Octupolar Tensors for Liquid Crystals

TL;DR

This paper provides an algebraic characterization of octupolar tensors in liquid crystals, revealing possible intra-octupolar transitions and offering a quantitative framework for understanding molecular orientations.

Contribution

It introduces a closed-form algebraic expression for the dome-shaped and separatrix surfaces of octupolar tensors, advancing the understanding of phase transitions in liquid crystals.

Findings

Derived algebraic expressions for key tensor surfaces.

Identified potential intra-octupolar transition points.

Enhanced the theoretical framework for liquid crystal phases.

Abstract

A third-order three-dimensional symmetric traceless tensor, called the \emph{octupolar} tensor, has been introduced to study tetrahedratic nematic phases in liquid crystals. The octupolar \emph{potential}, a scalar-valued function generated on the unit sphere by that tensor, should ideally have four maxima capturing the most probable molecular orientations (on the vertices of a tetrahedron), but it was recently found to possess an equally generic variant with \emph{three} maxima instead of four. It was also shown that the irreducible admissible region for the octupolar tensor in a three-dimensional parameter space is bounded by a dome-shaped surface, beneath which is a \emph{separatrix} surface connecting the two generic octupolar states. The latter surface, which was obtained through numerical continuation, may be physically interpreted as marking a possible \emph{intra-octupolar} transition. In this paper, by using the resultant theory of algebraic geometry and the E-characteristic polynomial of spectral theory of tensors, we give a closed-form, algebraic expression for both the dome-shaped surface and the separatrix surface. This turns the envisaged intra-octupolar transition into a quantitative, possibly observable prediction. Some other properties of octupolar tensors are also studied.

Equations

$$\sum _{i=1}^n{t}_{ii{i}_3\cdots i_m}=0\text{ for all }{i}_3,\dots ,i_m=1,2,\dots ,n,$$

$$[\mathcal{T}{Q}^m{]}_{i_1{i}_2\cdots i_m}=\sum _{j_1=1}^n\sum _{j_2=1}^n\cdots \sum _{j_m=1}^n{t}_{j_1{j}_2\cdots j_m}{q}_{i_1{j}_1}{q}_{i_2{j}_2}\cdots q_{i_m{j}_m}.$$

$$\sum _{i=1}^n\sum _{j_1=1}^m\sum _{j_2=1}^m{t}_{j_1{j}_2{j}_3\cdots j_m}{q}_{i{j}_1}{q}_{i{j}_2}=\sum _{i=1}^n{t}_{ii{j}_3\cdots j_m}=0$$

$$\sum _{i=1}^n[\mathcal{T}{Q}^m{]}_{ii{i}_3\cdots i_m}$$

$$\mathcal{A}=\left[\begin{array}[]{ccc|ccc|ccc}a_{111}&a_{112}&a_{113}&a_{112}&a_{122}&a_{123}&a_{113}&a_{123}&a_{133}\\ a_{112}&a_{122}&a_{123}&a_{122}&a_{222}&a_{223}&a_{123}&a_{223}&a_{233}\\ a_{113}&a_{123}&a_{133}&a_{123}&a_{223}&a_{233}&a_{133}&a_{233}&a_{333}\\ \end{array}\right]\in\mathbb{R}^{[3,3]}$$

$$\{\begin{array}{r}{a}_{111}+a_{122}+a_{133}=0,\\ a_{112}+a_{222}+a_{233}=0,\\ a_{113}+a_{223}+a_{333}=0.\end{array}$$

$${\alpha }_0=a_{123},{\alpha }_1=a_{111},{\alpha }_2=a_{222},{\alpha }_3=a_{333},{\beta }_1=a_{122},{\beta }_2=a_{233},{\beta }_3=a_{113}.$$

$$\mathcal{A}=$$

$$\Phi (\mathbf{x})$$

$${\alpha }_3=\Phi (0,0,1)\ge 0.$$

$$\{\begin{array}{rl}& {\alpha }_1{x}_1^2+2{\alpha }_0{x}_2{x}_3+{\beta }_1{x}_2^2+2{\beta }_3{x}_1{x}_3-({\alpha }_1+{\beta }_1)x_3^2-2({\alpha }_2+{\beta }_2)x_1{x}_2=\lambda x_1,\\ & {\alpha }_2{x}_2^2+2{\alpha }_0{x}_1{x}_3+2{\beta }_1{x}_1{x}_2+{\beta }_2{x}_3^2-({\alpha }_2+{\beta }_2)x_1^2-2({\alpha }_3+{\beta }_3)x_2{x}_3=\lambda x_2,\\ & {\alpha }_3{x}_3^2+2{\alpha }_0{x}_1{x}_2+2{\beta }_2{x}_2{x}_3+{\beta }_3{x}_1^2-2({\alpha }_1+{\beta }_1)x_1{x}_3-({\alpha }_3+{\beta }_3)x_2^2=\lambda x_3,\\ & x_1^2+x_2^2+x_3^2=1.\end{array}$$

