appeared first in the paper of Wunderlich [W-38] in 1938!

To prove that there is no other classes, we will strongly use Lemma 1. The
singularity of the described type occurs when two of the three lines, tangent
to Darboux quadric at a point and meeting each its own polar line, coincide
and the point is not a vertex or limit circle of any pencil.

###### Proposition 4

Consider the family of lines such that

1) they meet two fixed lines $L_{1}$ and $L_{2}$ and

2) they are tangent to the Darboux quadric.

If the tangent points are on a circle then either $L_{1}$ intersects $L_{2}$,
or both $L_{1}$ and $L_{2}$ are tangent to the Darboux quadric, or one meets
the dual of the other at some point on the Darboux quadric.

Moreover, in the case of skew $L_{1},L_{2}$ tangent to the Darboux quadric at
two points $p_{1},p_{2}$, the curve of touching points splits into 2 circles,
their planes containing the line $p_{1}p_{2}$ and bissecting the angles
between two planes $P_{1},P_{2}$, where $P_{i}$ is the plane through $L_{i}$
and $p_{1}p_{2}$.

Proof: Let $[a:b:c:f:g:h]$ be Plucker coordinates of a line $L$ touching the
Darboux quadric. Then, by Proposition 3, they satisfy equations (2) and (3).
Therefore $a^{2}+b^{2}+c^{2}\neq 0$ and we can normalize these coordinates to
$a^{2}+b^{2}+c^{2}=f^{2}+g^{2}+h^{2}=1$. One easily calculates the point
$p=(x,y,z)=(cg-bh,ah-cf,bf-ag)$ where the line $L$ touches the quadric. Due to
normalization, one can rewrite this as

$a=hy-gz,\ \ \ b=fz-hx,\ \ \ c=gx-fy.$ (4)

To simplify calculations, we can bring the Plücker coordinates of $L_{1}$ to
simple form by Möbius transformation. We suppose that $L_{1}$ and $L_{2}$ are
skew since the case of $L_{1},L_{2}$ intersecting is obvious.

If $L_{1}$ is hyperbolic we can choose a representative as
$L_{1}=[0:0:1:0:0:0]$. Then one has $m\neq 0$ since the lines $L_{1},L_{2}$ do
not intersect. (We used the fact that the Plücker coordinates of intersecting
lines are orthogonal with respect to the bilinear symmetric form defining the
Plücker quadric.) Let us set $m=1$. Moreover, the coordinates of $L_{2}$
satisfy the Plücker equation $ku+lv+w=0$ thus giving $w$. Since $L$ intersect
$L_{1}$ and $L_{2}$ we have $h=0$ and $ka+lb+c+uf+gv+wh=0$ respectively. The
above two equations cut a curve from the three-dimesional variety of lines
touching the quadric.

The touching points $p$ trace a curve on the Darboux quadric. This curve can
be computed as follows. Equations (2) and (4) imply $fx+gy+hz=0$, with $h=0$
we have $g=-fx/y$. Now normalization $f^{2}+g^{2}+h^{2}=1$ gives
$(x^{2}+y^{2})f^{2}=y^{2}$, which means $f$ is not identically zero. Therefore
intersection condition $ka+lb+c+uf+gv+wh=0$ is equivalent to $kxz+lyz+uy-
vx-x^{2}-y^{2}=0$. This equation cuts the curve of tangent points on the
Darboux quadric (i.e. unit sphere centered at the origin). If this curve is in
a plane $z=Ax+By+C$ then by direct calculation one gets $C^{2}=1$. One can
choose $C=1$ and then, by further calculations, we get $A=-k$, $B=-l$, $l=u$
$k=-v$. Thus $L_{2}=[u:v:0:-v:u:1]$ lies in the plane tangent to the Darboux
quadric at $(0,0,-1)$, which is the the point where the dual to $L_{2}$ spears
the Darboux quadric. Note that the plane equation is $z=vx-uy+1$.

If $L_{1}$ is hyperbolic we choose $L_{1}=[0:0:0:0:0:1]$. Then $w\neq 0$ and
we can normalize $L_{2}=[u:v:1:k:l:m]$ and the rest of the reasoning goes in a
similar way.

If $L_{1}$ is parabolic we choose $L_{1}=[0:-1:0:1:0:0]$. For skew
$L_{1},L_{2}$ we have $-l+u\neq 0$. Normalizing $l-u=1$ gives $l=u+1$. Further
we rewrite (4) as

$f=bz-cy,\ \ \ g=cx-az,\ \ \ h=ay-bx$

and, proceeding as before, obtain by calculation that $L_{2}$ satisfy (2),
which means it is tangent to the Darboux quadric.

To check the last claim, we normalize the configuration so that $L_{1}$ and
$L_{2}$ are tangent to the Darbox quadric at points $(0,0,\pm 1)$, look for a
plane containing a circle of tangent points in the form $y=Ax$, find two
solutions for $A$, and verify the geometry by calculation. $\Box$

###### Definition 2

Motivated by Lemma 1, we will refer to the circle of tangent points described
in Proposition 4 whose polar point is not a intersection of $L_{1},L_{2}$ as
singular circle.

Remark 1. It is easy to see that the curve of touching points is of order
four. In the hypothesis of the Proposition for the case of skew $L_{1}$,
$L_{2}$ not tangent to the Darboux quadric, this curve also splits. One
component is a circle and the real trace of other is the point, where dual to
the elliptic line meets the hyperbolic.

###### Corollary 5

For hexagonal circular 3-webs formed by three pencils with non-planar polar
lines, any two $L_{1},L_{2}$ of polar lines are either dual or obey
geometrical restriction described by Proposition 4. Moreover, the polar point
of a singular circle, defined by two polar lines, belongs to the third one.

Proof: Let $L_{1},L_{2}$ be skew but not dual and such that the curve of
points, where lines $L$ meeting both $L_{1},L_{2}$ touch the Darboux quadric,
is not planar. Then by Lemma 1 any such line meets also the third polar line
$L_{3}$. Since $L_{1},L_{2}$ are skew $L_{3}$ cannot intersect neither of
$L_{1},L_{2}$. Thus $L_{1},L_{2},L_{3}$ belong to one ruling of some quadric
$Q$ and the touching lines $L$ form the other ruling. But then
$L_{1},L_{2},L_{3}$ are also tangent to the Darboux quadric. This contradicts
our initial assumption. $\Box$

Let $(p,q),(r,s)\in\mathbb{R}^{2}$ be vertices of an elliptic pencil, then the
pencil circles form the family

$I(x,y):=\frac{[(p-x)(r-x)+(q-y)(s-y)]^{2}}{[(p-x)^{2}+(q-y)^{2}][(r-x)^{2}+(s-y)^{2}]}=const.$

The circles are the integral curves of the ODE

$\omega_{e}:=d(I)=f(x,y)dx+g(x,y)dy=0.$ (5)

The circles of the hyperbolic pencil with limit circles at $(p,q),(r,s)$ are
orthogonal to the circles of the above elliptic one. They are the integral
curves of the ODE

$\omega_{h}:=g(x,y)dx-f(x,y)dy=0.$ (6)

Now we work out the cases with parabolic pencils. Let $(p,q)\in\mathbb{R}^{2}$
be the vertex of a parabolic pencil and $[r:1-r]\in\mathbb{P}$ a direction
orthogonal to the line tangent to all circles of the pencil. Then the pencil
circles form the family

$\tilde{I}(x,y):=\frac{(x-p)^{2}+(y-q)^{2}}{r(p-x)+(r-1)(y-q)}=const.$

The circles are the integral curves of the ODE

$\omega_{p}:=d(\tilde{I})=0.$ (7)

For the exceptional direction $(1,-1)$, the pencil circles family is

$\bar{I}(x,y):=\frac{(x-p)^{2}+(y-q)^{2}}{(x-p)-(y-q)}=const$

and the corresponding EDO

$\omega_{\bar{p}}:=d(\bar{I})=0.$ (8)

Observe that differential forms $\sigma_{i}$, describing a 3-web of circles
formed by 3 pencils, are algebraic. Thus Lemma 1 remains valid also over
complex numbers in passing from $\mathbb{RP}^{3}$ to $\mathbb{CP}^{3}$. By
circles here we understand sections of the complex Darboux quadric by complex
planes. The complexification simplifies the proof of the following theorem.

###### Theorem 6

[S-05] Any hexagonal circular 3-webs formed by three pencils with non-coplanar
polar lines is Möbius equivalent to one from the Blaschke-Wunderlich-
Balabanova-Erdoǧan-Lazareva list.

Proof: We consider all types of non-coplanar polar line triples
$L_{1},L_{2},L_{3}$.

$\bullet$ Three hyperbolic pencils.
By Corollary 5, all lines intersect at one point $p$. This point cannot be
outside the Darboux quadric. In fact, applying a suitable Möbius
transformation, we send this point to an infinite one, say $p_{x}=[1:0:0:0]$
and the plane of $L_{1},L_{2}$ to the plane $Z=0$. Then, by Corollary 5, the
polar line $L_{3}$ joins $p_{x}$ and $p_{z}=[0:0:1:0]$, the polar point of the
plane $Z=0$. Thus $L_{3}$ cannot be hyperbolic as supposed. The point $p$
cannot be on the Darboux quadric either: we can send it to $p=(1,0,0)$ and the
plain of $L_{1},L_{2}$ to the plane $Z=0$. Now Corollary 5 implies that
$L_{3}$ joins $p$ and $p_{z}$. Thus $L_{3}$ is parabolic and not hyperbolic.

Therefore $p$ is inside the Darboux quadric and one can send it to the origin
$(0,0,0)$. Corollary 5 implies that any of the lines $L_{1},L_{2},L_{3}$
contains the point dual to the plane of the other two lines. Thus
$L_{1},L_{2},L_{3}$ are orthogonal and we have the web shown on the left of
Figure 1. Note that there is only one such web up to Möbius transformation.

$\bullet$ One elliptic and two hyperbolic pencils.
We can suppose that $L_{3}$ is an elliptic polar line and that $L_{3}$ is the
infinite line in the plane $Z=0$. Due to Corollary 5 the polar lines
$L_{1},L_{2}$, being hyperbolic, must intersect. Then the plane of
$L_{1},L_{2}$ is dual to some point on $L_{3}$ and therefore contains the line
$X=Y=0$ dual to $L_{3}$ and can be assumed to be the $Y=0$.

Suppose that none of $L_{1},L_{2}$ is dual to $L_{3}$. If both $L_{1},L_{2}$
intersect $L_{3}$ then the intersection point is $p_{x}=[1:0:0:0]$. Applying
Corollary 5 to the pair $L_{1},L_{3}$, we conclude that $L_{2}$, joining
$p_{x}$ and the polar point of the plane of $L_{1},L_{3}$ does not meet the
Darboux quadric and is not hyperbolic as supposed. Thus we can suppose that
$L_{1},L_{3}$ are skew and that $L_{1}$, by Corollary 5, contains $(0,0,-1)$.
Since $L_{1}$ is not dual to $L_{3}$, we infer by Corollary 5 that $L_{2}$
contains the dual point of the singular circle for $L_{1},L_{3}$. This point
lies in the tangent plane to the Darboux quadric at $(0,0,1)$, therefore
$L_{2}$, being non-parabolic, cannot meet $L_{3}$ and cannot also contain
$(0,0,1)$. Then it passes through $(0,0,-1)$, which contradicts the geometry
restriction imposed by Corollary 5.

