∎

11institutetext: L. C. B. da Silva 22institutetext: Department of Physics of
Complex Systems, Weizmann Institute of Science, Rehovot 7610001, Israel.
22email<EMAIL_ADDRESS>33institutetext: T. Bar 44institutetext:
Department of Physics of Complex Systems, Weizmann Institute of Science,
Rehovot 7610001, Israel. 55institutetext: E. Efrati 66institutetext:
Department of Physics of Complex Systems, Weizmann Institute of Science,
Rehovot 7610001, Israel. 66email<EMAIL_ADDRESS>
# Compatible director fields in $\mathbb{R}^{3}$

Luiz C. B. da Silva  Tal Bar  Efi Efrati

###### Abstract

The geometry and interactions between the constituents of a liquid crystal,
which are responsible for inducing the partial order in the fluid, may locally
favor an attempted phase that could not be realized in $\mathbb{R}^{3}$. While
states that are incompatible with the geometry of $\mathbb{R}^{3}$ were
identified more than 50 years ago, the collection of compatible states
remained poorly understood and not well characterized. Recently, the
compatibility conditions for three-dimensional director fields were derived
using the method of moving frames. These compatibility conditions take the
form of six differential relations in five scalar fields locally
characterizing the director field. In this work, we rederive these equations
using a more transparent approach employing vector calculus. We then use these
equations to characterize a wide collection of compatible phases.

###### Keywords:

Geometric frustration Incompatibility Liquid crystal Frobenius

##  1 Introduction

Liquid crystalline textures are often described by extrinsic quantities. For
example, a distorted nematic phase will be described using the polar and
azimuthal angles describing the orientation of the unit director field
relative to some fixed Cartesian frame. In contrast, the short range
interactions that give rise to the liquid crystalline phase and its resulting
texture prescribe the relative positioning and orientation of the constituents
locally. To properly describe these locally preferred relative positioning and
orientation of the constituents one must resort to an intrinsic description,
i.e., a description by quantities available to an observer residing within the
material manifold (and oblivious of the lab frame of reference). Examples of
such local intrinsic quantities include the bend, splay, and twist scalars
that appear in the Frank free energy density GP95 . As these intrinsic
quantities are obtained from derivatives of the local director orientation,
the values they assume are not independent of each other. The relations
restricting the values of these intrinsic fields are called their
compatibility conditions.

Bent-core liquid crystals locally favor a phase of constant positive bend and
vanishing splay and twist NE18 ; S18 . Such a phase was shown to be
incompatible with the geometry of $\mathbb{R}^{3}$ Meyer1976 . In place of the
incompatible locally favored phase, bent-core liquid crystals assume the
“closest” compatible phase, where the distance is measured using the system
free energy selingerAnnRev21 . Such “close-by” compatible phases include the
twist-bend (heliconical) phase C+14  and the splay-bend phase CK19 , but also
include many other less uniform (and less known) states. Identifying these
states requires a comprehensive characterization of the collection of
compatible states. Moreover, a clear characterization of the collection of
compatible states in terms of local variables would advance our ability to
solve inverse design problems where one seeks the local distortion fields that
give rise to a specific configuration GAE19 .

For two dimensions, it was shown that the compatibility condition consists of
a single first-order partial differential equation in the splay and bend of
the director field. It was also shown that when the bend and splay fields are
indeed compatible they suffice to uniquely define a director field, up to
rigid motions NE18 . This result relied on the ability to construct a two-
dimensional orthogonal coordinate system whose parametric curves are
everywhere locally tangent to the director and to the director normal. This
construction greatly simplifies the geometric treatment, yet it can not be
generalized to three dimensions. Consequently, for three-dimensional director
fields, we are required to formulate the problem in terms of the spatial
variation of a local orthogonal tetrad, an approach made formal through
Cartan’s method of moving frames dSE21 ; PA21 . While this approach proved
very potent and concise, it is less accessible compared to other approaches
(such as, for example, vector calculus). In what follows, we repeat the
derivation of the compatibility conditions carried out in dSE21 ; PA21  using
vector calculus (as carried out in V19  for the special case of uniform
phases). We then explore possible solutions to the general systems of
equations in selected cases exhibiting some simplifying symmetries.

The remaining of this manuscript is divided as follows. In Section 2, we
present the fundamental equations defining the local representation of the
director gradients, and discuss the relation of this approach to Cartan’s
moving frame method. In Section 3, we derive the corresponding compatibility
conditions, which then provide necessary and sufficient conditions for the
existence of a director field with a prescribed set of deformation modes. In
Section 4, we find specific solutions to these equations under the simplifying
assumptions of additional symmetries. In other words, we provide several
examples of how to use the compatibility conditions to find compatible
deformation modes. Finally, in the concluding remarks section 5, we present
alternative ways of interpreting the compatibility conditions.

##  2 Local representation of the director gradient

Consider a liquid-crystalline phase in a domain $U\subset\mathbb{R}^{3}$
described by a unit vector field $\mathbf{n}$ with Cartesian components
$n_{j}$. We shall refer to $\hat{\mathbf{n}}$ as the director field. Let
$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ be two orthogonal unit vectors that
span the space normal to $\hat{\mathbf{n}}$, and $p_{j}$ and $q_{j}$ be their
corresponding Cartesian components.111As a concrete example, if
$\hat{\mathbf{n}}$ does not coincide with a coordinate direction, say the
$x$-direction $\hat{x}$ in the domain, we can set
$\hat{\mathbf{p}}=\hat{\mathbf{n}}\times\hat{x}/\|\hat{\mathbf{n}}\times\hat{x}\|$
and $\hat{\mathbf{q}}=\hat{\mathbf{n}}\times\hat{\mathbf{p}}$. However, it may
often prove useful to choose $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ with
some extra properties which make them more appropriate to the context.
Following Machon and Alexander MA16  and Selinger S18 , we denote (sum on
repeated indices)

$J^{n}_{ij}\equiv\partial_{i}n_{j}=-n_{i}(b_{p}p_{j}+b_{q}q_{j})+\frac{s}{2}(\delta_{ij}-n_{i}n_{j})+\frac{t}{2}\epsilon_{ijk}n_{k}+\Delta_{ij},$
(1)

where $\partial_{i}=\frac{\partial}{\partial x^{i}}$ and $x^{i}$ is the
$i^{th}$ Cartesian coordinate ($i=\\{1,2,3\\}$) and the deformation modes
appearing in $J^{n}$ are the components of the bend vector
$\mathbf{b}=b_{p}\hat{\mathbf{p}}+b_{q}\hat{\mathbf{q}}=\hat{\mathbf{n}}\times\nabla\times\hat{\mathbf{n}}=-\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{n}}$
whose modulus gives the bend $b=\|\mathbf{b}\|$, the splay
$s=\nabla\cdot\hat{\mathbf{n}}$, the twist
$t=\hat{\mathbf{n}}\cdot\nabla\times\hat{\mathbf{n}}$, and the Cartesian
components of the so-called biaxial splay $\Delta_{ij}$. (See Ref. S18  for a
further discussion concerning the interpretation of the deformation modes.)

The biaxial splay $\Delta_{ij}$ corresponds to the traceless and symmetric
part of the gradient of $\hat{\mathbf{n}}$. Only recently has the biaxial
splay been studied as a local property of the constituents of a liquid
crystal, similarly to the local bend, splay and twist selingerPRE22 . When it
is expressed in the basis
$\mathcal{F}=\\{\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$, the
biaxial splay takes the form

$(\Delta_{ij})_{\mathcal{F}}=\left(\begin{array}[]{ccc}0&0&0\\\
0&\Delta_{1}&\Delta_{2}\\\ 0&\Delta_{2}&-\Delta_{1}\\\
\end{array}\right),\quad\text{i.e.,}\quad\Delta_{ij}=\Delta_{1}p_{i}p_{j}+\Delta_{2}(p_{i}q_{j}+q_{i}p_{j})-\Delta_{1}q_{i}q_{j}.$

The variation of the components of $\hat{\mathbf{n}}$ above is given in terms
of the components of $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$. Thus, to
obtain a complete system of equations we need to also provide
$J_{ij}^{p}=\partial_{i}p_{j}$ and $J_{ij}^{q}=\partial_{i}q_{j}$. The
restriction that $\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}$ are
triply orthogonal unit vector fields allows us to express these gradients
using only three additional scalar fields. We thus obtain a closed system of
three first order equations prescribing the spatial variation of the triad
$\hat{\mathbf{n}}$, $\hat{\mathbf{p}}$, and $\hat{\mathbf{q}}$:

$\displaystyle\partial_{i}n_{j}$ $\displaystyle=$
$\displaystyle\left[-n_{i}b_{p}+\left(\frac{s}{2}+\Delta_{1}\right)p_{i}-\left(\frac{t}{2}-\Delta_{2}\right)q_{i}\right]\,p_{j}$
(2)
$\displaystyle+\left[-n_{i}b_{q}+\left(\frac{t}{2}+\Delta_{2}\right)p_{i}+\left(\frac{s}{2}-\Delta_{1}\right)q_{i}\right]\,q_{j},$
$\displaystyle\partial_{i}p_{j}$ $\displaystyle=$
$\displaystyle\left[b_{p}n_{i}-\left(\frac{s}{2}+\Delta_{1}\right)p_{i}-\left(-\frac{t}{2}+\Delta_{2}\right)q_{i}\right]n_{j}+[\alpha
n_{i}+\beta p_{i}+\gamma q_{i}]q_{j},$ (3) $\displaystyle\partial_{i}q_{j}$
$\displaystyle=$
$\displaystyle\left[b_{q}n_{i}-\left(\frac{t}{2}+\Delta_{2}\right)p_{i}-\left(\frac{s}{2}-\Delta_{1}\right)q_{i}\right]n_{j}+[-\alpha
n_{i}-\beta p_{i}-\gamma q_{i}]p_{j}.$ (4)

These are the fundamental reconstruction equations for a director field in
$\mathbb{R}^{3}$, and form the basis for all the following calculations. We
may alternatively express $J^{n}$, $J^{p}$, and $J^{q}$ in matrix form through

$J^{n}=R^{T}M^{n}R,\,\quad J^{p}=R^{T}M^{p}R,\quad\mbox{and}\quad
J^{q}=R^{T}M^{q}R,\,\quad$ (5)

where

$R=\left(\begin{array}[]{ccc}n_{1}&n_{2}&n_{3}\\\ p_{1}&p_{2}&p_{3}\\\
q_{1}&q_{2}&q_{3}\\\ \end{array}\right),\quad
M^{n}=\left(\begin{array}[]{ccc}0&-b_{p}&-b_{q}\\\\[6.0pt]
0&\dfrac{s}{2}+\Delta_{1}&\dfrac{t}{2}+\Delta_{2}\\\\[6.0pt]
0&-\dfrac{t}{2}+\Delta_{2}&\dfrac{s}{2}-\Delta_{1}\\\ \end{array}\right),$
$M^{p}=\left(\begin{array}[]{ccc}b_{p}&0&\alpha\\\\[6.0pt]
-\dfrac{s}{2}-\Delta_{1}&0&\beta\\\\[6.0pt]
\dfrac{t}{2}-\Delta_{2}&0&\gamma\\\ \end{array}\right),\,\mbox{ and }\quad
M^{q}=\left(\begin{array}[]{ccc}b_{q}&-\alpha&0\\\\[6.0pt]
-\dfrac{t}{2}-\Delta_{2}&-\beta&0\\\\[6.0pt]
-\dfrac{s}{2}+\Delta_{1}&-\gamma&0\\\ \end{array}\right).$ (6)

######  Remark 1 (On the relation to Cartan’s method of moving frames)

For any given vectors $\mathbf{v}$ and $\mathbf{w}$, we have
$(\mathbf{v}\cdot\nabla\hat{\mathbf{n}})\cdot\mathbf{w}=v_{i}J_{ij}^{n}w_{j}$.
Comparison with the approach via the moving frame method dSE21  shows that
$(\mathbf{n}_{i}\cdot\nabla\mathbf{n}_{k})\cdot\mathbf{n}_{j}=\eta_{k}^{j}(\mathbf{n}_{i})$
and, therefore, the second and third columns of $M^{n}$ are respectively given
by
$(\,\eta_{1}^{2}(\hat{\mathbf{n}})\quad\eta_{1}^{2}(\hat{\mathbf{p}})\quad\eta_{1}^{2}(\hat{\mathbf{q}})\,)^{T}$
and
$(\,\eta_{1}^{3}(\hat{\mathbf{n}})\quad\eta_{1}^{3}(\hat{\mathbf{p}})\quad\eta_{1}^{3}(\hat{\mathbf{q}})\,)^{T}$.
Analogously, the first and third columns of $M^{p}$ are respectively given by
$(\,\eta_{2}^{1}(\hat{\mathbf{n}})\quad\eta_{2}^{1}(\hat{\mathbf{p}})\quad\eta_{2}^{1}(\hat{\mathbf{q}})\,)^{T}$
and
$(\,\eta_{2}^{3}(\hat{\mathbf{n}})\quad\eta_{2}^{3}(\hat{\mathbf{p}})\quad\eta_{2}^{3}(\hat{\mathbf{q}})\,)^{T}$,
while the first and second columns of $M^{q}$ are respectively given by
$(\,\eta_{3}^{1}(\hat{\mathbf{n}})\quad\eta_{3}^{1}(\hat{\mathbf{p}})\quad\eta_{3}^{1}(\hat{\mathbf{q}})\,)^{T}$
and
$(\,\eta_{3}^{2}(\hat{\mathbf{n}})\quad\eta_{3}^{2}(\hat{\mathbf{p}})\quad\eta_{3}^{2}(\hat{\mathbf{q}})\,)^{T}$.
(Note that $\eta_{i}^{j}=-\eta_{j}^{i}$, which implies that $M^{n}$ and
$M^{p}$, or $M^{n}$ and $M^{q}$, or $M^{p}$ and $M^{q}$, share one similar
column up to a sign.) Under this identification, the compatibility conditions
derived in the next section as the solvability condition for the set of first
order partial differential equations (2), (3) and (4), coincide with the
structure equations of the corresponding moving frame.