$${\alpha }_1+{\beta }_1=0\text{ and }{\beta }_2=0.$$

$${\alpha }_1=0.$$

$$\mathcal{A}=\left[\begin{array}[]{ccc|ccc|ccc}0&-\alpha_{2}&\beta_{3}&-\alpha_{2}&0&\alpha_{0}&\beta_{3}&\alpha_{0}&0\\ -\alpha_{2}&0&\alpha_{0}&0&\alpha_{2}&-\alpha_{3}-\beta_{3}&\alpha_{0}&-\alpha_{3}-\beta_{3}&0\\ \beta_{3}&\alpha_{0}&0&\alpha_{0}&-\alpha_{3}-\beta_{3}&0&0&0&\alpha_{3}\\ \end{array}\right],$$

$$\Phi (\mathbf{x};{\alpha }_0,{\alpha }_2,{\alpha }_3,{\beta }_3)={\alpha }_2{x}_2^3+{\alpha }_3{x}_3^3+6{\alpha }_0{x}_1{x}_2{x}_3+3{\beta }_3{x}_1^2{x}_3-3{\alpha }_2{x}_1^2{x}_2-3({\alpha }_3+{\beta }_3)x_2^2{x}_3$$

$${\alpha }_2\ge 0$$

$$\Phi (x_1,x_2,x_3;{\alpha }_0,{\alpha }_2,{\alpha }_3,{\beta }_3)=\Phi (x_1,-x_2,x_3;-{\alpha }_0,-{\alpha }_2,{\alpha }_3,{\beta }_3).$$

$$3{\alpha }_3^2-4{\alpha }_3{\beta }_3-4{\beta }_3^2-4{\alpha }_0^2\ge 0.$$

$$\{\begin{array}{rl}\max & \Phi (\mathbf{x})=\mathcal{A}{\mathbf{x}}^3\\ s.t.& {\mathbf{x}}^{\top }\mathbf{x}=1.\end{array}$$

$$L(\mathbf{x},\lambda )=-\frac{1}{3}\mathcal{A}{\mathbf{x}}^3+\frac{\lambda }{2}({\mathbf{x}}^{\top }\mathbf{x}-1).$$

$${\nabla }_{\mathbf{x}\mathbf{x}}^{2}L(\mathbf{x},\lambda )=\lambda I-2\mathcal{A}\mathbf{x},$$

$$\mathcal{A}{\mathbf{x}}^2=\lambda \mathbf{x}\text{ and }{\mathbf{x}}^{\top }\mathbf{x}=1,$$

$$P^{\top }{\nabla }_{\mathbf{x}\mathbf{x}}^{2}L(\mathbf{x},\lambda )P=\lambda (I+\mathbf{x}{\mathbf{x}}^{\top })-2\mathcal{A}\mathbf{x}.$$

$$[P^{\top}\nabla_{{\bf x}{\bf x}}^{2}L({\bf x},\lambda)P]_{\lambda=\alpha_{3},{\bf x}=(0,0,1)^{\top}}=\left[\begin{array}[]{ccc}\alpha_{3}-2\beta_{3}&-2\alpha_{0}&0\\ -2\alpha_{0}&3\alpha_{3}+2\beta_{3}&0\\ 0&0&0\\ \end{array}\right].$$

$${\mu }_1=0,{\mu }_{2,3}=2{\alpha }_3\pm \sqrt{({\alpha }_3+2{\beta }_3{)}^2+4{\alpha }_0^2},$$

$$3{\alpha }_3^2-4{\alpha }_3{\beta }_3-4{\beta }_3^2-4{\alpha }_0^2\ge 0.$$

$${\alpha }_3=1,{\alpha }_2\ge 0,\text{ and }{\alpha }_0^2+({\beta }_3+\frac{1}{2}{)}^2\le 1.$$

$$\mathcal{A}(\alpha_{0},\beta_{3},\alpha_{2})=\left[\begin{array}[]{ccc|ccc|ccc}0&-\alpha_{2}&\beta_{3}&-\alpha_{2}&0&\alpha_{0}&\beta_{3}&\alpha_{0}&0\\ -\alpha_{2}&0&\alpha_{0}&0&\alpha_{2}&-1-\beta_{3}&\alpha_{0}&-1-\beta_{3}&0\\ \beta_{3}&\alpha_{0}&0&\alpha_{0}&-1-\beta_{3}&0&0&0&1\\ \end{array}\right]$$

$$\Phi (\mathbf{x})={\alpha }_2{x}_2^3+x_3^3+6{\alpha }_0{x}_1{x}_2{x}_3+3{\beta }_3{x}_1^2{x}_3-3{\alpha }_2{x}_1^2{x}_2-3(1+{\beta }_3)x_2^2{x}_3$$

$$\mathcal{A}{\bf x}^{2}=\left(\begin{array}[]{c}-2\alpha_{2}x_{1}x_{2}+2\beta_{3}x_{1}x_{3}+2\alpha_{0}x_{2}x_{3}\\ -\alpha_{2}x_{1}^{2}+\alpha_{2}x_{2}^{2}+2\alpha_{0}x_{1}x_{3}-2(1+\beta_{3})x_{2}x_{3}\\ \beta_{3}x_{1}^{2}-(1+\beta_{3})x_{2}^{2}+x_{3}^{2}+2\alpha_{0}x_{1}x_{2}\end{array}\right)\equiv\left(\begin{aligned} F_{3}(x_{1},x_{2},x_{3})\\ F_{1}(x_{1},x_{2},x_{3})\\ F_{2}(x_{1},x_{2},x_{3})\end{aligned}\right).$$

Full text

Key words. Octupolar order tensors; Resultants; Traceless tensors; Liquid crystas; Intra-octupolar transitions.