So we can assume that $L_{2}$ is dual to $L_{3}$, i.e. it is the line $X=Y=0$.
Corollary 5 prevents $L_{1}$ to be skew with $L_{3}$. Thus it meets $L_{3}$
and therefore intersect $L_{2}$ in a point inside the Darboux quadric. We
obtain the hexagonal web equivalent to the one on the right of Figure 1.

$\bullet$ One hyperbolic and two elliptic pencils.
By Corollary 5, elliptic lines, say $L_{2},L_{3}$, intersect. We can assume
that the hyperbolic line $L_{1}$ is the coordinate axis $X=Y=0$.

First consider the case when two lines, $L_{1},L_{2}$, are dual. Then $L_{2}$
is the infinite line in the plane $Z=0$ and we can assume the intersection
line of $L_{2}$ and $L_{3}$ being the point $p_{y}=[0:1:0:0]$. Then $L_{3}$
cannot be skew with $L_{1}$ due to Corolllary 5: the polar of the singular
circle determined by $L_{1},L_{3}$ is finite and cannot lie on the infinite
line $L_{2}$. Thus $L_{3}$, being elliptic, intersect $L_{1}$ outside the
Darboux quadric and we get the web equivalent to the one shown on the left of
Figure 2.

Now suppose that no pair $L_{1},L_{i}$ is skew. Then $L_{2}$ and $L_{3}$
intersect $L_{1}$. Since the triple $L_{1},L_{2},L_{3}$ is not coplanar, all
three lines intersect at one point outside the Darboux quadric, which can be
taken as $p_{z}=[0:0:1:0]$. By Corollary 5, the line $L_{3}$ contains the dual
point of the singular circle determined by $L_{1},L_{2}$ and the line $L_{2}$
contains the dual point of the singular circle determined by $L_{1},L_{3}$.
This fixes $L_{1},L_{2},L_{3}$ up to rotation around $z$-axis and gives the
web shown in the center of Figure 1.

Finally, consider the case with skew but not dual $L_{1},L_{2}$. Applying
Möbius transformation, we send the intersection point of $L_{2}$ and $L_{3}$
to $p_{x}=[1:0:0:0]$ (preserving the position of $L_{1}$). Then $L_{3}$ joins
$p_{x}$ and the polar point of the singular circle determined by
$L_{1},L_{2}$. We obtain the web, equivalent to one on the left of Figure 3.

$\bullet$ Three elliptic pencils.
We treat this case using the complex version of Corollary 5. First we conclude
that, being nonplanar, all three polar lines intersect at one point, which we
can send to $p_{y}=[0:1:0:0]$. None of 3 real planes, containing a pair of
polar lines, can miss completely the real Darboux quadric. In fact, if the
real plane of $L_{1},L_{2}$ do not intersect the real Darboux quadric then the
polar point of the complex singular circle of $L_{1},L_{2}$ is inside the real
Darboux quadric and $L_{3}$, being elliptic, cannot contain this point. None
of 3 real planes, containing a pair of polar lines, can intersect the real
Darboux quadric. If the real plane of $L_{1},L_{2}$ cuts the real Darboux
quadric, the polar point of the real singular circle of $L_{1},L_{2}$ is
outside the real Darboux quadric and the real plane of $L_{1},L_{3}$ do not
meet the real Darboux quadric. Thus all 3 planes are tangent to the Darboux
quadric. A representative of such web is shown in the center of Figure 3.

$\bullet$ One parabolic and two hyperbolic pencils.
By Corollary 5 all 3 polar lines must intersect in one point. The intersection
point cannot be inside the Darboux quadric since one line is parabolic. It
also cannot be outside: we can send it to $p_{x}=[1:0:0:0]$ and the parabolic
line, joining $p_{x}$ and the point dual to the plane of hyperbolic lines,
will miss the Darboux quadric. Therefore this point is on the Darboux quadric.
One can move the plane of hyperbolic lines to $Z=0$. Then the parabolic line
contains $p_{z}=[0:0:1:0]$ and none of the hyperbolic lines can pass through
the point dual the singular circle of the other two lines. Thus there is no
hexagonal web with non-planar parabolic and two hyperbolic polar lines.

$\bullet$ One parabolic, one hyperbolic and one elliptic pencil.
By Corollary 5 the parabolic line meets the other two. If the hyperbolic line
intersects the elliptic at some point $p$ outside the Darboux quadric, we can
move the hyperbolic line to $X=Z=0$ and $p$ to $p_{y}=[0:1:0:0]$. Then the
point $p_{s}$, dual for the singular circle of hyperbolic and elliptic lines,
is infinite and the third polar line, joining $p_{y}$ and $p_{s}$ cannot be
parabolic.

Therefore hyperbolic and elliptic lines are skew and Corollary 5 fixes the
configuration up to Möbius transformation: if these lines are dual we get the
type shown in the center of Figure 2, otherwise the type on the right of
Figure 3.

$\bullet$ Two parabolic and one hyperbolic pencils.
By Corollary 5 the hyperbolic line meets both parabolic lines.

If the parabolic lines are skew then the corresponding two singular circles
have their polar points on the line dual to the one joining the points of
tangency of parabolic lines with the Darboux quadric. This dual line is
elliptic, therefore this configuration is not possible.

If the parabolic lines intersect outside the Darboux quadric then we can bring
the plane of their intersection to $Z=0$. Now the the hyperbolic line contains
$p_{z}=[0:0:1:0]$ by Corollary 5 and, intersecting the both elliptic lines,
must meet them at their common point. Then it misses the Darboux quadric and
is not hyperbolic.

Therefore the parabolic lines are tangent to the Darboux quadric at the same
point. Since the polar lines are not coplanar the hyperbolic line contains
this point. We can bring the hyperbolic line to $X=Y=0$. Then Corollary 5
implies that the parabolic lines are dual and we obtain the web type shown on
the right of Figure 2.

$\bullet$ Two parabolic and one elliptic pencils.
By Corollary 5 the elliptic line meets both parabolic lines.

If the parabolic lines are skew then the corresponding two singular circles
have their dual points on the line dual to the one joining the points of
tangency of parabolic lines with the Darboux quadric. The third polar line
must contain these point, it is elliptic and we get the web type presented in
Figure 4.

If the parapolic lines intersect outside the Darboux quadric then we can bring
the plane of their intersection to $Z=0$. By Corollray 5 the elliptic line
contains $p_{z}=[0:0:1:0]$ and, intersecting the both elliptic lines, must
meet them at their common point. Thus we get the type shown in Figure 4.

If the parabolic lines are tangent to the Darboux quadric at the same point
then the third line, being elliptic, cannot pass through this point. Therefore
it is coplanar with the parabolic lines.

$\bullet$ Three parabolic pencils.
Suppose that two polar lines $L_{1}$ and $L_{2}$ intersect. If the
intersection point is outside the Darboux quadric then we bring the plane of
$L_{1},L_{2}$ to $Z=0$. Since the third polar line $L_{3}$ do not lie in this
plain it contains $p_{z}=[0:0:1:0]$ by Corollary 5. Therefore it is skew with
at least one of $L_{1},L_{2}$. Let it be $L_{1}$. Then by Corollary 5 the line
$L_{2}$ contains both dual point of two singular circles of $L_{1},L_{3}$
which is obviously not possible.

If $L_{1},L_{2}$ touch the Darboux quadric at the same point then the line
$L_{3}$ is skew with at least one of $L_{1},L_{2}$. Again this is precluded by
Corollary 5.

Therefore $L_{1},L_{2},L_{3}$ are pairwise skew. Consider two singular circles
$C_{1},C_{2}$ of $L_{1},L_{2}$, their polar points $p_{1},p_{2}$ and the
family of lines $L$ touching the Darboux quadric and meeting both
$L_{1},L_{2}$. The third line $L_{3}$, being parabolic, cannot contain both
points $p_{1},p_{2}$. If $p_{1}\notin L_{3}$ then by Lemma 1 the family of
lines $L$ touching the Darboux quadric at points of $C_{1}$ must meet also
$L_{3}$. Therefore the lines $L$ constitute one ruling of a quadric touching
the Darboux quadric along $C_{1}$. Therefore $L_{1},L_{2},L_{3}$ belong to the
second ruling. If $p_{2}\in L_{3}$ then, sending the plane of $C_{1}$ to $Y=0$
and the points, where $L_{1},L_{2}$ touch the Darboux quadric, to $(0,0,\pm
1)$, we see that $p_{2}=[1:0:0:0]$ since $C_{2}$ is orthogonal to $C_{1}$ and
passes through $(0,0,\pm 1)$. Then $L_{3}$ must touch the Darboux quadric also
at one of the points $(0,0,\pm 1)$, which contradicts the initial assumption
that all 3 lines are skew. Thus $p_{2}\notin L_{3}$. Then the family of lines
$L$ touching the Darboux quadric at points of $C_{2}$ must meet also $L_{3}$.
This is not possible as the lines $L$, constituting a ruling of a quadric that
touch the Darboux quadric along $C_{1}$ cannot meet the orthogonal circle
$C_{2}$. $\Box$

##  5 Polar curve splits into conic and straight line

First we describe the types then we prove that the list is complete. Some
types are one-parametric families and we denote the parameter value by $c$.
The polar conic will be given either by an explicit parametrization of the
circle equations and we reserve $u$ for the parameter, or by indicating the
conic equations. The former representation gives also a parametrization of the
polar conic: to a circle

$\epsilon(x^{2}+y^{2})+\alpha x+\beta y+\gamma=0,$

(where $\epsilon=0$ or $\epsilon=1$, the case $\epsilon=0$ giving a line)
corresponds the polar point with the tetracyclic coordinates

$[\alpha:\beta:\gamma-\epsilon:-\gamma-\epsilon].$ (9)

The parameter for circles in the pencil will be denoted by $v$.

To check hexagonality, one computes the Blaschke curvature using the formula
in Apendix as follows. The polar conic gives a one-parameter family of circles
on the unit sphere. The stereographic projection from the ”south” pole

$X=\frac{2x}{1+x^{2}+y^{2}},\ \ \ \ Y=\frac{2y}{1+x^{2}+y^{2}},\ \ \ \
Z=\frac{1-x^{2}-y^{2}}{1+x^{2}+y^{2}},$

transforms this family into a family of circles in the plane parametrized by
the points of the conic. The ODE for circles, defined by the conic,

$P^{2}+A(x,y)P+B(x,y)=0,\ \ \ \ P=\frac{dy}{dx},$ (10)

is obtained by differentiating and excluding the coordinates of the conic
points. The slope $R$ comes from the pencil of circles.

For some types we add an additional geometric detail in the ”title” to
separate the types.

1\. Polar conic plane does not cut Darboux quadric, hyperbolic pencil.

The pencil with limit circles at the origin and in the infinite point gives
circles

$x^{2}+y^{2}=v,$

the polar conic is the circle

$X_{0}^{2}+Y_{0}^{2}+4cX_{0}Z_{0}=0,\ \ \ U_{0}=0,\ \ \ c>0,$

defining the family

$x^{2}+y^{2}-\frac{4c}{c^{2}u^{2}+1}x+\frac{4c^{2}u}{c^{2}u^{2}+1}y=1,$

the circles of the family enveloping the cyclic

$(x^{2}+y^{2})^{2}-(x^{2}+y^{2})(4cx+2)-4c^{2}y^{2}+4cx+1=0$

as shown on the left in Figure 5.

Figure 5: Types 1, 2, 3.

2\. Polar conic plane does not cut Darboux quadric, hyperbolic pencil, webs
symmetric by rotations.

