# Napoleonic triangles on the sphere

Serena Dipierro  ,  Lyle Noakes  and  Enrico Valdinoci  Department of
Mathematics and Statistics, University of Western Australia, 35 Stirling
Highway, WA6009 Crawley, Australia<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

As is well-known, numerical experiments show that Napoleon’s Theorem for
planar triangles does not extend to a similar statement for triangles on the
unit sphere $S^{2}$. Spherical triangles for which an extension of Napoleon’s
Theorem holds are called Napoleonic, and until now the only known examples
have been equilateral. In this paper we determine all Napoleonic spherical
triangles, including a class corresponding to points on a 2-dimensional
ellipsoid, whose Napoleonisations are all congruent. Other new classes of
examples are also found, according to different versions of Napoleon’s Theorem
for the sphere. The classification follows from successive simplifications of
a complicated original algebraic condition, exploiting geometric symmetries
and algebraic factorisations.

##  1\. Introduction

###  1.1. Napoleonic questions

A beautiful, theorem, often attributed to Napoleon Bonaparte, is that if
equilateral triangles are constructed either all outwards or all inwards on
the sides of a triangle in the Euclidean plane $E^{2}$, then the lines
connecting the centres of those equilateral triangles themselves form an
equilateral triangle.

######  Theorem 1.1 (Napoleon’s Theorem).

Given distinct $P_{0}$, $P_{1}$, $P_{2}\in E^{2}$, let $Q_{0}$, $Q_{1}$,
$Q_{2}\in E^{2}$ be such that, for every $i\in\mathbb{Z}/3\mathbb{Z}$, the
triangle $Q_{i}P_{i+1}P_{i+2}$ is equilateral with interior disjoint from the
interior of the triangle $P_{0}P_{1}P_{2}$. Then $R_{0}R_{1}R_{2}$ is
equilateral, where $R_{i}$ is the centroid of $Q_{i}P_{i+1}P_{i+2}$.

If instead “disjoint” is replaced by “not disjoint” the triangle
$R_{0}R_{1}R_{2}$ is also equilateral. $\square$

We refer to [MR635364, MR2928662] for some of the history of Napoleon’s
Theorem, including discussion of its attribution. For extensions of Napoleon’s
Theorem see [MR1847491, MR2410581, MR2558297] and the references therein (also
https://www.cut-the-knot.org/proofs/napoleon.shtml for a collection of related
work). Even in flat space, these kinds of elementary questions have attracted
continued attention. For example, the PDN Theorem [PETR-zbMATH02641104, DOUG-
MR2178, NEUM-MR6839], extending Napoleon’s Theorem from planar triangles to
polygons, was found independently by three first-rate mathematicians,
including Fields Medallist Jesse Douglas.

Our extension of Theorem 1.1 is a classification of Napoleonic triangles in
the unit $2$-sphere $S^{2}$ in Euclidean $3$-pace $E^{3}$. Here vertices are
points in $S^{2}$, and the distance $d(P,Q)\in[0,\pi]$ between $P,Q\in S^{2}$
is $\arccos\langle P,Q\rangle$, where $\langle~{},~{}\rangle$ is the Euclidean
inner product. Edges of spherical triangles are defined to be minimal great
circle arcs $PQ$. Then for vertices $P_{0},P_{1},P_{2}\in S^{2}$ where
$P_{i}\not=\pm P_{i+1}$, the spherical triangle $P_{0}P_{1}P_{2}$ is a union
of edges $P_{i}P_{i+1}$. It is convenient to consider the indices $i=0,1,2$ as
elements of $\mathbb{Z}/3\mathbb{Z}$ namely the integers reduced mod $3$.

A well-known quick proof of the planar Napoleon’s Theorem uses symmetries and
dilations, but this breaks down in spherical geometry, because most spherical
triangles are not Napoleonic in the following sense.

######  Definition 1.2.

A Napoleonisation of a spherical triangle $P_{0}P_{1}P_{2}$ is a spherical
triangle of the form $R_{0}R_{1}R_{2}$, where, for all
$i\in\mathbb{Z}/3\mathbb{Z}$, $R_{i}$ is the centroid of an equilateral
spherical triangle $Q_{i}P_{i+1}P_{i+2}$. We say $P_{0}P_{1}P_{2}$ is
Napoleonic when it has an equilateral Napoleonisation. $\square$

For there to exist a vertex $Q_{i}\in S^{2}$ with $Q_{i}P_{i+1}P_{i+2}$
equilateral, we must have $d(P_{i+1},P_{i+2})\leqslant 2\pi/3$ for all
$i\in\mathbb{Z}/3\mathbb{Z}$. So assume from now on that
$d(P_{i+1},P_{i+2})\leqslant 2\pi/3$. Notice also that

  * •

when $d(P_{i+1},P_{i+2})=2\pi/3$ the vertex $Q_{i}$ is unique, and
$Q_{i}P_{i+1}P_{i+2}$ are cogeodesic, namely they lie on a great circle.

  * •

when $0<d(P_{i+1},P_{i+2})<2\pi/3$ there are precisely two choices of $Q_{i}$.

######  Example 1.3.

Suppose that $P_{0},P_{1},P_{2}$ are cogeodesic: then they are all orthogonal
to some $N\in S^{2}$, and

$\sum_{i=0}^{2}d(P_{i},P_{i+1})=2\pi\Longrightarrow
d(P_{i},P_{i+1})=2\pi/3\hbox{ for all }i\in\mathbb{Z}/3\mathbb{Z}.$

So $P_{0}P_{1}P_{2}$ is already equilateral and, for all
$i\in\mathbb{Z}/3\mathbb{Z}$, we have $Q_{i}=P_{i}$ and $R_{i}=\pm N$. So the
$R_{i}$ are not distinct, $R_{0}R_{1}R_{2}$ is not considered as a triangle,
and $P_{0}P_{1}P_{2}$ is not Napoleonic. On the other hand, if
$P_{0}P_{1}P_{2}$ is equilateral, with $d(P_{i},P_{i+1})<2\pi/3$ then, by
symmetry, $P_{0}P_{1}P_{2}$ is Napoleonic. $\square$

Suppose from now on that, in addition to $d(P_{i},P_{i+1})\leqslant 2\pi/3$,
the $P_{i}$ are distinct and not cogeodesic. The interior of the spherical
triangle $P_{0}P_{1}P_{2}$ is the path-component of the spherical centroid (or
barycentre)

$\frac{P_{0}+P_{1}+P_{2}}{\|P_{0}+P_{1}+P_{2}\|}$

in $S^{2}\setminus P_{0}P_{1}P_{2}$.

The Napoleonisation $R_{0}R_{1}R_{2}$ is said to be outward (respectively
inward) when all $Q_{i}$ are directed away from (respectively towards) the
interior of the spherical triangle $P_{0}P_{1}P_{2}$. More detail is given in
§1.3 below. We call $P_{0}P_{1}P_{2}$ outward-Napoleonic or inward-Napoleonic
when its outward Napoleonisation or inward Napoleonisation is equilateral,
respectively. To begin we consider outward Napoleonisations.

Figure 1. Vertices (green) of the outward Napoleonisation of the spherical
triangle $P_{0}P_{1}P_{2}$, with $P_{0}:=(1,0,0)$,
$P_{1}:=\left(\frac{3}{4},\frac{1}{4},\frac{\sqrt{6}}{4}\right)$ and
$P_{2}:=\left(\frac{1}{2},\frac{1}{2},\frac{\sqrt{2}}{2}\right)$ (blue).

Figure 1 shows the vertices (green) $R_{0},R_{1},R_{2}$ of the outward
Napoleonisation for a particular choice of $P_{0},P_{1},P_{2}$ in the unit
sphere $S^{2}$. The spherical distances
$d(R_{0},R_{1}),d(R_{1},R_{2}),d(R_{2},R_{0})$ are

$0.474066,0.490357,0.448728.$

So $R_{0}R_{1}R_{2}$ is not equilateral, namely $P_{0}P_{1}P_{2}$ is not
outward-Napoleonic.

Next we give some explicit examples of Napoleonic constructions on the sphere.

###  1.2. Napoleonic examples and counterexamples

As noted in [TREUDEN] and again in §1.1, examples of non-Napoleonic spherical
triangles are easily found using Mathematica. Examples can also be found using
mathematical reasoning as follows.

######  Example 1.4.

Let $\delta>0$, to be taken as small as we wish. We denote by $o(1)$
quantities which are infinitesimal as $\delta\searrow 0$.

Let $P_{0}:=(1,0,0)$ and
$P_{1}:=\left(-\frac{1}{2},\frac{\sqrt{3}}{2},0\right)$. Let also $P_{2}\in
S^{2}$ be such that

$\|P_{0}-P_{2}\|<\delta\qquad{\mbox{ and }}\qquad\langle
P_{2},\,(0,1,0)\rangle>0.$

Notice that there is one, and only one, possible choice of $Q_{2}\in S^{2}$
such that the spherical triangle $P_{0}P_{1}Q_{2}$ is equilateral, namely
$Q_{2}:=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2},0\right)$ (which is actually
the largest possible equilateral spherical triangle in $S^{2}$). There are
however two spherical triangles $P_{0}P_{1}Q_{2}$, one containing $(0,0,1)$
and one containing $(0,0,-1)$. Let us pick the first one (the other choice
leading to similar calculations). In this case, the centroid $R_{2}$ (taken as
the intersection of the medians) of the spherical triangle $P_{0}P_{1}Q_{2}$
is $(0,0,1)$.

We also observe that the spherical distance $\vartheta_{0}$ between $P_{1}$
and $P_{2}$ is such that

$\cos\vartheta_{0}=\langle
P_{1},P_{2}\rangle=\left\langle\left(-\frac{1}{2},\frac{\sqrt{3}}{2},0\right),\,P_{2}\right\rangle>\left\langle\left(-\frac{1}{2},0,0\right),\,P_{2}\right\rangle\geqslant-\frac{1}{2},$

therefore there are two possible choices for $Q_{0}\in S^{2}$ such that the
spherical triangle $P_{1}P_{2}Q_{0}$ is equilateral. Anyway, since the choice
of $Q_{2}$ was unique, $Q_{0}$ needs to be at an $o(1)$-distance from $Q_{2}$.
Consequently, the centroid $R_{0}$ of $P_{1}P_{2}Q_{0}$ is also $o(1)$-close
to either $(0,0,1)$ or $(0,0,-1)$.

Also, since $P_{0}$ and $P_{2}$ lie at an $o(1)$-distance from each other,
either equilateral spherical triangle $P_{0}P_{2}Q_{1}$ (there are two
possible choices of $Q_{1}\in S^{2}$) is contained in an $o(1)$-neighborhood
of $P_{0}$, therefore its centroid $R_{1}$ is also contained in a
$o(1)$-neighborhood of $P_{0}$.

