Nomenclature

I,	second-order identify tensor;	eiejek,	orthonormal base in three dimensions;
ε,	permutation tensor (i.e., LeviCivita tensor) with components εijk in three dimensions;	tr A, AT,	trace and transpose, respectively, of second-order tensor A;
Q,	orthogonal tensor with components qij;	det A,	determinant of second-order tensor A;
〈Q〉A,	second-order tensor A under orthogonal transformation;	,	second-order tensor with components εklikklj in three dimensions;
	real number field with dimension n;	εA,	third-order tensor with components εijlalk in three dimensions;
	Hall tensor with components kijk;	,	m×n matrix on real number field.

1 Introduction

The tensor function representation theory is an essential topic in continuum mechanics, which focuses on the tensor invariants under coordinate transformations. Since tensor invariants often reveal more intrinsic information of materials than tensor components, the complete and irreducible representation for invariant tensor functions plays a key role in modeling nonlinear constitutive equations in both theoretical and applied physics. Such representation prescribes the number and the type of scalar variables required in the constitutive equations. These representations are efficient in the process of describing the physical behavior of anisotropic materials, because the invariant conditions are dominant and no other simple methods are able to determine such information. There are plenty of research works on this topic from the last century^[1-11]. For instance, the minimal integrity basis and irreducible function basis of a second-order tensor in three dimensions were well studied by Wang^[7], Smith^[4], Boehler^[1], Pennisi and Trovato^[3], and Zheng^[8]. Furthermore, Boehler et al.^[2] investigated polynomial invariants for the elasticity tensor. Some recent works were devoted to minimal integrity bases and irreducible function bases for third-order and fourth-order tensors^[12-13].

The tensor function representation theory is also closely related to the classical invariant theory in algebraic geometry^[14-17]. One of the most famous approaches for computing the complete invariant basis was first introduced by Hilbert^[15]. In 2017, Olive et al.^[18] studied the minimal integrity basis with 297 invariants of the fourth-order elasticity tensors successfully via the approaches from the algebraic geometry.

The Hall effect is an important magnetic effect observed in electric conductors and semiconductors^[19]. It was discovered in 1879 by and named after Edwin Hall^[20]. When an electric current density J is flowing through a plate and the plate is simultaneously immersed in a magnetic field H with a component transverse to the current, the electric field strength E is proportional to the current density and the magnetic field strength,

where the third-order tensor with components k_ijk is called the Hall tensor^[19]. We note that the representation of the Hall tensor under any orthonormal basis satisfies k_ijk = -k_jik for all i, j, k=1, 2, 3, since the Onsager relation for transport processes with time reversal is valid. The Hall tensor is essential for describing the electromagnetic induction. Therefore, it is significant to investigate the minimal integrity basis and irreducible function basis of the Hall tensor. In physics, there are other tensors which are third-order three-dimensional tensors whose first two indices are antisymmetric, for example, the tensors in the Faraday effect^[19].

This paper is devoted to the invariants of the Hall tensor and organized as follows. We first briefly review some basic definitions and the relationship between the integrity basis and the function basis of a tensor in Section 2. Then, we build a connection between the invariants of a Hall tensor and that of a second-order tensor, which is important for the subsequent contents. Furthermore, we propose a minimal isotropic integrity basis with 10 isotropic invariants of the Hall tensor in Section 3. In Section 4, we prove that the minimal integrity basis with 10 invariants of the Hall tensor is also its irreducible function basis. Finally, we draw some concluding remarks and raise one further question in Section 5.

2 Preliminaries

Denote as an mth-order tensor represented by a_i1i2…im under some orthonormal coordinates. A scalar-valued tensor function f() is called an isotropic invariant of if it is invariant under any orthogonal transformations, including rotations and reflections, i.e.,

or equivalently expressed by

where Q is an orthogonal tensor (QQ=QQ=I) with components q_ij. If f() is only invariant under rotations, i.e., f(〈Q〉)=f() for any orthogonal tensor Q with det Q=1, it is called a hemitropic invariant of tensor . Furthermore, if f() is a polynomial, it is called a polynomial invariant of . In the subsequence, invariants always stand for polynomial invariants unless specific remarks are made there.