The pencil with limit circles at the origin and at the infinite point gives
circles

$x^{2}+y^{2}=v,$

the polar conic is the circle

$X_{0}^{2}+Y_{0}^{2}=\frac{4Z_{0}^{2}}{c^{2}},\ \ \ U_{0}=0,$

defining the family

$x^{2}+y^{2}+\frac{2\cos(u)}{c}x+\frac{2\sin(u)}{c}y=1,$

the circles of the family enveloping the cyclic

$c^{2}(x^{2}+y^{2})^{2}-(2c^{2}+4)(x^{2}+y^{2})+c^{2}=0,$

which splits into two concentric circles as shown in the center of Figure 5.

3\. Polar conic plane cuts Darboux quadric, hyperbolic pencil.

The pencil with limit circles at the origin and at the infinite point gives
circles

$x^{2}+y^{2}=v,$

the polar conic is the circle

$x_{0}^{2}+y_{0}^{2}=4cx_{0},\ \ \ z_{0}=0,\ \ \ c>0,$

defining the family

$x^{2}+y^{2}+\frac{4c}{c^{2}u^{2}+1}x-\frac{4c^{2}u}{c^{2}u^{2}+1}y=-1,$

the circles of the family enveloping the cyclic

$(x^{2}+y^{2})^{2}+(x^{2}+y^{2})(4cx+2)-4c^{2}y^{2}+4cx+1=0,$

as shown on the right of Figure 5.

Figure 6: Types 4, 5, 6.

4\. Polar conic plane cuts Darboux quadric, hyperbolic pencil, webs symmetric
by rotations.

The pencil with limit circles at the origin and at the infinite points gives
circles

$x^{2}+y^{2}=v,$

the polar conic is the circle

$x_{0}^{2}+y_{0}^{2}=\frac{4}{c^{2}},\ \ \ z_{0}=0,$

defining the family

$x^{2}+y^{2}-\frac{2\cos(u)}{c}x-\frac{2\sin(u)}{c}y=-1,$

the circles of the family enveloping the cyclic

$c^{2}(x^{2}+y^{2})^{2}+(2c^{2}-4)(x^{2}+y^{2})+c^{2}=0,$

which splits into two concentric circles as shown on the left in Figure 6.

5\. Polar conic plane cuts Darboux quadric, elliptic pencil, webs symmetric by
homotheties.

The pencil with vertexes at the origin and at the infinite point gives lines

$y=vx,$

the polar conic is

$y_{0}^{2}+cz_{0}^{2}=c,\ \ \ x_{0}=0,\ \ c>1,$

defining the family of circles

$x^{2}+(y+u)^{2}=(1-1/c)u^{2},$

the circles enveloping the lines

$x^{2}=(c-1)y^{2}$

as shown in the center in Figure 6. The webs of the family are symmetric by
homotheties with the center in the origin.

6\. Polar conic plane cuts Darboux quadric, elliptic pencil, polar conic and
dual to conic line intersect in 2 points on the Darboux quadric.

The pencil with vertexes at the origin and at the infinite point gives lines

$y=vx,$

the polar conic is

$2cy_{0}=z_{0}^{2}-1,\ \ \ x_{0}=0,\ \ \ c>0,$

defining the family of circles

$x^{2}+y^{2}+\frac{(u-1)}{cu}y+\frac{u-1}{u+1}=0,$

the circles enveloping the cyclic

$c^{2}(x^{2}+y^{2})^{2}+(4cy-2c^{2})(x^{2}+y^{2})+(2y+c)^{2}=0$

as shown on the right in Figure 6.

Figure 7: Types 7, 8, 9.

7\. Polar conic plane cuts Darboux quadric, elliptic pencil.

The pencil with vertexes at the origin and at the infinite point gives lines

$y=vx,$

the polar conic is

$2cy_{0}z_{0}+1=z_{0}^{2},\ \ \ x_{0}=0,\ \ \ c>0,$

defining the family of circles

$x^{2}+y^{2}-\frac{c+u}{cu}y+\frac{c+u}{c-u}=0,$

as shown on the left in Figure 7.

8\. Polar conic plane cuts Darboux quadric, parabolic pencil.

The pencil with the vertex at the infinite point gives lines

$y=v,$

the polar conic

$x_{0}^{2}-z_{0}=1,\ \ \ y_{0}=0,$

gives the family of circles

$\left(x-\frac{1}{u}\right)^{2}+y^{2}=\frac{u^{2}-1}{u^{2}}.$

The circles envelopes the ellipse

$x^{2}+2y^{2}=2$

as shown in the center in Figure 7.

9\. Polar conic plane cuts Darboux quadric, parabolic pencil, webs symmetric
by translations.

The pencil with the vertex at the infinite point gives lines

$y=v,$

the polar conic is

$x_{0}^{2}+(1-\sqrt{2})z_{0}^{2}-2\sqrt{2}z_{0}=1+\sqrt{2},\ \ \ y_{0}=0,$

defining the family of circles

$(x+u)^{2}+y^{2}=1,$

the circles touching the lines

$y=\pm 1$

as shown on the right in Figure 7.

10\. Polar conic plane tangent to Darboux quadric, hyperbolic pencil.

The pencil with limit circles at $(1,0)$ and $(-1,0))$ gives circles

$x^{2}+y^{2}+1=vx,$

the polar conic is

$x_{0}^{2}+(1-c)y_{0}^{2}=c,\ \ \ z_{0}=-1,\ \ \ c>0,\ \ \ c\neq 1,$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the conic

$cx^{2}+\frac{c}{c-1}y^{2}=1$

with foci $(1,0)$ and $(-1,0))$ as shown on the left in Figure 8.

Figure 8: Type 10, 11, 12.

11\. Polar conic plane tangent to Darboux quadric, hyperbolic pencil, polar
line meets polar conic.

The pencil with limit circles at $(1,0)$ and $(-1,0))$ gives circles

$x^{2}+y^{2}+1=vx,$

the polar conic is

$y_{0}^{2}-cx_{0}y_{0}+cy_{0}+x_{0}=0,\ \ \ z_{0}=-1,\ \ \ c\geq 0,$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the parabola

$(cx-y)^{2}-(2c^{2}+4)x-2cy+c^{2}=0$

with focus at $(1,0)$ as shown in the center in Figure 8.

12\. Polar conic plane tangent to Darboux quadric, hyperbolic pencil, polar
line contains the point where polar conic plane touches the Darboux quadric.

The pencil with limit circles at the origin and infinity gives circles

$x^{2}+y^{2}=v,$

the polar conic is

$x_{0}^{2}+y_{0}^{2}-2y_{0}=0,\ \ \ z_{0}=-1$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the parabola

$x^{2}+2y=1$

with focus at $(0,0)$ as shown on the right in Figure 8.

13\. Polar conic plane tangent to Darboux quadric, hyperbolic pencil, web
symmetric by rotations.

The pencil with limit circles at the origin and infinity gives circles

$x^{2}+y^{2}=v,$

the polar conic is

$x_{0}^{2}+y_{0}^{2}=1,\ \ \ z_{0}=-1$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the circle

$x^{2}+y^{2}=1$

as shown on the left in Figure 9.

Figure 9: Type 13, 14, 15.

14\. Polar conic plane tangent to Darboux quadric, elliptic pencil.

The pencil with vertexes at $(1,0)$ and $(-1,0))$ gives circles

$\frac{(x^{2}+y^{2}-1)^{2}}{(x^{2}+y^{2}-2x+1)(x^{2}+y^{2}+2x+1)}=v,$

the polar conic is

$cx_{0}^{2}+(c+1)y_{0}^{2}+1=0,\ \ \ z_{0}=-1,$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the conic

$\frac{x^{2}}{c}+\frac{y^{2}}{c+1}=-1$

with foci $(1,0)$ and $(-1,0))$ as shown in the center in Figure 9.

15\. Polar conic plane tangent to Darboux quadric, parabolic pencil.

The pencil with vertex at the origin gives circles

$x^{2}+y^{2}=2vy,$

the polar conic is

$x_{0}^{2}+y_{0}^{2}=1,\ \ \ z_{0}=-1$

defining the family of lines

$x_{0}x+y_{0}y=1,$

the lines of the family enveloping the circle

$x^{2}+y^{2}=1$

as shown on the right in Figure 9.

###### Theorem 7

The webs of different types in the above classification list are not Möbius
equivalent. The webs of the same type with different normal forms are not
Möbius equivalent.

Proof: The webs from different types are not Möbius equivalent: discrete
geometric invariants indicated in the discriptions, such as 1) presence of
infinitesimal symmetry and 2) the mutual position of polar line, polar conic
and Darboux quadric, effectively separate the types.

To see that the different normal forms within a family are not Möbius
equivalent, one computes the subgroup $G_{p}$ of $PSO(3,1)$ respecting the
positions of the polar conic plane and the polar line in the chosen
normalization and checks that the $G_{p}$-orbits of the canonical forms are
different.
For the first 4 types, $G_{p}$ is generated by $R_{z}$ and by reflections in
the coordinate planes.
For the 5th, 6th, 7th type, $G_{p}$ is generated by $B_{z}$ and by reflections
in the coordinate planes.
For the 10th, 11th and 14th types, $G_{p}$ is discrete and generated by
reflections in the planes $x=0$ and $y=0$. $\Box$

As in the case of 3 pencils, Lemma 1 effectively selects candidates among
3-webs that can be hexagonal. The singularities of the type described by the
Lemma arise when either 1) a line, joining 2 different points on the polar
conic, touches the Darboux quadric at a point $p_{t}$, while the polar conic
plane is not tangent the Darboux quadric or 2) a line, joining a point on the
polar line with a point on the polar conic, touches the Darboux quadric at a
point $p_{t}$ while the tangent plane to the Darboux quadric at $p_{t}$ is not
tangent to the polar conic.

In the former case, the points $p_{t}$ trace a circle, which is the
intersection of the polar conic plane with the Darboux quadric. Then, by Lemma
1, the polar of the polar conic plane lies on the pencil polar line.

In the latter case, consider a point $p_{r}$ running over the polar line. For
non-planar polar set, $p_{r}$ meets the polar conic plane $\pi_{c}$ only at
one point. Therefore the one-parameter family of cones tangent to the Darboux
quadric and having their vertexes at $p_{r}$ cuts the plane $\pi_{c}$ in a
one-parameter family of conics $c_{r}$. Each conic $c_{r}$ intersect the polar
conic $c_{p}$ at 4 (possibly complex or multiple) points. If the polar conic
$c_{p}$ is not a member of the family $\\{c_{r}\\}$, at least one of these 4
intersection points is moving along $c_{p}$ as $p_{r}$ runs over the polar
line. In fact, if intersection points are stable then $\\{c_{r}\\}$ is a
pencil of conics containing $c_{p}$.

Choose one such moving intersection point $p_{i}$. Lines $l_{r}$ tangent to
$c_{r}$ at $c_{i}$ form a one-parameter family, or congruence of lines. All
the objects in this construction are considered as complex but the polar conic
and the polar line must have equations with real coefficients and the real
part of the polar conic cannot lie completely inside the Darboux quadric.

###### Proposition 8

If the polar curve of a hexagonal circular 3-web is non-planar and splits into
a line and a smooth conic then for the web complexification hold true

1) the polar of the polar conic plane lies on the polar line, if this plane is
not tangent to the Darboux quadric and

2) the congruence of lines $l_{r}$ is a pencil with the vertex on the polar
conic, if the polar conic is not a member of the family $\\{c_{r}\\}$.

Proof: The first claim follows directly from Lemma 1. For conic planes missing
the Darboux quadric, the complex version works.