From these considerations, when $R_{0}$ is $o(1)$-close to $(0,0,1)=R_{2}$,
for small $\delta$ we arrive at

$\|R_{1}-R_{2}\|\geqslant\|P_{0}-R_{2}\|-o(1)=\|(1,0,0)-(0,0,1)\|-o(1)=\sqrt{2}-o(1)>o(1)=\|R_{0}-R_{2}\|,$

while, when $R_{0}$ is $o(1)$-close to $(0,0,-1)$, we see that

$\displaystyle\|R_{1}-R_{2}\|\leqslant\|P_{0}-R_{2}\|+o(1)=\|(1,0,0)-(0,0,1)\|+o(1)=\sqrt{2}+o(1)<2$
$\displaystyle\qquad=\|(1,0,0)-(-1,0,0)\|=\|R_{0}-R_{2}\|,$

showing that the spherical triangle $R_{0}R_{1}R_{2}$ is not equilateral.
$\square$

######  Remark 1.5.

Example 1.4 can be also understood by perturbing the degenerate triangle
$P_{0}P_{1}P_{0}$ where $P_{0}:=(1,0,0)=:P_{2}$ and
$P_{1}:=\left(-\frac{1}{2},\frac{\sqrt{3}}{2},0\right)$. In this case
$Q_{1}=(1,0,0)$ and
$Q_{0}=Q_{2}=\left(-\frac{1}{2},\frac{\sqrt{3}}{2},0\right)$. Furthermore,
$R_{1}=(1,0,0)$ and $R_{0}$, $R_{2}\in\\{(0,0,1),(0,0,-1)\\}$. We thus obtain
that $\|R_{1}-R_{2}\|=\sqrt{2}$ while $\|R_{0}-R_{2}\|\in\\{0,2\\}$, showing
that the spherical triangle $R_{0}R_{1}R_{2}$ is not equilateral. Although
$P_{0}P_{1}P_{0}$ does not have $3$ distinct vertices, this defect is removed
by suitably small perturbations. $\square$

By symmetry, both the outward and inward Napoleonisations of an equilateral
triangle are also equilateral, with $R_{0}=R_{1}=R_{2}$ for the inward
Napoleonisation. So equilateral spherical triangles are both outward-
Napoleonic and inward-Napoleonic.

Some Napoleonic spherical triangles are not equilateral, indeed not even
isosceles.

######  Example 1.6.

Let

$P_{0}:=(1,0,0),\qquad
P_{1}:=\left(-\frac{7}{50},\frac{\sqrt{2451}}{50},0\right)\qquad{\mbox{and}}\qquad
P_{2}:=\left(-\frac{17}{50},-\frac{423\sqrt{2451}}{40850},\frac{46\sqrt{4902}}{4085}\right).$

Then $P_{0}P_{1}P_{2}$ is not isosceles, because

(1.1) $\langle P_{0},P_{1}\rangle=-\frac{7}{50},\qquad\langle
P_{0},P_{2}\rangle=-\frac{17}{50}\qquad{\mbox{and}}\qquad\langle
P_{1},P_{2}\rangle=-\frac{23}{50}.$

Define $Q_{0},Q_{1},Q_{2}\in S^{2}$ by

(1.2)
$\begin{split}&Q_{0}:=\left(0,-\frac{23}{\sqrt{2451}},-31\sqrt{\frac{2}{2451}}\right),\\\
&Q_{1}:=\left(-\frac{17}{50},-\frac{1019}{50\sqrt{2451}},-\frac{148}{5}\sqrt{\frac{2}{2451}}\right)\\\
{\mbox{and}}\qquad&Q_{2}:=\left(-\frac{7}{50},-\frac{7}{50}\sqrt{\frac{57}{43}},-\frac{3}{5}\sqrt{\frac{114}{43}}.\right).\end{split}$

Then

$\displaystyle\langle P_{0},P_{1}\rangle=\langle P_{0},Q_{2}\rangle=\langle
P_{1},Q_{2}\rangle=-\frac{7}{50},$ $\displaystyle\langle
P_{0},P_{2}\rangle=\langle P_{0},Q_{1}\rangle=\langle
P_{2},Q_{1}\rangle=-\frac{17}{50}$ and $\displaystyle\langle
P_{1},P_{2}\rangle=\langle P_{1},Q_{0}\rangle=\langle
P_{2},Q_{0}\rangle=-\frac{23}{50},$

and so $P_{0}P_{1}Q_{2}$, $P_{0}P_{2}Q_{1}$ and $P_{1}P_{2}Q_{0}$ are
equilateral, with centroids respectively

$\displaystyle
R_{2}:=\frac{P_{0}+P_{1}+Q_{2}}{\|P_{0}+P_{1}+Q_{2}\|}=\left(\frac{\sqrt{6}}{5},\frac{3}{5}\sqrt{\frac{38}{43}},-\sqrt{\frac{19}{43}}\right),$
$\displaystyle
R_{1}:=\frac{P_{0}+P_{2}+Q_{1}}{\|P_{0}+P_{2}+Q_{1}\|}=\left(\frac{2}{5}\sqrt{\frac{2}{3}},-\frac{286}{15}\sqrt{\frac{2}{817}},-\frac{5}{3\sqrt{817}}\right)$
and $\displaystyle
R_{0}:=\frac{P_{1}+P_{2}+Q_{0}}{\|P_{1}+P_{2}+Q_{0}\|}=\left(-\frac{2}{5}\sqrt{6},\frac{8}{15}\sqrt{\frac{2}{817}},-\frac{17}{3\sqrt{817}}\right).$

As a result,

(1.3) $\langle R_{0},R_{1}\rangle=\langle R_{0},R_{2}\rangle=\langle
R_{1},R_{2}\rangle=-\frac{1}{3}.$

This shows that $P_{0}P_{1}P_{2}$ is outward-Napoleonic and not isosceles, as
illustrated in Figure 2.$\square$

Figure 2. Different views of the Napoleonic construction related to (1.1) and
(1.2) (the triangle $P_{0}P_{1}P_{2}$ being in blue and the points $R_{0}$,
$R_{1}$, $R_{2}$ in green).

Unlike the planar case, $R_{0}R_{1}R_{2}$ does not always have the same
barycentre as $P_{0}P_{1}P_{2}$. In the present instance, the barycentre of
$P_{0}P_{1}P_{2}$ is

$\left(\frac{13}{10\sqrt{7}},\frac{197}{10}\sqrt{\frac{3}{5719}},23\sqrt{\frac{6}{5719}}\right)\approx(0.491354,0.451198,0.744978),$

while the barycentre of $R_{0}R_{1}R_{2}$ is

$\left(-\frac{1}{5}\sqrt{\frac{2}{3}},-\frac{107}{15}\sqrt{\frac{2}{817}},-\frac{79}{3\sqrt{817}}\right)\approx(-0.163299,-0.352936,-0.921287).$

$\square$

These examples show that:

  * •

unlike planar triangles, not all spherical triangles are Napoleonic,

  * •

as for planar triangles, all equilateral spherical triangles are Napoleonic,

  * •

there exist non-isosceles outward-Napoleonic spherical triangles,

  * •

the barycentre of a spherical Napoleonisation is not necessarily the same as
the barycentre of the original triangle, even when the latter is Napoleonic.

###  1.3. Classification of Outward and Inward Napoleonic Spherical Triangles

For $i\in\mathbb{Z}/3\mathbb{Z}$ let $P_{i}\in S^{2}$ be distinct,
satisfying111The condition $\langle P_{i+1},P_{i+2}\rangle\geqslant-1/2$ is
needed to ensure that there exist equilateral spherical triangles
$Q_{i}P_{i+1}P_{i+2}$, and $\langle P_{i+1},P_{i+2}\rangle\not=-1/2$ ensures
that the interiors of the equilateral triangles are well-defined (otherwise
$Q_{i},P_{i+1},P_{i+2}$ would be cogeodesic).

(1.4) $\langle P_{i+1},P_{i+2}\rangle>-1/2.$

Define

(1.5) $d_{i}:=\sqrt{1+2\langle P_{i+1},P_{i+2}\rangle}\in(0,\sqrt{3}),$

determined by the spherical distance between $P_{i+1}$ and $P_{i+2}$. Note
that $\chi:=\langle P_{i},P_{i+1}\times P_{i+2}\rangle$ is nonzero and
independent of $i\in\mathbb{Z}/3\mathbb{Z}$. We relabel the $P_{i}$ if
necessary, to ensure that $\chi>0$. Then in §2 we see that $Q_{i}$ is directed
towards the interior of $P_{0}P_{1}P_{2}$ precisely when $\langle
Q_{i},P_{i+1}\times P_{i+2}\rangle>0$ (otherwise outwards).

By symmetry, equilateral spherical triangles are both outward-Napoleonic and
inward-Napoleonic. Our main result is that these are the only inward-
Napoleonic spherical triangles, and that the non-equilateral outward-
Napoleonic spherical triangles are given by a quadratic condition (with all
outward-Napoleonisations congruent).

######  Theorem 1.7.

Suppose that the spherical triangle $P_{0}P_{1}P_{2}$ is not equilateral. Then
$P_{0}P_{1}P_{2}$ is not inward-Napoleonic, and is outward-Napoleonic if and
only if

(1.6) $d_{0}^{2}+d_{1}^{2}+d_{2}^{2}+d_{0}d_{1}+d_{0}d_{2}+d_{1}d_{2}=2.$

In such a case, the equilateral Napoleonisation $R_{0}R_{1}R_{2}$ has side
$\pi-\arccos(1/3)$.

So condition (1.6) is satisfied by $P_{0}P_{1}P_{2}$ 222Indeed, in Example 1.6
one has $d_{0}=\frac{\sqrt{2}}{5}$, $d_{1}=\frac{2\sqrt{2}}{5}$ and
$d_{2}=\frac{3\sqrt{2}}{5}$, which satisfy condition (1.6). in Example 1.6,
where the dimensions (1.3) of the Napoleonisation are also given by Theorem
1.7. Condition (1.6) is illustrated in Figure 3 and, after rotating
coordinates

$\left(\begin{matrix}X\\\ Y\\\
Z\end{matrix}\right)=\left(\begin{matrix}1/\sqrt{3}&1/\sqrt{3}&1/\sqrt{3}\\\
-\sqrt{2/3}&1/\sqrt{6}&1/\sqrt{6}\\\
0&-1/\sqrt{2}&1/\sqrt{2}\end{matrix}\right)\left(\begin{matrix}d_{0}\\\
d_{1}\\\ d_{2}\end{matrix}\right)$

defines part of333The conditions $d_{i}\in(0,\sqrt{3})$ must still be
satisfied. a standard ellipsoid of revolution

$2X^{2}+\frac{Y^{2}}{2}+\frac{Z^{2}}{2}=2.$

Figure 3. The ellipsoid of revolution
$\\{d_{0}^{2}+d_{1}^{2}+d_{2}^{2}+d_{0}d_{1}+d_{0}d_{2}+d_{1}d_{2}=2\\}$ and
its intersection with the region where the $d_{i}\in(0,\sqrt{3})$.

The paper is organised as follows. In §2 we introduce a mathematical framework
for studying Napoleonic spherical triangles. This allows us to make in §3 the
arguments required for the proof of Theorem 1.7, namely the classification of
Napoleonic triangles in $S^{2}$. This completes the proofs of our mathematical
results. In §4 we make some more general contextual remarks. Finally in
§LABEL:CONC we list some conclusions.