For any second-order tensor, the hemitropic invariants and the isotropic invariants are equivalent, since it keeps unaltered under the central inversion -I^[9]. Nevertheless, any isotropic polynomial invariant of a third-order tensor has to be the summation of several even order degree polynomials.

We then briefly review the definitions and properties of (minimal) integrity bases and (irreducible) function bases of a tensor.

Definition 1 (integrity basis) Let Ψ = {ψ₁, ψ₂, …, ψ_r} be a set of isotropic or hemitropicv, (or hemitropic, respectively) invariants of a tensor .

(ⅰ) Ψ is said to be polynomial irreducible if none of ψ₁, ψ₂, …, ψ_r can be expressed by a polynomial of the remainders;

(ⅱ) Ψ is called an isotropic (hemitropic, respectively) integrity basis if any isotropic (or hemitropic, respectively) invariant of is expressible by a polynomial of ψ₁, ψ₂, …, ψ_r;

(ⅲ) Ψ is called an isotropic (or hemitropic, respectively) minimal integrity basis if it is polynomial irreducible and an isotropic(or hemitropic, respectively) integrity basis.

Definition 2 (function basis) Let Ψ = {ψ₁, ψ₂, …, ψ_r} be a set of isotropic or hemitropic, respectively) invariants of a tensor .

(ⅰ) An invariant in Ψ is said to be functionally irreducible if it cannot be expressed by a single-valued function of the remainders. Ψ is said to be functionally irreducible if all of ψ₁, ψ₂, …, ψ_r are functionally irreducible;

(ⅱ) Ψ is called an isotropic (or hemitropic, respectively) function basis if any isotropic (or hemitropic, respectively) invariant of is expressible by a function of ψ₁, ψ₂, …, ψ_r;

(ⅲ) Ψ is called an isotropic (or hemitropic, respectively) irreducible function basis if it is functionally irreducible and is an isotropic(or hemitropic, respectively) function basis.

It is straightforward to verify by definitions that an isotropic (or hemitropic, respectively) integrity basis is an isotropic (or hemitropic, respectively) function basis, but the converse is incorrect in general. Thus, the number of invariants in an isotropic (or hemitropic, respectively) irreducible function basis is no greater than the number of invariants in an isotropic (or hemitropic, respectively) minimal integrity basis. For instance, the number of irreducible function basis of a third-order traceless symmetric tensor is 11 which is less than 13, the number of invariants in its minimal integrity basis^[13].

Particularly, it has been proved that the number of invariants of each degree in an isotropic (or hemitropic, respectively) minimal integrity basis is always fixed^[18]. Nevertheless, it is still unclear whether the number of invariants of an irreducible function basis is fixed.

3 Minimal integrity basis of Hall tensor

Let be a Hall tensor represented by k_ijk under an orthonormal basis e_ie_je_k. Define a second-order tensor A accordingly, with components a_ij under this orthonormal basis, by the tensor product operation,

or equivalently,

where ε is the third-order Levi-Civita tensor. Conversely, the Hall tensor can also be expressed with this second-order tensor by

or equivalently,

There are nine independent components in a Hall tensor due to the anti-symmetry of the first two indices of the components in a Hall tensor. Without loss of generality, denote the nine independent components of the Hall tensor as k₁₂₁, k₁₂₂, k₁₂₃, k₁₃₁, k₁₃₂, k₁₃₃, k₂₃₁, k₂₃₂, and k₂₃₃.

Then, under a right-handed coordinate, the representation of the associated second-order tensor can be written in a matrix form mathematically,

The following theorem reveals the relationship between the invariants of the Hall tensor and the ones of its associated second-order tensor.

Theorem 1 Let be a Hall tensor with components k_ijk. We use A() to denote the associated second-order tensor of .

(ⅰ) Any isotropic invariant of is an isotropic invariant of A().

(ⅱ) Any isotropic invariant of A() with even degree is an isotropic invariant of , and any isotropic invariant of A() with odd degree is a hemitropic invariant of .