To derive the second claim, observe that the line $p_{r}p_{i}$ touches the
Darboux quadric at a singular point treated by Lemma 1. Thus the tangency
point must trace a circle on the Darboux quadric and the polar point $p_{0}$
of the circle must lie on the polar conic. Then the plane, tangent to the
Darboux quadric and passing through $p_{r}p_{i}$, contains $p_{0}$. This plane
cuts the plane $\pi_{c}$ of the polar conic along a line $l_{r}$, passing
through the $p_{i}$ and tangent to the corresponding conic $c_{r}$. Hence
$\\{l_{r}\\}$ is the pencil with vertex at $p_{0}$. $\Box$

###### Theorem 9

If the polar curve of a hexagonal circular 3-web splits into a smooth conic
and a straight line, not lying in a plane of the conic, then the web is Möbius
equivalent to one from the above presented list.

Proof: The polar conic plane either completely misses the Darboux quadric, or
cuts it in a circle, or is tangent to it. The polar line is either hyperbolic,
or elliptic, or parabolic. Thus we have 9 cases to consider, each case
defining a set of webs (possibly empty).

$\bullet$ Polar conic plane misses the Darboux quadric.

Applying a suitable Möbius transformation, we can send the polar conic plane
to infinity. Then by Proposition 8 the polar line contains the origin of the
affine chart and therefore meets the Darboux quadric at 2 points. Thus the
polar line is hyperbolic. Applying a rotation around the origin, we map these
two points to $(0,0,\pm 1)$. In the affine coordinates $x=\frac{X}{Z},\
y=\frac{Y}{Z}$ on the polar conic plane $U=0$, the conics $c_{r}$ are

$x^{2}+y^{2}=r,$

where $r$ is considered as a complex parameter. These conics are real only for
real non-negative $r$. If the polar conic coincides with one of the conics
$c_{r}$ then we get Type 2, symmetric by $R_{z}$.

If the polar conic is not one of $c_{r}$ then, by Proposition 8, the points,
where lines of pencil with the vertex at some point
$p_{0}=(x_{0},y_{0})\in\mathbb{C}^{2}$ touch the circles $c_{r}$, run over the
polar conic. The line of the pencil $y=y_{0}+k(x-x_{0})$ corresponding to the
parameter $k\in\mathbb{C}$, is tangent to the conic $c_{r}$ for

$r=\frac{x_{0}^{2}k^{2}-2x_{0}y_{0}k+y_{0}^{2}}{k^{2}+1}$

at the point

$p(k)=(x(k),y(k))=\left(\frac{k(x_{0}k-y_{0})}{k^{2}+1},\frac{y_{0}-x_{0}k}{k^{2}+1}\right).$

The points $p(k)$ run over the complex conic

$x^{2}+y^{2}=x_{0}x+y_{0}x.$

This conic is real if and only if $x_{0}$ and $y_{0}$ are real. It is smooth
if and only if $p_{0}\neq(0,0)$. We got the first web of the list, Type 1.
This conic is the circle, passing through the origin $O=(0,0)$ and $p_{0}$ and
having its center at the midpoint of the segment $Op_{0}$. (We obtained a
theorem of scholar geometry.) Using $R_{z}$ we normalize $p_{0}$ to
$y_{0}=0,x_{0}>0$.

For infinite $p_{0}$, different from the cyclic points, the line $l(k)$ of the
pencil $y=ax+k$, $a\in\mathbb{C}$ touches a unique conic of the family
$\\{c_{r}\\}$ at the point

$p(k)=\left(-\frac{ak}{a^{2}+1},\frac{k}{a^{2}+1}\right).$

The points $p(k)$ are collinear: $x(k)+ay(k)=0$ and we cannot obtain a smooth
polar conic in this way. Finally, if $p_{0}$ is cyclic, the lines from the
pencil touch the conics $c_{r}$ at the very point $p_{0}$ and we get no conic
at all.

$\bullet$ Polar conic plane cuts the Darboux quadric, hyperbolic pencil.

We send the conic plane to $Z=0$. Now the polar line contains the infinite
point $[0:0:1:0]$ by Proposition 8. The subgroup of the Möbius group,
preserving the plane $Z=0$, is generated by rotations around $Z$-axis and the
boosts $B_{x}$ and $B_{y}$. Using this subgroup, we normalize the polar line
to $x=y=0$. The conics $c_{r}$ are circles in the plane $z=0$ with the center
at the origin. Repeating the arguments that we used above for conic planes
missing the Darboux quadric, we get Types 3 and 4.

$\bullet$ Polar conic plane cuts the Darboux quadric, elliptic pencil.

We normalize the conic plane to $X=0$. Then by Proposition 8 the point
$[1:0:0:0]$ lies on the polar line. The polar line, being elliptic, meets the
plane $X=0$ at a point $p$ outside the unit circle. Möbius transformations,
preserving the plane $X=0$, are generated by rotations around $X$-axis and the
boosts $B_{y}$ and $B_{z}$. Using these transformations, we send the point $p$
to $[0:1:0:0:0]$. Now the polar line is $U=Z=0$ and the conics $c_{r}$ have
equations

$z^{2}+\frac{y^{2}}{r}=1$

in the affine coordinates. If the polar conic coincides with one of these
conics then $r$ is real and positive and we get Type 5.

Othervise, by the second claim of Proposition 8, the lines $l_{r}$ meet at
some point $p_{0}\in c_{p}$. If $p_{0}=(0,y_{0},z_{0})$ is finite, a line of
the pencil of lines $z=z_{0}+k(y-y_{0})$ with vertex at $p_{0}$ is tangent to
the conic $c_{r}$ with

$r=\frac{y_{0}^{2}k^{2}-2y_{0}z_{0}k+z_{0}^{2}-1}{k^{2}}.$

Thus the points, where the lines $l_{r}$ touch $c_{r}$, are parametrized by
$k$ via

$y=\frac{y_{0}^{2}k^{2}-2y_{0}z_{0}k+z_{0}^{2}-1}{k(y_{0}k-z_{0})},\ \ \
z=\frac{1}{z_{0}-y_{0}k}.$

Excluding $k$, we get the conic

$y_{0}z^{2}-z_{0}yz+y-y_{0}=0.$

This conic is real only for real $y_{0},z_{0}$ and is smooth if and only if
$y_{0}(z^{2}_{0}-1)\neq 0.$ If $z^{2}_{0}<1$ we normalize to $z_{0}=0$
applying $B_{z}$ and get Type 6. If $z^{2}_{0}>1$ we send $p_{0}$ to infinity
applying $B_{z}$.

For infinite point $p_{0}=[0:Y_{0}:Z_{0}:0]$ we conclude that $Y_{0}\neq 0$
and $Z_{0}\neq 0$, otherwise the tangency points $p_{i}$ do not trace a conic.
Thus we can set $p_{0}=[0:y_{0}:1:0]$, where $y_{0}\neq 0$. In the affine
coordinates $u=\frac{U}{Z}$, $y=\frac{Y}{Z}$ in the plane $X=0$, the conic
$c_{r}$ is

$1+\frac{y^{2}}{r}=u^{2}.$

A line from the pencil $y=ku+y_{0}$ is tangent to the conic $c_{r}$ if and
only if

$r=k^{2}-y_{0}^{2}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are parameterized by $k$
via

$y=\frac{y_{0}^{2}-k^{2}}{y_{0}},\ \ \ u=-\frac{k}{y_{0}}.$

This is a parametrization of the polar conic of Type 7

$y_{0}u^{2}+y=y_{0},$

which becomes

$y_{0}U^{2}-y_{0}Z^{2}+YZ=0$

in homogeneous coordinates.

$\bullet$ Polar conic plane cuts the Darboux quadric, parabolic pencil.

We normalize the conic plane to $Y=0$. Then the point $[0:1:0:0]$ is on the
polar line. The plane $Y=0$ is stable under rotations $R_{y}$ along the
$y$-axis and under boosts $B_{x}$ and $B_{z}$. Rotating, if necessary, around
the $y$-axis, we bring the polar line to $z=-1,$ $x=0.$ In the affine
coordinates on the plane $Y=0$, the conics $c_{r}$ have equations

$x^{2}+\frac{r-1}{r}z^{2}-\frac{2}{r}z-\frac{r+1}{r}=0.$

If the polar conic coincides with one of $c_{r}$ then $r$ is real. The chosen
position of polar conic plane and polar line is stable by action of the group
generated by $B_{z}$. Applying it, one can normalize to $r=\frac{1}{\sqrt{2}}$
and we get Type 9.

Otherwise consider first the pencil of lines $z=z_{0}+k(x-x_{0})$ with finite
vertex at $p_{0}=(x_{0},0,z_{0})$. A line from the pencil is tangent to the
conic $c_{r}$ with

$r=\frac{x_{0}^{2}k^{2}-2x_{0}(z_{0}+1)k+(z0+1)^{2}}{(x_{0}^{2}-1)k^{2}-2x_{0}z_{0}k+z_{0}^{2}-1}.$

Thus the points, where the lines $l_{r}$ touch $c_{r}$, are parametrized by
$k$ via

$x=\frac{k(x_{0}k-z_{0}-1)}{k^{2}-x_{0}k+z_{0}+1},\ \ \
z=\frac{(x_{0}^{2}-1)k^{2}-x_{0}(2z_{0}+1)k+z_{0}(z_{0}+1)}{k^{2}-x_{0}k+z_{0}+1}.$

Excluding $k$, we get the conic

$(z_{0}+1)x^{2}-x_{0}xz+z^{2}-x_{0}x+(1-z_{0})z-z_{0}=0.$ (11)

This conic is real only for real $x_{0},z_{0}$ and is smooth if and only if
$z_{0}\neq-1.$ The chosen position of polar conic plane and polar line is
stable by action of the group generated by $B_{z}$ and $B_{x}-R_{y}$. Consider
the orbits of points in the plane $Y=0.$ The orbit dimension is two for points
outside the union of the line $U+Z=0$ and the circle $X^{2}+Z^{2}=U^{2}$, and
is one on this union except for their common point. Thus the finite
representatives of the orbits are $[0:0:0:1]$, $[0:0:1:1]$ and $[0:0:-1:1].$
For the first two points, the conics (11) lie inside the Darboux quadric and
there is no real circles. For the point $[0:0:-1:1]$ the conic (11) is not
smooth.

For infinite vertexes $p_{0}$, the pencil $z=k$ gives a non-smooth conic. Thus
the pencil can be chosen as $x=az+k$. A line from the pencil is tangent to the
conic $c_{r}$ with

$r=\frac{(k-a)^{2}}{k^{2}-a^{2}-1}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=\frac{k-a}{a^{2}-ak+1},\ \ \ z=\frac{k^{2}-ak-1}{a^{2}-ak+1}.$

Excluding $k$, we get the conic

$x^{2}-axz-ax-z-1=0,$

which is real only for real $a$. Taking into account the action of the group
generated by $B_{z}$ and $B_{x}-R_{y}$, we set $a=0$ and get Type 8.

$\bullet$ Polar conic plane tangent to Darboux quadric, hyperbolic pencil.

If the hyperbolic line does not contain the point where the polar conic plane
touches the Darboux quadric then we sent this point to $(0,0,-1)$ and the
polar line to $Y=Z=0$. In the affine coordinates on the plane $z=-1$, the
conics $c_{r}$ have equations

$(x-r)^{2}+(1-r^{2})y^{2}+1-r^{2}=0.$

If the polar conic coincides with one of $c_{r}$ then the web curvature

$K_{B}=\frac{4(r^{2}-1)^{2}(x^{2}-1)(rx^{2}+ry^{2}+(r^{2}-3)x+r)(x^{2}+y^{2}-2rx+1)^{4}}{x^{4}y^{3}(x^{2}+1-2rx)^{6}}$

vanishes only for $r=\pm 1$, the conic $c_{r}$ being non-smooth for these
values.