##  3 Compatibility equations

One of the main challenges in constructing the compatibility conditions, that
prescribe the necessary relations between the intrinsic local fields
describing a director in $\mathbb{R}^{3}$, is that it is a priori unknown how
many such fields are required to uniquely describe such a field. The
fundamental equations above, Eqs. (2), (3), and (4), make use of nine scalar
(and pseudoscalar) local intrinsic quantities, thus bounding the number of the
intrinsic descriptors required to uniquely define the texture at some
neighborhood. In the next section, we will show that the number of intrinsic
fields necessary to fully describe a director field in $\mathbb{R}^{3}$ may be
further reduced to only five. However, for simplicity and in order to properly
construct the compatibility conditions, we first assume that all the intrinsic
fields that appear in Eqs. (2), (3), and (4) are known, as are the components
of their derivatives along the vectors $\hat{\mathbf{n}}$, $\hat{\mathbf{p}}$,
and $\hat{\mathbf{q}}$. We then consider Eqs. (2), (3), and (4) as partial
differential equations (PDE’s) describing the spatial evolution of the
orthonormal triplet $\hat{\mathbf{n}}$, $\hat{\mathbf{p}}$, and
$\hat{\mathbf{q}}$. Given some initial value for the orthonormal triplet at a
point in a domain (as well as all the fields that appear in the PDE’s), we
could obtain the values the triplet obtains in its neighborhood by integrating

$\nabla\hat{\mathbf{n}}=J^{n},\quad\nabla\hat{\mathbf{p}}=J^{p},\quad\text{and}\quad\nabla\hat{\mathbf{q}}=J^{q}.$

Such a system of PDE’s is meaningful and could be solved only if its
solvability conditions are satisfied. In the present case, these necessary and
sufficient solvability conditions on a simply connected domain read (see
Theorem 10.9 of Apostol  or Sect. 8 of Chapter 2 of LeviCivita ):

$\nabla\times\nabla n_{j}=0,\quad\nabla\times\nabla
p_{j}=0,\quad\mbox{and}\quad\nabla\times\nabla q_{j}=0,\quad j\in\\{1,2,3\\}.$
(7)

Had all the components of the gradients $J^{n}$, $J^{p}$, and $J^{q}$ been
independent, these equations would have formed 27 differential equations in 27
unknown fields. However, the expressions above teach us that the gradients may
be fully expressed in terms of only 9 fields:
$b_{p},b_{q},s,t,\Delta_{1},\Delta_{2},\alpha,\beta,\gamma$. Moreover, because
$\hat{\mathbf{n}},\hat{\mathbf{p}}$, and $\hat{\mathbf{q}}$ are unit vectors,
in each of their curl equations three of the nine components vanish
identically. Due to the mutual orthogonality of these vectors, of the
remaining eighteen equations only nine are independent. Thus, the
compatibility conditions of the system consist of nine equations in nine
fields. We next derive these equations. To find the compatibility equations,
we define the commutation tensor $C^{n}_{k\ell}$:

$C^{n}_{k\ell}=\epsilon_{ijk}\partial_{i}\partial_{j}n_{\ell}=\epsilon_{ijk}\partial_{i}J_{j\ell}.$
(8)

Note that $\nabla\times\nabla
n_{\ell}=(C^{n}_{1\ell},C^{n}_{2\ell},C^{n}_{3\ell})$ must vanish in Euclidean
space. Analogously, we can define
$C_{k\ell}^{p}=\epsilon_{ijk}\partial_{i}J_{j\ell}^{p}$ and
$C_{k\ell}^{q}=\epsilon_{ijk}\partial_{i}J_{j\ell}^{q}$. The compatibility
conditions require all the components of the three commutation tensors to
vanish. However, contracting the obtained commutation tensors with the vectors
of the frame $\\{\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ yields
more concise and transparent equations of the form:
$C^{n}_{k\ell}\,n_{k}n_{\ell}=0$, $C^{n}_{k\ell}\,n_{k}p_{\ell}=0$, etc.

The contractions using $C^{n}_{k\ell}$ will provide six non-trivial equations,
Eqs. (16)–(21) at the end of this section, while the contractions using
$C^{p}_{k\ell}$ will provide three additional non-trivial equations, Eqs.
(22)–(24) at the end of this section. The resulting nine first order equations
in nine fields could be further simplified. We use three of the equations to
express $\alpha$, $\beta$, and $\gamma$, which describe the deformation modes
of $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$, in terms of
$\\{b_{p},b_{q},t,s,\Delta_{1},\Delta_{2}\\}$ and their gradients.
Substituting these expressions in the compatibility equations yields _six_
compatibility conditions in terms of the _six_ deformation modes of
$\hat{\mathbf{n}}$: $\\{b_{p},b_{q},t,s,\Delta_{1},\Delta_{2}\\}$, three
equations of first order and three of second order.

In the expression for $\nabla\hat{\mathbf{n}}$, Eq. (2), there are six
deformation modes. However, there exists a gauge freedom in the choice of
$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$, which implies we need only five
scalar fields. Indeed, we may choose a frame
$\\{\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ where:

  1. (i)

$\hat{\mathbf{p}}=\frac{1}{b}\mathbf{b}$, which implies $b_{q}=0$.
Geometrically, $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ have the same
direction as the principal normal and binormal vector fields of the integral
lines of the director $\hat{\mathbf{n}}$; or

  2. (ii)

$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ are the eigenvectors of the biaxial
splay, which implies $\Delta_{2}=0$.

Therefore, taking the gauge freedom into account we may say the compatibility
conditions consist of _six_ equations in _five_ deformation modes. The gauge
invariant physical deformation modes can be taken to be the splay, $s$, bend
(magnitude), $b$, twist, $t$, biaxial splay (magnitude),
$\Delta=\sqrt{(\Delta_{1})^{2}+(\Delta_{2})^{2}}$, and the relative angle
between the principal direction of the biaxial splay and the bend vector,
$\phi$, which satisfies
$\mathbf{b}\cdot\mathcal{D}\mathbf{b}=b^{2}\Delta\cos(2\phi)$, where
$\mathcal{D}$ denotes the biaxial splay acting as an operator on the plane
normal to $\hat{\mathbf{n}}$.

###### Remark 2

Sometimes, it may prove useful to employ other gauge choices for
$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$. For example $\hat{\mathbf{p}}$ and
$\hat{\mathbf{q}}$ may be chosen such that they minimize rotation along the
integral lines of the director. This choices implies that $\alpha=0$ B75 .
(The equations of motion of the frame
$\\{\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ along the integral
curves of the director are given by the first set of equations in (11).)

###  3.1 Obtaining the compatibility equations

From Eq. (2), the vector $\nabla\times\nabla
n_{\ell}=(C^{n}_{1\ell},C^{n}_{2\ell},C^{n}_{3\ell})$ takes the form

$\displaystyle\nabla\times\nabla n_{\ell}$ $\displaystyle=$
$\displaystyle{\nabla
p_{\ell}\times[-b_{p}\hat{\mathbf{n}}+(\frac{s}{2}+\Delta_{1})\hat{\mathbf{p}}+(-\frac{t}{2}+\Delta_{2})\hat{\mathbf{q}}]}$
(9) $\displaystyle+{\nabla
q_{\ell}\times[-b_{q}\hat{\mathbf{n}}+(\frac{t}{2}+\Delta_{2})\hat{\mathbf{p}}+(\frac{s}{2}-\Delta_{1})\hat{\mathbf{q}}]}$
$\displaystyle+p_{\ell}\nabla\times[-b_{p}\hat{\mathbf{n}}+(\frac{s}{2}+\Delta_{1})\hat{\mathbf{p}}+(-\frac{t}{2}+\Delta_{2})\hat{\mathbf{q}}]$
$\displaystyle+q_{\ell}\nabla\times[-b_{q}\hat{\mathbf{n}}+(\frac{t}{2}+\Delta_{2})\hat{\mathbf{p}}+(\frac{s}{2}-\Delta_{1})\hat{\mathbf{q}}].$

As it is easier to compute dot products than to compute cross products, we are
going to exploit the vector calculus identity

$\nabla\times(\mathbf{A}\times\mathbf{B})=(\nabla\cdot\mathbf{B})\mathbf{A}-(\nabla\cdot\mathbf{A})\mathbf{B}+\mathbf{B}\cdot\nabla\mathbf{A}-\mathbf{A}\cdot\nabla\mathbf{B}$

in order to write

$\displaystyle\nabla\times\hat{\mathbf{n}}$ $\displaystyle=$
$\displaystyle\nabla\times(\hat{\mathbf{p}}\times\hat{\mathbf{q}})=(\nabla\cdot\hat{\mathbf{q}})\hat{\mathbf{p}}-(\nabla\cdot\hat{\mathbf{p}})\hat{\mathbf{q}}+\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{p}}-\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{q}},$
$\displaystyle\nabla\times\hat{\mathbf{p}}$ $\displaystyle=$
$\displaystyle\nabla\times(\hat{\mathbf{q}}\times\hat{\mathbf{n}})=(\nabla\cdot\hat{\mathbf{n}})\hat{\mathbf{q}}-(\nabla\cdot\hat{\mathbf{q}})\hat{\mathbf{n}}+\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{q}}-\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{n}},$
$\displaystyle\nabla\times\hat{\mathbf{q}}$ $\displaystyle=$
$\displaystyle\nabla\times(\hat{\mathbf{n}}\times\hat{\mathbf{p}})=(\nabla\cdot\hat{\mathbf{p}})\hat{\mathbf{n}}-(\nabla\cdot\hat{\mathbf{n}})\hat{\mathbf{p}}+\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{n}}-\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{p}}.$

Now, from Eqs. (2), (3), and (4) we have

$\nabla\cdot\hat{\mathbf{n}}=s,\quad\nabla\cdot\hat{\mathbf{p}}=b_{p}+\gamma,\quad\nabla\cdot\hat{\mathbf{q}}=b_{q}-\beta,$
(10)
$\begin{array}[]{l}\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{n}}=-b_{p}\,\hat{\mathbf{p}}-b_{q}\,\hat{\mathbf{q}}\\\\[3.0pt]
\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{p}}=b_{p}\,\hat{\mathbf{n}}+\alpha\,\hat{\mathbf{q}}\\\\[3.0pt]
\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{q}}=b_{q}\,\hat{\mathbf{n}}-\alpha\,\hat{\mathbf{p}}\\\
\end{array},\quad\begin{array}[]{l}\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{p}}=\beta\,\hat{\mathbf{q}}-(\frac{s}{2}+\Delta_{1})\,\hat{\mathbf{n}}\\\\[3.0pt]
\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{q}}=-\beta\,\hat{\mathbf{p}}-(\frac{t}{2}+\Delta_{2})\,\hat{\mathbf{n}}\\\\[3.0pt]
\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{n}}=(\frac{s}{2}+\Delta_{1})\,\hat{\mathbf{p}}+(\frac{t}{2}+\Delta_{2})\,\hat{\mathbf{q}}\\\
\end{array},$ (11)

and

$\begin{array}[]{l}\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{q}}=-(\frac{s}{2}-\Delta_{1})\,\hat{\mathbf{n}}-\gamma\,\hat{\mathbf{p}}\\\\[3.0pt]
\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{n}}=(\frac{s}{2}-\Delta_{1})\,\hat{\mathbf{q}}-(\frac{t}{2}-\Delta_{2})\,\hat{\mathbf{p}}\\\\[3.0pt]
\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{p}}=\gamma\,\hat{\mathbf{q}}-(-\frac{t}{2}+\Delta_{2})\,\hat{\mathbf{n}}\\\
\end{array}.$ (12)

Thus, we can express the curl of each vector field in the frame as

$\displaystyle\nabla\times\hat{\mathbf{n}}$ $\displaystyle=$ $\displaystyle
t\,\hat{\mathbf{n}}+b_{q}\,\hat{\mathbf{p}}-b_{p}\,\hat{\mathbf{q}},$ (13)
$\displaystyle\nabla\times\hat{\mathbf{p}}$ $\displaystyle=$
$\displaystyle\beta\hat{\mathbf{n}}+[(\frac{t}{2}-\Delta_{2})-\alpha]\hat{\mathbf{p}}+(\frac{s}{2}+\Delta_{1})\hat{\mathbf{q}},$
(14) $\displaystyle\nabla\times\hat{\mathbf{q}}$ $\displaystyle=$
$\displaystyle\gamma\hat{\mathbf{n}}-(\frac{s}{2}-\Delta_{1})\,\hat{\mathbf{p}}+[(\frac{t}{2}+\Delta_{2})-\alpha]\,\hat{\mathbf{q}}.$
(15)

On the other hand, using the identity

$(\mathbf{A}\times\mathbf{B})\times\mathbf{C}=(\mathbf{A}\cdot\mathbf{C})\mathbf{B}-(\mathbf{B}\cdot\mathbf{C})\mathbf{A},$

we have (for any scalar function $f$)

$\displaystyle\hat{\mathbf{n}}\times\nabla f$ $\displaystyle=$
$\displaystyle(\hat{\mathbf{p}}\times\hat{\mathbf{q}})\times\nabla
f=(\hat{\mathbf{p}}\cdot\nabla f)\hat{\mathbf{q}}-(\hat{\mathbf{q}}\cdot\nabla
f)\hat{\mathbf{p}},$ $\displaystyle\hat{\mathbf{p}}\times\nabla f$
$\displaystyle=$
$\displaystyle(\hat{\mathbf{q}}\times\hat{\mathbf{n}})\times\nabla
f=(\hat{\mathbf{q}}\cdot\nabla f)\hat{\mathbf{n}}-(\hat{\mathbf{n}}\cdot\nabla
f)\hat{\mathbf{q}},$ $\displaystyle\hat{\mathbf{q}}\times\nabla f$
$\displaystyle=$
$\displaystyle(\hat{\mathbf{n}}\times\hat{\mathbf{p}})\times\nabla
f=(\hat{\mathbf{n}}\cdot\nabla f)\hat{\mathbf{p}}-(\hat{\mathbf{p}}\cdot\nabla
f)\hat{\mathbf{n}}.$

We may now substitute for $\hat{\mathbf{n}}\times\nabla p_{\ell}$,
$\hat{\mathbf{n}}\times\nabla q_{\ell}$, $\hat{\mathbf{p}}\times\nabla
p_{\ell}$, $\hat{\mathbf{p}}\times\nabla q_{\ell}$,
$\hat{\mathbf{q}}\times\nabla p_{\ell}$, $\hat{\mathbf{q}}\times\nabla
q_{\ell}$ and $\nabla\times\hat{\mathbf{n}}$, $\nabla\times\hat{\mathbf{p}}$,
$\nabla\times\hat{\mathbf{q}}$ in the commutation tensor $C^{n}_{k\ell}$.
Finally, we may compute the six contractions $C^{n}_{k\ell}\,q_{k}p_{\ell}$,
$C^{n}_{k\ell}\,p_{k}p_{\ell}$, $C^{n}_{k\ell}\,n_{k}p_{\ell}$,
$C^{n}_{k\ell}\,q_{k}q_{\ell}$, $C^{n}_{k\ell}\,p_{k}q_{\ell}$, and
$C^{n}_{k\ell}\,n_{k}q_{\ell}$ (which must all vanish in $\mathbb{R}^{3}$) in
order to obtain the following compatibility equations:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}+\Delta_{1})_{,n}-b_{p,p}-b_{p}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta_{1}-(\Delta)^{2}+2\alpha\Delta_{2}+\beta
b_{q},$ (16) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(-\frac{t}{2}+\Delta_{2})_{,n}-b_{p,q}-b_{p}b_{q}-s(-\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1}+\gamma
b_{q},$ (17) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(-\frac{t}{2}+\Delta_{2})_{,p}+(\frac{s}{2}+\Delta_{1})_{,q}+tb_{p}-2\beta\Delta_{1}-2\gamma\Delta_{2},$
(18) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{t}{2}+\Delta_{2})_{,n}-b_{q,p}-b_{p}b_{q}-s(\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1}-\beta
b_{p},$ (19) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}-\Delta_{1})_{,n}-b_{q,q}-b_{q}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}+s\Delta_{1}-(\Delta)^{2}-2\alpha\Delta_{2}-\gamma
b_{p},$ (20) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}-\Delta_{1})_{,p}+(\frac{t}{2}+\Delta_{2})_{,q}+tb_{q}-2\beta\Delta_{2}+2\gamma\Delta_{1}.$
(21)

Here, $f_{,n}$, $f_{,p}$, and $f_{,q}$ indicate the derivative of $f$ in the
direction of $\hat{\mathbf{n}}$, $\hat{\mathbf{p}}$, and $\hat{\mathbf{q}}$,
respectively.