Proof (ⅰ) An isotropic invariant f() of the Hall tensor is also a polynomial function of its associated second-order tensor A(), denoted by g(A) := f(εA). Now, we need to show that g(A) is an isotropic invariant of A(). Let Q be any orthogonal tensor. According to the definition of isotropic invariants, we have

Make use of the equality 〈Q〉ε = (det Q)ε. Then,

An isotropic invariant of a third-order tensor must be an even function,

Hence, g(〈Q〉A)=g(A), i.e., g(A) is an isotropic invariant of A.

(ⅱ) Denote an invariant of A() as g(A). It is also a polynomial of the Hall tensor , denoted by f() := . For any orthogonal tensor Q, since g(A) is an invariant, we know

Recall 〈Q〉ε = (det Q)ε. Then,

Hence, when g(A) is an invariant of even degree, we have f(〈Q〉)=f() for any orthogonal tensor Q. That is, f() is an isotropic invariant of the Hall tensor .

When g(A) is an invariant of odd degree, only for orthogonal tensor Q satisfying det Q =1, it holds that f(〈Q)=f(), which means that f() is a hemitropic invariant of the Hall tensor . The proof is completed.

Hence, we can construct an integrity basis for a Hall tensor from the integrity basis of its associated second-order tensor. For the associated second-order tensor A(), we split it into A() = T + W, where T is symmetric with components t_ij = and W is skew-symmetric with components w_ij = . It is well known that 7 invariants tr T, tr T², tr T³, tr W², tr TW², tr T² W², and tr T² W² T W form a minimal integrity basis of A() and also an irreducible function basis as well. We denote the invariants of A() as follows:

The following theorem shows the way to obtain a minimal integrity basis of from this particular minimal integrity basis of A().

Theorem 2 Let be a Hall tensor with components k_ijk, and A() be its associated second-order tensor with components a_ij. Denote K₂ := I₁², J₄ := I₁I₃, K₄ := I₁J₃, J₆ := I₃², K₆ := J₃², and L₆ :=I₃J₃. Then, the invariant set

(1)

is a minimal integrity basis of .

Proof By Theorem 1, any isotropic invariant of is also an invariant of A(), and thus can be expressed by a polynomial p(I₁, I₂, J₂, I₃, J₃, I₄, I₆). Moreover, any isotropic invariant of an even order tensor consists of several even degree monomials. Each even degree monomial containing I₁, I₃, and J₃ should be a polynomial of I₁², I₁I₃, I₁J₃, I₃², I₃J₃, and J₃². Therefore, the isotropic invariant p(I₁, I₂, J₂, I₃, J₃, I₄, I₆) can also be written into a polynomial of the invariants in (1). That is, (1) is an integrity basis of .

Next, we need to verify the polynomial irreducibility of this integrity basis. A natural observation is that these isotropic invariants are homogenous polynomials of the 9 independent components in the Hall tensor . A similar approach as the method proposed by Chen et al.^[12] is employed in this part.

(i) There are exactly 3 degree-2 isotropic invariants I₂, J₂, and K₂ in this integrity basis. Take I₂ for example. If it is not polynomial irreducible with the other 9 invariants in this basis, it has to be a linear combination of the other 2 degree-2 invariants J₂ and K₂. Therefore, if I₂, J₂ and K₂ are polynomial irreducible, the unique triple of (c₁, c₂, c₃) such that

(2)

is c₁ = c₂ = c₃ =0. Note that (2) holds for an arbitrary Hall tensor. Thus, when we generate n points y₁, y₂, …, y_n∈⁹ where ⁹ is the real number field with dimension 9, c₁, c₂, and c₃ must be the solution to the linear system of equations,

(3)

The coefficient matrix of System (3) is denoted by M₂, and denote r(M₂) as the rank of the coefficient matrix M₂. Then, r(M₂) shows the number of polynomial irreducible invariants in these three isotropic invariants. Take n=3 and

By numerical calculations, we can determine r(M₂)=3. Hence, the only solution to (3) is c₁ = c₂ = c₃ =0, which implies that these three invariants of degree 2 are polynomial irreducible.