1 Introduction

Liquid crystals are well-known for visualization applications in flat panel electronic displays. But beyond that, various optical and electronic devices, such as laser printers, light-emitting diodes, field-effect transistors, and holographic data storage, were invented with the development of bent-core (banana-shaped) liquid crystals [8, 9]. A third-order three dimensional symmetric traceless tensor was introduced in [6] to characterize condensed phases exhibited by bent-core molecules [16, 11]. Based on such a tensor, the orientationally ordered octupolar (tetrahedratic) phase has been both predicted theoretically [13, 2] and confirmed experimentally [22]. After that, the octupolar order parameters of liquid crystals have been widely studied [1, 20, 7]. Generalized liquid crystal phases are also considered in [17, 12], which feature octupolar order tensors among so many others.

Ideally, the octupolar order in a molecular ensemble of generalized liquid crystals is expected to exhibit four directions in space pointing towards the vertices of a tetrahedron, along which specific molecular axes are most likely to be oriented. Fel [6] first proposed to use a third-order three dimensional symmetric tensor $\mathcal{A}$, which here is called the octupolar tensor, to describe the tetrahedratic symmetry of the octupolar order. According to the Buckingham’s formula [3, 20] for the expansion in Cartesian tensors of the orientational probability density function for uniaxial nematics, the octupolar tensor $\mathcal{A}$ is traceless, i.e., the trace of each slice matrix of $\mathcal{A}$ vanishes. As shown in [7], by a judicious choice of the Cartesian coordinate system, such a tensor can be represented by three independent parameters, which there were called $\alpha_{0},\beta_{3}$ and $\alpha_{2}$.

Virga [21] and Gaeta and Virga [7], in their studies of third-order octupolar tensors in two and three space dimensions, respectively, also introduced the octupolar potential, a scalar-valued function on the unit sphere obtained from the octupolar tensor. In particular, Gaeta and Virga [7] showed that the irreducible admissible region for the octupolar potential is bounded by a surface in the three-dimensional parameter space which has the form of a dome and, more importantly, that there are indeed two generic octupolar states, divided by a separatrix surface in paramter space. Physically, the latter surface was interpreted as representing a possible intra-octupolar transition.

In this paper, by using the resultant theory of algebraic geometry and the E-characteristic polynomial of the spectral theory of tensors, we give a closed-form, algebraic expression for both the dome and the separatrix. This turns the intra-octupolar transition envisioned in [7] into a quantitative, possibly observable prediction. Some other properties of octupolar tensors are also studied.

In Section 2, we prove that the traceless property of octupolar tensors is preserved under orthogonal transformations. This property motives us to choose a proper Cartesian coordinate system to reduce the independent elements of the octupolar tensor $\mathcal{A}$ from seven to three. By assuming that the North pole $(0,0,1)^{\top}$ is a maximum point of the octupolar potential on the unit sphere, we identify an irreducible, admissible region for only three independent parameters of $\mathcal{A}$, which we shall also call $\alpha_{0},\beta_{3}$ and $\alpha_{2}$ to ease comparison with [7].

Using the multi-polynomial resultant from algebraic geometry [5], we derive the resultant $\mathrm{Res}(\mathcal{A}{\bf x}^{2})$ in Section 3. $\mathrm{Res}(\mathcal{A}{\bf x}^{2})=0$ implies that $\mathcal{A}$ has zero E-eigenvalues. After that, we construct the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ of $\mathcal{A}$; this latter is a fourteen-degree polynomial with only even degree terms and its constant term is the square of $\mathrm{Res}(\mathcal{A}{\bf x}^{2})$. When $\mathrm{Res}(\mathcal{A}{\bf x}^{2})\neq 0$, $\phi_{\mathcal{A}}(\lambda)=0$ if and only if $\lambda$ is an E-eigenvalue of $\mathcal{A}$.

In Section 4, by assuming that the North pole $(0,0,1)^{\top}$ is the global maximum point of the octupolar potential on the unit sphere, the admissible region is further reduced. The boundary of such a reduced admissible region was referred to as the dome in [7]. By exploring the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ of $\mathcal{A}$, we give the algebraic expression for the dome explicitly.

We said already that two generic octupolar states were identified in in [7] on the two sides of a separatrix surface in parameter space. It should be added here that in in [7] both the dome and the separatix were determined by numerical continuation: no closed-form was given for either of them. In Section 5, we also obtain the explicit, algebraic expression for the separatrix.

Finally, some concluding remarks are drawn in Section 6.