The contractions $C^{n}_{k\ell}\,n_{k}n_{\ell}$,
$C^{n}_{k\ell}\,p_{k}n_{\ell}$, and $C^{n}_{k\ell}\,q_{k}n_{\ell}$ vanish
trivially and provide no further compatibility equations. It remains to
compute $C_{k\ell}^{p}=\epsilon_{ijk}\partial_{i}J_{j\ell}^{p}$ and
$C_{k\ell}^{q}=\epsilon_{ijk}\partial_{i}J_{j\ell}^{q}$. Proceeding similarly
as we did for $C^{n}_{k\ell}$, it turns out that the computation of
$C_{k\ell}^{p}$ results in three additional independent equations:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha_{,p}-\beta_{,n}-(b_{q}+\beta)(\frac{s}{2}+\Delta_{1})+(b_{p}-\gamma)(\frac{t}{2}+\Delta_{2})+\alpha(b_{p}+\gamma),$
(22) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha_{,q}-\gamma_{,n}+(b_{p}-\gamma){(\frac{s}{2}-\Delta_{1})}-(b_{q}+\beta)(-\frac{t}{2}+\Delta_{2})+\alpha(b_{q}-\beta),$
(23) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\beta_{,q}-\gamma_{,p}-\beta^{2}-\gamma^{2}-t\alpha-\frac{s^{2}}{4}-\frac{t^{2}}{4}+(\Delta)^{2}.$
(24)

The compatibility equations (16)–(24) thus constitute the necessary and
sufficient conditions for Eqs. (2), (3), and (4) to be meaningful and allow
path independent integration. They correspond to Eqs. (19) of Ref. dSE21  once
we identify $\alpha=\omega_{1}$, $\beta=\kappa_{2}^{1}$, and
$\gamma=-\kappa_{3}^{2}$.

Note that the variation of the director as expressed in
$\nabla\hat{\mathbf{n}}$ depends only on six deformation modes (or
alternatively, the five gauge invariant quantities). However, integration of
the equations for $\nabla\hat{\mathbf{n}}$ also requires knowledge of
$\alpha$, $\beta$, and $\gamma$, which describe the spatial variation of
$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$. It is therefore natural to ask
whether these variables are independent from the fields describing the
deformation modes of $\hat{\mathbf{n}}$, or could be eliminated from the
equations. As we next show, the latter is indeed the case; as the first six
compatibility conditions contain no derivatives of the fields $\alpha$,
$\beta$, and $\gamma$ we may use three of these equations to express them
using the six fields $\\{b_{p},b_{q},t,s,\Delta_{1},\Delta_{2}\\}$ and their
first derivatives. When substituted back to the equations, we obtain six
differential equations in the six unknown fields describing the deformations
of $\hat{\mathbf{n}}$; three equations of first order and three equations of
second order.

If $\Delta_{1}^{2}+\Delta_{2}^{2}\not=0$, then combining Eqs. (18) and (21)
(multiplied by $\Delta_{1}$ and $\Delta_{2}$, respectively) allows us to
express $\beta$ as

$\displaystyle\beta$ $\displaystyle=$
$\displaystyle\frac{\Delta_{1}t_{,p}-\Delta_{2}s_{,p}+\Delta_{1}s_{,q}+\Delta_{2}t_{,q}}{4\Delta^{2}}$
(25)
$\displaystyle-\frac{\Delta_{1}\Delta_{2,p}-\Delta_{2}\Delta_{1,p}}{2\Delta^{2}}+\frac{(\Delta^{2})_{,q}}{4\Delta^{2}}+t\,\frac{b_{p}\Delta_{1}+b_{q}\Delta_{2}}{2\Delta^{2}}.$

Now, a different linear combination of these two equations (essentially
multiplied by $\Delta_{2}$ and $-\Delta_{1}$, respectively) allows us to
express $\gamma$ as

$\displaystyle\gamma$ $\displaystyle=$
$\displaystyle\frac{\Delta_{2}t_{,p}+\Delta_{1}s_{,p}-\Delta_{1}t_{,q}+\Delta_{2}s_{,q}}{4\Delta^{2}}$
(26)
$\displaystyle+\frac{\Delta_{2}\Delta_{1,q}-\Delta_{1}\Delta_{2,q}}{2\Delta^{2}}-\frac{(\Delta^{2})_{,p}}{4\Delta^{2}}+t\,\frac{b_{p}\Delta_{2}-b_{q}\Delta_{1}}{2\Delta^{2}}.$

Finally, summing Eq. (16) multiplied by $\Delta_{2}$, Eq. (17) multiplied by
$-\Delta_{1}$, Eq. (19) multiplied by $-\Delta_{1}$, and Eq. (20) multiplied
by $-\Delta_{2}$, and using the above expressions for $\beta$ and $\gamma$
allow us to express $\alpha$ as

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\frac{\Delta_{2}b_{p,p}-\Delta_{2}b_{q,q}-\Delta_{1}b_{q,p}-\Delta_{1}b_{p,q}}{4\Delta^{2}}-\frac{b_{q}\Delta_{1,p}-b_{p}\Delta_{2,p}+b_{p}\Delta_{1,q}+b_{q}\Delta_{2,q}}{8\Delta^{2}}$
(27)
$\displaystyle+\frac{\Delta_{2}\Delta_{1,n}-\Delta_{1}\Delta_{2,n}}{2\Delta^{2}}-{\frac{b_{p}(s_{,q}+t_{,p})+b_{q}(t_{,q}-s_{,p})}{16\Delta^{2}}}$
$\displaystyle-\frac{tb^{2}}{8\Delta^{2}}+\frac{(b_{p}^{2}-b_{q}^{2})\Delta_{2}-2b_{p}b_{q}\Delta_{1}}{4\Delta^{2}}.$

On the other hand, if the biaxial splay vanishes,
$\Delta_{1}^{2}+\Delta_{2}^{2}=0$, and $b_{p}^{2}+b_{q}^{2}\not=0$, then we
may use Eq. (16) and Eq. (19) to express $\beta$ as

$\displaystyle\beta$ $\displaystyle=$
$\displaystyle\frac{b_{q}s_{,n}-b_{p}t_{,n}}{2b^{2}}{+\frac{b_{q}b_{p,p}-b_{p}b_{q,p}}{b^{2}}}+\frac{(s^{2}-t^{2})b_{q}-2stb_{p}}{4b^{2}}$
(28)

and we may use Eqs. (17) and (20) to express $\gamma$ as

$\displaystyle\gamma$ $\displaystyle=$
$\displaystyle-\frac{b_{p}s_{,n}+b_{q}t_{,n}}{2b^{2}}+\frac{b_{q}b_{p,q}-b_{p}b_{q,q}}{b^{2}}-\frac{(s^{2}-t^{2})b_{p}+2stb_{q}}{4b^{2}}.$
(29)

Finally, when the biaxial splay vanishes, we can write $\alpha$ as a function
of the deformation modes by substituting for $\beta$ and $\gamma$ in Eq. (24)
or, alternatively, we can get rid of $\alpha$ by choosing $\hat{\mathbf{p}}$
and $\hat{\mathbf{q}}$ such that $\alpha=0$ (see Remark 2).

##  4 Selected compatible phases: specific solutions

The set of equations (16)-(24) forms the compatibility conditions for the
fundamental equations (2), (3), and (4). Their satisfaction assures the
fundamental equations could be integrated to obtain a solution for the
director $\hat{\mathbf{n}}$. Finding compatible phases thus amounts to finding
solutions to Eqs. (16)-(24). There are several approaches for this task; the
first and most systematic approach is to treat the compatibility conditions as
PDEs for the unknown local intrinsic fields. The second approach identifies
the compatibility conditions as algebraic equations in the components of the
triad $\hat{\mathbf{n}}$, $\hat{\mathbf{p}}$, and $\hat{\mathbf{q}}$ and
solves for them directly. However, each of these approaches requires the
calculation of additional compatibility conditions (this time treating the
local intrinsic fields as the unknowns, see concluding remarks for further
discussion). We thus do not follow these approaches but rather follow a third,
simpler approach. We seek solutions to simplified versions of the
compatibility conditions obtained by imposing restrictions on the values of
the deformation modes, such as imposing some of them are constant. Given
enough constraints, the compatibility conditions simplify and could be clearly
interpreted allowing us to characterize the resulting phases. In this method,
the existence of the sought solutions is obtained by explicit construction.
Conversely, in the absence of an explicit construction, we cannot assure a
phase with the sought simplified properties exists. We begin by presenting the
recently studied case where all the deformation modes are constant. We then
discuss the case where only one deformation mode is not constant, and finally,
we investigate director fields with only two non-vanishing deformation modes.

###  4.1 Directors with uniform distortion fields

Among all director fields, the simplest ones are the so-called uniform
distortion directors, i.e., those director fields with all deformation modes
constant in space. In this case, it is known that V19  (see also dSE21 ):

  1. (a)

if $\Delta=0$, then $b=s=t=0$, i.e., there is only the trivial solution;

  2. (b)

if $\Delta\not=0$, then $s=0$, $t=\pm 2\Delta$, and
$\phi=\frac{(2k+1)\pi}{4}$, $k\in\\{0,1,2,3\\}$, where $\phi$ is again the
angle formed by the bend vector and the principal direction of the biaxial
splay. In addition, the bend vector $\mathbf{b}$ bisects the principal
directions of the biaxial splay: if $t=2\Delta$, then $k=0$ or $k=2$ and if
$t=-2\Delta$ then $k=1,3$.

Geometrically, the two families of uniform phases correspond to a foliation of
space by parallel helices (one family of solution is the mirror image of the
other), see Fig. 1, Left and Center. The geometry of the helices, namely,
their values of the curvature $\kappa$ and the torsion $\tau$, depends on the
two free parameters, biaxial splay $\Delta$ and bend $b$:

$\kappa=b\quad\mbox{ and
}\quad\tau=\alpha=\mp\frac{b^{2}}{4\Delta}+\frac{(b_{p}^{2}-b_{q}^{2})\Delta_{2}-2b_{p}b_{q}\Delta_{1}}{4\Delta^{2}}.$

Remember that a helix of radius $r$ and pitch $p$,
$\vartheta\mapsto(r\cos\vartheta,r\sin\vartheta,p\vartheta)$, has curvature
$\kappa=\frac{r}{r^{2}+p^{2}}$ and torsion $\tau=\frac{p}{r^{2}+p^{2}}$. Thus,
a helix of curvature $\kappa$ and torsion $\tau$ has radius
$r=\frac{\kappa}{\kappa^{2}+\tau^{2}}$ and pitch
$p=\frac{\tau}{\kappa^{2}+\tau^{2}}$.

Figure 1: Uniform distortion and splay-only phases. (Left) The heliconical
uniform distortion phase in which the director rotates at a constant rate
along a fixed direction (not necessarily perpendicular to the director), while
displaying translational invariance in the plane perpendicular to the screw
axis V19 . The director is thus tangent to a helical conic surface from which
it inherits its name. Integral curves of the director (solid black line on the
right) provide a foliation of space by parallel helices of curvature
$\kappa=b$ and torsion $\tau=-b^{2}/{2\Delta}$. (Right) A phase with all
deformation modes equal to zero except for the splay. A splay-only phase is
often termed a hedgehog phase. The integral lines of the hedgehog director can
be described as the unit normal field of a family of concentric spheres. (In
the figures, the colors of the director vectors are just a guide to the eyes.)

###  4.2 Directors with a single non-uniform distortion field

We now proceed to investigate whether it is possible to have a director field
in Euclidean space where all but one deformation mode is constant. The
theorems below show that, except for a few special cases, this is not
generally possible, thus revealing an interesting rigidity of uniform
distortion fields.

######  Theorem 4.1 (Rigidity of non-uniform directors. I)

Let $\hat{\mathbf{n}}$ be a director field in Euclidean space with deformation
modes $b_{p},b_{q},s,t,\Delta_{1}=\Delta$, and $\Delta_{2}=0$, i.e.,
$\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ are the eigendirections of the
biaxial splay. If $b_{p},b_{q},s$, and $t$ are constant, then the modulus of
the biaxial splay $\Delta$ must be also constant.

######  Theorem 4.2 (Rigidity of non-uniform directors. II)

Let $\hat{\mathbf{n}}$ be a director field in Euclidean space with deformation
modes $b_{p}=b,b_{q}=0,s,t,\Delta_{1}$, and $\Delta_{2}$, i.e.,
$\hat{\mathbf{p}}$ is the normalized bend vector. If
$\Delta_{1},\Delta_{2},s$, and $t$ are constant, but $t\not=-2\Delta_{1}$ and
$t\not=0$, then the bend $b$ must be also constant.

######  Theorem 4.3 (Rigidity of non-uniform directors. III)

Let $\hat{\mathbf{n}}$ be a director field in Euclidean space with deformation
modes $b_{p},b_{q},t,s$ and vanishing biaxial splay $\Delta_{1}=\Delta_{2}=0$.

  1. (a)

If $b_{p},b_{q}$, and $t$ are constant and $b_{p}^{2}+b_{q}^{2}$ do not
vanish, then the splay $s$ must be also constant.

  2. (b)

If $b_{p},b_{q}$, and $s$ are constant, then the twist $t$ must be also
constant.

The proofs of Theorems 4.1, 4.2, and 4.3 can be found in the appendices.

####  4.2.1 Non-uniform splay phases

Let us understand the exceptions to Theorems 4.1, 4.2, and 4.3. The first
possibility for a simple non-uniform deformation mode director field
corresponds to the case where $b=0,t=0,\Delta=0$, but $s$ is non-constant. An
example of such deformation mode is the hedgehog director field, see Fig. 1,
Right. Geometrically, the director $\hat{\mathbf{n}}$ corresponds to the
normals of a family of concentric spheres. The value of the splay at a point
located at a sphere of radius $R$ is $s=\pm 2/R$. It turns out that this is
the only non-trivial solution. To see that, we first need to take into account
the following geometric facts:

  1. (a)

if $t=0$, then there exists a foliation of (a possibly finite portion of)
space by surfaces such that the unit normal of each leaf is given by the
director $\hat{\mathbf{n}}$ Aminov ;

  2. (b)

if $t=0$, then the shape operator $A$ of each leaf of the foliation in (a) is
given by (see Subsect. 4.3)

$A=-\mathrm{d}\hat{\mathbf{n}}=-\frac{s}{2}I-\left(\begin{array}[]{cc}\Delta_{1}&\Delta_{2}\\\
\Delta_{2}&-\Delta_{1}\\\ \end{array}\right),$

where $I$ is the identity operator acting on the plane orthogonal to
$\hat{\mathbf{n}}$;

  3. (c)

if $t=0$ and $b=0$, then the foliation in (a) is given by a family of parallel
surfaces Aminov . In this case, we can parametrize the region of
$\mathbb{R}^{3}$ where $\hat{\mathbf{n}}$ is defined as

$\mathbf{R}(u,v,w)=\mathbf{r}(u,v)+w\hat{\mathbf{n}}(u,v).$

Each surface $w=\mbox{const.}$ is parallel to a prescribed surface
$\Sigma^{2}$ parametrized by $(u,v)\mapsto\mathbf{r}(u,v)$.

Now, assume there exists a phase with $t=0$, $b=0$, and $\Delta=0$. As
$\hat{\mathbf{n}}$ has $t=0$ and $b=0$, then $\hat{\mathbf{n}}$ is the field
of unit normals of a family of parallel surfaces. As, in addition, $\Delta=0$,
then the shape operator of each leave is just $A=-\frac{s}{2}I$, which implies
the leaves are totally umbilical surfaces (planes or spheres). Therefore,
$\hat{\mathbf{n}}$ is either the trivial solution (foliation by parallel
planes) or a hedgehog (foliation by concentric spheres, see Fig. 1, Right.).