##  2\. Spherical Triangles

For $P_{0}\not=\pm P_{1}\in S^{2}$ and any $Q\in S^{2}$ write
$Q=a_{0}P_{0}+a_{1}P_{1}+bP_{0}\times P_{1}$. Then

(2.1) $1=\langle Q,Q\rangle=a_{0}^{2}+a_{1}^{2}+2a_{0}a_{1}\langle
P_{0},P_{1}\rangle+b^{2}(1-\langle P_{0},P_{1}\rangle^{2}).$

From $a_{0}+a_{1}\langle P_{0},P_{1}\rangle=\langle Q,P_{0}\rangle$ and
$a_{0}\langle P_{0},P_{1}\rangle+a_{1}=\langle Q,P_{1}\rangle$ we obtain

(2.2) $a_{0}=\frac{\langle Q,P_{0}\rangle-\langle P_{0},P_{1}\rangle\langle
Q,P_{1}\rangle}{1-\langle P_{0},P_{1}\rangle^{2}},~{}a_{1}=\frac{\langle
Q,P_{1}\rangle-\langle P_{0},P_{1}\rangle\langle Q,P_{0}\rangle}{1-\langle
P_{0},P_{1}\rangle^{2}}.$

Let $P_{0}QP_{1}$ be isosceles with $\langle P_{0},Q\rangle=\langle
P_{1},Q\rangle$. Then (2.2) reduces to

(2.3) $a_{0}=a_{1}=a:=\frac{\langle P_{0},Q\rangle}{1+\langle
P_{0},P_{1}\rangle}.$

Now take $P_{0}P_{1}Q_{2}$ to be equilateral. By (2.1), (2.3),

(2.4) $b=\epsilon_{2}\frac{\sqrt{1+2\langle P_{0},P_{1}\rangle}}{1+\langle
P_{0},P_{1}\rangle}$

where $\langle P_{0},P_{1}\rangle\geqslant-1/2$ in order for the equilateral
triangle444When $\langle P_{0},P_{1}\rangle=-1/2$ the equilateral spherical
triangle $P_{0}P_{1}Q_{2}$ is the great circle through $P_{0},P_{1}$. to
exist, and $\epsilon_{2}=\pm 1$. Then

(2.5) $Q_{2}=\frac{1}{1+\langle P_{0},P_{1}\rangle}(\langle
P_{0},P_{1}\rangle(P_{0}+P_{1})+\epsilon_{2}\sqrt{1+2\langle
P_{0},P_{1}\rangle}P_{0}\times P_{1}).$

Suppose now that $P_{0}P_{1}$ is an edge of a spherical triangle
$P_{0}P_{1}P_{2}$ where $\langle P_{i},P_{i+1}\rangle\in(-1/2,1)$ for all
$i\in\mathbb{Z}/3\mathbb{Z}$, and $\chi:=\langle P_{i},P_{i+1}\times
P_{i+2}\rangle>0$. From formula (2.5) we see that $Q_{2}$ is directed towards
the interior of $P_{0}P_{1}P_{2}$ precisely when the sign $\epsilon_{2}$ of
$\langle Q_{2},P_{0}\times P_{1}\rangle$ is positive. So $\epsilon_{2}=-1$ in
Figure 4. More generally, $Q_{i+2}$ is directed towards (away from) the
interior of $P_{0}P_{1}P_{2}$ when the sign $\epsilon_{i+2}$ of $\langle
Q_{i+2},P_{i}\times P_{i+1}\rangle$ is positive (negative).

Figure 4. Constructing an equilateral spherical triangle $P_{0}P_{1}Q_{2}$,
given the spherical triangle $P_{0}P_{1}P_{2}$.

The centroid of the equilateral Euclidean triangle $P_{0}P_{1}Q_{2}$ is

$\frac{1}{3(1+\langle P_{0},P_{1}\rangle))}((1+2\langle
P_{0},P_{1}\rangle)(P_{0}+P_{1})+\epsilon_{2}\sqrt{1+2\langle
P_{0},P_{1}\rangle}P_{0}\times P_{1}).$

So the equilateral spherical triangle $P_{0}P_{1}Q_{2}$ has centroid

$R_{2}:=\frac{(1+2\langle
P_{0},P_{1}\rangle)(P_{0}+P_{1})+\epsilon_{2}\sqrt{1+2\langle
P_{0},P_{1}\rangle}P_{0}\times P_{1}}{\|(1+2\langle
P_{0},P_{1}\rangle)(P_{0}+P_{1})+\epsilon_{2}\sqrt{1+2\langle
P_{0},P_{1}\rangle}P_{0}\times P_{1}\|}=$ $\frac{\sqrt{1+2\langle
P_{0},P_{1}\rangle}(P_{0}+P_{1})+\epsilon_{2}P_{0}\times
P_{1}}{\sqrt{3}(1+\langle P_{0},P_{1}\rangle)}.$

Similarly, an equilateral spherical triangle $P_{1},P_{2}Q_{0}$ has centroid

$R_{0}=\frac{\sqrt{1+2\langle
P_{1},P_{2}\rangle}(P_{1}+P_{2})+\epsilon_{0}P_{1}\times
P_{2}}{\sqrt{3}(1+\langle P_{1},P_{2}\rangle)}.$

So $3(1+\langle P_{0},P_{1}\rangle)(1+\langle P_{1},P_{2}\rangle)(1+\langle
P_{2},P_{0}\rangle)\langle R_{2},R_{0}\rangle=(1+\langle
P_{2},P_{0}\rangle)S_{1}$ where

$S_{1}:=\langle\sqrt{1+2\langle
P_{0},P_{1}\rangle}(P_{0}+P_{1})+\epsilon_{2}P_{0}\times
P_{1},\sqrt{1+2\langle
P_{1},P_{2}\rangle}(P_{1}+P_{2})+\epsilon_{0}P_{1}\times P_{2}\rangle=$
$\sqrt{1+2\langle P_{0},P_{1}\rangle}\sqrt{1+2\langle
P_{1},P_{2}\rangle}\langle P_{0}+P_{1},P_{1}+P_{2}\rangle+$
$\epsilon_{0}\sqrt{1+2\langle P_{0},P_{1}\rangle}\langle
P_{0}+P_{1},P_{1}\times P_{2}\rangle+\epsilon_{2}\sqrt{1+2\langle
P_{1},P_{2}\rangle}\langle P_{0}\times P_{1},P_{1}+P_{2}\rangle+$
$\epsilon_{2}\epsilon_{0}(1+\langle P_{2},P_{0}\rangle)\langle P_{0}\times
P_{1},P_{1}\times P_{2}\rangle=$ $\sqrt{1+2\langle
P_{0},P_{1}\rangle}\sqrt{1+2\langle P_{1},P_{2}\rangle}(1+\langle
P_{0},P_{1}\rangle+\langle P_{2},P_{0}\rangle+\langle P_{1},P_{2}\rangle)+$
$(\epsilon_{0}\sqrt{1+2\langle
P_{0},P_{1}\rangle}+\epsilon_{2}\sqrt{1+2\langle P_{1},P_{2}\rangle})\langle
P_{0},P_{1}\times P_{2}\rangle+\epsilon_{2}\epsilon_{0}(\langle
P_{0},P_{1}\rangle\langle P_{1},P_{2}\rangle-\langle P_{2},P_{0}\rangle)=$
$\alpha\sqrt{1+2\langle P_{0},P_{1}\rangle}\sqrt{1+2\langle
P_{1},P_{2}\rangle}+\chi(\epsilon_{0}\sqrt{1+2\langle
P_{0},P_{1}\rangle}+\epsilon_{2}\sqrt{1+2\langle
P_{1},P_{2}\rangle})+\epsilon_{2}\epsilon_{0}(\langle
P_{0},P_{1}\rangle\langle P_{1},P_{2}\rangle-\langle P_{2},P_{0}\rangle),$

where $\alpha:=1+\langle P_{0},P_{1}\rangle+\langle P_{2},P_{0}\rangle+\langle
P_{1},P_{2}\rangle$ is symmetric with respect to permutations of
$P_{0},P_{1},P_{2}$, and the scalar triple product $\chi:=\langle
P_{0},P_{1}\times P_{2}\rangle$ is symmetric with respect to cyclic
permutations of $i=0,1,2\in\mathbb{Z}/3\mathbb{Z}$.

Setting $d_{i}^{2}:=1+2\langle P_{i+1},P_{i+2}\rangle$ with
$d_{i}\in(0,\sqrt{3})$, and
$\gamma:=3(d_{0}^{2}+1)(d_{1}^{2}+1)(d_{2}^{2}+1)$, our calculation gives the
useful formula

(2.6) $\displaystyle\frac{\gamma\langle R_{i+2},R_{i}\rangle}{d_{i+1}^{2}+1}$
$\displaystyle=$ $\displaystyle 4(\alpha
d_{i+2}d_{i}+\chi(\epsilon_{i}d_{i+2}+\epsilon_{i+2}d_{i}))+\epsilon_{i+2}\epsilon_{i}((d_{i+2}^{2}-1)(d_{i}^{2}-1)-2(d_{i+1}^{2}-1)).$

By (2.2),

$P_{2}=\frac{\langle P_{2},P_{0}\rangle-\langle P_{0},P_{1}\rangle\langle
P_{1},P_{2}\rangle}{1-\langle P_{0},P_{1}\rangle^{2}}P_{0}+\frac{\langle
P_{1},P_{2}\rangle-\langle P_{0},P_{1}\rangle\langle
P_{2},P_{0}\rangle}{1-\langle P_{0},P_{1}\rangle^{2}}P_{1}+bP_{0}\times P_{1}$

where by (2.1), $b^{2}(1-\langle P_{0},P_{1}\rangle^{2})^{2}=$

$1-\langle P_{0},P_{1}\rangle^{2}-\langle P_{1},P_{2}\rangle^{2}-\langle
P_{2},P_{0}\rangle^{2}+2\langle P_{2},P_{0}\rangle\langle
P_{0},P_{1}\rangle\langle P_{1},P_{2}\rangle.$

Therefore, and because $\chi>0$,

$\chi=b(1-\langle P_{0},P_{1}\rangle^{2})=\sqrt{1-\langle
P_{0},P_{1}\rangle^{2}-\langle P_{1},P_{2}\rangle^{2}-\langle
P_{2},P_{0}\rangle^{2}+2\langle P_{2},P_{0}\rangle\langle
P_{0},P_{1}\rangle\langle P_{1},P_{2}\rangle}.$

Summarising, we can say

(2.7) $\displaystyle 2\alpha$ $\displaystyle=$ $\displaystyle
d_{0}^{2}+d_{1}^{2}+d_{2}^{2}-1,$ (2.8) $\displaystyle 4\chi^{2}$
$\displaystyle=$ $\displaystyle
2(1-\alpha)(1+2\alpha)+d_{0}^{2}d_{1}^{2}+d_{1}^{2}d_{2}^{2}+d_{2}^{2}d_{0}^{2}+d_{0}^{2}d_{1}^{2}d_{2}^{2},$