2 Preliminaries

The octupolar tensor in liquid crystals is a traceless tensor. Now, we give the formal definition of traceless tensors and prove that the traceless property of a symmetric tensor are invariant under orthogonal transformations [7]. For convenience, we denote as $\mathbb{R}^{[m,n]}$ the real-valued space of $m$th order $n$ dimensional symmetric tensors.

Definition 2.1**.**

Let $\mathcal{T}=[t_{i_{1}i_{2}\cdots i_{m}}]\in\mathbb{R}^{[m,n]}$. If

[TABLE]

then $\mathcal{T}$ is called a traceless tensor.

Theorem 2.2**.**

Let $\mathcal{T}=[t_{i_{1}i_{2}\cdots i_{m}}]\in\mathbb{R}^{[m,n]}$ be a traceless tensor and $Q=[q_{ij}]\in\mathbb{R}^{n\times n}$ be an orthogonal matrix. We denote by $\mathcal{T}Q^{m}\in\mathbb{R}^{[m,n]}$ the symmetric tensor whose elements are

[TABLE]

Then, $\mathcal{T}Q^{m}$ is also a traceless tensor.

Proof.

It is straightforward to see that the new tensor $\mathcal{T}Q^{m}$ is real-valued and symmetric. Now, we consider its slice matrices. As the sum of all matrix eigenvalues, the trace of a symmetric matrix is invariant under an orthogonal transformation. Hence, we get

[TABLE]

for all $j_{3},\ldots,j_{m}=1,2,\ldots,n$. By some calculations,

[TABLE]

where the last equality is valid because of (1). Hence, the new tensor $\mathcal{T}Q^{m}$ is also traceless. ∎

By exploting the rotational invariance of traceless tensors, we now represent the octupolar tensor as

[TABLE]

by choosing a proper Cartesian coordinate system. The traceless property of $\mathcal{A}$ means that

[TABLE]

Hence, there are seven independent elements in $\mathcal{A}$. Let

[TABLE]

Using the traceless property (3), we convert (2) into

[TABLE]

The associated octupolar potential as defined in [7] is

[TABLE]

On the unit sphere $\mathbb{S}^{2}\equiv\{{\bf x}=(x_{1},x_{2},x_{3})^{\top}:x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\}$, the polynomial $\Phi({\bf x})$ has at least one maximum point. Without loss of generality, we assume such a maximum point being the North pole $(0,0,1)^{\top}$ and

[TABLE]

From the spectral theory of tensors [18], we know that $\alpha_{3}$ is a Z-eigenvalue of $\mathcal{A}$ with an associated Z-eigenvector $(0,0,1)^{\top}$. All Z-eigenvectors and Z-eigenvalues must satisfy the following system:

[TABLE]

Hence, requiring $(0,0,1)^{\top}$ to be a solution, we obtain

[TABLE]

Moreover, because $\Phi(-x_{1},0,0)=-\Phi(x_{1},0,0)$, we could rotate the Cartesian coordinate system so that $\Phi(1,0,0)=0$ and we get

[TABLE]

Now, the octupolar tensor in (2) reduces to

[TABLE]

which features four independent elements, namely, $\alpha_{0}$, $\alpha_{2}$, $\alpha_{3}$, and $\beta_{3}$. Correspondingly, the octupolar potential (2) is

[TABLE]

for all ${\bf x}\in\mathbb{S}^{2}$. Without loss of generality, we can assume

[TABLE]

as a consequence of the following proposition.

Proposition 2.3**.**

For the octupolar potential (13), we have

[TABLE]

Next, we turn to the assumption that the North pole $(0,0,1)^{\top}$ is the maximum point of the octupolar potential with value $\alpha_{3}$.

Theorem 2.4**.**

Suppose that the North pole $(0,0,1)^{\top}$ is the maximum point of the octupolar potential $\Phi({\bf x})$ on $\mathbb{S}^{2}$. Then, we have that

[TABLE]

If the strict inequality holds in (6), then $(0,0,1)^{\top}$ is a (local) maximum point of $\Phi({\bf x})$ on $\mathbb{S}^{2}$.

Proof.

We consider the spherical optimization problem:

[TABLE]

Its Lagrangian is

[TABLE]

The Hessian of the Lagrangian is

[TABLE]

which is positive semidefinite on the tangent space ${\bf x}^{\bot}\equiv\{{\bf y}\in\mathbb{R}^{3}:{\bf x}^{\top}{\bf y}=0\}$ if ${\bf x}$ is a (local) maximum point of $\Phi({\bf x})$ on $\mathbb{S}^{2}$ [15]. Let $P\equiv I-{\bf x}{\bf x}^{\top}\in\mathbb{R}^{3\times 3}$ be the projection matrix onto ${\bf x}^{\bot}$. Then, the matrix $P^{\top}\nabla_{{\bf x}{\bf x}}^{2}L({\bf x},\lambda)P$ is positive semidefinite. By use of the first-order necessary condition,

[TABLE]

we have that

[TABLE]

Because the North pole $(0,0,1)^{\top}$ is a maximum point with the associated multiplier $\lambda=\alpha_{3}$, we arrive at

[TABLE]