A line from the pencil $y=y_{0}+k(x-x_{0})$ with finite vertex at
$p_{0}=(x_{0},y_{0},-1)$ is tangent to the conic $c_{r}$ with

$r=\frac{(x_{0}^{2}+1)k^{2}-2x_{0}y_{0}k+y_{0}^{2}+1}{2k(x_{0}k-y_{0})}.$

Thus the points, where the lines $l_{r}$ touch $c_{r}$, are parametrized by
$k$ via

$x=\frac{x_{0}(x_{0}^{2}-1)k^{3}-y_{0}(3x_{0}^{2}-1)k^{2}+x_{0}(3y_{0}^{2}+1)k-y_{0}(y_{0}^{2}+1)}{k((x_{0}^{2}-1)k^{2}-2x_{0}y_{0}k+(y_{0}^{2}-1))},$
$y=\frac{2(x_{0}k-y_{0})}{(x_{0}^{2}-1)k^{2}-2x_{0}y_{0}k+(y_{0}^{2}-1)}.$

Excluding $k$, we get the cubic

$(x_{0}^{2}-1)y^{3}+(y_{0}^{2}-1)x^{2}y-2x_{0}y_{0}xy^{2}+2y_{0}x^{2}+2y_{0}y^{2}-2x_{0}y_{0}x+(x_{0}^{2}-y_{0}^{2})y=0.$
(12)

This cubic splits into a smooth real conic and a line in 3 cases:
1) for $y_{0}=0$ equation (12) factors as
$y(x^{2}+(1-x_{0}^{2})y^{2}-x_{0}^{2})=0$ and we get Type 10,
2) for $x_{0}=1$, $y_{0}\neq 0$ equation (12) factors as
$(x-1)((y_{0}^{2}-1)xy-2y_{0}y^{2}+2y_{0}x+(y_{0}^{2}-1)y)=0$, 3) for
$x_{0}=-1$, $y_{0}\neq 0$ equation (12) factors as
$(x+1)((y_{0}^{2}-1)xy+2y_{0}y^{2}+2y_{0}x-(y_{0}^{2}-1)y)=0$.
The cases 2) and 3) give Type 11, the substitution $x\to-x$ reducing one to
the other.

For infinite vertexes $p_{0}$, the pencil $x=k$ gives a non-smooth conic. Thus
the pencil can be chosen as $y=ax+k$. A line from the pencil is tangent to the
conic $c_{r}$ with

$r=\frac{k^{2}+a^{2}+1}{2ak}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=\frac{k(k^{2}-a^{2}+1)}{a(k^{2}-a^{2}-1)},\ \ \
y=\frac{2k}{a^{2}-k^{2}+1}.$

Excluding $k$, we get the cubic

$y^{3}+a^{2}x^{2}y+2axy^{2}+2ax+(1-a^{2})y=0.$

The cubic splits into a line and a conic only for $a=0$ or for $a^{2}\pm
2ia-1=0$. The former case gives non-real and non-smooth conic $y^{2}+1=0$, the
latter - the non-real conic $xy\pm i(y^{2}+2)=0$.

If the hyperbolic line contains the point where the polar conic plane touches
the Darboux quadric then we sent this point to $(0,0,-1)$ and the polar line
to $X=Y=0$. In the affine coordinates on the plane $z=-1$, the conics $c_{r}$
are concentric circles. The case of concentric circles was considered above.
We get Type 12 and Type 13.

$\bullet$ Polar conic plane tangent to Darboux quadric, elliptic pencil.
If one vertex of the elliptic pencil coincides with the point where the polar
conic plane touches the Darboux quadric then the polar curve is planar.

Thus we can sent the tangent point to $(0,0,-1)$ and the vertexes to $(\pm
1,0,0)$. The conics $c_{r}$ have equations

$(r^{2}+1)x^{2}+(y+r)^{2}=r^{2}+1.$

If the polar conic coincides with one of $c_{r}$ then the web curvature

$K_{B}=-\frac{64(r^{2}+1)^{2}x(y^{2}+1)(rx^{2}+ry^{2}+(3+r^{2})y-r)(2ry-x^{2}-y^{2}+1)^{4}}{(x^{2}-y^{2}-1)^{4}(r^{2}-x^{2}+1)^{6}}$

vanishes only for $r=\pm i$, the conic $c_{r}$ being non-smooth for these
values.

A line from the pencil $y=y_{0}+k(x-x_{0})$ is tangent to the conic $c_{r}$
with

$r=\frac{(x_{0}^{2}-1)k^{2}-2x_{0}y_{0}k+(y_{0}^{2}-1)}{2(x_{0}k-y_{0})}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=\frac{2k(x_{0}k-y_{0})}{(x_{0}^{2}+1)k^{2}-2x_{0}y_{0}k+y_{0}^{2}+1},$
$y=-\frac{x_{0}(x_{0}^{2}-1)k^{3}-y_{0}(3x_{0}^{2}-1)k^{2}+x_{0}(3y_{0}^{2}+1)k-y_{0}(y_{0}^{2}+1)}{(x_{0}^{2}+1)k^{2}-2x_{0}y_{0}k+y_{0}^{2}+1}.$

Excluding $k$, we get the cubic

$(y_{0}^{2}+1)x^{3}-2x_{0}y_{0}x^{2}y+(x_{0}^{2}+1)xy^{2}-2x_{0}x^{2}-2x_{0}y^{2}+(x_{0}^{2}-y_{0}^{2})x+2x_{0}y_{0}y=0.$

This cubic splits into a conic and a line in 4 cases:
1) for $y_{0}=\pm i$ the cubic equation factors as

$(\pm i-y)(\pm 2ix_{0}x^{2}-(1+x_{0}^{2})xy\mp i(1+x_{0}^{2})x+2x_{0}y)=0,$

2) for $y_{0}=\pm ix_{0}$ the cubic equation factors as

$(x\pm iy)((1-x_{0}^{2})x^{2}\mp i(x_{0}^{2}+1)xy-2x_{0}x\pm
2ix_{0}y+2x_{0}^{2})=0,$

3) for $x_{0}=0$, the cubic equation factors as
$x((y_{0}^{2}+1)x^{2}+y^{2}-y_{0}^{2})=0$,
4) for $x=\pm 1$ the cubic equation factors as $(x\mp
1)((y_{0}^{2}+1)x^{2}+2y^{2}\mp 2y_{0}xy\pm(y_{0}^{2}-1)x-2y_{0}y)=0.$
In the cases 1) and 2) the conic is not real, the case 3) gives Type 14, for
the case 4) the web curvature does not vanishes.

For infinite vertexes $p_{0}$, the pencil $x=k$ gives a non-smooth conic. Thus
the pencil can be chosen as $y=ax+k$. A line from the pencil is tangent to the
conic $c_{r}$ with

$r=\frac{a^{2}-k^{2}+1}{2k}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=-\frac{2ak}{k^{2}+a^{2}+1},\ \ \ y=\frac{k(k^{2}-a^{2}+1)}{k^{2}+a^{2}+1}.$

Excluding $k$, we get the cubic

$a^{2}x^{3}-2ax^{2}y+xy^{2}+(1-a^{2})x+2ay=0.$

The cubic splits into a line and a conic in 2 cases:
1) for $a=0$ the cubic equation factors as $x(y^{2}+1)=0$ and the conic is
non-smooth
2) for $a=\pm i$ the cubic equation factors as $(x\pm iy)(x^{2}\pm ixy-2)=0,$
and the conic is not real.

$\bullet$ Polar conic plane tangent to Darboux quadric, parabolic pencil.
For non-planar polar curves, the points, where the polar line and polar conic
plane touch the Darboux quadric, are different. Thus we can normalize the
polar conic plane to $z=-1$ and the polar line to $x=0$, $z=1$. This
configuration is preserved by $B_{z}$. The conics $c_{r}$ have equations

$\frac{r}{4}x^{2}+y=\frac{1}{r}.$

If the polar conic coincides with one of $c_{r}$ then the web curvature

$K_{B}=-\frac{4096r^{5}xy^{2}(ry+2x^{2}+2y^{2})(ry-x^{2}-y^{2})^{4}}{(x^{2}-y^{2})^{4}(r^{2}-4x^{2})^{6}}$

vanishes only for $r=0$, the conic $c_{r}$ being non-smooth for this value.

A line from the pencil $y=y_{0}+k(x-x_{0})$ is tangent to the conic $c_{r}$
with

$r=\frac{k^{2}+1}{y_{0}-x_{0}k}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=\frac{2k(x_{0}k-y_{0})}{k^{2}+1},$
$y=\frac{x_{0}k^{3}-y_{0}k^{2}-x_{0}k+y_{0})}{k^{2}+1}.$

Excluding $k$, we get the cubic

$x^{3}+xy^{2}-2x_{0}x^{2}-2x_{0}y^{2}+(x_{0}^{2}-y_{0}^{2})x+2x_{0}y_{0}y=0.$

This cubic splits into a conic and a line in 2 cases:
1) for $y_{0}=\alpha x_{0}$, where $\alpha^{2}\pm 2i\alpha-1=0$, the cubic
equation factors as

$(x\pm iy)(\pm ixy-x^{2}+2x_{0}x\mp 2ix_{0}y-2x_{0}^{2})=0,$

2) for $x_{0}=0$ the cubic equation factors as $x(x^{2}+y^{2}-y_{0}^{2})=0$.
Ii the case 1) the conic is not real, the case 2) gives Type 15 after
rescaling by $B_{z}$.

For infinite vertexes $p_{0}$, the pencil $y=k$ gives a non-smooth conic. Thus
the pencil can be chosen as $x=ay+k$. A line from the pencil is tangent to the
conic $c_{r}$ with

$r=-\frac{a^{2}+1}{ak}.$

The points, where the lines $l_{r}$ touch $c_{r}$, are

$x=\frac{2k}{a^{2}+1},\ \ \ y=\frac{k(1-a^{2})}{a(1+a^{2})}.$

Excluding $k$, we get the line $(a^{2}-1)x+2y=0.$

Thus we have considered all the cases with non-planar polar curve, the theorem
is proved. $\Box$

##  6 Hexagonal circular 3-webs with polar curves of degree three

All the known hexagonal circular 3-webs are ”algebraic” in the sense that
their polar curves are algebraic, namely, the irreducible components of polar
curves are cubics, conics, and straight lines.

###### Theorem 10

There is no hexagonal circular 3-web whose polar curve is a rational normal
curve.

Proof: Suppose there is such a web with rational normal curve $N$ as the polar
curve. Consider a family of (possibly complex) bisecant lines $L$ of $N$ that
are tangent to the Darboux quadric. The points of tangency form a curve $C$.
As the curve $N$, being of degree three, cannot have trisecant lines, the
curve $C$ must be a circle whose polar point $p_{0}$ lies on $N$ by Lemma 1.
Consider the projection $\pi$ from $p_{0}$ to the plane of the circle $C$. The
image $\pi(N)$ of $N$ is a conic, the image $\pi(C)$ of $C$ is the circle $C$
itself. Any busecant line $L$ of the considered family meets the circle $C$ at
a point $p_{L}$ and the curve $N$ at two points $p_{1},p_{2}$. Then the image
$\pi(L)$ touches $C$ at $\pi(p_{L})=p_{L}$ and intersects $\pi(N)$ at
$\pi(p_{1}),\pi(p_{2})$. Consider one of the four (possibly complex) points
$q_{i}$ of intersection of the conics $\pi(N)$ and $C$. The curves $\pi(N)$
and $C$ cannot be tangent at $q_{i}$ since for $q_{i}=p_{L}$ we have
$\pi(p_{1})=\pi(p_{2})=p_{L}$ but $p_{1},p_{2},p_{0}$ are not collinear. If
$p_{L}$ is not coinciding with any of $p_{1},p_{2}$ then the 4 points
$\pi^{-1}(q_{i}),p_{1},p_{2},p_{0}$ on $N$ are coplanar but $N$ is of degree
3. Therefore one of the pints $p_{1},p_{2}$ coincides with $p_{L}$ and the
plane of $C$ intesect the curve $N$ of degree three at 4 points $q_{i}$. This
contradiction finishes the proof. $\Box$

The above proven theorem gives classification of a natural class of webs under
study.