##  3\. Proof of Theorem 1.7

Note that $P_{0}P_{1}P_{2}$ is equilateral precisely when $d_{0}=d_{1}=d_{2}$.
This case is excluded in the present section, namely we suppose that
$d:=(d_{0},d_{1},d_{2})\in D:=(0,\sqrt{3})^{3}\setminus\Delta$, where
$\Delta:=\\{(\delta,\delta,\delta):\delta\in(0,\sqrt{3})\\}$. Set555Because of
these choices, if $\langle P_{2},P_{0}\rangle=\langle P_{1},P_{0}\rangle$ then
$\langle R_{2},R_{0}\rangle=\langle R_{1},R_{0}\rangle$. Similarly, if
$\langle P_{0},P_{1}\rangle=\langle P_{2},P_{1}\rangle$ then $\langle
R_{0},R_{1}\rangle=\langle R_{2},R_{1}\rangle$, and if $\langle
P_{1},P_{2}\rangle=\langle P_{0},P_{2}\rangle$ then $\langle
R_{1},R_{2}\rangle=\langle R_{0},R_{2}\rangle$. At this stage however, our
attention is not limited to the case where $P_{0}P_{1}P_{2}$ is isosceles.
$\epsilon_{0}=\epsilon_{1}=\epsilon_{2}=\epsilon:=\pm 1$. So the equilateral
triangles constructed on the edges of $P_{0}P_{1}P_{2}$ are directed either
all outwards ($\epsilon=-1$) or all inwards ($\epsilon=+1$). By (2.6),

(3.1) $\displaystyle\frac{\gamma\langle R_{i+2},R_{i}\rangle}{d_{i+1}^{2}+1}$
$\displaystyle=$ $\displaystyle 4(\alpha
d_{i+2}d_{i}+\epsilon\chi(d_{i+2}+d_{i}))+(d_{i+2}^{2}-1)(d_{i}^{2}-1)-2(d_{i+1}^{2}-1).$

A short calculation then shows that $\langle R_{2},R_{0}\rangle=\langle
R_{0},R_{1}\rangle$ precisely when

$4\left(\alpha
d_{0}(1-d_{1}d_{2})+\epsilon\chi(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})\right)(d_{2}-d_{1})=-2(d_{2}^{2}-d_{1}^{2})(d_{0}^{2}+d_{1}^{2}+d_{2}^{2}-1).$

The right hand side is $-4\alpha(d_{2}^{2}-d_{1}^{2})$, and so our condition
is that either $d_{1}=d_{2}$ or

(3.2)
$\displaystyle\alpha(d_{0}+d_{1}+d_{2}-d_{0}d_{1}d_{2})+\epsilon\chi(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})$
$\displaystyle=$ $\displaystyle 0.$

This equation is invariant with respect to cyclic permutations of the
subscripts and, because $d\notin\Delta$, equation (3.2) is necessary and
sufficient for $R_{0}R_{1}R_{2}$ to be equilateral. We observe that:

  1. (1)

For $d\in D$ we have $d_{0}+d_{1}+d_{2}-d_{0}d_{1}d_{2}>0$ because otherwise,
by the AM-GM inequality,

$d_{0}d_{1}d_{2}\geqslant d_{0}+d_{1}+d_{2}\geqslant
3(d_{0}d_{1}d_{2})^{1/3}\Longrightarrow d_{0}d_{1}d_{2}\geqslant 3\sqrt{3},$

contradicting $d_{0},d_{1},d_{2}\in(0,\sqrt{3})$.

  2. (2)

For any given $d\in D$, equation (3.2) cannot hold for both $\epsilon=1$ and
$\epsilon=-1$. Otherwise,

$\alpha(d_{0}+d_{1}+d_{2}-d_{0}d_{1}d_{2})=0=\chi(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})\Longrightarrow$
$d_{0}^{2}+d_{1}^{2}+d_{2}^{2}=1=d_{0}d_{1}+d_{1}d_{2}+d_{2}d_{0}\Longrightarrow
d_{0}+d_{1}+d_{2}=\sqrt{3}\Longrightarrow$ $\langle
d,(1,1,1)\rangle^{2}=3=\langle
d,d\rangle\langle(1,1,1),(1,1,1)\rangle\Longrightarrow d=(1,1,1)/\sqrt{3}$

by Cauchy-Schwarz, contradicting $d\notin\Delta$.

  3. (3)

So (3.2) holds neither for any $d\in D\cap S^{2}_{+}$, nor for any $d\in D$
satisfying

$d_{0}d_{1}+d_{1}d_{2}+d_{2}d_{0}=1.$

  4. (4)

$\epsilon$ has the sign of

$(1-d_{0}^{2}-d_{1}^{2}-d_{2}^{2})(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0}).$

Hence, a necessary condition for $P_{0}P_{1}P_{2}$ to be either outward-
Napoleonic or inward-Napoleonic is

(3.3)
$\alpha^{2}(d_{0}+d_{1}+d_{2}-d_{0}d_{1}d_{2})^{2}-\chi^{2}(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})^{2}=0$

with $\alpha,\chi^{2}$ given by (2.7), (2.8) as symmetric polynomials in
$d_{0},d_{1},d_{2}$.

Substituting for $\alpha,\chi^{2}$, a calculation shows that the left-hand
side of (3.3) is $\gamma/12$ times

(3.4)
$(d_{0}^{2}+d_{1}^{2}+d_{2}^{2}-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})(d_{0}^{2}+d_{1}^{2}+d_{2}^{2}+d_{0}d_{1}+d_{1}d_{2}+d_{2}d_{0}-2),$

where the first factor is nonzero because $d\notin\Delta$. So
$P_{0}P_{1}P_{2}$ is either outward-Napoleonic or inward-Napoleonic when

$(d_{0}+\frac{d_{1}}{2}+\frac{d_{2}}{2})^{2}+\frac{3}{4}(d_{1}+\frac{d_{2}}{3})^{2}+\frac{2}{3}d_{2}^{2}=2.$

When this happens we find that, using
$d_{0}^{2}+d_{1}^{2}+d_{2}^{2}-1=1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0}$,

$(1-d_{0}^{2}-d_{1}^{2}-d_{2}^{2})(1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0})=-(1-d_{0}^{2}-d_{1}^{2}-d_{2}^{2})^{2}<0$

by observations (2), (3). Then $\epsilon=-1$ by observation (4), namely
$P_{0}P_{1}P_{2}$ is outward-Napoleonic (it cannot be inward-Napoleonic).

More can be said: using
$1-d_{0}d_{1}-d_{1}d_{2}-d_{2}d_{0}=d_{0}^{2}+d_{1}^{2}+d_{2}^{2}-1=2\alpha$
and $\epsilon=-1$ in (3.2), we find that

$2\chi=d_{0}+d_{1}+d_{2}-d_{0}d_{1}d_{2}.$

Then, substituting for $\chi,\alpha$, we find

$(d_{0}^{2}+1)(d_{2}^{2}+1)+(4(\alpha
d_{2}d_{0}-\chi(d_{2}+d_{0}))+(d_{2}^{2}-1)(d_{0}^{2}-1)-2(d_{1}^{2}-1)=$
$2(d_{0}d_{2}-1)(d_{0}^{2}+d_{1}^{2}+d_{2}^{2}+d_{0}d_{1}+d_{1}d_{2}+d_{2}d_{0}-2)=0.$

Comparing the left-hand side with equation (3.1) for $i=0$,

$\langle R_{2},R_{0}\rangle=\langle R_{0},R_{1}\rangle=\langle
R_{1},R_{2}\rangle=-1/3.$

So the Napoleonisation $R_{0}R_{1}R_{2}$ belongs to the congruency class of
spherical equilateral triangles of perimeter $3(\pi-\arccos(1/3))$. This
proves Theorem 1.7. $\square$

##  4\. Context

The study of triangular patterns arises in several situations and we present
here some of them, both in the Euclidean and spherical case, to provide
context to the Napoleonic construction.

To start with, the theory of crystalline growth provides a number of examples
of triangular structures, which occur often in nature, especially for mineral
crystals. These structures can arise as sections of trigonal crystal shapes in
ruby, quartz, brucite, calcite, agate, jasper, etc., and in groups of sodium
nitrate and nickel arsenide, see [SSC, page 11, page 46, Figure A.4]. They
also occur as sections of tetragonal systems of silicon, see [PAUL], and in
hexagonal patterns of beryl, cancrinite, apatite, sugilite, etc., see [ATLAS],
and of oxygen atoms in hexagonal lattices of ice, see [PETRE, Figure 2.3] and
[DEER, pages 4-5].

With regards to organic molecules, the case of the compound theophylline is
also noteworthy, since, differently from most organic compounds, it may form
crystals with a triangular form, see [UNUS] (and in particular Figure 1 there
for optical microscope images showing the growth of triangular theophylline
crystals on a glass slide). In terms of applications, theophylline is a
bronchodilator and is used as a medication to treat asthma and obstructive
pulmonary diseases. Triangular crystals also occur in the crystalline growth
of multilayers silver nanocrystals, see [Courty2007].

Another interesting situation arises in the theory of superfluids and
superconductors, in which vortices have been found to form triangular lattice
arrangements, as a configuration that minimises the total energy, see
[zbMATH03188431, BEN, CORR] (remarkably, in this setting, the original idea of
[Abrikosov] that the energy is minimised by a square lattice turns out to be
inaccurate).

Figure 5. Aerial views of Canberra, Damascus, Gramichelle, Hamina, Naples, New
Dheli, Perth, Palmanova, Paris, Rome, Washington (screenshots from Google
Earth).

Further context comes in urban design and town planning, since several cities
exhibit triangle-shaped blocks, see Figures 5 and 6. We refer to [HES] for
more information about hexagonal and triangular plannings in urban design. A
nice scenario in which Napoleonic constructions arise in the context of city
planning is therefore the case of an initial block given by a (possibly
irregular) triangle, which gets developed by three adjacent blocks formed by
equilateral triangles. If, for instance, we assume that each of these new
blocks is endowed with a communication antenna located at the centre of the
block, the fundamental Napoleonic question is whether or not these antennas
form an equilateral666Indeed, one can suppose that a rewarding strategy in
signal transmission consists in “covering the largest possible surface with
the least possible mutual distances among the antennas”: the formulation of
this isoperimetric problem in the class of triangular patterns of antennas is
optimised by equilateral configurations (in turn, the isoperimetric problem
for triangles can be explicitly solved, for example, by writing the area of a
general triangle using Heron’s formula and using the AM-GM inequality to see
that the optimiser is equilateral; another proof can be performed by purely
geometrical arguments by considering an ellipse with foci centred at two
vertices and passing through the third one). triangle.

Figure 6. Detroit city layout plan following the 1805 fire that destroyed most
of the city (public domain image from Wikipedia).

Besides the interest in mathematical theories, it is worth recalling that
triangular patterns of antennas are broadly used in modern technologies. For
instance, a triangulation network is implemented in North America for
satellite communication, and satellite geodesy exploited an equilateral
triangle on the Earth with operational basis established in 1963 at Aberdeen
Proving Ground 39∘28’24.4”N 76∘08’27.0”W, Maryland, Chandler Air Force Station
43∘53’52”N 95∘56’45”W, Minnesota, and Greenville Air Force Base 33∘28’58”N
90∘59’08”W, Mississippi: these stations form a triangle with about 1500 km on
a side, see Appendix E of the North American Datum report
https://www.ngs.noaa.gov/PUBS_LIB/NAD_National_Academy_of_Science.pdf and
Figure 7 (actually, the geodesic distance between 39∘28’24.4”N 76∘08’27.0”W
and 43∘53’52”N 95∘56’45”W is about 1700 km, the one between 43∘53’52”N
95∘56’45”W and 33∘28’58”N 90∘59’08”W is about 1250 km, and the one between
33∘28’58”N 90∘59’08”W and 39∘28’24.4”N 76∘08’27.0”W is about 1500 km). Also,
in the Laser Interferometer Space Antenna (LISA), three spacecraft are
arranged in an equilateral triangle, see [LISA]. See also [MYSTE] for other
different applications of equilateral triangles.