As easily seen, this projected Hessian has eigenvalues

[TABLE]

which are all required to be non-negative [15]. Hence, we obtain the following inequality

[TABLE]

If the strict inequality holds, i.e., if $3\alpha_{3}^{2}-4\alpha_{3}\beta_{3}-4\beta_{3}^{2}-4\alpha_{0}^{2}>0$, then the North pole $(0,0,1)^{\top}$ is a (local) maximum point of $\Phi({\bf x})$ on $\mathbb{S}^{2}$ [15]. ∎

We first consider the case that $\alpha_{3}=0$. From Theorem 2.4, we know that $\alpha_{0}=\beta_{3}=0$. If $\alpha_{2}>0$, then $\Phi(0,1,0)=\alpha_{2}>\alpha_{3}=\Phi(0,0,1)$. This contradicts that $(0,0,1)^{\top}$ is a maximum point. Hence $\alpha_{2}=0$ and the octupolar tensor $\mathcal{A}$ is the trivial zero tensor.

In the remainder of this paper, we shall consider the case that $\alpha_{3}$ is positive. Without loss of generality, by Proposition 2.3 and Theorem 2.4, we can choose

[TABLE]

Then, the octupolar tensor

[TABLE]

has only three independent elements and the associated octupolar potential is given by

[TABLE]

for all ${\bf x}\in\mathbb{S}^{2}$.

3 The E-characteristic polynomial

Qi [18] introduced E-eigenvalues and E-eigenvectors for a symmetric tensor and showed that they are invariant under orthonormal coordinate changes. E-eigenvalues and E-eigenvectors were further studied in [19, 14, 4]. Furthermore, the coefficients of the E-characteristic polynomial of a tensor are orthonormal invariants of that tensor [10]. If the E-eigenvalues and the associated E-eigenvectors of a real-valued symmetric tensor are real, we call them the Z-eigenvalues and the Z-eigenvectors of the tensor, respectively.

Using the notion of resultant from algebraic geometry [5], we write now compute explicitly the resultant $\mathrm{Res}(\mathcal{A}{\bf x}^{2})$, where

[TABLE]

Since $F_{1}$, $F_{2}$, and $F_{3}$ are homogeneous polynomials with degree two in the variables $x_{1}$, $x_{2}$, and $x_{3}$, the multi-polynomial system

[TABLE]

has a trivial solution $x_{1}=x_{2}=x_{3}=0$. However, we are interested in its non-trivial solutions and we assume that the multi-polynomial system (14) has a non-zero common (complex) root.

According to Theorem 2.3 in Chapter 3 of [5], there is a unique irreducible polynomial $\mathrm{Res}(F_{1},F_{2},F_{3})$ such that $\mathrm{Res}(x_{1}^{2},x_{2}^{2},x_{3}^{2})=1$ and the system $F_{1}=F_{2}=F_{3}=0$ has a non-trivial solution over $\mathbb{C}$ if, and only if, $\mathrm{Res}(F_{1},F_{2},F_{3})=0$. We follow the approach in Chapter 3, $\S 4$ of [5]. Since the degree of each of the $F_{i}$ is $d_{i}=2$, we set

[TABLE]

We divide monomials ${\bf x}^{{\bm{\upsilon}}}=x_{1}^{\upsilon_{1}}x_{2}^{\upsilon_{2}}x_{3}^{\upsilon_{3}}$ of total degree $|{\bm{\upsilon}}|\equiv\upsilon_{1}+\upsilon_{2}+\upsilon_{3}=d$ into three sets:

[TABLE]

Clearly, there are ${d+2\choose 2}=15$ monomials ${\bf x}^{{\bm{\upsilon}}}$ with $|{\bm{\upsilon}}|=d$ and each lies in one of sets $S_{1}$, $S_{2}$, and $S_{3}$, which are mutually disjoint. We next write the system of equations

[TABLE]

Its coefficient matrix,

[TABLE]

in the unknowns ${\bf x}^{{\bm{\upsilon}}}$ with $|{\bm{\upsilon}}|=d$ is important in the sense that

[TABLE]

Now, we consider to the extraneous factor. A monomial ${\bf x}^{{\bm{\upsilon}}}$ of total degree $d=4$ is reduced if $x_{i}^{2}$ divides ${\bf x}^{{\bm{\upsilon}}}$ for exactly one $i$. Let $D^{\prime}$ be the determinant of the submatrix of $D$ obtained by deleting all rows and columns corresponding to reduced monomials, i.e.,

[TABLE]

By Theorem 4.9 in Chapter 3 of [5], to within a sign, the resultant reads as

[TABLE]

Theorem 3.1** ([5]).**

There exists a vector ${\bf x}\neq{\bf 0}$ such that $\mathcal{A}{\bf x}^{2}={\bf 0}$ if, and only if, $\mathrm{Res}(\mathcal{A}{\bf x}^{2})=0$, where the formula for $\mathrm{Res}(\mathcal{A}{\bf x}^{2})$ is given by (3).

By the same approach, we compute the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ of the octupolar tensor (12), which is a resultant of the following system of homogeneous polynomial equations

[TABLE]

Using the software Mathematica, we obtain the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$.