####  4.2.2 Non-uniform bend phases

Figure 2: Example of a non-uniform bend phase with $t=0$ and $s,\Delta$
constant. In the figures, the arrows representing the director are colored by
the value of the bend $b$. (Left) The non-uniform bend and uniform splay $2d$
phase obtained by considering $\hat{\mathbf{n}}$ orthogonal to the one-
parameter family of circles of radius $\rho$ constant given by
$\lambda\mapsto\mathbf{r}_{\lambda}(\theta)=(\lambda+\rho\cos\theta,\rho\sin\theta)$
(dashed lines). The integral lines of $\hat{\mathbf{n}}$ (full lines) have a
bend equal to $b=\frac{1}{\rho}\tan\theta$ while the splay $s=1/\rho$ is
constant. (Right) The $3d$ phase obtained by translating the $2d$ phase on the
left in the direction orthogonal to the plane of the $2d$ phase. The director
$\hat{\mathbf{n}}$ is orthogonal to a family of cylinders of the same radius
and parallel axes but with distinct centers. Since $\hat{\mathbf{n}}$ is
normal to the foliation by the cylinders
$\rho\mapsto\mathbf{R}_{\lambda}(\theta,z)=(\lambda+\rho\cos\theta,\rho\sin\theta,z)$,
the twist vanishes $t=0$. In addition, as the leaves have uniform geometry,
their shape operator $A_{n}=-d\hat{\mathbf{n}}$ is constant, see Eq. (57).
Therefore, the biaxial splay and splay must be constant. (The cylinders have
distinct colors to ease the visualization.)

Let us now investigate the exceptions to Theorem 4.2. We may have a phase with
non-uniform bend provided that $s,\Delta_{1}$, and $\Delta_{2}$ are constant,
$\Delta_{1}^{2}+\Delta_{2}^{2}\not=0$, and $t=0$ or $t=-2\Delta_{2}$.

a) Non-uniform bend phases with zero twist: First assume that $t=0$ and
$\Delta_{1},\Delta_{2}$, and $s$ are constant, with
$\Delta_{1}^{2}+\Delta_{2}^{2}\not=0$, and $b_{q}=0$, but $b_{p}=b$ non-
uniform. We are going to show that we can construct $\hat{\mathbf{n}}$ as the
unit normal of a foliation of a finite domain of space by cylinders of the
same radius and with parallel axes, as illustrated in Fig. 2.

Here, the compatibility equations (16)–(21) become

$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,p}-b^{2}-\frac{s^{2}}{4}-s\Delta_{1}-(\Delta)^{2}+2\alpha\Delta_{2},$ (30)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,q}-s\Delta_{2}-2\alpha\Delta_{1},$ (31) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\beta\Delta_{1}-2\gamma\Delta_{2},$ (32) $\displaystyle 0$
$\displaystyle=$ $\displaystyle-s\Delta_{2}-2\alpha\Delta_{1}-\beta b,$ (33)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{s^{2}}{4}+s\Delta_{1}-(\Delta)^{2}-2\alpha\Delta_{2}-\gamma
b,$ (34) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\beta\Delta_{2}+2\gamma\Delta_{1}.$ (35)

From Eqs. (32) and (35), we conclude that

$t,\nabla\Delta_{1},\nabla\Delta_{2},\nabla
s,b_{q}=0\Rightarrow\beta,\gamma=0.$ (36)

Now, adding Eq. (33) multiplied by $\Delta_{1}$ and Eq. (34) multiplied by
$\Delta_{2}$, we can obtain for $\alpha$

$t,\nabla\Delta_{1},\nabla\Delta_{2},\nabla
s,b_{q}=0\Rightarrow\alpha=-\frac{\Delta_{2}}{2\Delta^{2}}\left(\frac{s^{2}}{2}+\Delta^{2}\right),$
(37)

which is constant. Let us exploit the remaining compatibility equations, Eq.
(23), which gives

$0=b(\frac{s}{2}-\Delta_{1})\Rightarrow s=2\Delta_{1},$ (38)

where we are assuming $b\not=0$. (Otherwise, the phase would be uniform.) On
the other hand, Eq. (24) gives

$0=-\frac{s^{2}}{4}+\Delta_{1}^{2}+\Delta_{2}^{2}\Rightarrow\Delta_{2}=0.$
(39)

In particular, $\alpha=0$. As mentioned in the previous subsection, the
condition $t=0$ implies that the director corresponds to the field of unit
normals to leaves of a foliation by surfaces whose shape operators are given
by $A=-\frac{s}{2}I-\mathcal{D}$, where $I$ is the identity operator and
$\mathcal{D}$ is the traceless and symmetric operator acting on the plane
orthogonal to $\hat{\mathbf{n}}$ such that $\mathcal{D}_{11}=\Delta_{1}$ and
$\mathcal{D}_{12}=\Delta_{2}$. From what we deduced so far, the director
$\hat{\mathbf{n}}$ corresponds to the field of unit normals of surfaces with
shape operator

$A=-\mathrm{d}\hat{\mathbf{n}}=-\frac{s}{2}I-\left(\begin{array}[]{cc}\Delta_{1}&\Delta_{2}\\\
\Delta_{2}&-\Delta_{1}\\\ \end{array}\right)=\left(\begin{array}[]{cc}-s&0\\\
0&0\\\ \end{array}\right).$ (40)

Therefore, the leaves of the foliation are cylinders of radius $R=1/s$ and
axis $\hat{\mathbf{q}}$. Substituting $\alpha,\beta,\gamma=0$,
$s=2\Delta_{1}$, and $\Delta_{2}=0$ in Eq. (4) implies
$\nabla\hat{\mathbf{q}}=0$ and, therefore, it follows that the cylinders’ axes
are all parallel.

Finally, substituting $s=2\Delta_{1}$ in Eq. (30) gives the following
evolution equation for $b$ along $\hat{\mathbf{p}}$

$-b_{,p}=b^{2}+4\Delta^{2}=b^{2}+s^{2}.$ (41)

Noting that the system displays translational invariance along the direction
of $\hat{\mathbf{q}}$, Eq. (41) can be interpreted as the compatibility
condition for a $2d$ director with uniform splay, and non-uniform bend NE18 ;
NEcomment .

b) Non-uniform bend phases with non-zero twist: Assume that
$\Delta_{1},\Delta_{2},s$ are constant, $t=-2\Delta_{2}\not=0$, and $b_{q}=0$
(so, $b_{p}=b$). We are going to show that $s=2\Delta_{1}$, which implies that
$\hat{\mathbf{n}}$ is a cholesteric with pitch axis
$\hat{\mathbf{P}}=\hat{\mathbf{q}}$.

We say that $\hat{\mathbf{n}}$ is a cholesteric with pitch axis
$\hat{\mathbf{P}}$ if $\hat{\mathbf{P}}\cdot\nabla\hat{\mathbf{n}}$ has no
component along $\hat{\mathbf{P}}$ B+14 . Following the notation in B+14 , we
may define a new parameter $q$ from
$\hat{\mathbf{P}}\cdot\nabla\hat{\mathbf{n}}$. The Eqs. (11) and (12) imply
the director $\hat{\mathbf{n}}$ is (i) a cholesteric with pitch axis
$\hat{\mathbf{P}}=\hat{\mathbf{p}}$ if and only if $s=-2\Delta_{1}$ and (ii) a
cholesteric with pitch axis $\hat{\mathbf{P}}=\hat{\mathbf{q}}$ if and only if
$s=2\Delta_{1}$. In general, a director field may have two, one, or no
cholesteric pitch axes depending on whether the cholestericity
$\mathcal{R}=t^{2}-2\sigma$ is positive, zero, or negative, respectively.
Here, $\sigma$ denotes the saddle-splay B+14  which can be expressed in terms
of the local deformation modes by S18 :

$\sigma=\nabla\cdot[s\,\hat{\mathbf{n}}-\hat{\mathbf{n}}\cdot\nabla\hat{\mathbf{n}}]=\frac{1}{2}\left(s^{2}+t^{2}\right)-2\Delta^{2}.$
(42)

Using that our non-uniform bend phase satisfies $t=-2\Delta_{2}$ and
$s=2\Delta_{1}$ implies that $\sigma=0$ and, therefore, the cholestericity is
positive. Thus, there must exist a second cholesteric pitch. Indeed, in
addition to $\hat{\mathbf{P}}_{1}=\hat{\mathbf{q}}$, the director
$\hat{\mathbf{n}}$ also has
$\hat{\mathbf{P}}_{2}=\frac{1}{\Delta^{2}}(\Delta_{2}\hat{\mathbf{p}}-\Delta_{1}\hat{\mathbf{q}})$
as a cholesteric pitch axis. Note that neither $\hat{\mathbf{P}}_{1}$ nor
$\hat{\mathbf{P}}_{2}$ provides a foliation (since $\hat{\mathbf{P}}_{i}$
would be the field of unit normals, the director $\hat{\mathbf{n}}$ would be
tangent to the leaves): both pitch axes have twist
$\hat{\mathbf{P}}_{1}\cdot\nabla\times\hat{\mathbf{P}}_{1}$ and
$\hat{\mathbf{P}}_{2}\cdot\nabla\times\hat{\mathbf{P}}_{2}$ given by
$-2\Delta_{2}(1+\frac{b^{2}}{4\Delta^{2}})$, where we used that
$\alpha=\frac{\Delta_{2}}{2\Delta^{2}}b^{2}$ as we are going to show in Eq.
(52) below.

In this case, the second pitch axis, $\hat{\mathbf{P}}_{2}$, has an
interesting property, its integral curves are straight lines. Indeed, this is
the same as showing that the $\hat{\mathbf{P}}_{2}$-integral curves have no
curvature, i.e., $\hat{\mathbf{P}}_{2}\cdot\nabla\hat{\mathbf{P}}_{2}=0$.
Using Eqs. (11) and (12) with $t=-2\Delta_{2}$ and $s=2\Delta_{1}$ constant,
we have

$\displaystyle\hat{\mathbf{P}}_{2}\cdot\nabla\hat{\mathbf{P}}_{2}$
$\displaystyle=$
$\displaystyle\frac{1}{\Delta^{2}}\left[\Delta_{2}^{2}\,\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{p}}+\Delta_{1}^{2}\,\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{q}}-\Delta_{1}\Delta_{2}(\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{q}}+\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{p}})\right]$
(43) $\displaystyle=$
$\displaystyle\frac{1}{\Delta^{2}}[\Delta_{2}^{2}(\beta\hat{\mathbf{q}}-2\Delta_{1}\hat{\mathbf{n}})-\gamma\Delta_{1}^{2}\,\hat{\mathbf{p}}-\Delta_{1}\Delta_{2}(-\beta\hat{\mathbf{p}}+\gamma\hat{\mathbf{q}}-2\Delta_{2}\hat{\mathbf{n}})]$
$\displaystyle=$
$\displaystyle\frac{\beta\Delta_{2}-\gamma\Delta_{1}}{\Delta^{2}}(\Delta_{1}\,\hat{\mathbf{p}}+\Delta_{2}\,\hat{\mathbf{q}}).$

Now, using Eq. (51), to be proved below, we obtain
$\beta\Delta_{2}-\gamma\Delta_{1}=0$ and, therefore,
$\hat{\mathbf{P}}_{2}\cdot\nabla\hat{\mathbf{P}}_{2}=0$ as stated.

If we choose
$\\{\hat{\mathbf{n}},\hat{\mathbf{P}}_{2},\hat{\mathbf{Q}}_{2}\equiv\hat{\mathbf{n}}\times\hat{\mathbf{P}}_{2}\\}$
as a new frame, then the corresponding deformation modes
$\\{\tilde{b}_{p},\tilde{b}_{q},\tilde{t}=t,\tilde{s}=s,\tilde{\Delta}_{1},\tilde{\Delta}_{2},\tilde{\alpha},\tilde{\beta},\tilde{\gamma}\\}$
are

$\tilde{b}_{p}=\frac{b\Delta_{2}}{\Delta},\tilde{b}_{q}=\frac{b\Delta_{1}}{\Delta},\tilde{\Delta}_{1}=-\Delta_{1},\tilde{\Delta}_{2}=\Delta_{2},\tilde{\alpha}=\alpha,\tilde{\beta}=0,\tilde{\gamma}=-\frac{b\Delta_{2}}{\Delta}.$
(44)

Remember that $t$ and $s$ do not depend on the choice of frame.

Now, let us show that $s=2\Delta_{1}$, which implies that $\hat{\mathbf{n}}$
constitutes a cholesteric phase. Assuming that $\Delta_{1},\Delta_{2},s$ are
constant, $t=-2\Delta_{2}\not=0$, and $b_{q}=0$, the compatibility equations
(16)–(21) become

$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,p}-b^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta_{1}-(\Delta)^{2}+2\alpha\Delta_{2},$
(45) $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,q}-s(-\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1},$ (46) $\displaystyle 0$
$\displaystyle=$ $\displaystyle tb-2\beta\Delta_{1}-2\gamma\Delta_{2},$ (47)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-s(\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1}-\beta b,$ (48)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{s^{2}}{4}+\frac{t^{2}}{4}+s\Delta_{1}-(\Delta)^{2}-2\alpha\Delta_{2}-\gamma
b,$ (49) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\beta\Delta_{2}+2\gamma\Delta_{1}.$ (50)

Since $\Delta_{1}^{2}+\Delta_{2}^{2}\not=0$, the coefficients $\beta$ and
$\gamma$ are given by

$\beta=\frac{tb\Delta_{1}}{2\Delta^{2}}=-\frac{b\Delta_{1}\Delta_{2}}{\Delta^{2}}\,\mbox{
and
}\,\gamma=\frac{tb\Delta_{2}}{2\Delta^{2}}=-\frac{b\Delta_{2}^{2}}{\Delta^{2}}.$
(51)

Then, Eq. (48) implies

$\alpha=-\frac{\beta
b}{2\Delta_{1}}=-\frac{t}{4\Delta^{2}}\,b^{2}=\frac{\Delta_{2}}{2\Delta^{2}}\,b^{2}.$
(52)

Note that Eqs. (22) and (23) become

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha_{,p}-\beta_{,n}-\beta(\frac{s}{2}+\Delta_{1})+\alpha(b+\gamma),$
(53) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha_{,q}-\gamma_{,n}+(b-\gamma)(\frac{s}{2}-\Delta_{1})-2\beta\Delta_{2}-\alpha\beta.$
(54)

We can manipulate the compatibility equations and deduce that, see Eqs. (119)
and (120) in the appendix, we can write

$\displaystyle\Delta_{2}b_{,p}-\Delta_{1}b_{,q}$ $\displaystyle=$
$\displaystyle
2\Delta_{2}(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2})-b^{2}(\frac{t}{2}+\Delta_{2})-st\Delta_{1}$
(55) $\displaystyle=$
$\displaystyle-2\Delta_{2}(\frac{s^{2}}{4}+\Delta_{1}^{2})+2s\Delta_{1}\Delta_{2}=-2\Delta_{2}(\frac{s}{2}-\Delta_{1})^{2},$

where in the second equality we used that $t=-2\Delta_{2}$.