It is interesting to point out that the Napoleonic construction is also
related to the Fermat-Steiner problem, see [MR2022170, §7.1]. This problem was
posed in 1629 by Fermat and addressed by Torricelli sometime before 1640. The
question is to find a point whose total sum of distances from the vertices of
a given triangle is a minimum. That is, given a triangle with vertices $A$,
$B$ and $C$, find a point $X$ for which $\|X-A\|+\|X-B\|+\|X-C\|$ is a
minimum. Remarkably, the point $X$ can be obtained as the outcome of a
Napoleonic construction. Indeed, one constructs the three outward pointing
equilateral triangles on the sides of the triangle with vertices $A$, $B$ and
$C$ and connects each vertex of the original triangle to the new vertex of the
opposite equilateral triangle, the common intersection of these segments thus
identifying the desired point $X$ (this, if all the angles of the original
triangle are less than $\frac{2\pi}{3}$, otherwise $X$ coincides with the
vertex corresponding to the largest angle). The point $X$ found in this way is
sometimes called “Fermat point” (or “Torricelli point”, or “Steiner point”).

The Fermat-Steiner problem itself is a source of questions of practical
importance, such as the so-called “factory problem”: suppose that there are
three factories, labeled $A$, $B$, and $C$, which need to be supplied by
shipments of a certain material originating at a warehouse $X$, and assume
additionally that the cost of the shipment is proportional to the distance
between the warehouse and the factory receiving the shipment. Once again, the
total cost would be thus proportional to $\|X-A\|+\|X-B\|+\|X-C\|$ and
therefore the optimal location for $X$ is that of the Fermat point above.

Figure 7. Location of Aberdeen Proving Ground, Maryland, Chandler Air Force
Station, Minnesota, and Greenville Air Force Base, Mississippi (obtained via
Mathematica).

Equilateral triangular patterns naturally emerge from chaotic processes (see
[CHF, Figure 6.2]). Besides, other interesting occurrences of equilateral
triangles arise for the Kakeya needle problem: namely, equilateral triangles
are the convex figures of least area in which a line segment of given length 1
can be moved around and return to its original position with reversed
orientation, see [GARD, page 221] and the references therein (and Kakeya
problems are nowadays mainstream in the contemporary mathematics research, see
e.g. [MR2206765, MR3946717]).

Equilateral triangles also play a pivotal role in numerics and finite element
methods, since, for numerical accuracy and efficiency, the shape of the finite
elements is often chosen to resemble as much as possible the “reference
elements”, such as equilateral triangles or tetrahedra, see e.g. [CAD, pages
106 and 137] and [KURO, page 242].

Extensions of the Napoleonic construction to polygons (rather than triangles)
are also possible. The most famous result in this framework is the so-called
PDN-Theorem, named after Karel Petr, Jesse Douglas, and Bernhard Neumann
[PETR-zbMATH02641104, DOUG-MR2178, NEUM-MR6839] (the special case of
quadrilaterals being related to Van Aubel’s Theorem, see [VANB], and to
Thébault’s Second Problem, see [MR3961650]).

In all these settings, the case of the flat plane is somewhat special and it
is intriguing to understand the role played by a possibly curved environment:
the case of constant curvature being the natural one to start with, we study
here spherical Napoleonic triangles (we also aim at discussing hyperbolic
spaces in a forthcoming project).

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# Parallel Texture Structures with Cofactor Zeros in Lepton Sector

Weijian Wang  Department of Physics, North China Electric Power University,
Baoding 071003, P. R. China<EMAIL_ADDRESS>
###### Abstract

In this paper we investigate the parallel texture structures with cofactor
zeros in the charged lepton and neutrino sectors. The textures can not be
obtained from arbitrary leptonic matrices by making weak basis
transformations, which therefore have physical meaning. The 15 parallel
textures are grouped as 4 classes where each class has the same physical
implications. It is founded that one of them is not phenomenological viable
and another is equivalent to the texture zero structures extensively explored
in previous literature. Thus we focus on the other two classes of parallel
texture structures and study the their phenomenological implications. The
constraints on the physical variables are obtained for each class, which are
essential for the model selection and can be measured by future experiments.
The model realization is illustrated in a radiated lepton mass model.

##  I Introduction

The discovery of neutrino oscillations has provided us with convincing
evidences for massive neutrinos and leptonic flavor mixing with high degree of
accuracyneu1 (1, 2, 3). The recent measurement of large reactor mixing angle
$\theta_{13}$ has not only open the door for us to explore the leptonic CP
violation and the mass hierarchy in the future experiments, but also highlight
the flavor puzzle of neutrino mass and mixing pattern which appears to be
rather different from the distinct mass hierarchy and the small mixing angles
shown by quarks. Although a full theory is still missing, several ideas have
been proposed by reducing the number of free parameters of seesaw modelsseesaw
(4) and introducing the specific structures into mass matrices to explain the
observed leptonic mixing pattern. The models include texture zeroszero (5),
hybrid textureshybrid (6, 7), zero tracesum (8), zero determinantdet (9),
vanishing minorsminor (10, 11, 12), two traceless submatricestra (13), equal
elements or cofactorsco (14), hybrid $M_{\nu}^{-1}$ textureshyco (15). Among
these models, the textures with zero elements or zero minors are particularly
interesting because of their connection to the flavor symmetries and the
stable behavior of running renomalization group. The phenomenological analysis
of neutrino mass matrices with texture zeros or cofactor zeros in flavor basis
have been widely investigated in earlier literaturezero (5, 10, 11, 12).

However, there is no priori requirement that the analysis must be done in
flavor basis. The more general situation should be considered in the basis
where both charged lepton mass matrix $M_{l}$ and neutrino mass matrix
$M_{\nu}$ are non-diagonal. In this spirit, the parallel $Ans\ddot{a}tze$ has
been proposed where $M_{l}$ and $M_{\nu}$ have the same structure (We denote
it ”parallel texture structure”). A popular parallel texture structure appears
as the Fritzsch-like modelfritzsch (16) with texture zeros in mass matrix and
is firstly applied to understand the quark mixing pattern. Subsequently the
idea is generalized to the lepton sectorzzx (17, 18). A systematic search on
the parallel structures with texture zeros in lepton mass matrices are
reported in Ref.GCB (19). It is shown that some sets of the texture zeros have
no physical meaning by themselves, since they can be obtained by making
suitable weak basis (WB) transformation from arbitrary mass matrix and leaving
the gauge currents invariant. The minimal no trivial case is the four texture
zeros model. Recently, a similar investigation is done in the context of
parallel hybrid textures with one zero and two equal elements hyp (20).

In this work, we study the parallel structures with two cofactor zeros in both
$M_{\nu}$ and $M_{\nu}$. As shown in Ref. minor1 (11), the cofactor zeros in
$M_{\nu}$ are generated by the type-I seesaw formula
$M_{\nu}=-M^{D}M_{R}^{-1}M_{D}^{T}$ with texture zeros in $M_{D}$ and $M_{R}$.
The cofactor zeros in $M_{l}$, on the other hand, seems to be rather unusual
because no flavor symmetry directly leads to the cofactor zeros of Dirac mass
matrix $M_{l}$. However, we will show that if we adopt the recent viewpoint
proposed by Mama (21) that the radiated lepton mass originate from the one-
loop diagram, a seesaw-like formula is possible for charged lepton masses and
the cofactor zeros in $M_{l}$ can be realized. There exists $C_{6}^{2}=15$
logically possible patterns. Furthermore, we assume the mass matrices to be
Hermitian and all neutrinos are massive, which indicates det$M_{\nu}\neq 0$
and existence of $M_{\nu}^{-1}$. Thus the mass textures $M_{\nu}$ with
cofactor zeros are equivalent to the $M_{\nu}^{-1}$ with texture zeros. As the
texture zero case GCB (19), the 15 textures structures can be grouped into 4
classes with each class having the same physical implications. Among the 4
classes, we find that one of them is not viable phenomenologically and another
class is equal to the matrices with texture zeros. Therefore we focus on the
other two classes having not been studied before.

The paper is organized as follows. In Sec. II, we discuss the classification
of textures and relate them to the experimental results. In Sec.III, we
diagonalize the mass matrices, confront the numerical results with the
experimental data and discuss their predictions. In Sec. IV, the the
realization of cofactor zeros in $M_{l}$ is discussed. A summary is given in
Sec. IV.

##  II Formalism

###  II.1 Weak basis equivalent classes

We assume the neutrinos to be Majorana fermions. The most general WB
transformations leaving gauge currents invariant is given by

$M_{l}\rightarrow M_{l}^{\prime}=W^{\dagger}M_{l}W_{R}\quad\quad\quad
M_{\nu}\rightarrow M_{\nu}^{\prime}=W^{T}M_{\nu}W$ (1)

where $W$, $W_{R}$ are $3\times 3$ unitary matrices. Therefore the parallel
texture with cofactor zeros located at different positions can be related by
permutation matrix $P$ as the WB transformation

$M_{l}^{\prime}=P^{T}M_{l}P\quad\quad\quad M_{\nu}^{\prime}=P^{T}M_{\nu}P$ (2)

The permutation matrix $P$ changes the positions of cofactor zeros but
preserves the parallel structures for both charged lepton and neutrino mass
textures. It is noted that $P$ belongs to the group of 6 permutations and are
isomorphic to $S_{3}$. Then the four cofactor zeros texture can be classified
into 4 classes as following:

Class I:

$\begin{split}\left(\begin{array}[]{ccc}\bigtriangleup&\times&\bigtriangleup\\\
\times&\times&\times\\\
\bigtriangleup&\times&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\bigtriangleup&\bigtriangleup&\times\\\
\bigtriangleup&\times&\times\\\
\times&\times&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\bigtriangleup&\times\\\
\bigtriangleup&\bigtriangleup&\times\\\
\times&\times&\times\end{array}\right)\\\
\left(\begin{array}[]{ccc}\times&\times&\times\\\
\times&\bigtriangleup&\bigtriangleup\\\
\times&\bigtriangleup&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\times&\bigtriangleup\\\
\times&\times&\times\\\
\bigtriangleup&\times&\bigtriangleup\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\times&\times\\\
\times&\times&\bigtriangleup\\\
\times&\bigtriangleup&\bigtriangleup\end{array}\right)\end{split}$ (3)

Class II:

$\begin{split}\left(\begin{array}[]{ccc}\bigtriangleup&\times&\times\\\
\times&\times&\bigtriangleup\\\
\times&\bigtriangleup&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\times&\bigtriangleup\\\
\times&\bigtriangleup&\times\\\
\bigtriangleup&\times&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\bigtriangleup&\times\\\
\bigtriangleup&\times&\times\\\
\times&\times&\bigtriangleup\end{array}\right)\end{split}$ (4)

Class III:

$\begin{split}\left(\begin{array}[]{ccc}\bigtriangleup&\times&\times\\\
\times&\bigtriangleup&\times\\\
\times&\times&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\bigtriangleup&\times&\times\\\
\times&\times&\times\\\
\times&\times&\bigtriangleup\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\times&\times\\\
\times&\bigtriangleup&\times\\\
\times&\times&\bigtriangleup\end{array}\right)\end{split}$ (5)

Class IV:

$\begin{split}\left(\begin{array}[]{ccc}\times&\bigtriangleup&\bigtriangleup\\\
\bigtriangleup&\times&\times\\\
\bigtriangleup&\times&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\bigtriangleup&\times\\\
\bigtriangleup&\times&\bigtriangleup\\\
\times&\bigtriangleup&\times\end{array}\right)\quad\quad\left(\begin{array}[]{ccc}\times&\times&\bigtriangleup\\\
\times&\times&\bigtriangleup\\\
\bigtriangleup&\bigtriangleup&\times\end{array}\right)\end{split}$ (6)

where ”$\bigtriangleup$” at $(i,j)$ position denotes the zero cofactor
$C_{ij}=0$ while ”$\times$” stands for arbitrary element. Since $M_{l,\nu}$
with cofactor zeros is equivalent to $M_{l,\nu}^{-1}$ with zero elements, the
classification given above is the same as the texture zero ones shown in
Ref.GCB (19) except for replacing ”$\bigtriangleup$” with ”0”. Like the
texture zero cases , the class IV leads to the decoupling of a generation of
lepton from mixing and thus not experimentally viable. On the other hand, one
can easily check that the textures of class I correspond to the texture zero
ones, which has already studied in previous literaturezzx (17, 18, 19). As an
example, for the first matrix of class I, we have

$\begin{split}M_{l,\nu}=\left(\begin{array}[]{ccc}\bigtriangleup&\times&\bigtriangleup\\\
\times&\times&\times\\\
\bigtriangleup&\times&\times\end{array}\right)\Rightarrow
M_{l,\nu}^{-1}=\left(\begin{array}[]{ccc}0&\times&0\\\ \times&\times&\times\\\
0&\times&\times\end{array}\right)\Rightarrow
M_{l,\nu}=\left(\begin{array}[]{ccc}\times&\times&\times\\\ \times&0&0\\\
\times&0&\times\end{array}\right)\end{split}$ (7)

Therefore only class II and class III have no trivial physical implications.

###  II.2 Some useful notations

As we have mentioned, we only need to investigate two mass matrices
respectively belonging to to representations of class II and class III. In
this work, we choose

$\begin{split}M_{l,\nu}^{II}=\left(\begin{array}[]{ccc}\bigtriangleup&\times&\times\\\
\times&\times&\bigtriangleup\\\
\times&\bigtriangleup&\times\end{array}\right)\quad\quad
M_{l,\nu}^{III}=\left(\begin{array}[]{ccc}\bigtriangleup&\times&\times\\\
\times&\bigtriangleup&\times\\\
\times&\times&\times\end{array}\right)\end{split}$ (8)

The charged leptonic mass texture $M_{l}$ is a complex Hermitian matrix and
the Majorana neutrino mass texture $M_{\nu}$ is a complex symmetric matrix.
They are diagonalized by unitary matrix $V_{l}$ and $V_{\nu}$

$M_{l}=V_{l}M_{l}^{D}V_{l}^{\dagger}\quad\quad
M_{\nu}=V_{\nu}M_{\nu}^{D}V_{\nu}^{T}$ (9)

where $M_{l}^{D}=Diag(m_{e},m_{\mu},m_{\tau})$,
$M_{\nu}^{D}=Diag(m_{1},m_{2},m_{3})$. The Pontecorvo-Maki-Nakagawa-Sakata
matrixPMNS (22) $U_{PMNS}$ is given by

$U_{PMNS}=V_{l}^{\dagger}V_{\nu}$ (10)

and can be parameterized as

$U_{PMNS}=UP_{\nu}=\left(\begin{array}[]{ccc}c_{12}c_{13}&c_{13}s_{12}&s_{13}e^{-i\delta}\\\
-s_{12}c_{23}-c_{12}s_{13}s_{23}e^{i\delta}&c_{12}c_{23}-s_{12}s_{13}s_{23}e^{i\delta}&c_{13}s_{23}\\\
s_{23}s_{12}-c_{12}c_{23}s_{13}e^{i\delta}&-c_{12}s_{23}-c_{23}s_{12}s_{13}e^{i\delta}&c_{13}c_{23}\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
0&e^{i\alpha}&0\\\ 0&0&e^{i(\beta+\delta)}\end{array}\right)$ (11)

where the abbreviations $s_{ij}=\sin\theta_{ij}$ and $c_{ij}=\cos\theta_{ij}$
are used. The $\alpha$ and $\beta$ in $P_{\nu}$ denote two Majorana CP-
violating phases and $\delta$ in U denotes the Dirac CP-violating phase. In
order to facilitate our calculation, it is better to start from $M_{l}^{-1}$
rather than $M_{l}$. From (9), we get

$M_{l}^{-1}=V_{l}(M_{l}^{D})^{-1}V_{l}^{\dagger}$ (12)

So the $V_{l}$ can not only diagonalize the $M_{l}$ but also $M_{l}^{-1}$.
Furthermore, we treat the Hermitian matrix $M_{l}^{-1}$ to be factorisable.
i.e

$M_{l}^{-1}=K_{l}(M_{l}^{-1})^{r}K_{l}^{\dagger}$ (13)

where $K_{l}$ is the unitary phase matrix and can be parameterized as
$K_{l}=diag(1,e^{i\phi_{1}},e^{i\phi_{2}})$. The $(M_{l}^{-1})^{r}$ becomes a
real symmetric matrix which can be diagonalized by real orthogonal matrix
$O_{l}$. Then we have

$V_{l}=K_{l}O_{l}$ (14)

and

$U_{PMNS}=O_{l}^{T}K_{l}^{\dagger}V_{\nu}$ (15)

From (9), (10) and (15), the neutrino mass matrix $M_{\nu}$ is given by

$M_{\nu}=K_{l}VP_{\nu}M_{\nu}^{D}P_{\nu}V^{T}K_{l}^{\dagger}$ (16)

where $V\equiv O_{l}U$. From (16) The restriction of two cofactor zeros on
$M_{\nu}$

$M_{\nu(pq)}M_{\nu(rs)}-M_{\nu(tu)}M_{\nu(vw)}=0\quad\quad
M_{\nu(p^{\prime}q^{\prime})}M_{\nu(r^{\prime}s^{\prime})}-M_{\nu(t^{\prime}u^{\prime})}M_{\nu(v^{\prime}w^{\prime})}=0$
(17)

induces two equations

$m_{1}m_{2}K_{3}e^{2i\alpha}+m_{2}m_{3}K_{1}e^{2i(\alpha+\beta+\delta)}+m_{3}m_{1}K_{2}e^{2i(\beta+\delta)}=0$
(18)
$m_{1}m_{2}L_{3}e^{2i\alpha}+m_{2}m_{3}L_{1}e^{2i(\alpha+\beta+\delta)}+m_{3}m_{1}L_{2}e^{2i(\beta+\delta)}=0$
(19)

where

$K_{i}=(V_{pj}V_{qj}V_{rk}V_{sk}-V_{tj}V_{uj}V_{vk}V_{wk})+(j\leftrightarrow
k)$ (20)
$L_{i}=(V_{p^{\prime}j}V_{q^{\prime}j}V_{r^{\prime}k}V_{s^{\prime}k}-V_{t^{\prime}j}V_{u^{\prime}j}V_{v^{\prime}k}V_{w^{\prime}k})+(j\leftrightarrow
k)$ (21)

with $(i,j,k)$ a cyclic permutation of (1,2,3). After solving Eq.(18) and
(19), we arrive at

$\frac{m_{1}}{m_{2}}e^{-2i\alpha}=\frac{K_{3}L_{1}-K_{1}L_{3}}{K_{2}L_{3}-K_{3}L_{2}}$
(22)
$\frac{m_{1}}{m_{3}}e^{-2i\beta}=\frac{K_{2}L_{1}-K_{1}L_{2}}{K_{3}L_{2}-K_{2}L_{3}}e^{2i\delta}$
(23)

With the help of Eq.(22) and (23), we obtain the magnitudes of mass radios

$\rho=\Big{|}\frac{m_{1}}{m_{3}}e^{-2i\beta}\Big{|}$ (24)
$\sigma=\Big{|}\frac{m_{1}}{m_{2}}e^{-2i\alpha}\Big{|}$ (25)

as well as the two Majorana CP-violating phases

$\alpha=-\frac{1}{2}arg\Big{(}\frac{K_{3}L_{1}-K_{1}L_{3}}{K_{2}L_{3}-K_{3}L_{2}}\Big{)}$
(26)
$\beta=-\frac{1}{2}arg\Big{(}\frac{K_{2}L_{1}-K_{1}L_{2}}{K_{3}L_{3}-K_{2}L_{3}}e^{2i\delta}\Big{)}$
(27)

The results of Eq. (24),(25), (26) and (27) imply that the two mass ratios
($\rho$ and $\sigma$) and two Majorana CP-violating phases ($\alpha$ and
$\beta$) are fully determined in terms of the real orthogonal matrix $O_{l}$
and $U$($\theta_{12},\theta_{23},\theta_{13}$ and $\delta$). The neutrino mass
ratios $\rho$ and $\sigma$ are related to the ratios of two neutrino mass-
squared ratios obtained from the solar and atmosphere oscillation experiments
as

$R_{\nu}\equiv\frac{\delta m^{2}}{\Delta
m^{2}}=\frac{2\rho^{2}(1-\sigma^{2})}{|2\sigma^{2}-\rho^{2}-\rho^{2}\sigma^{2}|}$
(28)

and to the three neutrino mass as

$m_{2}=\sqrt{\frac{\delta m^{2}}{1-\sigma^{2}}}\quad\quad m_{1}=\sigma
m_{2}\quad\quad m_{3}=\frac{m_{1}}{\rho}$ (29)

where $\delta m^{2}\equiv m_{2}^{2}-m_{1}^{2}$ and $\Delta m^{2}\equiv\mid
m_{3}^{2}-\frac{1}{2}(m_{1}^{2}+m_{2}^{2})\mid$. In the numerical analysis, we
use the latest global-fit neutrino oscillation experimental data, at 3$\sigma$
confidential level, which is listed in Ref.data (24)

$\begin{split}\sin^{2}\theta_{12}/10^{-1}=3.08^{+0.51}_{-0.49}\quad\sin^{2}\theta_{23}/10^{-1}=4.25^{+2.16}_{-0.68}\quad\sin^{2}\theta_{13}/10^{-2}=2.34^{+0.63}_{-0.57}\\\
\delta m^{2}/10^{-5}=7.54^{+0.64}_{-0.55}eV^{2}\quad\quad\bigtriangleup
m^{2}/10^{-3}=2.44^{+0.22}_{-0.22}eV^{2}\end{split}$ (30)

for normal hierarchy (NH) and

$\begin{split}\sin^{2}\theta_{12}/10^{-1}=3.08^{+0.51}_{-0.49}\quad\sin^{2}\theta_{23}/10^{-1}=4.25^{+2.22}_{-0.74}\quad\sin^{2}\theta_{13}/10^{-2}=2.34^{+0.61}_{-0.61}\\\
\delta m^{2}/10^{-5}=7.54^{+0.64}_{-0.55}eV^{2}\quad\quad\bigtriangleup
m^{2}/10^{-3}=2.40^{+0.21}_{-0.23}eV^{2}\end{split}$ (31)

for inverted hierarchy(IH). There is no constraint on the Dirac CP-violating
phase $\delta$ at $3\sigma$ level, however, the recent global fit tends to
give $\delta\approx 1.40\pi$. In neutrino oscillation experiments, CP
violation effect is usually reflected by the Jarlskog rephasing invariant
quantityJas (23) defined as

$J_{CP}=s_{12}s_{23}s_{13}c_{12}c_{23}c_{13}^{2}\sin\delta$ (32)

The Majorana nature of neutrino can be determined if any signal of
neutrinoless double decay is observed, implying the violation of leptonic
number violation. The decay ratio is related to the effective of neutrino
$m_{ee}$, which is written as

$m_{ee}=|m_{1}c_{12}^{2}c_{13}^{2}+m_{2}s_{12}^{2}c_{13}^{2}e^{2i\alpha}+m_{3}s_{13}^{2}e^{2i\beta}|$
(33)

Although a $3\sigma$ result of $m_{ee}=(0.11-0.56)$ eV is reported by the
Heidelberg-Moscow CollaborationHM (25), this result is criticized in Ref NND2
(26) and shall be checked by the forthcoming experiment. It is believed that
that the next generation $0\nu\beta\beta$ experiments, with the sensitivity of
$m_{ee}$ being up to 0.01 eVNDD (27), will open the window to not only the
absolute neutrino mass scale but also the Majorana-type CP violation. Besides
the $0\nu\beta\beta$ experiments, a more severe constraint was set from the
recent cosmology observation. Recently, an upper bound on the sum of neutrino
mass $\sum m_{i}<0.23$ eV is reported by Plank CollaborationPlanck (28)
combined with the WMAP, high-resolution CMB and BAO experiments.