###### Corollary 11

Suppose that the polar curve of a hexagonal circular 3-web is algebraic of
degree three. Then the polar curve is either planar or the web is Möbius
equivalent to one described by Theorems 6 and 9.

##  7 Hexagonal circular 3-webs with Möbius symmetry

The Möbius group in $\mathbb{RP}^{3}$ can be realized as
$PGL_{2}(\mathbb{C})$, or equivalently, as the group of fractional-linear
transformations of $z=x+iy$, where $(x,y)$ are cartesian coordinates in the
plane. A generator of any 1-dimensional subalgebra can be brought to the
Jordan normal form by adjoint action. The generator can be chosen either as

$\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right),\ \ \ \mbox{or}\ \ \
\left(\begin{array}[]{cc}\lambda&0\\\ 0&-\lambda\end{array}\right),$

where $\lambda=\alpha+i\beta$ is some complex number with
$Re(\lambda)=\alpha$, $Im(\lambda)=\beta$.

One way to obtain symmetric hexagonal 3-webs is provided by the Wunderlich
construction (see [W-38] and Introduction).

Another easy way to produce hexagonal circular 3-webs is to choose two orbits
of polar points such that one is a conic and the other is a coplanar straight
line, these two orbits forming a polar curve. This construction may degenerate
if there are 3 orbits which are coplanar straight lines.

For translations $(x,y)\mapsto(x,y+u)$, the orbit of a polar point for a
circle $(x-a)^{2}+y^{2}=r$, parametrized by $u$ as follows
$[-2a:-2u:a^{2}+u^{2}-r-1:-a^{2}-u^{2}+r-1]$, is a conic in the plane
$-X+a(Z+U)=0$. For a nonvertical line $y=ax$, the orbit is a line
$Z+U=X+aY=0$, parametrized by $u$ via $[a:-1:-u:u]$. Thus we immediately get
the following hexagonal 3-webs symmetric by translations.

T1. 3 families of parallel lines.

Möbius orbits of such webs form a two-parametric family. The polar curve
splits into 3 coplanar lines. Observe that any web of this family has a
3-dimensional symmetry group.

T2. Polar curve splits into conic and coplanar line.

There is only one Möbius class of such webs, any representative is formed by
horizontal lines $y=const$ and by the orbit of a circle which can be chosen as
the unitary one centered at the origin.

T3. Wunderlich’s type.

There are several types, depending on the position of generating curves.

  1. 1.

Nondegenerate type. Webs are formed by vertical lines and by two different
orbits of circles. The family of orbits is two-parametric: by translation and
rescaling we can fix one orbit.

  2. 2.

Coinciding circle orbits. There is only one Möbius class of this type. It was
already obtained as Type 9 in the classification of Theorem 9.

  3. 3.

Generated by a circle and a line. The family of orbits is one-parametric: by
translation and rescaling we can fix the generating circle.

For dilatations $(x,y)\mapsto(ux,uy)$, the orbit of a polar point for a circle
$(x-a)^{2}+(y-b)^{2}=r$, parametrized by $u$ via
$[-2au:-2bu:a^{2}+b^{2}-r-u^{2}:-(a^{2}+b^{2}-r)-u^{2}]$, is a conic in the
plane $bX-aY=0$ if $(a,b)\neq(0,0)$ and $a^{2}+b^{2}\neq r$, the hyperbolic
line $X=Y=0$ if $(a,b)=(0,0)$, and the parabolic line $bX-aY=U-Z=0$ if
$a^{2}+b^{2}=r$. For a line $ax+by+c=0$ with $c\neq 0$, the orbit is the
parabolic line $bX-aU=Z+U=0$, parametrized by $u$ via $[au:bu:c:-c]$. The
lines with $c=0$ are invariant. Invoking the classification of the webs with 3
pencils, we list the following hexagonal 3-webs symmetric by dilatations.

D1=T1. 3 families of parallel lines.

D2. 2 coplanar parabolic lines and hyperbolic line intersecting them.

Family of parallel lines, the parabolic pencil with circles tangent to a line
of the family through the origin, and the family of concentric circles with
the center at the origin. There is only one Möbius class of such webs.

D3. 2 dual parabolic lines and hyperbolic line through their common point.

2 orthogonal families of parallel lines and the family of concentric circles
with the center at the origin. There is only one Möbius class of such webs, we
have already presented it on the right of Figure 2.

D4. Polar curve splits into conic and coplanar hyperbolic line.

Family of circles obtained by the dilatations from one not passing trough the
origin and not having its center at the origin and the family of concentric
circles with the center at the origin. Möbius orbits of such webs form a one-
parameter family.

D5. Polar curve splits into conic and tangent parabolic line.

Family of circles obtained by dilatations from one not passing trough the
origin and not having its center at the origin and the family of parallel
lines orthogonal to the orbit of centers of the circles. Möbius orbits of such
webs form a one-parameter family.

D6. Wunderlich’s type.

  1. 1.

Nondegenerate type, polar curve with 2 conics and elliptic line. Webs are
formed by pencil of lines centered at the origin and by two different orbits
of circles at general position. The family of Möbius orbits is 3-parametric:
one can choose the centers of circles on a fixed circle centered at the origin
and normalize by rotations.

  2. 2.

Polar curve with conic and elliptic line. Webs are formed by pencil of lines
centered at the origin and orbit of a circle. The family of Möbius orbits is
1-parametric, it is Type 5 in the classification of Theorem 9.

  3. 3.

Polar curve with conic, hyperbolic and elliptic line. Webs are formed by
pencil of lines centered at the origin, orbit of a circle, and the family of
concentric circles with the center at the origin. The family of Möbius orbits
is 1-parametric.

  4. 4.

Polar curve with conic, parabolic and elliptic line. Webs are formed by pencil
of lines centered at the origin, orbit of a circle, and a family of parallel
lines. The family of Möbius orbits is 2-parametric.

  5. 5.

Polar curve with hyperbolic line, elliptic line dual to hyperbolic, and
parabolic line intersecting them. Webs are formed by the pencil of lines
centered at the origin, the family of concentric circles with the center at
the origin, family of parallel lines. There is only one Möbius type. This web
is shown in the center of Figure 2.

  6. 6.

Polar curve with 2 parabolic lines touching Darboux quadric at the same point
and coplanar elliptic line. Webs are formed by the pencil of lines centered at
the origin and 2 families of parallel lines. The family of Möbius orbits is
1-parametric.

  7. 7.

Polar curve with 2 parabolic lines and elliptic line whose dual joins the
touching points of parabolic lines with Darboux quadric. Webs are formed by
the pencil of lines centered at the origin, a family of parallel lines and a
parabolic pencil with the vertex at the origin. The family of Möbius orbits is
1-parametric. A representative of this web is shown in Figure 4.

For rotations $(x,y)\mapsto(x\cos(t)+y\sin(t),-x\sin(t)+y\cos(t))$, the orbit
of a polar point for a circle $(x-a)^{2}+y^{2}=r$, parameterized by $t$ via
$[-2a\cos(t):-2a\sin(t):a^{2}-r-1:-a^{2}+r-1]$, is a circle in the plane
$(a^{2}-r+1)Z+(a^{2}-r-1)U=0$ if $a\neq 0$. For a line $ax+by+c=0$ with $c\neq
0$, the orbit of its polar is also a circle
$[a\cos(t)-b\sin(t):a\sin(t)+b\cos(t):c:-c]$ in the plane $Z+U=0$. The circles
with $a=0$ are invariant, the orbit of a polar point of a line with $c=0$ is
an elliptic line. Thus one can easily list the following hexagonal 3-webs
symmetric by rotations.

R1. Polar curve is a circle symmetric by rotations around $z$-axis and
coplanar elliptic line.

The family of Möbius orbits is 1-parametric.

R2. Wunderlich’s type.

One foliation is formed by orbits of points by rotations and the other two are
images of two circles by rotation, these circles can coincide or ”degenerate”
into straight lines. Examples of such webs with coinciding generators are
shown in Figure 5 (center), Figure 6 (left), and Figure 9 (left). Considering
The reader easily describes different types and compute corresponding Möbius
orbit dimension of the webs.

###### Theorem 12

Hexagonal circular 3-web with 1-dimensional Möbius symmetry is Möbius
equivalent to one of the above described T-,D-, or R-types.

Proof: A circular 3-web, symmetric by a given 1-parametric subgroup of the
Möbius group and not obtained by the Wunderlich construction, is fixed by a
choice of three curves, each being either a circle or a straight line, one
from each foliation. Two circles can coincide: an orbit of circle still gives
a (singular) 2-web. Moreover, one can move around these curves by the
stabilizer of the infinitesimal generator of the subgroup.

$\bullet$ Webs symmetric by translations $\partial_{y}$.
Consider a point such that none of the leaf tangents is parallel to the field
$\partial_{y}$. If a generating curve $C_{1}$ of some foliation is a circle
then there are 2 circles from the orbit of $C_{1}$ passing through this point.
Thus there are two locally defined direction fields
$\partial_{x}+P_{\pm}\partial_{y}$ tangent to these 2 circles. Globally they
are not separable: one direction swaps for the other upon running along
$C_{1}$.

Consider one of the foliations and the corresponding direction field
$\partial_{x}+P\partial_{y}$. Since the foliation is symmetric, the slope $P$
does not depend on $y$. Since all the leaves are circles of the same
curvature, the foliation has a first integral

$\frac{({P^{\prime}})^{2}}{(P^{2}+1)^{3}},\ \ \ \mbox{where}\ \ \
P^{\prime}=\frac{dP}{dx}.$

The differential equation

$\frac{({P^{\prime}})^{2}}{(P^{2}+1)^{3}}=A^{2}=const$

has general solution

$P(x)=\frac{A(x-x_{0})}{\sqrt{1-A^{2}(x-x_{0})^{2}}},$

the zero value of $A$ corresponding to the foliation by straight parallel
lines. Let the slopes of the other two web foliations be $Q$ and $R$, and the
connection form be $\gamma=\alpha(x)dx+\beta(x)dy$. Hexagonality condition
$d\gamma=0$ implies $\beta(x)=const$, which amounts to

$k=\frac{P^{\prime}}{(P-Q)(P-R)}+\frac{Q^{\prime}}{(Q-P)(Q-R)}+\frac{R^{\prime}}{(R-P)(R-Q)}=const.$
(13)

If all foliations are formed by circles and not by straight lines then the
above identity is not possible. In fact, each of the slopes $P,Q,R$,
considered as function of complex $x$, has two ramification points, for
example, $P$ has singularities at $x_{\pm}=x_{0}\pm 1/A$. If at least one of
the 6 ramification points is not coinciding with one of the others, then,
supposing it be $x_{+}$ of $P$ and expanding the expression (13) for
$x=x_{+}+t^{2}$ by $t$ at $t=0$, one sees that it has a simple pole and
therefore cannot be constant. Therefore each ramification point of any slope
coincides with a ramification point of another slope.