Theorem 3.2**.**

The E-characteristic polynomial of the octupolar tensor (12) is

[TABLE]

where

[TABLE]

and

[TABLE]

The E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ is a polynomial of degree $14$, with no odd-degree terms. By comparing the expression of $c_{12}$ and (3), we have the following corollary:

Corollary 3.3**.**

The constant term of the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ is

[TABLE]

This corollary is in agreement with Theorem 3.5 of [10] in this case.

Theorem 3.4**.**

Suppose that $\mathcal{A}{\bf x}^{2}={\bf 0}$ has only the trivial solution ${\bf x}={\bf 0}$. Then, the system (18) has a non-trivial solution $(x_{0},\ldots,x_{3})$ over $\mathbb{C}$ if, and only if, $\phi_{\mathcal{A}}(\lambda)=0$.

Proof.

Because $\mathcal{A}{\bf x}^{2}={\bf 0}$ has only the trivial solution ${\bf x}={\bf 0}$, we know that the system of $\mathcal{A}{\bf x}^{2}={\bf 0}$ and ${\bf x}^{\top}{\bf x}=0$ has only a zero solution. By Proposition 2.6 in Chapter 3 of [5], we obtain the desired conclusion. ∎

Corollary 3.5**.**

If $\mathrm{Res}(\mathcal{A}{\bf x}^{2})\neq 0$, then all E-eigenvalues of the octupolar tensor $\mathcal{A}$ are non-zero.

Proof.

Since $\mathrm{Res}(\mathcal{A}{\bf x}^{2})\neq 0$, the system $\mathcal{A}{\bf x}^{2}={\bf 0}$ has only the trivial solution by Theorem 3.1. Let us compute $\phi_{\mathcal{A}}(0)=[\mathrm{Res}(\mathcal{A}{\bf x}^{2})]^{2}>0$. By Theorem 3.4, $\lambda=0$ is not an E-eigenvalue of $\mathcal{A}$. ∎

4 Dome: the reduced admissible region

Assume that the North pole $(0,0,1)^{\top}$ is the global maximum point of the octupolar potential $\Phi({\bf x})$ on the unit sphere $\mathbb{S}^{2}$. That is, we assume that the octupolar tensor $\mathcal{A}$ has the largest Z-eigenvalue $\lambda=1$ with an associated Z-eigenvector $(0,0,1)^{\top}$. In the admissible region (11), there is a reduced region such that the maximal Z-eigenvalue of $\mathcal{A}$ is $1$. The boundary of this reduced admissible region is called the dome [7]: its apex is at $\alpha_{0}=0,\beta_{3}=-\frac{1}{2},\alpha_{2}=\frac{\sqrt{2}}{2}$, and it meets the plane $\alpha_{2}=0$ along the circle $\alpha_{0}^{2}+\beta_{3}^{2}+\beta_{3}=0$.

Now, we are in a position to give an explicit formula for the dome. We consider the E-characteristic polynomial $\phi_{\mathcal{A}}(\lambda)$ in Theorem 3.2. Clearly, $\lambda=1$ is a root of $\phi_{\mathcal{A}}(\lambda)$. Since the dome is the locus where the maximal Z-eigenvalue is $\lambda=1$, we substitute $\lambda=1$ into $\phi_{\mathcal{A}}(\lambda)/(\lambda^{2}-1)=0$ and we obtain the following equation

[TABLE]

where

[TABLE]

[TABLE]

and

[TABLE]

Because of (11), we know that $c_{1}(\alpha_{0},\beta_{3},\alpha_{2})\geq 0$ and the equality holds on the boundary of the admissible region. Hence, $c_{1}(\alpha_{0},\beta_{3},\alpha_{2})=0$ is a trivial solution of (19).

As for $c_{2}(\alpha_{0},\beta_{3},\alpha_{2})$, this is a quadratic function in $\alpha_{2}^{2}$ which attains its minimum value $4\alpha_{0}^{2}(4\alpha_{0}^{2}-3(1+2\beta_{3})^{2})^{2}\geq 0$. If $\alpha_{0}=0$, then $c_{2}(\alpha_{0},\beta_{3},\alpha_{2})=(-8\alpha_{2}^{2}+(1+2\beta_{3})^{3})^{2}$. Hence, when

[TABLE]

we have $c_{2}(\alpha_{0},\beta_{3},\alpha_{2})=0$. If $4\alpha_{0}^{2}-3(1+2\beta_{3})^{2}=0$, then $c_{2}=64(\alpha_{2}^{2}+(1+2\beta_{3})^{3})^{2}$. Hence, when

[TABLE]

we also have $c_{2}(\alpha_{0},\beta_{3},\alpha_{2})=0$. However, under either (21) or (22), there are two E-eigenvectors corresponding to the E-eigenvalue $1$. The first one is the North pole $(0,0,1)^{\top}$ and the other one is always a complex vector according to by our direct computations. Hence, we omit these two lines.