(ⅱ) For the invariants of degree 4, we need to consider the following linear equation:

(4)

where c₁, c₂, …, c₉ are scalars. If the unique (c₁, c₂, …, c₉) such that (4) holds for any Hall tensor is (0, 0, ..., 0), all the 3 degree-4 invariants I₄, J₄, and K₄ are polynomial irreducible. We generate n points y₁, y₂, …, y_n∈⁹ and consider the following linear system:

(5)

The coefficient matrix of (5) is denoted by M₄. Take n=9 and

We can verify that the rank of M₄ is r(M₄)= 9, which implies that these three degree-4 invariants cannot be polynomial represented by other invariants of degree-4 and degree-2.

(ⅲ) Similarly, in the case of degree 6, the verification linear equation is

(6)

Thus, we generate n points y₁, y₂, …, y_n∈⁹. Consider a linear system similar with the system (5). Its coefficient matrix is denoted as M₆, and its rank is denoted by r(M₆). Take n=23 and

Then, r(M₆)= 23, which implies that these four invariants with degree 6 are polynomial irreducible in the integrity basis.

Therefore, we have shown that (1) is a minimal integrity basis of .

In the above discussion, we fix the inducing initial, i.e., a particular minimal integrity basis of the second-order tensor. Nevertheless, the minimal integrity basis is generally not unique. We can also start from another minimal integrity basis of the second-order tensor, denoted by

Construct another integrity basis of the Hall tensor in the same way, where

Since this integrity basis has already got the same number of invariants as the minimal integrity basis (1), it must also be a minimal integrity basis. Therefore, we have the following corollary.

Corollary 1 Let be a Hall tensor with components k_ijk, and A() be its associated second-order tensor with components a_ij. Let be any minimal integrity basis of the second-order tensor A(). Denote with , and . Then, Ψ is a minimal integrity basis of the Hall tensor .

4 Irreducible function basis

Since a minimal integrity basis for a tensor is also a function basis, the number of invariants in an irreducible function basis consisting of polynomial invariants is no more than that of a minimal integrity basis. Moreover, the number of invariants in a minimal integrity basis of a tensor can be very big. For example, the number of minimal integrity basis of an elasticity tensor is 297^[18]. However, from an experimental point of view, it will be easier to detect all the values of the invariants in an irreducible function basis of a tensor. Hence, it is meaningful to study the irreducible function basis of a tensor. For a symmetric third-order tensor, one of its irreducible function base contains 11 invariants, while its minimal integrity basis contains 13 invariants^[13].

In this section, we shall show that the minimal integrity basis given in Section 3 is also an irreducible function basis of the Hall tensor . According to the method proposed by Pennisi and Trovato^[3] in 1987, to show that a given function basis of a tensor is functionally irreducible, for each invariant in this basis, we need to find two different sets of independent variables in the tensor, denoted by V and V^', such that this invariant takes different values in V and V^', while all the remainders are the same in V and V^'. The following theorem is proved in this spirit.

Theorem 3 The set {I₂, J₂, K₂, I₄, J₄, K₄, I₆, J₆, K₆, L₆} is an irreducible function basis of the Hall tensor.

Proof. It can be verified by definitions that an integrity basis of a tensor is a function basis of the tensor. We have proved in Section 3 that these ten invariants form a minimal integrity basis of the Hall tensor. Thus, this basis is also a function basis.

Denote V={k₁₂₁, k₁₂₂, k₁₂₃, k₁₃₁, k₁₃₂, k₁₃₃, k₂₃₁, k₂₃₂, k₂₃₃} and V^'={k₁₂₁^', k₁₂₂^', k₁₂₃^', k₁₃₁^', k₁₃₂^', k₁₃₃^', k₂₃₁^', k₂₃₂^', k₂₃₃^'} as two different sets of independent variables of the Hall tensor . Then, we shall find ten pairs of {V, V^'} to show that all the ten isotropic invariants in (1) are functionally irreducible.

(ⅰ) For I₂, in V, let k₁₂₁ = k₁₂₂ = k₁₂₃ = k₁₃₂ = k₁₃₃ = k₂₃₁ = k₂₃₃ = 0, k₁₃₁= -1, and k₂₃₂= 1.