The ramification points correspond to the group orbits (lines $x=const$)
tangent to the foliation circles. If only two circles are tangent along some
orbit $x=const$ then, applying Lemma 1, we conclude that either this orbit
belongs to the third foliation and the web is of Wunderlich’s type or the two
generating circles coincide. In the latter case the ramification points of the
third foliation again must coincide with the common ramification points of the
first two. This means that that all orbits of generating circles coincide and
we do not have a 3-web.

Therefore either all leaves of all foliations are straight lines or one of the
foliation, say the one corresponding to $R$, is formed by straight lines with
slope $R=const$ and the other two are formed by orbits of the same circle and
$Q=-P$. Then it is immediate that $k=0$ and $R=0$. We get the type $T2$. The
webs of Wunderlich’s type, which are hexagonal, are excluded from
consideration by the coordinate choice.

$\bullet$ Webs symmetric by dilatations $x\partial_{x}+y\partial_{y}$.
Stabilizer of the 1-dimensional algebra spanned by
$x\partial_{x}+y\partial_{y}$ is generated by rotation
$y\partial_{x}-x\partial_{y}$ and dilatation $x\partial_{x}+y\partial_{y}$.
The orbit of the polar of the generating curve is either a conic, or a
hyperbolic line, or a parabolic line. Parabolic lines touch the Darboux
quadric at one of the stationary point of the dilatation.

We use modified polar coordinates rectifying the symmetry

$u=\arctan\left(\frac{y}{x}\right),\ \ \ v=\frac{1}{2}\ln(x^{2}+y^{2}).$ (14)

A curve $x\mapsto(x,y(x))$ is a circle if and only if

$y^{\prime\prime\prime}=\frac{3y^{\prime}(y^{\prime\prime})^{2}}{1+(y^{\prime})^{2}},$

therefore integral curves of a symmetric vector field
$\partial_{u}+P(u)\partial_{v}$ are circles if and only if

$P^{\prime\prime}=\frac{3P}{P^{2}+1}(P^{\prime})^{2}-P(P^{2}+1).$ (15)

The integral curves are straight lines if and only if $y^{\prime\prime}=0$,
which is equivalent to $P^{\prime}=P^{2}+1$. Hence (apply
$z\mapsto\frac{1}{z}$) the integral curves are circles of a parabolic pencil
with the vertex at the origin if and only if $P^{\prime}=-(P^{2}+1)$ thus
giving $P(u)=\tan(u-u_{0})$ and $P(u)=-\tan(u-u_{0})$ respectively. The second
order equation (15) has a first integral

$\frac{(P^{\prime})^{2}}{(P^{2}+1)^{3}}-\frac{1}{P^{2}+1}=A=const,$

allowing also to integrate it

$P(u)=\frac{\sqrt{A+1}\tan(u-u_{0})}{\sqrt{1-A\tan^{2}(u-u_{0})}}.$ (16)

To the hyperbolic pencil with the circles centered at the origin corresponds
the solution $P\equiv 0$.

For hexagonal webs the connection form is $\gamma=\alpha(u)du+\beta(u)dv$.
Hexagonality condition $d\gamma=0$ implies $\beta(u)=const$, giving again (13)
(where the slopes of the other foliations are $Q$ and $R$). Observe that the
choice of local coordinates $u,v$ excludes webs of Wunderlich’s type and we
have to show that hexagonal are only the types D1, D2, D3, D4, D5. This can be
done as follows. The types D1, D2, D3 are webs with three pencils, the case
being settled earlier. Therefore we study the webs whose polar curve includes
at least one conic.

The approach used for translation works also in this case if we pass to
complex webs: the singular points of solutions $P,Q,R$ become complex if the
corresponding generating circle on the Darboux quadric separates the stable
points $(0,0,\pm 1)$ of the dilatation. Considering behavior of the expression
$k$ at singular points of $P,Q,R$ we see that necessarily a singular point of
slope for one foliation must coincide with a singular point for another.
Singular points are the points (possibly complex) where the group orbit
touches the generating circle, or lies either on a line of one foliation or on
the common tangent to circles of parabolic pencil with the vertex at $(0,0)$.

Applying Lemma 1 as in the case of translation, we infer that, for webs of
non-Wunderlich’s type, having two foliations with conic polar curves and
coinciding singular points, these polar curves must coincide and the third
polar must be a line.

Consider the points where the circles from the common orbit are tangent. Lemma
1 implies that for non-Wunderlich’s type the leaves of the third foliations
must be also tangent to the circles and we get either type D4 or D5.

Finally, if two components of the polar curve are lines and the third is a
conic then the lines must be parabolic for non-Wunderlich’s type. In fact,
considering the points where circles of the hyperbolic pencil are tangent to
circles corresponding to conic, we conclude by Lemma 1 that the third
foliation lines are also tangent to the circles at these points but then the
singular points of the ”conic” solution to (15) can not be compensated.

Thus the polar lines are parabolic. The expression $k$ cannot be constant if
the poles of $Q,R$, corresponding to these lines, do not coincide with
singular points of $P$, which represent a conic in the polar curve. Therefore
the generating circle of this ”conic” solution does not separate the
stationary point of the dilatation and the singular points are real. Then by
Lemma 1 the web cannot be hexagonal.

$\bullet$ Webs symmetric by rotations $y\partial_{x}-x\partial_{y}$.
Stabilizer of the 1-dimensional subalgebra spanned by
$y\partial_{x}-x\partial_{y}$ is generated by rotations
$y\partial_{x}-x\partial_{y}$ and dilatations $x\partial_{x}+y\partial_{y}$.
The polar orbit for a generating curve is either a circle or an elliptic line.

We again use polar coordinates (14) to describe symmetric vector fields
$\partial_{v}+P(v)\partial_{u}$. The integral curves of such vector field are
circles (in coordinates $x,y$, of course) if and only if

$P^{\prime\prime}=\frac{3P}{P^{2}+1}(P^{\prime})^{2}+P(P^{2}+1).$ (17)

The integral curves are straight lines if and only if
$P^{\prime}=-P(P^{2}+1)$. The integral curves are circles passing through the
origin if and only if $P^{\prime}=P(P^{2}+1)$. Solutions of these two
equations are $P(v)=\frac{1}{\sqrt{e^{2(v-v_{0})}-1}}$ and
$P(v)=\frac{1}{\sqrt{e^{-2(v-v_{0})}-1}}$ respectively. The second order
equation (17) has a first integral

$\frac{(P^{\prime})^{2}}{(P^{2}+1)^{3}}+\frac{1}{P^{2}+1}=A^{2}=const,$

allowing also to integrate it

$P(v)=\frac{\sqrt{A^{2}-1}\tanh(v-v_{0})}{\sqrt{1-A^{2}\tanh^{2}(v-v_{0})}},\
\ A^{2}\neq 1.$ (18)

The value $A^{2}=1$ gives the special solutions with generating circles
passing through the stationary points of the symmetry, i.e. elliptic pencil
with the lines passing through the origin with $P\equiv 0$. Analysis of the
behavior of $k$ at singular points and use of Lemma 1, similar to the ones
performed above, show that only the type R1 is hexagonal, the Wunderlich types
being excluded by the choice of variables. (In fact, multiplying by $i$ the
independent variable of the differential equation (15) reduces it to (17).)

$\bullet$ Loxodromic symmetry.
Finally we show that there is no hexagonal 3-webs symmetric by loxodromic
vector field $y\partial_{x}-x\partial_{y}+\kappa(x\partial_{x}+y\partial_{y})$
for any $\kappa\neq 0$. Note that the Wunderlich construction does not give
circular webs as the symmetry orbits are spirals. We use the following
coordinates

$s=\frac{\kappa}{2}\ln(x^{2}+y^{2})-\arctan\left(\frac{y}{x}\right),\ \ \
t=\kappa\arctan\left(\frac{y}{x}\right)+\frac{1}{2}\ln(x^{2}+y^{2}),$

the variable $t$ being invariant by the symmetry. Integral curves of a
symmetric vector field $\partial_{t}+P(t)\partial_{s}$ are circles (in
coordinates $x,y$) if and only if

$P^{\prime\prime}=\frac{3P}{P^{2}+1}(P^{\prime})^{2}+\frac{(P^{2}+1)(\kappa
P+1)(P-\kappa)}{(\kappa^{2}+1)^{2}}.$ (19)

This equation has a first integral

$\frac{(P^{\prime})^{2}}{(P^{2}+1)^{3}}+\frac{2\kappa
P-\kappa^{2}+1}{(\kappa^{2}+1)(P^{2}+1)}=A=const.$

This integral does not allow to integrate (19) in elementary functions but
allows to study the behavior of solutions at singular points. A singular point
emerges when a symmetry orbit is tangent to a circle (or a line) of
corresponding foliation.

Consider possible singularity types. If we exclude from consideration the
stationary points of symmetry then the symmetry vector field touches a generic
circle at two points and the tangency is simple. The corresponding solution
$P$ has two singularities if the tangency points belong to different orbits
and only one if the points lie on the same orbit. There are circles, for which
two tangency points merge to give only one singularity. If the generating
curve is a line or a circle through the origin then the solution has only one
singularity. Finally, there are two constant solutions, namely $P=-1/\kappa$
corresponding to the invariant hyperbolic line $X=Y=0$ and $P=k$ corresponding
to the invariant elliptic line $Z=U=0$.

If a solution $P$ has two singularities at $t_{1},t_{2}$ then $A\neq 0$ and
the singularity is of the same type as for the non-loxodromic cases:

$P(t)=\frac{1}{\sqrt[4]{4A}\sqrt{t-t_{i}}}+\\{analytic\ function\ of\
\sqrt{t-t_{i}}\\},$

If a solution $P$ is generated by a circle tangent to the symmetry trajectory
$t=t_{0}$ and the tangency is of second order then there is only one
singularity of the following type at $t=t_{0}$:

$P(t)=c(t-t_{0})^{-\frac{2}{3}}+...,$

where $c\neq 0$ and the omitted terms are not essential for our analysis. The
condition of double tangency is equivalent to $A=0$. The corresponding
generating circle $(x-a)^{2}+y^{2}=r^{2}$ verifies the relation
$r^{2}=\frac{(k^{2}+1)a^{2}}{k^{2}}.$ For an orbit $t=t_{0}$ there is at most
one such circle.

Let $P,Q,R$ be solutions giving a hexagonal 3-web. These solutions, as well as
the coordinates $s,t$, are defined only locally but we can prolong them along
a symmetry orbit, along a leaf of some of the 3 foliations, or along any
curve, as long as we do not meet a singular point of one of $P,Q,R$. The
condition of hexagonality (13) remains satisfied along any such prolongations.
Therefore we cannot meet a singularity of only one of $P,Q,R$, they emerge
necessarily at least in pairs.

Suppose there is a symmetric hexagonal 3-web. Consider a non-singular point.
There are 3 leaves passing through it. Each leaf can be considered as the
generating curve of the respective foliation. At least one of the
corresponding solutions $P,Q,R$ has a singular point. Let us run along the
respective leaf until we meet a singularity $t_{1}$ of $P,Q$ or $R$. Then at
least two of $P,Q,R$ are singular at $t_{1}$ and the orbit $t=t_{1}$ is
tangent to at least two web leaves at a singular point $p_{s}$. Since the
symmetry orbit is not a circle, Lemma 1 implies that either exactly two leaves
at $p_{s}$ are tangent and therefore coincide or all three leaves are tangent
at $p_{s}$.