##  III Phenomenological implications of parallel cofactor zero textures

###  III.1 Class II

In this section, we study the phenomenological implications of class II. The
factorisable formation of inverse charged leptonic matrix ($M_{l}^{-1})^{r}$
are parameterized as

$\begin{split}(M_{l}^{-1})^{r}_{II}=\left(\begin{array}[]{ccc}0&a&c\\\
a&b&0\\\ c&0&d\end{array}\right)\end{split}$ (34)

and can be diagonalized by an orthogonal matrix $O_{l}$

$O_{l}^{T}(M_{l}^{-1})^{r}_{II}O_{l}=diag(m_{e}^{-1},-m_{\mu}^{-1},m_{\tau}^{-1})$
(35)

where the coefficients $a,c,d$ are real and positive; The $m_{e},m_{\mu}$ and
$m_{\tau}$ denote the masse eigenvalues of charged leptons for three
generations. The minus sign in (35) has been introduced to facilitate the
analytical calculation and has no physical meaning since the charged lepton is
Dirac fermions. Using the invariant Tr$(M_{l}^{-1})^{r}$,
Det$(M_{l}^{-1})^{r}$ and Tr$(M_{l}^{-1})^{r^{2}}$ the nozero elements of
$(M_{l}^{-1})^{r}$ can be expressed in terms of three mass eigenvalues
$m_{e},m_{\mu}$, $m_{\tau}$ and $d$

$a=\sqrt{-\frac{(m_{e}^{-1}-m_{\mu}^{-1}-d)(m_{e}^{-1}+m_{\tau}^{-1}-d)(-m_{\mu}^{-1}+m_{\tau}^{-1}-d)}{m_{e}^{-1}-m_{\tau}^{-1}+m_{\tau}^{-1-2d}}}$
(36) $b=m_{e}^{-1}-m_{\mu}^{-1}+m_{\tau}^{-1}-d$ (37)
$c=\sqrt{\frac{(d-m_{e}^{-1})(d+m_{\mu}^{-1})(d-m_{\tau}^{-1})}{m_{e}^{-1}-m_{\tau}^{-1}+m_{\tau}^{-1}}-2d}$
(38)

where the parameter $d$ is allowed in the range of $0<d<m_{\tau}^{-1}$ and
$m_{e}^{-1}-m_{\tau}^{-1}<d<m_{e}^{-1}$. Then the $O_{l}$ can be easily
constructed as

$\begin{split}O_{l}=\left(\begin{array}[]{ccc}\frac{(b-m_{e}^{-1})(d-m_{e}^{-1})}{N_{1}}&\frac{(b+m_{\mu}^{-1})(d+m_{\mu}^{-1})}{N_{2}}&\frac{(b-m_{\tau}^{-1})(d-m_{\tau}^{-1})}{N_{3}}\\\
-\frac{a(d-m_{e}^{-1})}{N_{1}}&-\frac{a(d+m_{\mu}^{-1})}{N_{2}}&-\frac{a(d-m_{\tau}^{-1})}{N_{3}}\\\
-\frac{c(b-m_{e}^{-1})}{N_{1}}&-\frac{c(b+m_{\mu}^{-1})}{N_{2}}&-\frac{c(b-m_{\tau}^{-1})}{N_{3}}\end{array}\right)\end{split}$
(39)

where the $a,b$ and $c$ in(39) is given in (36), (37) and (38); The $N_{1}$,
$N_{2}$ and $N_{3}$ are the normalized coefficients given by

$N_{1}^{2}=(b-m_{e}^{-1})^{2}(d-m_{e}^{-1})^{2}+a^{2}(d-m_{e}^{-1})^{2}+c^{2}(b-m_{\tau}^{-1})^{2}$
(40)
$N_{2}^{2}=(b+m_{\mu}^{-1})^{2}(d+m_{\mu}^{-1})^{2}+a^{2}(d+m_{\mu}^{-1})^{2}+c^{2}(b+m_{\mu}^{-1})^{2}$
(41)
$N_{3}^{2}=(b-m_{\tau}^{-1})^{2}(d-m_{\tau}^{-1})^{2}+a^{2}(d-m_{\tau}^{-1})^{2}+c^{2}(b-m_{\tau}^{-1})^{2}$
(42)

Substitute the $O_{l}$ we obtained into (39) to(24), (25), (26), (27) and
(28), the ratio of mass squared difference can be expressed via eight
parameters. i.e three mixing angles $(\theta_{12},\theta_{23},\theta_{13})$,
one Dirac CP-violating phase $\delta$, three charged lepton mass
$(m_{e},m_{\mu},m_{\tau})$ and a parameter $d$. Here we choose the three
charged leptonic masses at the electroweak scale($\mu\simeq M_{Z}$) i.ezzx2
(29)

$m_{e}=0.486570154MeV\quad\quad m_{\mu}=102.7181377MeV\quad\quad
m_{\tau}=1746.17MeV$ (43)

In the numerical analysis, We randomly vary the three mixing angles
$(\theta_{12},\theta_{23},\theta_{13})$ in their $3\sigma$ range and parameter
$d$ in its proper range. Up to now, no bound was set on Dirac CP-violating
phase $\delta$ at 3 $\sigma$ level, so we vary it randomly in the range of
$[0,2\pi]$. Using Eq. (28), the mass-squared difference ratio $R_{\nu}$ is
determined. Then the input parameters is empirically acceptable when the
$R_{\nu}$ falls inside the the $3\sigma$ range of experimental data, otherwise
they are excluded. Finally, we get the value of neutrino mass and Majorana CP-
violating phase $\alpha$ and $\beta$ though Eq.(24), (25). Since we have
already obtained the absolute neutrino mass $m_{1,2,3}$, the further
constraint from cosmology should be considered. In this work, we set the upper
bound on the sum of neutrino mass $\Sigma m_{i}$ less than 0.23 eV.

We present the numerical results of class II in Fig.1 for the NH and in Fig.2
for the IH. One can see from the figures that different mass spectra exhibit
different correlations between physical variables. For the NH case, the Dirac
CP-violating phase $\delta$ is highly restricted in the range of
$60^{\circ}\sim 70^{\circ}$ and leads to the Jarlskog rephasing invariant
$|J_{CP}|>0.02$ which is promising to be detected in the future long baseline
neutrino oscillation experiments. On the other hand, there exists a bound of
$\theta_{23}>48^{\circ}$. Although accepted at 3$\sigma$ level, this result is
phenomenologically ruled out at $2\sigma$ level since recent experiments tend
to give $\theta_{23}<\pi/4$. We obtain the bound on the lightest neutrino mass
$M_{1}$, $0.025eV<m_{1}<0.075eV$ and the effective Majorana neutrino mass
$m_{ee}$ $0.04eV<m_{ee}<0.10eV$ which reaches the accuracy of future
neutrinoless double beta decay $(0\nu\beta\beta)$ experiments. The correlation
between $\alpha$ and $\beta$ is also illustrated that the small range is
allowed at $3\sigma$ level. For the IH case, the three mixing angles
$\theta_{12}$, $\theta_{23}$, and $\theta_{13}$ are fully covered the
$3\sigma$ range while the constrained Dirac CP-violated phase $\delta$ lies in
the range of $70^{\circ}\sim 290^{\circ}$, leading to the
$|J_{CP}|\sim(0)-(0.04)$. Interestingly, $m_{ee}$ and the lightest neutrino
mass $m_{3}$ exhibit a strong dependence on $\delta$. Such correlations are
essential for the model selection and could be tested by experiments. There
also exists a bound of $0.005eV<m_{ee}<0.095eV$ which could be in principle
tested by future $0\nu\beta\beta$ experiments. The Majorana phase $\alpha$ is
covered in the whole range of $-90^{\circ}\sim 90^{\circ}$ while $\beta$ is
constrained in the range of $-25^{\circ}\sim 25^{\circ}$.

Figure 1: The correlation plots for class II(NH).  Figure 2: The correlation
plots for class II(IH).  Figure 3: The correlation plots for class III(NH).

###  III.2 Class III

Let’s consider another class of textures which is phenomenologically
interesting. In the factorisable case, the real matrix $(M_{l}^{-1})^{r}$ is
parameterized as

$\begin{split}(M_{l}^{-1})^{r}_{III}=\left(\begin{array}[]{ccc}0&a&b\\\
a&0&c\\\ b&c&d\end{array}\right)\end{split}$ (44)

where $a,b,c$ and $d$ are real number. Without loss of generality, the
parameter $b,c$ are set to be positive.The matrix can be diagonalized by the
orthogonal matrix $O_{l}$

$O_{l}^{T}(M_{l}^{-1})^{r}_{III}O_{l}=diag(m_{e}^{-1},-m_{\mu}^{-1},m_{\tau}^{-1})$
(45)

Different from class II, we choose $a$ as the free parameter since the trace
of $(M_{l}^{-1})^{r}$ has already fixed $d$ to be

$d=m_{e}^{-1}-m_{\mu}^{-1}+m_{\tau}^{-1}$ (46)

Using the invariant Det$(M_{l}^{-1})^{r}$ and Tr$(M_{l}^{-1})^{r^{2}}$, the
parameters $b,c$ can be expressed by three charged leptonic mass
eigenvalues$(m_{e},m_{\mu},m_{\tau})$ and $a$

$(b\pm
c)^{2}=-(-m_{e}^{-1}m_{\mu}^{-1}+m_{e}^{-1}m_{\tau}^{-1}-m_{\mu}^{-1}m_{\tau}^{-1})-a^{2}\pm\frac{a^{2}(m_{e}^{-1}-m_{\mu}^{-1}+m_{\tau}^{-1})-m_{e}^{-1}m_{\mu}^{-1}m_{\tau}^{-1}}{a}$
(47)