Then, in V^', let k₁₂₁^' = k₁₂₂^' = k₁₂₃^' = k₁₃₂^' = k₁₃₃^' = k₂₃₁^' = k₂₃₃^' = 0, k₁₃₁^'= -2, and k₂₃₂^'= 2.

We have I₂ = 2 and I₂^' = 8, while other invariants { J₂, K₂, I₄, J₄, K₄, I₆, J₆, K₆, L₆} and {J₂^', K₂^', I₄^', J₄^', K₄^', I₆^', J₆^', K₆^', L₆^'} are all equal to 0. This means that I₂ is functionally irreducible in the function basis (1).

(ⅱ) For J₂, in V, let k₁₂₁ = k₁₂₂ = k₁₂₃ = k₁₃₂ = k₁₃₃ = k₂₃₁ = k₂₃₃ = 0, k₁₃₁= 1, and k₂₃₂= 1. Then, in V^', let all the variables be 0.

We have J₂ = 2 and J₂^' = 0, while the other invariants {I₂, K₂, I₄, J₄, K₄, I₆, J₆, K₆, L₆} and {I₂^', K₂^', I₄^', J₄^', K₄^', I₆^', J₆^', K₆^', L₆^'} are all equal to 0. This means that J₂ is functionally irreducible in the function basis (1).

(ⅲ) For K₂, in V, let k₁₂₁ = k₁₂₂ = k₁₃₁ = k₁₃₃ = k₂₃₂ = k₂₃₃ = 0, , k₁₃₂= 0, and k₂₃₁= .

In V^', let k₁₂₁^' = k₁₂₂^' = k₁₃₁^' = k₁₃₃^' = k₂₃₂^' = k₂₃₃^' = 0, k₁₂₃^'= 1, , and k₂₃₁^'= 1.

We have K₂ = 0. It is not equal to K₂^' = (2 -, while the other invariants are I₂ = I₂^'= 2 + and J₂ = J₂^'= I₄ = I₄^' = J₄ = J₄^' = K₄ = K₄^' = I₆ = I₆^' = J₆ = J₆^' = K₆ = K₆^' = L₆ = L₆^' = 0. This means that K₂ is functionally irreducible.

(ⅳ) For I₄, in V, let k₁₂₁=-2, k₁₂₂=0, k₁₂₃=1, k₁₃₁=1, k₁₃₂=1, k₁₃₃=0, k₂₃₁=0, k₂₃₂=1, and k₂₃₃=2.

In V^', let k₁₂₁^'=-, k₁₂₂^'=-, k₁₂₃^'=1, k₁₃₁^'=0, k₁₃₂^'=1, k₁₃₃^'=-, k₂₃₁^'=0, k₂₃₂^'=0, and k₂₃₃^'=.

We have I₄ = 5. It is not equal to I₄^' = 7, while I₂ = I₂^'= 2, J₂ = J₂^'= 10, K₆ = K₆^'= 9, and the others are all equal to 0. This means that I₄ is functionally irreducible.

(ⅴ) For J₄, assume s=4+ and t=4-. In V, let k₁₂₁ = k₁₂₂ = k₁₃₁ = k₁₃₃ = k₂₃₂ = k₂₃₃ = 0, and

In V^', let k₁₂₁^' = k₁₂₂^' = k₁₃₁^' = k₁₃₃^' = k₂₃₂^' = k₂₃₃^' = 0,

and k₂₃₁^'= 0.

We have . Meanwhile,

and the others are all equal to 0. This shows that J₄ is functionally irreducible.

(ⅵ) For K₄, in V, let

In V^', let

We have

Meanwhile,

and the others are all equal to 0. This shows that K₄ is functionally irreducible.

(ⅶ) For I₆, in V, let k₁₂₁= -1, k₁₂₂=-1, k₁₂₃= 1, k₁₃₁= 1, k₁₃₂= 1, k₁₃₃= -1, k₂₃₁=0, k₂₃₂=1, and k₂₃₃=1.