Then, we turn attention to the equation $c_{3}(\alpha_{0},\beta_{3},\alpha_{2})=0$, which has multiple roots in $\alpha_{2}$ for fixed $\alpha_{0}$ and $\beta_{3}$. For example, when $\alpha_{0}=0$ and $\beta_{3}=-0.8$, $\alpha_{2}^{(1)}=\alpha_{2}^{(2)}=\frac{2}{\sqrt{17}}\approx 0.4851$ and $\alpha_{2}^{(3)}=\frac{4\sqrt{7}}{5\sqrt{5}}\approx 0.9466$ are roots of the equation, whereas when $\alpha_{0}=0.1$ and $\beta_{3}=-0.8$, the roots are $\alpha_{2}^{(1)}=0.3765$, $\alpha_{2}^{(2)}=0.5862$, and $\alpha_{2}^{(3)}=9459$. Which value of $\alpha_{2}$ then describes the dome? If the largest Z-eigenvalue of $\mathcal{A}(\alpha_{0},\beta_{3},\alpha_{2})$ is larger that $1$, then the triple $(\alpha_{0},\beta_{3},\alpha_{2})$ is above the dome. By direct numerical explorations, we found that, for given $\alpha_{0}$ and $\beta_{3}$, the dome lies on the smallest non-negative value of $\alpha_{2}$ such that $c_{3}(\alpha_{0},\beta_{3},\alpha_{2})=0$, i.e.,

[TABLE]

The contour profile of the dome as given by (23) is illustrated in Figure 1.

Finally, we say more on the apex and the base of the dome. At the apex of the dome $(\alpha_{0},\beta_{3},\alpha_{2})=(0,-\tfrac{1}{2},\frac{\sqrt{2}}{2})$ and the E-characteristic polynomial of the octupolar tensor $\mathcal{A}(0,-\tfrac{1}{2},\frac{\sqrt{2}}{2})$ is $\phi_{\mathcal{A}}(\lambda)=19683\lambda^{6}(\lambda^{2}-1)^{4}.$There is a quadruple root $\lambda^{2}=1$ of such a tensor corresponding to four Z-eigenvectors:

[TABLE]

The polar plots of the octupolar potential $\Phi({\bf x})$ is illustrated in Figure 2(a).

At the base of the dome, $\alpha_{2}=0$ and $\alpha_{0}^{2}+\beta_{3}^{2}+\beta_{3}=0$, which represents a circle of center in $\alpha_{0}=0$, $\beta_{3}=\frac{1}{2}$ and radius $\frac{1}{2}$; there, the E-characteristic polynomial reduces to $\phi_{\mathcal{A}}(\lambda)=-64\lambda^{8}(\lambda^{2}-1)^{6}$. Hence, $\lambda^{2}=1$ is a triple root of $\phi_{\mathcal{A}}(\lambda)$. Specifically, $\mathcal{A}(0,0,0)$ has four Z-eigenvectors, namely,

[TABLE]

corresponding to the positive Z-eigenvalue $\lambda=1$. In this case, the polar plot of the octupolar potential $\Phi({\bf x})$ is illustrated in Figure 2(b). The two plots in Figure 2 illustrate the typical appearance of the octupolar potential in the two generic orientational states of generalized liquid crystals described by an octupolar order tensor.

5 Separatrix

Gaeta and Virga [7] showed by numerical continuation that there is a separatrix surface between the two different generic states of the octupolar potential $\Phi$: in one generic state, $\Phi$ has four maxima and three (positive) saddles; in the other generic state, $\Phi$ has three maxima and two (positive) saddles.444Since $\Phi(\mathbf{x})$ is odd in $\mathbf{x}$, both maxima (and positive saddles) of the octupolar potential are accompanied by an equal number of minima (and negative saddles) in the antipodal positions on the unit sphere. Here, we determine explicitly the separatrix.

We recall the spherical optimization problem (7). When passing through the separatrix, one maxima $\lambda=\Phi({\bf x})$ of the octupolar potential $\Phi({\bf x})$ on the unit sphere $\mathbb{S}^{2}$ disappears. Hence, the Hessian $\nabla_{{\bf x}{\bf x}}^{2}L({\bf x},\lambda)$ in (8) of the Lagrangian (7) is singular on the tangent space ${\bf x}^{\bot}$, i.e., the projected Hessian $P^{\top}\nabla_{{\bf x}{\bf x}}^{2}L({\bf x},\lambda)P$ in (10) has two zero eigenvalues. Clearly, $\mu_{1}=0$ is an eigenvalue of the projected Hessian (10) with the associated eigenvector ${\bf x}$. Suppose that $\mu_{2}$ and $\mu_{3}$ are the other eigenvalues of the projected Hessian (10). Let $\sigma\equiv\mu_{2}\mu_{3}=\mu_{1}\mu_{2}+\mu_{1}\mu_{3}+\mu_{2}\mu_{3}$. Then, by linear algebra, $\sigma$ equals the sum of all $2$-by-$2$ principal minors of the projected Hessian. Using (10), we have that

[TABLE]