With the help of Eq. (46) and Eq. (47), one can construct the diagonalized
matrix $O_{l}$ to be

$\begin{split}(M_{l}^{-1})^{r}_{III}=\left(\begin{array}[]{ccc}\frac{O(11)}{N_{1}}&\frac{O(12)}{N_{2}}&\frac{O(13)}{N_{3}}\\\
\frac{O(21)}{N_{1}}&\frac{O(22)}{N_{2}}&\frac{O(23)}{N_{3}}\\\
\frac{O(31)}{N_{1}}&\frac{O(32)}{N_{2}}&\frac{O(33)}{N_{3}}\end{array}\right)\end{split}$
(48)

where

$\begin{split}O(11)=&am_{e}(bm_{e}+ca^{-1})+bm_{e}(m_{e}^{-1}a^{-1}-m_{e}a)\\\
O(12)=&-am_{\mu}(-bm_{\mu}+ca^{-1})-bm_{\mu}(-m_{\mu}^{-1}a^{-1}+m_{\mu}a)\\\
O(13)=&am_{\tau}(bm_{\tau}+ca^{-1})+bm_{\tau}(m_{\tau}^{-1}a^{-1}-m_{\tau}a)\\\
&O(21)=bm_{e}+ca^{-1}\\\ &O(22)=-bm_{\mu}+ca^{-1}\\\
&O(23)=bm_{\tau}+ca^{-1}\\\ &O(31)=m_{e}^{-1}a^{-1}-m_{e}a\\\
&O(32)=-m_{\mu}^{-1}a^{-1}+m_{\mu}a\\\ &O(33)=m_{\tau}^{-1}a^{-1}-m_{\tau}a\\\
\end{split}$ (49)

and the normalized coefficients is given by

$\begin{split}N_{1}^{2}=O(11)^{2}+O(21)^{2}+O(31)^{2}\\\
N_{2}^{2}=O(12)^{2}+O(22)^{2}+O(32)^{2}\\\
N_{3}^{2}=O(13)^{2}+O(23)^{2}+O(33)^{2}\end{split}$ (50)

From the condition that $b,c$ are real and positive, we have the free
parameter $a$ allowed in the range of

$-\Big{(}\frac{m_{e}^{-1}m_{\mu}^{-1}m_{\tau}}{m_{e}^{-1}-m_{\mu}^{-1}+m_{\tau}^{-1}}\Big{)}^{\frac{1}{2}}<a<0$
(51)

or

$\Big{(}\frac{m_{e}^{-1}m_{\mu}^{-1}m_{\tau}}{m_{e}^{-1}-m_{\mu}^{-1}+m_{\tau}^{-1}}\Big{)}^{\frac{1}{2}}<a<(m_{e}^{-1}m_{\mu}^{-1}+m_{\mu}m_{\tau}-m_{e}^{-1}m_{\tau}^{-1})^{\frac{1}{2}}$
(52)

Now we repeat the previous analysis. The class III with inverted hierarchy are
now found to be unacceptable by current experimental data. We present the
allowed region for class III with normal mass hierarchy in Fig.3. It is
observed that no bound is set on $\theta_{12}$ and $\theta_{13}$. However the
Dirac CP-violating phase $\delta$ is restricted in two regions. We denote them
respectively as R1:
$0^{\circ}<\delta<60^{\circ}(300^{\circ}<\delta<360^{\circ})$ and R2:
$90^{\circ}<\delta<150^{\circ}(210^{\circ}<\delta<270^{\circ})$. Each shows
different predictions. In R1, $\theta_{23}$ varies in its $3\sigma$ range. We
obtain a highly suppressed $m_{ee}\simeq 0$eV which is beyond the accuracy of
future $0\nu\beta\beta$ experiments and implies the underlying cancelation of
three neutrino masses in $m_{ee}$. There also exists the lower bound on the
lightest neutrino mass $m_{1}>0.02eV$. On the other hand, in R2 we find
$\theta_{23}<45^{\circ}$ which is supported by $2\sigma$ experimental
constraint. We also obtain $|J_{CP}|>0.03$. Moveover, the bound on $m_{ee}$ is
founded in the range of $0.075$eV $\sim 0.2$eV and can be potentially detected
by future experiment.

Figure 4: The one-loop diagram for generating radiated charged lepton masses

##  IV Cofactor zeros in charged lepton matrices

One reminds the type-I seesaw mechanism as $M_{\nu}=-M_{D}M_{R}M_{D}^{T}$.
Then the cofactor zeros of $M_{\nu}$ are attributed to the texture zeros in
$M_{D}$ and $M_{R}$. Generally, this can be easily realized by Abelian $Z_{n}$
flavor symmetryzn (30, 11). Can the cofactor zeros in $M_{l}$ arise using the
same way? At the tree level, it is obviously impossible. At the loop level,
the answer is yes! Here we adopt the model proposed by Mama (21), consisting
of the SM extended by adding three Dirac singlet neutral fermion
$N_{k}(k=1,2,3)$, a doublet scalar $(\eta^{+},\eta^{0})$ and a charged singlet
$\sigma^{+}$. In Ma’s model, the particles transforms under the proper
$U(1)_{D}$ gauge symmetry and $A_{4}$ flavor symmetry. Here, we choose the
$Z_{2}^{(A)}$ instead of $A_{4}$ flavor symmetry under which $N_{k}$,
$(\eta^{+},\eta^{0})$ and $\sigma^{+}$ are odd. To forbid the tree level Dirac
lepton mass, another $Z_{2}^{(B)}$ symmetry is imposed such that $l_{R}$ and
$\sigma^{+}$ are odd while others are even. Actually, the flavor symmetry we
propose is the same as the one in Refma1 (31) where the Dirac neutrino mass is
generated at one-loop level. The allowed Yukawa interactions are
$y_{ij}\overline{N}_{iR}(l_{jL}\eta^{+}-\nu_{jL}\eta^{0})$ and
$h_{ij}\overline{l}_{iR}N_{jL}\sigma^{-}$. The $Z_{2}^{(B)}$ is allowed to be
softly broken by the trilinear term
$\mu(\eta^{+}\phi^{0}-\eta^{0}\phi^{+})\sigma^{-}$ with the SM vacuum
expectation $v=\langle\phi^{0}\rangle$. The one-loop charged lepton mass is
thus generated as shown in Fig. 4, the result being

$(M_{l})_{ij}=\frac{\sin
2\theta}{32\pi^{2}}\sum\limits_{k}y_{ik}M_{k}\Big{[}\frac{m_{1}^{2}}{m_{1}^{2}-M_{k}^{2}}\ln\Big{(}\frac{m_{1}^{2}}{M_{k}^{2}}\Big{)}-\frac{m_{2}^{2}}{m_{2}^{2}-M_{k}^{2}}\ln\Big{(}\frac{m_{2}^{2}}{M_{k}^{2}}\Big{)}\Big{]}h_{kj}^{{\dagger}}$
(53)

The $m_{1,2}$ and $\theta$ denote the eigenvalues and the mixing angle of mass
squared texture

$\begin{split}\left(\begin{array}[]{cc}m_{\sigma}^{2}&\mu v\\\ \mu
v&m_{\eta}^{2}\end{array}\right)\end{split}$ (54)

with

$m_{1,2}^{2}=\frac{1}{2}\big{[}m_{\eta}^{2}+m_{\sigma}^{2}\mp\sqrt{(m_{\eta}^{2}-m_{\sigma}^{2})^{2}+4\mu^{2}v^{2}}\big{]}$
(55)

and

$\sin
2\theta=\frac{2\mu^{2}v^{2}}{\sqrt{(m_{\eta}^{2}-m_{\sigma}^{2})^{2}+4\mu^{2}v^{2}}}$
(56)

For $M_{k}\gg m_{1,2}$, (53) is simplified as

$(M_{l})_{ij}\simeq\frac{\sin
2\theta}{32\pi^{2}}m_{1}^{2}\sum\limits_{k}F\Big{(}\frac{m_{1}^{2}}{m_{2}^{2}},\frac{M_{k}^{2}}{m_{1}^{2}}\Big{)}y_{ik}\frac{1}{M_{k}}h^{{\dagger}}_{kj}$
(57)

with

$F(x,y)\equiv x\ln\Big{(}\frac{y}{x}\Big{)}+\ln y$ (58)

Following the same strategy of Ref.gu (32), the
$F\Big{(}\frac{m_{1}^{2}}{m_{2}^{2}},\frac{M_{k}^{2}}{m_{1}^{2}}\Big{)}$ is
treated as a constant at leading order if three $M_{k}$ are assumed to be
nearly degenerated. Then we get

$M_{l}\sim m_{1}y(M_{N})_{diag}^{-1}h^{{\dagger}}$ (59)

On the other hand, if we assume $m_{\eta}\simeq m_{\sigma}\simeq M_{k}$ and
note $\mu v\ll M_{k}^{2}$ , then

$(M_{l})_{ij}\simeq\frac{\mu
v}{16\pi^{2}}\sum\limits_{k}y_{ik}\frac{1}{M_{k}}h^{{\dagger}}_{kj}\sim\mu
vy(M_{N})_{diag}^{-1}h^{{\dagger}}$ (60)

The expression also appears in ma3 (33) where the Majorana neutrino mass is
generated at one-loop level. From (59) and (60), the charged leptons acquire
the radiated masses via the seesaw-like mechanism and masses of heavy Dirac
neutral particles $N_{k}$ play the role of seesaw scale.

Consider now the weak basis where the mass matrix of $M_{N}$ is not diagonal.
It is obvious that, working in the context of the seesaw-like mechanism with a
diagonal Dirac matrices $y$ and $h$, the vanishing cofactors in the charged
lepton mass matrix are equivalent to texture zeros in the heavy Dirac fermion
mass matrix $M_{N}$. As having done in neutrino sector, the texture zeros in
$y$, $h$, and $M_{N}$ are easily achieved by introducing extra $Z_{n}$ flavor
symmetries. Form eq. (60), it is clear that the seesaw-like scale $M_{N}$ is
reduced to TeV by the smallness of factor $\mu v$/$16\pi^{2}$ originated from
softly broken $Z^{(B)}_{2}$.

##  V Conclusion and discussion

In this work, we have studied the parallel structures with cofactor zeros in
lepton mass matrices. These matrices can not obtained from arbitrary Hermitian
texture by making WB transformations. Using the permutation transformation,
the 15 possible textures are grouped into 4 classes where the matrices in each
class lead to the same physical implications. Among the 4 classes, one of them
is not compatible with experimental results and another is equivalent to the
texture zero structures explored extensively in previous literature. We focus
on the other two classes (class II and class III). Using the new results from
the neutrino oscillation and cosmology experiments, a systematic and
phenomenological analysis are proposed for each class and mass hierarchy. We
have demonstrated that some predictions for the atmosphere mixing angle
$\theta_{23}$, the Dirac CP-violating phase $\delta$ and the Majorana
effective neutrino mass $m_{ee}$ are rather interesting and deserve to be
explored in the future experiments. We also demonstrate how the cofactor zeros
arise in a seesaw-like model where charged lepton mass are generated at one-
loop level. We expect that a cooperation between phenomenological study and
the flavor symmetry study will finally help us real the structure of leptonic
texture.

###### Acknowledgements.

The author would like to thank Z. Z. Xing for the useful discussion during
this work and J. Zhang for his kind help in plotting Fig. 4.

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