Now, adding Eq. (53) multiplied by $-\Delta_{2}$ and Eq. (54) multiplied by
$\Delta_{1}$, and using the relation above for
$\Delta_{2}b_{,p}-\Delta_{1}b_{,q}$, finally give

$0=\frac{4b\Delta_{2}^{2}}{\Delta_{2}}\left(\frac{s}{2}-\Delta_{1}\right)^{2}\Rightarrow
s=2\Delta_{1}.$ (56)

###  4.3 Vanishing twist phases

It is known that the necessary and sufficient condition for the director
$\hat{\mathbf{n}}$ be orthogonal to the leaves of a foliation of (a possibly
finite domain of) space by surfaces is the vanishing of the twist
$t=\hat{\mathbf{n}}\cdot\nabla\times\hat{\mathbf{n}}=0$ Aminov . (As is often
the case with if-and-only-if theorems, one of the implications is easy to
prove. Indeed, if $\hat{\mathbf{n}}=\nabla f/\|\nabla f\|$, i.e.,
$\hat{\mathbf{n}}$ is a unit normal for the level sets of $f$, then performing
the explicit calculations shows that
$\hat{\mathbf{n}}\cdot\nabla\times\hat{\mathbf{n}}=0$.)

Let us investigate how to relate the geometry of the leaves to the deformation
modes of the director. For a director $\hat{\mathbf{n}}$ at a point $p$, the
tangent plane of the leaf passing through $p$ is given by
$T_{p}=\\{\mathbf{v}:\mathbf{v}\cdot\hat{\mathbf{n}}=0\\}$. We may then
associate with $\hat{\mathbf{n}}$ the family of shape operators that measure
the variation of $\hat{\mathbf{n}}$ along $T_{p}$

$\begin{array}[]{rcccl}A_{n}&:&T_{p}&\to&T_{p}\\\
&&\mathbf{v}&\mapsto&-(\mathbf{v}\cdot\nabla)\hat{\mathbf{n}}.\\\
\end{array}\quad$ (57)

Note that $A_{n}(\mathbf{v})$ does belong to $T_{p}$:
$\hat{\mathbf{n}}\cdot\hat{\mathbf{n}}=1\Rightarrow(\mathbf{v}\cdot\nabla\hat{\mathbf{n}})\cdot\hat{\mathbf{n}}=0$.

Taking into account Eqs. (11) and (12), the matrix coefficients of $A_{n}$ in
the basis $\\{\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ is precisely

$A_{n}=\left(\begin{array}[]{cr}-(\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{n}})\cdot\hat{\mathbf{p}}&-(\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{n}})\cdot\hat{\mathbf{p}}\\\
-(\hat{\mathbf{p}}\cdot\nabla\hat{\mathbf{n}})\cdot\hat{\mathbf{q}}&-(\hat{\mathbf{q}}\cdot\nabla\hat{\mathbf{n}})\cdot\hat{\mathbf{q}}\end{array}\right)=\left(\begin{array}[]{cr}-(\frac{s}{2}+\Delta_{1})&-(-\frac{t}{2}+\Delta_{2})\\\
-(\frac{t}{2}+\Delta_{2})&-(\frac{s}{2}-\Delta_{1})\end{array}\right).$ (58)

Note that the family of operators $A_{n}$ are indeed the shape operators of
the leaves of a foliation if, and only if, each $A_{n}$ is symmetric:
$A_{n}=A_{n}^{T}\Leftrightarrow t=0$.

Note that $s=-\mbox{Tr}(A_{n})$. In addition, using the relation between the
saddle-splay, twist, splay, and biaxial splay given by Eq. (42), we can write

$\sigma=2\det(A_{n}).$ (59)

Thus, if we assume $t=0$, then the Mean $H$ and Gaussian $K$ curvatures of the
family of shape operators $A_{n}$ associated with the director
$\hat{\mathbf{n}}$ satisfy $s=-2H$ and $\sigma=2K$. Moreover, if
$\\{\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ are the eigenvectors of the biaxial
splay, then they also correspond to the principal directions of each surface
in the foliation orthogonal to $\hat{\mathbf{n}}$.

####  4.3.1 Zero bend and zero twist phases

From Eq. (11), we see that $\nabla\times\hat{\mathbf{n}}=0$ if, and only if,
$t=0$ and $b=0$. Then, we can locally write $\hat{\mathbf{n}}=\nabla f$ for
some scalar function $f$. Since the gradient of $f$ is of unit length, the
length of an integral line of $\hat{\mathbf{n}}$ connecting a point of the
level set $\\{f=c_{1}\\}$ to a point of the level set $\\{f=c_{2}\\}$ is
always $|c_{2}-c_{2}|$. It follows that the distance between the level sets
$\\{f=c_{1}\\}$ and $\\{f=c_{2}\\}$ is precisely $|c_{2}-c_{2}|$. An example
of such a phase is given by the splay-only hedgehog phase, see Fig. 1, Right.

We can state that $\hat{\mathbf{n}}$ corresponds to the unit normals of a
foliation by equidistant surfaces if, and only if, $t=0$ and $b=0$. In this
case, we can parametrize the region where $\hat{\mathbf{n}}$ is defined as

$(u,v,w)\mapsto\mathbf{R}(u,v,w)=\mathbf{r}(v,w)+u\,\hat{\mathbf{n}}(v,w),$

where $\mathbf{r}$ parametrizes some chosen level set of $f$. As we will prove
in Subsect. 4.6, if $b=0$, the values of the deformations modes are then
determined by the values they assume on the points of $\mathbf{r}(v,w)$. Using
Eqs. (84) and (86) from Subsect. 4.6, the mean and Gaussian curvatures of a
surface of the foliation at a distance $u$ from $\mathbf{r}(v,w)$ are given by

$H=\frac{H_{0}-K_{0}u}{1-2H_{0}u+K_{0}u^{2}}\quad\mbox{and}\quad
K=\frac{K_{0}}{1-2H_{0}u+K_{0}u^{2}},$ (60)

where $H_{0}$ and $K_{0}$ are the mean and Gaussian curvatures of
$\mathbf{r}$. (See also DoCarmo76 , Exercise 11, Chapter 3.)

###  4.4 Zero splay and zero twist phases

Figure 3: Example of a zero twist and zero splay phase. In the figures, the
arrows representing the director are colored by the value of the bend $b$.
(Left and Right) Distinct views of the phase obtained from the unit normal
vector field of a foliation of a portion of space by (vertically translated)
helicoids, i.e., the levels sets of $f(x,y,z)=z-p\arctan(y/x)$. The integral
lines of the director field $\hat{\mathbf{n}}=\nabla f/\|\nabla f\|$ are
helices and they can be parametrized as
$c(u)=(\rho_{0}\cos(-\omega\,u+\theta_{0}),\rho_{0}\sin(-\omega\,u+\theta_{0}),(\omega\rho_{0}^{2}/p)\,u+p\,\theta_{0})$,
where $\omega=(p/\rho_{0})(\rho_{0}^{2}+p^{2})^{-\frac{1}{2}}$ and the initial
condition $c(0)$ is taken as a point of $f(x,y,z)=0$. The bend is the
curvature of the helices: $b={p^{2}}/{\rho_{0}(p^{2}+\rho_{0}^{2})}$.
Consequently, the phase has non-uniform bend. In addition, note that the
closer to the common axis of the helicoids, i.e., the smaller the value of
$\rho_{0}$, the higher the bend. (In the figures, the helicoids have distinct
colors to ease the visualization.)

As the splay relates to the mean curvature of the leaves orthogonal to the
director, $s=-2H$, we see that if $s=0$, then $\hat{\mathbf{n}}$ is the unit
normal of a foliation of a portion of space by minimal surfaces.

As an example, we may consider the level sets of the scalar function
$f(x,y,z)=z-p\arctan(y/x)$. Here, each leaf of the foliation obtained from $f$
is a (vertically translated) helicoid of pitch $p$, which is a well-known
minimal surface, see Fig. 3. The director is given by $\hat{\mathbf{n}}=\nabla
f/\|\nabla f\|$, where $\nabla f=(py/(x^{2}+y^{2}),-px/(x^{2}+y^{2}),1)$. Note
that the director $\hat{\mathbf{n}}$ and the gradient $\nabla f$ have the same
integral lines up to reparametrization. A curve
$c(\theta)=(x(\theta),y(\theta),z(\theta))$ is an integral line of $\nabla f$
if

$\dot{x}(\theta)=\frac{p\,y(\theta)}{x(\theta)^{2}+y(\theta)^{2}},\quad\dot{y}(\theta)=-\frac{p\,x(\theta)}{x(\theta)^{2}+y(\theta)^{2}},\quad\mbox{and}\quad\dot{z}(\theta)=1.$

The solution $c(\theta)$ is a helix given by

$c(\theta;\rho_{0},\theta_{0})=(\rho_{0}\cos(-\frac{1}{\rho_{0}^{2}}p\theta+\theta_{0}),\rho_{0}\sin(-\frac{1}{\rho_{0}^{2}}p\theta+\theta_{0}),\theta+p\theta_{0}),$

where the initial condition $c(0)$ is taken as a point on a chosen helicoid,
e.g., $\Sigma_{0}:z=p\arctan\frac{y}{x}$.(222The solution of
$\dot{z}(\theta)=1$ is $z(\theta)=\theta+\zeta_{0}$. The fact that $c(0)$ must
be a point of a helicoid implies $\zeta_{0}$ depends on $\theta_{0}$. If
$c(0)\in\Sigma_{0}$, then $\zeta_{0}=p\theta_{0}$.) In this case, $\rho_{0}$
and $\theta_{0}$ parametrize $\Sigma_{0}$. Finally, reparametrizing
$c(\theta)$ by its arc-length gives an integral line of the director. Note
that $c(\theta)$ is a helix of radius $\rho_{0}$ and pitch
$P=-\frac{\rho_{0}}{p}$. Thus, the integral lines of the director
$\hat{\mathbf{n}}$ have a handedness opposite to that of the helices foliating
the family of helicoids orthogonal to $\hat{\mathbf{n}}$.

###  4.5 Zero biaxial splay and zero twist phases

Figure 4:  Example of a zero twist and zero biaxial splay phase. In the
figures, the arrows representing the director are colored by the value of the
bend $b$. (Left) The non-uniform bend and uniform splay $2d$ phase obtained by
considering $\hat{\mathbf{n}}$ orthogonal to the one-parameter family of
circles of radius $\rho$ constant given by
$\lambda\mapsto\mathbf{r}_{\lambda}(\theta)=(\lambda+\rho\cos\theta,\rho\sin\theta)$
(dashed lines). The integral lines of $\hat{\mathbf{n}}$ (full lines) have a
bend equal to $b=\frac{1}{\rho}\tan\theta$ while the splay $s=2/\rho$ is
constant. (Right) The $3d$ phase obtained by rotating the $2d$ phase on the
left around $\mathbf{r}_{\lambda}$. The director $\hat{\mathbf{n}}$ is
orthogonal to a family of spheres with distinct centers but the same radii.
Since $\hat{\mathbf{n}}$ can be used as the unit normal of foliation of a
finite domain of space by the spheres
$\rho\mapsto\mathbf{R}_{\lambda}(\theta,\psi)=(\lambda+\rho\cos\theta,\rho\sin\theta\cos\psi,\rho\sin\theta\sin\psi)$,
the twist vanishes $t=0$. In addition, as the leaves are totally umbilical
surfaces, their shape operator $A_{n}=-\mathrm{d}\hat{\mathbf{n}}$ are a
multiple of the identity, see Eq. (57). Therefore, the biaxial splay vanishes
as well. (The spheres have distinct colors to ease the visualization.) Figure
5: Example of a zero twist and zero biaxial splay phase. In the figures, the
arrows representing the director are colored by the value of the bend $b$.
(Left) The non-uniform bend and non-uniform splay $2d$ phase obtained by
considering $\hat{\mathbf{n}}$ orthogonal to the one-parameter family of
circles
$\lambda\mapsto\mathbf{r}_{\lambda}(\theta)=(\lambda-\lambda\cos\theta,\lambda\sin\theta)$
(dashed lines). The integral lines of $\hat{\mathbf{n}}$ (full lines) are
circles or a horizontal line and, therefore, $\hat{\mathbf{n}}\cdot\nabla
b=0$. On the other hand, the splay varies only along the direction of
$\hat{\mathbf{n}}$: $\hat{\mathbf{p}}\cdot\nabla s=0$. The derivatives
$\hat{\mathbf{p}}\cdot\nabla b$ and $\hat{\mathbf{n}}\cdot\nabla s$ must
comply with the $2d$ compatibility equation
$0=b^{2}+s^{2}+\hat{\mathbf{n}}\cdot\nabla s+\hat{\mathbf{p}}\cdot\nabla b$
NE18 ; NEcomment . (Right) The $3d$ phase obtained by rotating the $2d$ phase
on the left around $\mathbf{r}_{\lambda}$. The director $\hat{\mathbf{n}}$ is
orthogonal to a family of spheres with distinct centers and distinct radii.
Since $\hat{\mathbf{n}}$ can be used as the unit normal of foliation by the
spheres
$\lambda\mapsto\mathbf{R}_{\lambda}(\theta,\psi)=(\lambda-\lambda\cos\theta,\lambda\sin\theta\cos\psi,\lambda\sin\theta\sin\psi)$,
the twist vanishes $t=0$. As the leaves are totally umbilical surfaces, their
shape operator $A_{n}=-\mathrm{d}\hat{\mathbf{n}}$ are a multiple of the
identity, see Eq. (57). Therefore, the biaxial splay vanishes as well. (The
spheres have distinct colors to ease the visualization.)

If the biaxial splay vanishes, then the shape operator of the leaves
orthogonal to the director is diagonal with two identical eigenvalues.
Therefore, the leaves must be totally umbilical and, thus, they are either
portions of spheres or planes DoCarmo76 , Prop. 4 of Chap. 3. It follows that
the splay is constant along the leaves, i.e., $\hat{\mathbf{p}}\cdot\nabla
s=0$ and $\hat{\mathbf{q}}\cdot\nabla s=0$.

In Fig. 4 we illustrate a phase where the director $\hat{\mathbf{n}}$ is
orthogonal to the leaves of a foliation of a finite domain of space by spheres
of constant radius centered along a line, while Fig. 5 illustrates a phase
where the director $\hat{\mathbf{n}}$ is orthogonal to the leaves of a
foliation by spheres centered along a line but with distinct radii. Finally,
note that concentric circles yield the hedgehog phase as illustrated in Fig.
1, Right.

It turns out that the process of obtaining phases with zero twist and zero
biaxial splay from the unit normals of a foliation of an open set of
$\mathbb{R}^{3}$ by pieces of spheres is generic. Indeed, given a point $p$ in
space, consider the integral line $c(u)$ of $\hat{\mathbf{n}}$ passing through
$p$. Then, every point of $c(u)$ is intersected by a unique sphere whose
center and radius may be denoted by $\mathbf{r}(u)$ and $\rho(u)$. (If some of
the leaves are planes, we should allow for $1/\rho$ to have zeros.) In Fig. 4
we have $\mathbf{r}(\lambda)=\lambda\hat{\mathbf{x}}$ and $\rho$ constant,
while in Fig. 5 we have $\mathbf{r}(\lambda)=\lambda\hat{\mathbf{x}}$ and
$\rho(\lambda)=\lambda$.