Moreover, if $\lambda\neq 0$ and ${\bf x}\neq 0$ satisfy $\mathcal{A}{\bf x}^{2}=\lambda{\bf x}$, then $\frac{\lambda}{\|{\bf x}\|}$ and $\frac{1}{\|{\bf x}\|}{\bf x}$ satisfy (9). Hence, we could omit the spherical constraint ${\bf x}^{\top}{\bf x}=1$ temporarily and just consider the system of homogeneous polynomial equations (5) and

[TABLE]

Using the approach introduced in Section 3, we obtain the resultant of (5) and (25)

[TABLE]

where

[TABLE]

[TABLE]

Below the dome, the contour plot of the separatrix is illustrated in Figure 3. Figure 3 shows a $6$-fold symmetry, which confirms equation (39) of Gaeta and Virga [7]. We now contrast the separatrix and the dome given by (26) and (23) to the same surfaces found numerically in [7]. To this purpose, let

[TABLE]

with $\rho\in[0,\frac{1}{2}]$ and $\chi\in(-\pi,\pi]$. In Figure 4, we illustrate the cross-sections for $\chi=-\frac{\pi}{2},-\frac{5\pi}{12},-\frac{\pi}{3},-\frac{\pi}{4},-\frac{\pi}{6}$ of both the dome and the separatrix, in dash-dot lines and solid lines, respectively. It can easily be seen that Figure 4 is consistent with Figure 8 in [7].

It readily follows from (27) that separatix represented by (26) intersects the plane $\alpha_{2}=0$ along the circles whose radii are the positive roots of the polynomial

[TABLE]

the smallest of which identifies the base of the dome. As also shown in [7], the octupolar potential possesses a monkey saddle when the parameters are chosen on this circle. Apart from the three points shown in Figure 3 where the separatrix touches this circle, all other points of the latter do not properly belong to the separatrix defined as the locus that separates regions with four and three maxima of the octupolar potential: they will be considered as spurious points, and so discarded in the following.

Next, to reduce (26) to a simpler form, we study the special case where we set $\chi=-\frac{\pi}{2}$ in (27), so as to describe a cross-section of the separatrix that reaches the base of the dome. The equation $c_{3}(\alpha_{0},\beta_{3},\alpha_{2})=0$ for the dome reduces to

[TABLE]

Clearly, $\rho+1>0$. Because $(1-\rho)(1+2\rho)^{2}$ is monotonically increasing in $\rho\in[0,\frac{1}{2}]$, we get $(1-\rho)(1+2\rho)^{2}\geq 1$. Hence, by $2\alpha_{2}^{2}-(1-\rho)(1+2\rho)^{2}=0$ and $\alpha_{2}\geq 0$, we have that the region in parameter space where $\alpha_{2}\geq\frac{1}{\sqrt{2}}$ lies above the dome. Moreover, from $\alpha_{2}^{2}(\rho-2)-2\rho+1=0$, we know that for $\chi=-\frac{\pi}{2}$ the cross-section of the dome is the curve

[TABLE]

For $\chi=-\frac{\pi}{2}$, the separatrix could be rewritten as

[TABLE]

For $\rho\in[0,\frac{1}{2}]$, $(1-\rho)^{2}(1+\rho)^{3}>0$. Moreover, $\alpha_{2}^{2}+4(1-\rho)(1-2\rho)\geq 0$ and the equality holds if, and only if, $(\rho,\alpha_{2})=(\frac{1}{2},0)$. Also, $\alpha_{2}^{4}+\alpha_{2}^{2}(6\rho+4)+\rho^{2}(2\rho+1)^{2}\geq 0$ and the equality holds if, and only if, $(\rho,\alpha_{2})=(0,0)$. From $3(3-\rho)\alpha_{2}^{2}-4\rho^{2}(1-2\rho)=0$, we finally obtain that for $\chi=-\frac{\pi}{2}$ the cross-section of the separatrix is the curve

[TABLE]

By (28) and (29), for $\chi=-\frac{\pi}{2}$, the corresponding cross-sections are tangent at $\rho=\frac{1}{2}$ with vertical tangent. Since

[TABLE]

in $\rho\in[0,\frac{1}{2}]$ and the equality holds if, and only if, $\rho=\frac{1}{2}$, for $\chi=-\frac{\pi}{2}$ the dome is above the separatrix.

By a similar discussion applied at the case $\chi=-\frac{\pi}{6}$, we also obtain the following curves as representations of the meridian cross-sections of the dome and separatrix, respectively,

[TABLE]

They intersect for $(\rho,\alpha_{2})=\left(\frac{1}{3},\frac{\sqrt{2}}{3\sqrt{3}}\right)$.

6 Conclusion

We studied the octupolar tensor arising from liquid crystal science. The traceless property of octupolar tensors was shown to be preserved under orthogonal transformations. The resultant and the E-characteristic polynomial of the octupolar tensor were constructed explicitly. Using the resultant theory of algebraic geometry and the E-characteristic polynomial of the spectral theory of tensors, we gave an explicit, algebraic expression for the dome and the separatrix, the two significant surfaces for the representation of the octupolar order in three space dimensions. It would be interesting to apply the same algebraic techniques to higher order tensors (or in higher space dimensions) to see whether the pattern of multi-generic states described explicitly in this paper does indeed persist.