In the former case suppose that the coinciding leaves are $C_{Q}$ and $C_{R}$.
Then the leaf $C_{P}$ at $p_{s}$ is different from $C_{Q}=C_{R}$ at $p_{s}$.
Let us go along $C_{P}$ keeping track of $Q,R$ until one of the two leaves
corresponding to $Q$ and $R$ touches $C_{P}$ at some $\bar{p}_{s}$. Such point
exists untill all the foliations are formed by straight lines. Then by Lemma 1
this leaf coincide with $C_{P}$ at $\bar{p}_{s}$, thus all three generating
curves of the web coincide and there is no 3-web. If all 3 generating curves
are straight lines then $C_{P}$ is necessary the line through the origin and
$P=\kappa$. Applying the map $z\mapsto 1/z$ we transform the lines $C_{Q}$ and
$C_{R}$ into circles and the above argument applies.

If all three leaves are tangent at $p_{s}$ then at least one leaf, say
$C_{P}$, is different from any of the other two. Let us go again along $C_{P}$
keeping track of $Q,R$ until one of the two leaves corresponding to $Q$ and
$R$ touches $C_{P}$. Let it be $C_{Q}$. Repeating the above used argument we
conclude that such point $\bar{p}_{s}$ exists and $C_{P}$ coincides at
$\bar{p}_{s}$ with $C_{Q}$. Then $C_{P}$ coincides with $C_{Q}$ also at
$p_{s}$. Now either all 3 generating curves $C_{P},C_{Q},C_{R}$ coincide at
$p_{s}$ and we do not have 3-web, or $C_{R}$ is different from $C_{P}=C_{Q}$
at $p_{s}$. Now we repeat the trick with prolongation, this time along
$C_{R}$, and conclude that $C_{P}=C_{Q}=C_{R}$ at $p_{s}$. Thus all three
generating curves coincide and we can get at most 2-web. $\Box$

##  8 Concluding remarks

###  8.1 Circular hexagonal 3-webs on surfaces

Pottmann, Shi, and Skopenkov [PSS-12] classified circular hexagonal 3-webs on
nontrivial Darboux cyclides: such surfaces carry up to 6 one-parameter
families of circles, 3 families can be picked up in 5 different ways to form a
hexagonal web.

###  8.2 Erdoǧan’s approach to Theorem 6

The first attempt to prove Theorem 6 appeared in [E-89]. The idea was to
choose a Möbius normalization sending one of the vertexes to infinity, to set
$y=0$, and to obtain ”sufficient number of equations” to fix the pencil
configurations. The author claimed that the curvature equation for $y=0$ is a
polynomial one of degree 6 in $x$, though the calculation itself was not
present. Nowadays, armed with a powerful computer (32GB of RAM is enough) and
a symbolic computation system like Maple, one can perform this computations
and check that the degree may be much higher (in fact, up to 18) for some
choices of polar line types.

Anyway, brute computer force does work: with the above mentioned equipment the
author of this paper managed to derive the classification results. The
treating has the following steps:

  1. 1.

Choosing an initial Möbius normalization.

  2. 2.

Computing the curvature.

  3. 3.

Isolating and factoring the highest homogeneous part of the curvature
equation.

  4. 4.

Möbius renormalization adjusted to the geometric information obtained in the
previous step and repeating from the step 2 until one makes the curvature
vanish.

###  8.3 Boundaries of regular domain for hexagonal 3-webs

To avoid heavy computation of the curvature in proving Theorem 6, the author
of [S-05] suggested to use the structure of web singular set (see Lemma 1).
The presented proof was not correct. The author argued that the web equation
$u_{3}=F(u_{1},u_{2})$, relating first integrals $u_{i}$ of the web foliations
$\mathcal{F}_{i}$, may be rewritten as $u_{3}=f(\alpha(u_{1})+\beta(u_{2}))$
and used this form on the curve of singular points $\Gamma_{1}$. This argument
is definitely wrong as the functions $\alpha,\beta$ typically have
singularities on $\Gamma_{1}$: a simple counterexample is the web equation
$u_{3}=u_{1}u_{2}$.

###  8.4 Hexagonal 3-subwebs

Consider the autodual tetrahedron with vertexes at $[1:0:0:0]$, $[0:1:0:0]$,
$[0:0:1:0]$ and $[0:0:0:1]$. Lines joining the vertexes of this tetrahedron
give a 6-web $A_{6}$ with 6 pencils of circles. Any 3-subweb of this 6-web is
hexagonal: any 3 lines are either coplanar or give a polar curve of a
hexagonal 3-web. By direct computation (better use computer!) one checks the
following claim.

###### Proposition 13

The rank of the 6-web $A_{6}$ is maximal, i.e. is equal to 10.

Another remarkable feature of this autodual 6-web is that the infinitesimal
operators, corresponding to the pencils in the sense of Proposition 3, form
the basis $R_{x},R_{y},R_{z},B_{x},B_{y},B_{z}$ of the Lie algebra of the
Möbius group so that the commutator of any two of them is either zero or an
operator of the basis.

There is also autodual 4-web $A_{4}$ whose polar curve is the union of 4
pencils corresponding to $R_{z},B_{z},B_{x}-R_{y},R_{x}+B_{y}$. This web has
similar properties: any its 3-subweb is hexagonal, the commutator of any two
of the four operators is either zero or an operator of the set, its rank is
maximal. One finds more examples with hexagonal subwebs among symmetric webs
of Wunderlich’s type (see Section 7).

###  8.5 Conjecture

The polar curve of a hexagonal circular 3-web is an algebraic curve such that
each its irreducible component is a planar curve of degree at most 3.

## Appendix: curvature equation

$(A^{2}-4B)(R^{2}+AR+B)[(AR+2B)A_{xx}+A(R^{2}-B)A_{xy}-BR(A+2R)A_{yy}+$
---
$-(A+2R)B_{xx}+(A^{2}-2R^{2}-2B)B_{xy}+R(A^{2}-2B+AR)B_{yy}+$
$(4B-A^{2})R_{xx}+A(A^{2}-4B)R_{xy}+B(4B-A^{2})R_{yy}]+$
$(A^{2}-4B)^{2}(A+2R)R_{x}^{2}+(A+2R)(A^{2}-4AR-4R^{2}-8B)B_{x}^{2}+$
$-(2A^{3}R^{2}+A^{2}R^{3}+7A^{2}BR+4ABR^{2}+4BR^{3}+4AB^{2}-4B^{2}R)A_{x}^{2}+$
$(B-R^{2})(2A^{3}R+A^{2}R^{2}+A^{2}B+4BR^{2}+4B^{2})A_{x}A_{y}-A(A^{2}-4B)^{2}(A+2R)R_{x}R_{y}+$
$(2A^{3}R-A^{4}+4A^{2}R^{2}-8AR^{3}-8R^{4}+8A^{2}B-16BR^{2}-8B^{2})B_{x}B_{y}+$
$BR(2A^{3}R+7A^{2}R^{2}+4AR^{3}+A^{2}B+4ABR-4BR^{2}+4B^{2})A_{y}^{2}+$
$-R(A^{4}-A^{3}R-6A^{2}R^{2}-4AR^{3}-8A^{2}B-8ABR+8B^{2})B_{y}^{2}+$
$B(A^{2}-4B)^{2}(A+2R)R_{y}^{2}+(A^{2}-4B)(A^{2}R-AR^{2}-AB-8BR)A_{x}R_{x}+$
$(3A^{3}R+13A^{2}R^{2}+8AR^{3}+A^{2}B+12ABR-4BR^{2}+12B^{2})A_{x}B_{x}+$
$2(A^{2}-4B)(A^{2}+AR+R^{2}-3B)B_{x}R_{x}+$
$(5A^{2}R^{3}-A^{4}R-A^{3}R^{2}+4AR^{4}+A^{2}BR+4ABR^{2}-4BR^{3}-4AB^{2}-4B^{2}R)[A_{x}B_{y}+A_{y}B_{x}]+$
$(A^{2}-4B)(A^{2}R^{2}+A^{2}B+2ABR-2BR^{2}-2B^{2})[A_{x}R_{y}+A_{y}R_{x}]+$
$-A(A^{2}-4B)(A^{2}+AR+R^{2}-3B)[B_{x}R_{y}+B_{y}R_{x}]$
$B(A^{2}-4B)(A^{2}R-AR^{2}-AB-8BR)A_{y}R_{y}+$
$+(4B-A^{2})(A^{3}R+A^{2}R^{2}-A^{2}B-6ABR-6BR^{2}+2B^{2})B_{y}R_{y}+$
$-R(2A^{4}R+5A^{3}R^{2}+3A^{2}R^{3}-A^{2}BR+4ABR^{2}+4BR^{3}+8AB^{2}+20B^{2}R)A_{y}B_{y}=0.$

## Acknowledgements

This research was supported by FAPESP grant # 2022/12813-5.

## References

  * [B-73] Balabanova, R. S. Hexagonal circular three-webs, Plovdiv. Univ. Nauchn. Trud. Mat., pp. 128-141, 11, No. 4, 1973.
  * [B-29] Blaschke, W. Vorlesungen über Differentialgeometrie und geometrische Grundlagen von Einsteins Relativitätstheorie. III: Differentialgeometrie der Kreise und Kugeln. Bearbeitet von G. Thomsen. (German) Berlin, J. Springer (Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen) (1929).
  * [BB-38] Blaschke, W., Bol, G. Geometrie der Gewebe, Topologische Fragen der Differentialgeometrie. J. Springer, Berlin, 1938.
  * [B-55] Blaschke, W. Einführung in die Geometrie der Waben. (German) Birkhäuser Verlag, Basel-Stuttgart, 1955.
  * [GS-24] Graf, H., Sauer, R. Über dreifache Geradensysteme in der Ebene, welche Dreiecksnetze bilden, Sitzungsb. Math.-Naturw. Abt. (1924), 119-156.
  * [E-74] Erdoǧan, H. I. Düzlemde $6$-gen doku teşkil eden čember demety 3-üzleri, Ph.D. Thesis, Istanbul Teknik Ueniversitesi, Istanbul, 1974.
  * [E-89] Erdoǧan, H. I. Triples of circle-pencils forming a hexagonal three-web in $E^{2}$ J. Geom. 35 (1989), no. 1-2, 39–65.
  * [L-77] Lazareva, V. B. Three-webs formed by families of circles on the plane. Differential geometry (Russian), pp. 49–64, 140, Kalinin. Gos. Univ., Kalinin, 1977.
  * [N-14] Nilov, F. K. On new constructions in the Blaschke-Bol problem. (Russian) Mat. Sb. 205 (2014), no. 11, 125–144; translation in Sb. Math. 205 (2014), no. 11-12, 1650–1667.
  * [PSS-12] Pottmann, H., Shi, L., Skopenkov, M. Darboux cyclides and webs from circles. Comput. Aided Geom. Design 29 (2012), no. 1, 77–97.
  * [S-05] Shelekhov, A. M. Classifications of regular 3-webs formed by pencils of circles. (Russian) Sovrem. Mat. Prilozh. No. 32, Geom. Geom. Tkan. (2005), 7–28; translation in J. Math. Sci. (N.Y.) 143 (2007), no. 6, 3607–3629.
  * [S-32] Strubecker, K. Über eine Klasse spezieller Dreiecksnetze aus Kreisen. (German) Monatsh. Math. Phys. 39 (1932), no. 1, 395–398.
  * [V-29] Volk, O. Über spezielle Kreisnetze. (German) Sitzungsberichte München 59, 125–134 (1929).
  * [W-38] Wunderlich, W. Über ein besonderes Dreiecksnetz aus Kreisen, Sitzungsber. Akad. Wiss. Wien 147 (1938), 385–399