Now, let us compute the deformation modes of phases with $\hat{\mathbf{n}}$
orthogonal to a foliation by spheres. As concentric circles necessarily yield
the well known hedgehog or the trivial nematic phases, we may assume that the
curve $\mathbf{r}$ describing the centers of the spheres does not degenerate
to a point. Consider $\mathbf{r}(\lambda)$ parametrized by its arc-length,
i.e., $\mathbf{r}^{\prime}(\lambda)\cdot\mathbf{r}^{\prime}(\lambda)=1$. Now,
consider the Frenet frame of $\mathbf{r}$,
$\\{\hat{\mathbf{T}},\hat{\mathbf{N}},\hat{\mathbf{B}}\\}$, and define the
frame

$\hat{\mathbf{n}}=\cos\theta\hat{\mathbf{T}}+\sin\theta\cos\psi\hat{\mathbf{N}}+\sin\theta\sin\psi\hat{\mathbf{B}},$
$\hat{\mathbf{p}}=-\sin\theta\hat{\mathbf{T}}+\cos\theta\cos\psi\hat{\mathbf{N}}+\cos\theta\sin\psi\hat{\mathbf{B}},\quad\mbox{and}\quad\hat{\mathbf{q}}=-\sin\psi\hat{\mathbf{N}}+\cos\psi\hat{\mathbf{B}}.$

Let us denote by $\kappa$ and $\tau$ the curvature and torsion of
$\mathbf{r}$.

If the director $\hat{\mathbf{n}}$ is orthogonal to the spheres
$\Sigma_{\lambda}=\mathbb{S}^{2}(\mathbf{r}(\lambda),\rho(\lambda))$ of radii
$\rho(\lambda)$ and centered at the points of $\mathbf{r}(\lambda)$, then we
can parametrize a neighborhood of space where $\hat{\mathbf{n}}$ is defined as

$\mathbf{R}(\lambda,\theta,\psi)=\mathbf{r}(\lambda)+\rho(\lambda)\hat{\mathbf{n}}(\lambda,\theta,\psi).$
(61)

We can write the vector fields in the frame
$\\{\hat{\mathbf{n}},\hat{\mathbf{p}},\hat{\mathbf{q}}\\}$ in terms of the
parametric velocity vectors
$\\{\partial\mathbf{R}/\partial\lambda,\partial\mathbf{R}/\partial\theta,\partial\mathbf{R}/\partial\psi\\}$
as

$\begin{array}[]{c}\hat{\mathbf{n}}=\frac{1}{\rho^{\prime}+\cos\theta}\frac{\partial\mathbf{R}}{\partial\lambda}+\frac{\sin\theta-\rho\kappa\cos\psi}{\rho^{\prime}+\cos\theta}\frac{\partial\mathbf{R}}{\partial\theta}+\frac{\kappa\cos\theta\sin\psi-\tau\sin\theta}{\sin\theta(\rho^{\prime}+\cos\theta)}\frac{\partial\mathbf{R}}{\partial\psi},\\\\[6.0pt]
\hat{\mathbf{p}}=\frac{1}{\rho}\frac{\partial\mathbf{R}}{\partial\theta},\quad\mbox{and}\quad\hat{\mathbf{q}}=\frac{1}{\rho\sin\theta}\frac{\partial\mathbf{R}}{\partial\psi}.\\\
\end{array}$ (62)

Therefore, we can compute the directional derivatives
$\hat{\mathbf{n}}\cdot\nabla$, $\hat{\mathbf{p}}\cdot\nabla$, and
$\hat{\mathbf{q}}\cdot\nabla$ in terms of the partial derivatives with respect
to the coordinates $(\lambda,\theta,\psi)$: recall that
$\frac{\partial\mathbf{R}}{\partial\lambda}\cdot\nabla=\frac{\partial}{\partial\lambda}$,
$\frac{\partial\mathbf{R}}{\partial\theta}\cdot\nabla=\frac{\partial}{\partial\theta}$,
and
$\frac{\partial\mathbf{R}}{\partial\psi}\cdot\nabla=\frac{\partial}{\partial\psi}$.
Now, using Eqs. (11) and (12), we can finally obtain that

$b_{p}=-\dfrac{\sin\theta}{\rho(\cos\theta+\rho^{\prime})},\,b_{q}=0,\,\frac{s}{2}+\Delta_{1}=\frac{1}{\rho},\,\frac{s}{2}-\Delta_{1}=\frac{1}{\rho},\,\frac{t}{2}-\Delta_{2}=0,\,\frac{t}{2}+\Delta_{2}=0.$
(63)

Thus, this process provides us with phases of zero twist, zero biaxial splay,
and splay $s=\frac{2}{\rho}$ uniform on each leaf as expected. Note that the
bend displays non-trivial variation both within the leafs and between leafs as
a function of $\rho(\lambda)$, yet retains the rotational symmetry in the
plane normal to $\hat{\mathbf{T}}$.

###  4.6 Vanishing bend phases

Next we come to study directors with no bend, often called Beltrami fields.
(Note that we do not assume $\nabla\cdot\hat{\mathbf{n}}=0$ as some authors do
in studying Beltrami fields CK20 .) We are going to show that this family of
directors depends on three functions to be prescribed on an initial surface
transversal to $\hat{\mathbf{n}}$: see Eqs. (84), (85), and (86) below. In
addition, these three functions must comply with two differential relations as
a consequence of Eqs. (66) and (69) below. (The solutions (84), (85), and (86)
are obtained from the compatibility equations (64), (65), (67), and (68), that
only contain derivatives in the direction of $\hat{\mathbf{n}}$.)

Setting $b_{p}=b_{q}=0$ in the compatibility equations (16)–(21) gives

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}+\Delta_{1})_{,n}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta_{1}-(\Delta)^{2}+2\alpha\Delta_{2},$
(64) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(-\frac{t}{2}+\Delta_{2})_{,n}-s(-\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1},$
(65) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(-\frac{t}{2}+\Delta_{2})_{,p}+(\frac{s}{2}+\Delta_{1})_{,q}-2\beta\Delta_{1}-2\gamma\Delta_{2},$
(66) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{t}{2}+\Delta_{2})_{,n}-s(\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1},$
(67) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}-\Delta_{1})_{,n}-\frac{s^{2}}{4}+\frac{t^{2}}{4}+s\Delta_{1}-(\Delta)^{2}-2\alpha\Delta_{2},$
(68) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-(\frac{s}{2}-\Delta_{1})_{,p}+(\frac{t}{2}+\Delta_{2})_{,q}-2\beta\Delta_{2}+2\gamma\Delta_{1}.$
(69)

The coefficients $\alpha$, $\beta$, and $\gamma$ are

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\frac{\Delta_{2}\Delta_{1,n}-\Delta_{1}\Delta_{2,n}}{2\Delta^{2}},$
(70) $\displaystyle\beta$ $\displaystyle=$
$\displaystyle\frac{\Delta_{1}t_{,p}-\Delta_{2}s_{,p}+\Delta_{1}s_{,q}+\Delta_{2}t_{,q}}{4\Delta^{2}}-\frac{\Delta_{1}\Delta_{2,p}-\Delta_{2}\Delta_{1,p}}{2\Delta^{2}}+\frac{(\Delta^{2})_{,q}}{4\Delta^{2}},$
(71)

and

$\displaystyle\gamma$ $\displaystyle=$
$\displaystyle\frac{\Delta_{2}t_{,p}+\Delta_{1}s_{,p}-\Delta_{1}t_{,q}+\Delta_{2}s_{,q}}{4\Delta^{2}}+\frac{\Delta_{2}\Delta_{1,q}-\Delta_{1}\Delta_{2,q}}{2\Delta^{2}}-\frac{(\Delta^{2})_{,p}}{4\Delta^{2}}.$
(72)

Our main goal here is to show how the description of Beltrami fields depends
on the information prescribed on an initial surface. In the present case, the
triplet of deformation modes $\\{t,s,\sigma\\}$ can be interpreted more
transparently and naturally and thus will be used instead of the triplet
$\\{t,s,\Delta\\}$. Indeed, as discussed in Subsect. 4.3, the splay plays the
role of a mean curvature, the saddle-splay plays the role of a Gaussian
curvature, while the twist measures the deviation of the director from being
orthogonal to the leaves of a foliation.

Summing Eq. (64) and Eq. (68) and then using Eq. (42), we obtain

$s_{,n}+s^{2}=\sigma.$ (73)

On the other hand, subtracting Eq. (67) from Eq. (65) gives

$t_{,n}+st=0.$ (74)

To find the evolution equation for $\sigma$, we may use the evolution equation
for $\Delta^{2}$. First, subtract Eq. (64) from Eq. (68):

$\Delta_{1,n}+s\Delta_{1}-2\alpha\Delta_{2}=0.$ (75)

Now, summing Eq. (67) and Eq. (65):

$\Delta_{2,n}+s\Delta_{2}+2\alpha\Delta_{1}=0.$ (76)

Finally, summing Eq. (75) multiplied by $\Delta_{1}$ and Eq. (76) $\Delta_{2}$
gives

$(\Delta^{2})_{,n}+2s\Delta^{2}=0.$ (77)

Thus, using Eq. (42) together with Eqs. (73) and (74) allow us to obtain

$\sigma_{,n}+s\sigma=0.$ (78)

###### Remark 3

By choosing $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$ such that $\alpha=0$
(see Remark 2), we obtain the evolution equation for $\Delta_{i}$ as
$\Delta_{i,n}+s\Delta_{i}=0$, $i=1,2$. In addition, if $\alpha=0$, then Eq.
(70) implies that ratio between $\Delta_{1}$ and $\Delta_{2}$ is constant. In
the case of Beltrami fields, we get $\alpha=0$ if we choose $\hat{\mathbf{p}}$
and $\hat{\mathbf{q}}$ as the eigenvectors of the biaxial splay:
$\Delta_{2}=0\Rightarrow\alpha=0$ by Eq. (70).

Let us integrate Eqs. (73), (74), and (78). First, introduce a coordinate
system $(u,v,w)$ such that $\hat{\mathbf{n}}=\frac{\partial}{\partial u}$. For
example, if we consider a surface $(v,w)\mapsto\mathbf{r}(v,w)$ transversal to
$\hat{\mathbf{n}}$, then we may define
$(u,v,w)\mapsto\mathbf{R}(u,v,w)=\mathbf{r}(v,w)+u\,\hat{\mathbf{n}}(v,w)$,
where $\hat{\mathbf{n}}(v,w)$ is the restriction of $\hat{\mathbf{n}}$ on the
points of $\mathbf{r}$. Then, $f_{,n}=\hat{\mathbf{n}}\cdot\nabla f=\partial
f/\partial u=f_{u}$.

Taking the derivative of Eq. (73) and using Eq. (78) imply
$s_{uu}+2ss_{u}=\sigma_{u}=-s\sigma$. Using Eq. (78) again to eliminate
$\sigma$ finally gives

$s_{uu}+3ss_{u}+s^{3}=0.$ (79)

Introducing $Y(s)=\mathrm{d}s/\mathrm{d}u$, the equation above is mapped onto
$Y^{\prime}(s)=-3s-s^{3}/Y(s)$, which is an equation of Chini type Kamke1948 ,
Eq. I$\cdot 55$, p. 303. The general solution for $s$ is

$s(u,v,w)=\frac{2A(u+B)}{A\,u^{2}+2AB\,u+AB^{2}+3}=\frac{2A(u+B)}{A(u+B)^{2}+3},$
(80)

where $A=A(v,w)$ and $B=B(v,w)$. For $t$ and $\sigma$, the solutions are

$t(u,v,w)=t_{0}(v,w)\exp(-\int_{0}^{u}s(x)\mathrm{d}x)=\frac{C}{A(u+B)^{2}+3}$
(81)

and

$\sigma(u,v,w)=\sigma_{0}(v,w)\exp(-\int_{0}^{u}s(x)\mathrm{d}x)=\frac{D}{A(u+B)^{2}+3},$
(82)

where $C=C(v,w)$ and $D=D(v,w)$.

The functions $A,B,C$, and $D$ are not entirely arbitrary. Indeed, we have a
few relations between the functions $A,B,C,D$ and $t,s$, and $\sigma$ at
$u=0$, i.e., along the transversal surface $\mathbf{r}$. Namely,
$s_{0}=2AB/(AB^{2}+3)$, $t_{0}=C/(AB^{2}+3)$, and $\sigma_{0}=D/(AB^{2}+3)$,
where $s_{0}=s_{0}(v,w)=s(0,v,w)$, $t_{0}=t_{0}(v,w)=t(0,v,w)$, and
$\sigma_{0}=\sigma_{0}(v,w)=\sigma_{0}(0,v,w)$. An additional equation comes
from the fact that $s_{0}$ and $\sigma_{0}$ must be connected by
$s_{u}+s^{2}=-\sigma$:

$s_{u}+s^{2}=\frac{2A}{A(B+u)^{2}+3},\sigma=\frac{D}{A(u+B)^{2}+3}\stackrel{{\scriptstyle
u=0}}{{\Rightarrow}}D=-2A.$ (83)

Now, dividing $s_{0}$ by $\sigma_{0}$ gives

$\frac{s_{0}}{\sigma_{0}}=\frac{2AB}{D}=-B.$

In addition, using the expression for $\sigma_{0}$, we have

$-2A=AB^{2}\sigma_{0}+3\sigma_{0}=As_{0}^{2}/\sigma_{0}+3\sigma_{0},$

which gives

$A=-\frac{3\sigma_{0}}{2+\frac{s_{0}^{2}}{\sigma_{0}}}=-\frac{3\sigma_{0}^{2}}{2\sigma_{0}+s_{0}^{2}}\Rightarrow
D=\frac{6\sigma_{0}^{2}}{2\sigma_{0}+s_{0}^{2}}.$

Finally, the remaining coefficient is

$C=t_{0}(AB^{2}+3)=t_{0}\left(-\frac{3\sigma_{0}^{2}}{2\sigma_{0}+s_{0}^{2}}\frac{s_{0}^{2}}{\sigma_{0}^{2}}+3\right)=\frac{6\,t_{0}\sigma_{0}}{2\,\sigma_{0}+s_{0}^{2}}.$

In short, the general solutions for the splay, twist, and saddle splay of a
Beltrami director field are

$s=\frac{2\frac{-3\sigma_{0}^{2}}{2\sigma_{0}+s_{0}^{2}}(u-\frac{s_{0}}{\sigma_{0}})}{\frac{3[2\sigma_{0}+s_{0}^{2}-(s_{0}-u\sigma_{0})^{2}]}{2\sigma+s_{0}^{2}}}=\frac{2\sigma_{0}(s_{0}-\sigma_{0}u)}{2\sigma_{0}+s_{0}^{2}-(s_{0}-\sigma_{0}u)^{2}}=\frac{s_{0}-\sigma_{0}u}{1+s_{0}u-\sigma_{0}\frac{u^{2}}{2}},$
(84)
$\,t=\frac{2t_{0}\sigma_{0}}{2\sigma_{0}+s_{0}^{2}-(s_{0}-\sigma_{0}u)^{2}}=\frac{t_{0}}{1+s_{0}u-\sigma_{0}\frac{u^{2}}{2}},$
(85)

and

$\sigma=\frac{2\sigma_{0}^{2}}{2\sigma_{0}+s_{0}^{2}-(s_{0}-\sigma_{0}u)^{2}}=\frac{\sigma_{0}}{1+s_{0}u-\sigma_{0}\frac{u^{2}}{2}}.$
(86)

As discussed after Eqs. (75) and (76), we may choose $\hat{\mathbf{p}}$ and
$\hat{\mathbf{q}}$ such that $\Delta_{2}=0$. Under this Gauge choice, there
are only three remaining deformation modes to investigate, either
$\\{s,t,\sigma\\}$ or $\\{s,t,\Delta\\}$. Remember that splay, twist, biaxial
splay, and saddle-splay are connected by Eq. (42).