In V^', let k₁₂₁^'= -1, k₁₂₂^'= -1, k₁₂₃^'= 1, k₁₃₁^'= -1, k₁₃₂^'= 1, k₁₃₃^'= -1, k₂₃₁^'=0, k₂₃₂^'= -1, and k₂₃₃^'= 1.

We have I₆ = - I₆^' =2. Meanwhile, I₂ = I₂^'= 2, J₂ = J₂^'= 6, I₄ = I₄^'= 4, and the others are all equal to 0. This means that I₆ is functionally irreducible.

(ⅷ) For J₆, in V, let k₁₂₁ = k₁₂₂ = k₁₃₁ = k₁₃₃ = k₂₃₂ = k₂₃₃ = 0, k₁₂₃= , k₁₃₂= 0, and k₂₃₁= .

In V^', let k₁₂₁^' = k₁₂₂^' = k₁₃₁^' = k₁₃₃^' = k₂₃₂^' = k₂₃₃^' = 0, k₁₂₃^'= , k₁₃₂^'= , and k₂₃₁^'= .

We have J₆ = 0 ≠ J₆^' = 144. Meanwhile, I₂ = I₂^'=6, and the others are all equal to 0. This means that J₆ is functionally irreducible.

(ⅸ) For K₆, in V, let k₁₂₁=, k₁₂₂=1, k₁₂₃=0, k₁₃₁=, k₁₃₂=0, k₁₃₃=1, k₂₃₁=0, k₂₃₂=, and k₂₃₃=.

In V^', let k₁₂₁^'=-, k₁₂₂^'=, k₁₂₃^'=0, k₁₃₁^'=, k₁₃₂^'=0, k₁₃₃^'=, k₂₃₁^'=0, k₂₃₂^'=, and k₂₃₃^'=-.

We have . Meanwhile, , and the others are all equal to 0. This means that K₆ is functionally irreducible.

(ⅹ) For L₆, in V, let k₁₂₁= -1, k₁₂₂=, k₁₂₃=-1, k₁₃₁=0, k₁₃₂=2, k₁₃₃=, k₂₃₁=3, k₂₃₂=0, and k₂₃₃=1.

In V^', let , , and k₂₃₃^'= 0.

We have . Meanwhile, , , and the others are all equal to 0. This means that L₆ is functionally irreducible.

Therefore, this particular minimal integrity basis {I₂, J₂, K₂, I₄, J₄, K₄, I₆, J₆, K₆, L₆} is also an irreducible function basis of the Hall tensor .

In the above proof, the examples V and V^' in the cases (ⅰ), (ⅱ), (ⅳ), and (ⅶ) are based on related sets in Pennisi and Trovato^[3], while the examples V and V^' in the case (v) are suggested by Dr. Yannan Chen.

5 Conclusions and further question

In this paper, we investigate isotropic invariants of the Hall tensor. For this purpose, we connect the invariants of the Hall tensor with the ones of its associated second-order tensor A(). A() can be split into a second-order symmetric tensor T and a second-order skew-symmetric tensor W. Then, { I₁ := tr T, I₂ := tr T², J₂ := tr W², I₃:= tr T³, J₃ := tr T W², I₄:= tr T² W², I₆ := tr T² W² T W} is the minimal integrity basis of A() as in the previous sections. It is also an irreducible function basis of A(). We prove in this paper the following statements:

(ⅰ) {I₁², I₂, J₂, I₄, I₁I₃, I₁J₃, I₆, I₃², J₃²,I₃J₃} is an isotropic minimal integrity basis of the Hall tensor .

(ii) {I₁², I₂, J₂, I₄, I₁I₃, I₁J₃, I₆, I₃², J₃²,I₃J₃} is also an isotropic irreducible function basis of the Hall tensor as well.

Apart from this particular selection, we can also begin with any minimal integrity basis of the second-order tensor and use the same approach to construct an invariant basis of the Hall tensor. We prove in the paper that such basis of the Hall tensor is a minimal integrity basis.

A further question is whether there exists an irreducible function basis consisting of less than ten polynomial invariants.

Acknowledgements

We are thankful to Dr. Yannan CHEN for his helpful discussion.