The solutions (84), (85), and (86) depend on three functions to be prescribed
on an initial surface transversal to $\hat{\mathbf{n}}$. However, these
functions are not entirely arbitrary as they must comply with Eqs. (66) and
(69).

Finally, note that splay, twist, and saddle splay may be singular at those
points $\mathbf{R}(u,v,w)$ where the denominator in Eqs. (84), (85), and (86)
vanishes: $d(u)=1+s_{0}u-\sigma_{0}u^{2}/2=0$. If $\sigma_{0}=0$, this happens
when $u=1/s_{0}$. On the other hand, if $\sigma_{0}>0$, then
$s_{0}^{2}+2\sigma_{0}>0$ and $d(u)=0$ has two real distinct solutions, one
positive and one negative. Indeed, $\sqrt{s_{0}^{2}+2\sigma_{0}}>|s_{0}|$ and
then

$\left\\{\begin{array}[]{cc}s_{0}\geq
0\Rightarrow&\frac{s_{0}+\sqrt{s_{0}^{2}+2\sigma_{0}}}{\sigma_{0}}>\frac{s_{0}}{\sigma_{0}}+\frac{s_{0}}{\sigma_{0}}\geq
0\mbox{ and
}\frac{s_{0}-\sqrt{s_{0}^{2}+2\sigma_{0}}}{\sigma_{0}}<\frac{s_{0}}{\sigma_{0}}-\frac{s_{0}}{\sigma_{0}}<0\\\\[4.0pt]
s_{0}<0\Rightarrow&\frac{s_{0}+\sqrt{s_{0}^{2}+2\sigma_{0}}}{\sigma_{0}}>\frac{s_{0}}{\sigma_{0}}-\frac{s_{0}}{\sigma_{0}}>0\mbox{
and
}\frac{s_{0}-\sqrt{s_{0}^{2}+2\sigma_{0}}}{\sigma_{0}}<\frac{s_{0}}{\sigma_{0}}+\frac{s_{0}}{\sigma_{0}}<0\\\
\end{array}\right..$ (87)

Thus, for $\sigma_{0}<0$, $d(u)$ may have zero, one, or two real solutions as
long as $s_{0}^{2}+2\sigma_{0}<0$, $=0$, or $>0$, respectively. When
$s_{0}^{2}+2\sigma_{0}=0$, the unique solution for $d(u)=0$ is
$u=s_{0}/\sigma_{0}$. Contrarily to the case $\sigma_{0}>0$, when
$s_{0}^{2}+2\sigma_{0}>0$ and $\sigma_{0}<0$, the two solutions have the same
sign.

As an example, consider the director field $\hat{\mathbf{n}}(x,y,z)=(\cos
f(z),\sin f(z),0)$, where $x,y,z$ are Cartesian coordinates and $f$ is a
smooth function. Here, $\nabla\times\hat{\mathbf{n}}=(-f^{\prime}\cos
f,-f^{\prime}\sin f,0)=-f^{\prime}\hat{\mathbf{n}}$ and then
$\mathbf{b}=\hat{\mathbf{n}}\times\nabla\times\hat{\mathbf{n}}=0$. Therefore,
$\hat{\mathbf{n}}$ is a Beltrami director field. The twist is
$t=\hat{\mathbf{n}}\cdot\nabla\times\hat{\mathbf{n}}=-f^{\prime}$, the splay
is $s=\nabla\cdot\hat{\mathbf{n}}=0$, and finally the saddle splay is
$\sigma=0$. Let $\Sigma$ be a surface transversal to $\hat{\mathbf{n}}$ as,
for example, the plane $y=0$. Take as initial conditions on $\Sigma$ the
values $s_{0}(v,w)=0$, $\sigma_{0}(v,w)=0$, and $t_{0}(v,w)=-f^{\prime}(w)$.
As expected, applying Eqs. (84), (85), and (86) to these initial conditions
gives $t=t_{0}=-f^{\prime}(w)$, $s=0$, and $\sigma=0$.

###  4.7 Stacked planar phases

We conclude by examining the special case where the director constitutes
stacked planar phases yet with a possible twist. In this case, space is
foliated by planes in which $\hat{\mathbf{n}}$ and $\mathbf{b}$ span the local
tangent planes. Setting $\hat{\mathbf{p}}\parallel\mathbf{b}$ we obtain that
$\hat{\mathbf{q}}=\mbox{const.}$, and consequently $J^{q}=0$. We may thus
deduce that

$\Delta_{1}=\frac{s}{2},\quad\Delta_{2}=-\frac{t}{2},\quad
b_{q}=0,\quad\alpha=0,\quad\beta=0,\quad\gamma=0.$ (88)

In this case, the compatibility conditions assume a particular simple form:

$s^{2}+b_{p}^{2}+s_{,n}+b_{p,p}=0$ (89)

and

$b_{p,q}=st+t_{,n}\,,\qquad s_{,q}=-b_{p}t-t_{,p}\,.$ (90)

Equation (89) is the known two-dimensional compatibility condition within each
of the leaves NE18 ; NEcomment , while equations (90) describe how the bend
and splay evolve between leaves as a function of the twist and its tangential
derivatives. We note that differentiating equation (89) along
$\hat{\mathbf{q}}$ and substituting equation (90) (along with the relations
$(\hat{\mathbf{q}}\cdot\nabla)\hat{\mathbf{p}}=t\hat{\mathbf{n}}$ and
$(\hat{\mathbf{q}}\cdot\nabla)\hat{\mathbf{n}}=-t\hat{\mathbf{p}}$) results in
a trivial relation. Thus the satisfaction of equation (89) in one of the
tangent planes is propagated along $\hat{\mathbf{q}}$ by equations (90) to all
of space. We conclude that it suffices to prescribe the values the twist
assumes in a three dimensional domain, and compatible splay and bend functions
along a single plane to uniquely determine the texture in the domain. The
cholesteric phase described at the end of the previous subsection on Beltrami
fields is a particular example in which the bend and splay vanish identically,
and the twist is uniform across each of the planar leaves.

##  5 Concluding remarks

In this work, we derived through vector calculus the compatibility conditions
for three-dimensional director fields in Euclidean space. The results
presented here agree with the results obtained for uniform distortion fields
V19 , and also coincide with the results obtained through the method of moving
frames dSE21 ; PA21 . Our strategy consisted in seeing Eqs. (2), (3), and (4)
as a system of PDEs defining an orthonormal triad containing the director.
This system is naturally subjected to integrability conditions, which gives us
the compatibility equations. Conversely, under appropriate hypotheses on the
domain, e.g., if we consider the system of PDEs on an open ball, the
compatibility equations provide sufficient conditions for the existence a
triad containing the director with prescribed deformation modes.

We have shown that every three dimensional director field is fully
characterized by five fields, namely $s,t,b,\Delta$, and $\phi$ (the splay,
twist, bend magnitude, biaxial splay magnitude, and the relative angle between
the bend vector and the principal directions of the biaxial splay,
respectively). The values these fields can assume in space are related to each
other through six differential relations which are termed the compatibility
conditions. Any set of local fields that satisfies these equations corresponds
to a realizable triad, and to a liquid crystalline texture with the matching
values of $s,t,b,\Delta$, and $\phi$. This leads naturally to the problem of
finding solutions to the set of compatibility conditions.

The most comprehensive way of finding solutions to equations (16)-(24) is to
consider them as partial differential equations for the fields $s,t,b,\Delta$,
and $\phi$, and seek their solvability conditions in the most general sense,
and possibly classify their solutions. To date, we know of no attempt to
directly attack this formidable task. Another alternative approach is to
consider the compatibility conditions as algebraic equations for the sought
triad. This approach yields the Cartesian components of the triad as functions
of the intrinsic deformation modes and the Cartesian components of their
gradients. By construction, these will satisfy the compatibility conditions.
However, calculated in this fashion there is no assurance that the resulting
triad matches the sought deformation modes, e.g., that the splay of the found
director indeed coincides with the prescribed value of $s$. Requiring that the
resulting triad be indeed orthonormal and that the values of its deformation
modes coincide with the prescribed ones yield another set of compatibility
conditions. This idea of interpreting the compatibility conditions as
algebraic equations for the components of the sought vector fields was first
implemented in $2d$, see Section 3 of NE18 , and later in $3d$, see Section 4
of PA21 . However, in neither cases were the most general compatibility
conditions obtained.

Given the generality of the compatibility equations, it may be naive to expect
to solve them in general without severe simplifying assumptions. In practice,
the compatibility conditions should be seen as complementing the Euler-
Lagrange equations associated with an energy functional written in terms of
the deformation modes. The full system composed of Euler-Lagrange equations
and compatibility conditions may then prove amenable to further analytical
progress or numerical solutions. The path taken here does not incorporate the
Euler-Lagrange equations but rather follows a less general, yet more
applicable route. We imposed constraints, namely that some of the deformation
modes were constant or vanishing, and then solved reduced compatibility
conditions for the remaining deformation modes. This allowed us to greatly
simplify the compatibility conditions and to obtain the corresponding
compatible phases. Table 1 summarizes these results.

Phases of uniform distortion modes
---
$s=0,t=\pm 2\Delta,b_{p}=\pm b_{q}$ |  $\hat{\mathbf{n}}$ is tangent to a foliation of space by parallel helices
Phases with a single non-uniform distortion mode
$t,b,\Delta=0,\nabla s\not=0$ |  $\hat{\mathbf{n}}$ orthogonal to a foliation by concentric spheres (hedgehog)
$t,b_{q},\Delta_{2}=0,s=2\Delta_{1}$,$\nabla b\not=0$ |  $\hat{\mathbf{n}}$ orthogonal to a foliation by cylinders with parallel axes
$t=-2\Delta_{2}$,$s=2\Delta_{1}$,$b_{q}=0$,$\nabla b\not=0$ |  Special biaxial cholesteric
Phases of vanishing twist
($\hat{\mathbf{n}}$ is orthogonal to a foliation of a domain by surfaces)
$t=0,b=0$ | leaves are parallel surfaces
$t=0,\Delta=0$ |  leaves are spheres$/$planes
$t=0,s=0$ | leaves are minimal surfaces
Phases of vanishing bend (Beltrami fields)
$b=0$ |  $\hat{\mathbf{n}}$ tangent to a foliation of by straight lines
Table 1: Families of director fields with restricted deformation modes.

In a few of the cases presented here, the compatibility conditions reduce to a
boundary value problem allowing to identify the relevant degrees of freedom
that fully determine the phase in space. These dimensionally reduced
descriptions of the phases allow us to understand “how many” distinct fields
could be constructed with the desired property. More importantly, they also
advance our understanding in approaching the relevant inverse design problems,
where one seeks the local fields to prescribe in order to obtain a specific
desired texture GAE19 .

The compatibility conditions not only identify which local deformation modes
could not exist in $\mathbb{R}^{3}$, but can also serve to guide the
construction of optimal compromises that approximate well these unattainable
phases. The heliconical phase that arises in lieu of a phase of a constant
bend and vanishing splay, twist, and saddle splay constitutes a uniform
compromise associated with extensive energy. For small enough domains one may
expect non-uniform phases that are associated with super-extensive energy
scaling ME21  to provide a better approximation for the unattainable phase. In
this case, an optimal compromise could be constructed by coinciding with the
desired incompatible deformation modes along a surface (or a curve) and use
the compatibility conditions to propagate the texture to away from the surface
(curve) along which the deformation modes were prescribed.

## Appendix

In what follows, we make use of the shorthand notations

$\langle\mathcal{D}\mathbf{b},\mathbf{b}\rangle=(b_{p}^{2}-b_{q}^{2})\Delta_{1}+2b_{p}b_{q}\Delta_{2}=b^{2}\Delta\cos(2\phi)$
(91)

and

$\langle\mathcal{J}\mathcal{D}\mathbf{b},\mathbf{b}\rangle=(b_{q}^{2}-b_{p}^{2})\Delta_{2}+2b_{p}b_{q}\Delta_{1}=b^{2}\Delta\sin(2\phi).$
(92)

Here, $\phi$ is the angle formed by the bend vector and the principal
direction of the biaxial splay, while $\mathcal{D}$ and $\mathcal{J}$ denote
the biaxial splay and the counterclockwise $\frac{\pi}{2}$-rotation acting as
linear operators on the plane normal to the director field, respectively.

##  Appendix A Proof of Theorem 4.1

If $b_{p},b_{q},t$, and $s$ are constant and $\Delta_{2}=0$, so
$\Delta_{1}=\Delta$, then Eqs. (16)–(21) become

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\Delta_{,n}-b_{p}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta-\Delta^{2}+\beta
b_{q},$ (93) $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{p}b_{q}+\frac{st}{2}-2\alpha\Delta+\gamma b_{q},$ (94) $\displaystyle 0$
$\displaystyle=$ $\displaystyle\Delta_{,q}+tb_{p}-2\beta\Delta,$ (95)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{p}b_{q}-\frac{st}{2}-2\alpha\Delta-\beta b_{p},$ (96) $\displaystyle 0$
$\displaystyle=$
$\displaystyle\Delta_{,n}-b_{q}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}+s\Delta-\Delta^{2}-\gamma
b_{p},$ (97) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\Delta_{,p}+tb_{q}+2\gamma\Delta.$ (98)

The coefficients $\alpha$, $\beta$, and $\gamma$ are given by

$\alpha=-\frac{tb^{2}}{8\Delta^{2}}-\frac{b_{q}\Delta_{,p}+b_{p}\Delta_{,q}}{8\Delta^{2}}-\frac{b_{p}b_{q}}{2\Delta},$
(99)
$\beta=\frac{(\Delta^{2})_{,q}}{4\Delta^{2}}+t\,\frac{b_{p}\Delta}{2\Delta^{2}}=\frac{\Delta_{,q}}{2\Delta}+t\,\frac{b_{p}}{2\Delta},$
(100)

and

$\gamma=-\frac{(\Delta^{2})_{,p}}{4\Delta^{2}}-t\,\frac{b_{q}\Delta}{2\Delta^{2}}=-\frac{\Delta_{,p}}{2\Delta}-t\,\frac{b_{q}}{2\Delta}.$
(101)

Now, using the above expressions for $\alpha$, $\beta$, and $\gamma$ and
summing Eqs. (93) and (97) give

$0=\frac{b_{p}\Delta_{,p}+b_{q}\Delta_{,q}}{2\Delta}+\frac{-2\Delta
b^{2}+2b_{p}b_{q}t-4\Delta^{3}-\Delta(s^{2}-t^{2})}{2\Delta}.$ (102)

Subtracting Eq. (96) from Eq. (94) gives

$0=\frac{-b_{q}\Delta_{,p}+b_{p}\Delta_{,q}}{2\Delta}+\frac{(b_{p}^{2}-b_{q}^{2})t+2\Delta
st}{2\Delta}.$ (103)

If, in addition, $b_{p}=b_{q}=0$, then $4\Delta^{2}+s^{2}-t^{2}=0$, from which
we conclude that $\Delta$ must be constant. From now on, assume that
$b^{2}\not=0$. Therefore, we get

$b^{2}\Delta_{,p}=4b_{p}\Delta^{3}+[2b_{p}b^{2}+(s^{2}-t^{2})b_{p}+2stb_{q}]\,\Delta-
b_{q}b^{2}t$ (104)

and

$b^{2}\Delta_{,q}=4b_{q}\Delta^{3}+[2b_{q}b^{2}+(s^{2}-t^{2})b_{q}-2stb_{p}]\,\Delta-
b_{p}b^{2}t.$ (105)

We shall now exploit Eq. (24). First, we have

$\displaystyle\beta_{,q}$ $\displaystyle=$
$\displaystyle\frac{\Delta\Delta_{,qq}-(\Delta_{,q})^{2}-tb_{p}\Delta_{,q}}{2\Delta^{2}}$
(106) $\displaystyle=$
$\displaystyle\frac{12b_{q}\Delta^{2}+2b_{q}b^{2}+(s^{2}-t^{2})b_{q}-2stb_{p}}{2b^{2}\Delta}\Delta_{,q}-\frac{\Delta_{,q}+tb_{p}}{2\Delta^{2}}\Delta_{,q}$
$\displaystyle=$
$\displaystyle\frac{4b_{q}\Delta}{b^{4}}\Big{(}4b_{q}\Delta^{3}+\Delta[2b_{q}b^{2}+(s^{2}-t^{2})b_{q}-2stb_{p}]-b_{p}b^{2}t\Big{)}$

and

$\displaystyle-\gamma_{,p}$ $\displaystyle=$
$\displaystyle\frac{\Delta\Delta_{,pp}-(\Delta_{,p})^{2}-tb_{q}\Delta_{,p}}{2\Delta^{2}}$
(107) $\displaystyle=$
$\displaystyle\frac{12b_{p}\Delta^{2}+2b_{p}b^{2}+(s^{2}-t^{2})b_{p}+2stb_{q}}{2b^{2}\Delta}\Delta_{,p}-\frac{\Delta_{,p}+tb_{q}}{2\Delta^{2}}\Delta_{,p}$
$\displaystyle=$
$\displaystyle\frac{4b_{p}\Delta}{b^{4}}\Big{(}4b_{p}\Delta^{3}+\Delta[2b_{p}b^{2}+(s^{2}-t^{2})b_{p}+2stb_{q}]-b_{q}b^{2}t\Big{)}.$

Finally, from Eq. (24), we obtain the following rational function in $\Delta$:

$\frac{12}{b^{2}\Delta^{2}}\Big{\\{}\Delta^{6}+\frac{2(s^{2}-t^{2})+5b^{2}}{12b^{2}}\Delta^{4}+\frac{b^{2}(3t^{2}-4b^{2}-5s^{2})-(s^{2}+t^{2})^{2}}{48}\Delta^{2}$
$-\frac{tb_{p}b_{q}}{2}\Delta^{3}+\frac{tb_{q}[(s^{2}-t^{2})b_{p}+2stb_{q}]-st^{2}b^{2}+3tb_{p}b_{q}b^{2}}{24}\Delta-\frac{b^{4}t^{2}}{96}\Big{\\}}=0.$
(108)

As the coefficient of the highest power never vanishes, it follows that
$\Delta$ must be constant provided that $s,t,b_{p}$, and $b_{q}$ are constant.

##  Appendix B Proof of Theorem 4.2

If $\Delta_{1},\Delta_{2},t$, and $s$ are constant and $b_{q}=0$ (so,
$b_{p}=b$), the compatibility equations (16)–(21) become

$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,p}-b^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta_{1}-(\Delta)^{2}+2\alpha\Delta_{2},$
(109) $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{,q}-s(-\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1},$ (110) $\displaystyle 0$
$\displaystyle=$ $\displaystyle tb-2\beta\Delta_{1}-2\gamma\Delta_{2},$ (111)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-s(\frac{t}{2}+\Delta_{2})-2\alpha\Delta_{1}-\beta b,$ (112)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{s^{2}}{4}+\frac{t^{2}}{4}+s\Delta_{1}-(\Delta)^{2}-2\alpha\Delta_{2}-\gamma
b,$ (113) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-2\beta\Delta_{2}+2\gamma\Delta_{1}.$ (114)

If $\Delta_{1}^{2}+\Delta_{2}^{2}\not=0$, then the coefficients $\alpha$,
$\beta$, and $\gamma$ are given by

$\beta=\frac{tb\Delta_{1}}{2\Delta^{2}},\,\gamma=\frac{tb\Delta_{2}}{2\Delta^{2}},$
(115)

and

$\alpha=-\frac{\Delta_{1}b_{,q}-\Delta_{2}b_{,p}}{4\Delta^{2}}-\frac{tb^{2}}{8\Delta^{2}}-\frac{\langle\mathcal{J}\mathcal{D}\mathbf{b},\mathbf{b}\rangle}{4\Delta^{2}}.$
(116)

Here, we used the notation introduced in Eq. (92). Substituting $\alpha$,
$\beta$, and $\gamma$ in Eqs. (109) and (110) gives

$0=\frac{\Delta_{2}^{2}-2\Delta^{2}}{2\Delta^{2}}b_{,p}-\frac{\Delta_{1}\Delta_{2}}{2\Delta^{2}}b_{,q}-b^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-s\Delta_{1}-\Delta^{2}-\frac{tb^{2}\Delta_{2}}{4\Delta^{2}}-\frac{\langle\mathcal{J}\mathcal{D}\mathbf{b},\mathbf{b}\rangle}{2\Delta^{2}}\Delta_{2}$
(117)

and

$0=-\frac{\Delta_{1}\Delta_{2}}{2\Delta^{2}}b_{,p}+\frac{\Delta_{1}^{2}-2\Delta^{2}}{2\Delta^{2}}b_{,q}-s(-\frac{t}{2}+\Delta_{2})+\frac{tb^{2}\Delta_{1}}{4\Delta^{2}}+\frac{\langle\mathcal{J}\mathcal{D}\mathbf{b},\mathbf{b}\rangle}{2\Delta^{2}}\Delta_{1}.$
(118)

Now, summing Eq. (117) multiplied by $(\Delta_{1}^{2}-2\Delta^{2})$ with Eq.
(118) multiplied by $\Delta_{1}\Delta_{2}$ gives

$b_{,p}=\frac{2\Delta^{2}-\Delta_{1}^{2}}{\Delta^{2}}(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2})-b^{2}-\frac{tb^{2}}{2\Delta^{2}}\Delta_{2}-\frac{s\Delta_{1}\Delta_{2}}{\Delta^{2}}(\frac{t}{2}+\Delta_{2})-\frac{s\Delta_{1}^{3}}{\Delta^{2}}.$
(119)

On the other hand, summing Eq. (117) multiplied by $\Delta_{1}\Delta_{2}$ with
Eq. (118) multiplied by $(\Delta_{2}^{2}-2\Delta^{2})$ gives

$b_{,q}=s\frac{2\Delta^{2}-\Delta_{2}^{2}}{\Delta^{2}}(\frac{t}{2}-\Delta_{2})+\frac{\Delta_{1}\Delta_{2}}{\Delta^{2}}(\frac{s^{2}}{4}-\frac{t^{2}}{4}+\Delta^{2})+\frac{s\Delta_{1}^{2}\Delta_{2}}{\Delta^{2}}+\frac{tb^{2}}{2\Delta^{2}}\Delta_{1}.$
(120)

Let us now exploit the remaining compatibility equations, Eqs. (22), (23), and
(24) . First, we can rewrite $\alpha$ as

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\frac{1}{4\Delta^{2}}\left[2\Delta_{2}\left(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2}\right)-b^{2}\left(\frac{t}{2}+\Delta_{2}\right)-st\Delta_{1}-\frac{tb^{2}}{2}+b^{2}\Delta_{2}\right]$
(121) $\displaystyle=$
$\displaystyle\frac{1}{4\Delta^{2}}\left[2\Delta_{2}\left(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2}\right)-st\Delta_{1}-tb^{2}\right].$

Then, Eq. (24) gives

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\gamma_{,p}+\beta_{,q}-\beta^{2}-\gamma^{2}-t\,\alpha-\frac{s^{2}}{4}-\frac{t^{2}}{4}+(\Delta)^{2}$
$\displaystyle=$
$\displaystyle\frac{t}{2\Delta^{2}}(\Delta_{1}b_{,q}-\Delta_{2}b_{,p})-\frac{t^{2}b^{2}}{4\Delta^{2}}-t\,\alpha-\frac{s^{2}}{4}-\frac{t^{2}}{4}+(\Delta)^{2}$
$\displaystyle=$
$\displaystyle-\frac{t}{2\Delta^{2}}\left[2\Delta_{2}\left(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2}\right)-b^{2}\left(\frac{t}{2}+\Delta_{2}\right)-st\Delta_{1}\right]-\frac{t^{2}b^{2}}{4\Delta^{2}}$
$\displaystyle-\frac{t}{4\Delta^{2}}\left[2\Delta_{2}\left(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2}\right)-st\Delta_{1}-tb^{2}\right]-\frac{s^{2}}{4}-\frac{t^{2}}{4}+\Delta^{2}$
$\displaystyle=$
$\displaystyle\frac{tb^{2}}{2\Delta^{2}}\left(\frac{t}{2}+\Delta_{2}\right)-\frac{3t\Delta_{2}}{2\Delta^{2}}\left(\frac{t^{2}}{4}-\frac{s^{2}}{4}-\Delta^{2}\right)+\frac{3st^{2}\Delta_{1}}{4\Delta^{2}}-\frac{s^{2}}{4}-\frac{t^{2}}{4}+\Delta^{2}.$

Therefore, if neither $t=-2\Delta_{2}$ nor $t=0$, then $b$ must be constant.

It remains to analyze the case where $\Delta_{1}^{2}+\Delta_{2}^{2}=0$. Here,
$\beta$ and $\gamma$ are given by

$\beta=-\frac{st}{2b}\mbox{ and }\gamma=-\frac{s^{2}-t^{2}}{4b}.$ (122)

The Eqs. (16), (17), and (18) are

$0=-b_{,p}-b^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4},\,0=-b_{,q}+\frac{st}{2},\,0=tb.$

From the last equation, we conclude that either $b=0$ or $t=0$. Since we are
assuming $t\not=0$, then $b=0$, which implies that $s=\pm t$ and $st=0$. We
then deduce that $t$ must vanish, which is a contradiction. Consequently,
there exists no director field with $\Delta^{2}_{1}+\Delta_{2}^{2}=0$, $b=0$,
$s=0$, but $t\not=0$.

##  Appendix C Proof of Theorem 4.3

(a) Let us assume that $b_{p},b_{q}$, and $t$ are constant, and
$\Delta_{1}^{2}+\Delta_{2}^{2}=0$. Then, Eqs. (16)–(21) become

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}s_{,n}-b_{p}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}+\beta
b_{q},$ (123) $\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{p}b_{q}+\frac{st}{2}+\gamma b_{q},$ (124) $\displaystyle 0$
$\displaystyle=$ $\displaystyle\frac{1}{2}s_{,q}+tb_{p},$ (125) $\displaystyle
0$ $\displaystyle=$ $\displaystyle-b_{p}b_{q}-\frac{st}{2}-\beta b_{p},$ (126)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}s_{,n}-b_{q}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-\gamma
b_{p},$ (127) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}s_{,p}+tb_{q}.$ (128)

If, in addition, we assume $b\not=0$, then we can write $\beta$ and $\gamma$
as

$\beta=\frac{b_{q}}{2b^{2}}s_{,n}+\frac{(s^{2}-t^{2})b_{q}-2stb_{p}}{4b^{2}},\,\gamma=-\frac{b_{p}}{2b^{2}}s_{,n}-\frac{(s^{2}-t^{2})b_{p}+2stb_{q}}{4b^{2}}.$
(129)

By summing Eqs. (123) and (127), we obtain

$s_{,n}=-2b^{2}+\frac{t^{2}}{2}-\frac{s^{2}}{2},$ (130)

while from Eq. (128) and (125) we obtain

$s_{,p}=2tb_{q}\mbox{ and }s_{,q}=-2tb_{p}.$ (131)

So far, we have not specified who $\hat{\mathbf{p}}$ and $\hat{\mathbf{q}}$
are. Choosing them to satisfy $\alpha=0$ (Remark 2), we finally obtain from
Eq. (24) that

$0=-\frac{t^{2}+b^{2}}{4b^{2}}s^{2}+\frac{3t^{2}-4b^{2}}{4}.$ (132)

Since the coefficient of $s^{2}$ does not vanish, it follows that $s$ must be
constant.

Finally, if $\Delta_{1}^{2}+\Delta_{2}^{2}=0$ and $b=0$, then we have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{s_{,n}}{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4},$ (133)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{st}{2},$ (134)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{s_{,q}}{2},$ (135)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle-\frac{st}{2},$ (136)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{s_{,n}}{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4},$ (137)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle-\frac{s_{,p}}{2}.$ (138)

Then, either $t=0$ or $s=0$. If $s=0$, then it also follows from Eq. (133)
that $t=0$ and, therefore, $\hat{\mathbf{n}}$ is constant. On the other hand,
if $t=0$ we deduce that $s$ only varies along the integral curves of
$\hat{\mathbf{n}}$ and satisfies $s_{,n}=-\frac{s^{2}}{2}$. This leads to a
solution where $\hat{\mathbf{n}}$ is either constant or a hedgehog, i.e.,
$\hat{\mathbf{n}}$ correspond to the unit normal fields of a foliation of
space by concentric spheres.

(b) Now, let us assume that $b_{p},b_{q}$, and $s$ are constant, and
$\Delta_{1}^{2}+\Delta_{2}^{2}=0$. Then, the compatibility equations (16)–(21)
become

$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{p}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}+\beta b_{q},$ (139) $\displaystyle
0$ $\displaystyle=$
$\displaystyle\frac{1}{2}t_{,n}-b_{p}b_{q}+\frac{st}{2}+\gamma b_{q},$ (140)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle\frac{1}{2}t_{,p}+tb_{p},$
(141) $\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}t_{,n}-b_{p}b_{q}-\frac{st}{2}-\beta b_{p},$ (142)
$\displaystyle 0$ $\displaystyle=$ $\displaystyle-
b_{q}^{2}-\frac{s^{2}}{4}+\frac{t^{2}}{4}-\gamma b_{p},$ (143) $\displaystyle
0$ $\displaystyle=$ $\displaystyle\frac{1}{2}t_{,q}+tb_{q}.$ (144)

If, in addition, we assume that $b\not=0$, then we can write $\beta$ and
$\gamma$ as

$\beta=-\frac{b_{p}}{2b^{2}}t_{,n}+\frac{(s^{2}-t^{2})b_{q}-2stb_{p}}{4b^{2}},\,\gamma=-\frac{b_{q}}{2b^{2}}t_{,n}-\frac{(s^{2}-t^{2})b_{p}+2stb_{q}}{4b^{2}}.$
(145)

Finally, using $\beta$ and $\gamma$ above and summing Eqs. (139) and (143)
give

$0=b^{2}+\frac{t^{2}-s^{2}}{4},$ (146)

which implies that the twist must be constant.

Finally, if $\Delta_{1}^{2}+\Delta_{2}^{2}=0$ and $b=0$, then Eq. (139)
becomes

$0=-\frac{s^{2}}{4}+\frac{t^{2}}{4},$

which implies $t=\pm s$. Therefore, $t$ must be constant. This conclude the
proof of theorem 4.3.

###### Acknowledgements.

This work was funded by the Israel Science Foundation grant no. $1444/21$.
L.C.B.dS. acknowledges the support provided by the Morá Miriam Rozen Gerber
fellowship for Brazilian postdocs and the Faculty of Physics Postdoctoral
Excellence Fellowship.

## Conflict of interest

The authors declare that they have no conflict of interest.

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