A Direct Approach to the Polar Representation of Plane Tensors

Abstract

We show we can derive the so called polar representation of 2D symmetric second and fourth order real tensors essentially just relying on the spectral theorem for unitary tensors in the complex field. The use of a coordinate-free approach allows us to clearly and methodically detect rotation-invariant quantities and to readily (and directly) deduce the representation theorems.

1 Introduction

Representation theorems for tensors, especially for elasticity-like tensors, have been quite investigated in the last decades, in particular in connection with symmetry classes [6, 7, 20, 21]. The basic tools are of geometric nature and rely on group representations and harmonic decomposition [15, 16].

In this paper, we are interested in plane problems. Among the various ways to represent the plane elasticity tensor, a remarkable one is the so-called polar method (or polar formalism). The polar method was introduced by Verchery in the 70s [32] to describe the behavior of 2D, anisotropic, linear elastic materials. This technique relies upon an algebraic, complex variable change, which was essentially introduced by Kolosov [12], Muskhelishvili [19], and by Green and Zerna [9].

The Verchery transform works as follows (we refer to [29] for a complete presentation). Fix a Cartesian frame $(O; x_{1}, x_{2})$. Let $\mathbf {x}:=(x_{1},x_{2})$, $z:=x_{1} + \mathrm {i}\mkern 1mux_{2}$, $k:= e^{\mathrm {i}\mkern 1mu\frac{\pi}{4}}$, and introduce the following transformation:

$$ \mathsf{x}^{1} := \frac{1}{\sqrt{2}} \overline{k}z, \qquad \mathsf{x}^{2} := \overline{\mathsf{x}^{1}}, $$

which can be recast into $\mathbf{\mathsf{x}} = \mathbf{M}\mathbf {x}$ where $\mathbf{\mathsf{x}} := (\mathsf{x}^{1},\mathsf{x}^{2})$, and $\mathbf{M}$ is a complex unitary matrix. A 2D, fourth order tensor $\mathbb{T}$ of the type of (hyper)elasticity has six independent components. Denoting by $\mathbf{t}$ the tuple collecting the Cartesian components of $\mathbb{T}$ and by $\mathsf{t}$ the tuple collecting the Verchery-transformed ones, we get after rather lengthy algebraic computations $\mathsf{t} = \mathbf{M}_{4} \mathbf{t}$ for some $6\times 6$ complex, unitary matrix $\mathbf{M}_{4}$.

If we further consider a new Cartesian frame $(O; x_{1}', x_{2}')$ rotated by an angle $\theta$ with respect to the previous one, and we pose $r:=e^{-\mathrm {i}\mkern 1mu\theta}$, $z' := r z$, we get the “rotated” Verchery-transformed quantities

$$ (\mathsf{x}^{1})' := r\mathsf{x}^{1}, \qquad (\mathsf{x}^{2})' := \overline{r}\mathsf{x}^{2}, $$

so that we can write after some manipulation $\mathbf{\mathsf{x}}' = \mathbf{R}\mathbf{\mathsf{x}}$ for a diagonal complex matrix $\mathbf{R}$. After again some lengthy computations, one obtains a similar result for elasticity tensor:

$$ \mathsf{t}' = \mathbf{R}_{4} \mathsf{t} $$

for some $6\times 6$ diagonal complex matrix $\mathbf{R}_{4}$. Scrutiny of $\mathbf{R}_{4}$ suggests the definition of invariant quantities, which are called polar parameters. Unraveling the Verchery transform, one finally obtains the expressions of the Cartesian components of $\mathbb{T}$ in terms of the polar parameters.

The representation of 2D elasticity-like tensors via the polar method paved the way to a deeper understanding of such objects. Polar parameters are in fact intrinsically linked to the symmetries of the tensor. In particular, the possible symmetries emerge as one or more of the polar parameters assume a special value (often, zero).

It is remarkable that the polar representation allows a refinement of the classical picture given by the geometric approaches. In fact, it has been shown that a geometrical symmetry (an invariance to some rotation/reflection) may actually contain different mechanical (algebraic) symmetries, indistinguishable from the geometric viewpoint, but which lead to different mechanical behaviors. It is the case of the so-called $R_{0}$-orthotropy [27, 28], which is a special case of orthotropy. Unlike the case of ordinary orthotropy, if, say, the stiffness tensor exhibits $R_{0}$-orthotropy, its inverse (the compliance tensor) need not. As a further achievement, the polar method allows us to show the existence, in 2D, of rari-constant materials ¹ [30].

Besides the theoretical interest, the 2D setting is, at least from the structural viewpoint, of paramount importance for practical applications. In modern engineering and design, thin structures, both intrinsically flat or curved (plates or shells, in jargon) occupy a central role. The polar formalism has been successfully adopted in many cases to describe the anisotropy of stratified plates [8, 17, 22–24, 31], shells [3, 13], and even for plates having a continuously-variable stiffness [5, 11, 18].

In this paper, we show we can deduce the polar representation in a clear, coordinate-free way. In so doing, we further provide a simple geometric interpretation of the algebraic Verchery polar method, somehow reconciling the two viewpoints. Some classical results (as those announced in [4]) are re-obtained in a simple and natural way, with a clear geometrical interpretation.

The paper is structured as follows. In Sect. 2 we provide a self-contained review of complex inner product spaces, which are the underlying setting of this work. Despite many results are classical in the real setting, attention must be paid in the complex field, since the inner product is not bilinear, nor commutative (see [1, 10] for further details). In Sect. 3 we derive the polar representation for vectors, 2D second order tensors, and finally for 2D elasticity-like tensors.

2 Preliminaries

Throughout the paper, let $V$ be a finite dimensional inner product vector space over the field ℂ of complex numbers. For $z\in \mathbb{C}$, let $\overline{z}$ denote the complex conjugate of $z$. $V$ is endowed with a Hermitian inner product, i.e. a sesquilinear operator $\langle \cdot ,\cdot \rangle :V\times V \to \mathbb{C}$ such that

$$ \begin{aligned} \langle \mathbf {u}, \mathbf {v}\rangle = \overline{\langle \mathbf {v}, \mathbf {u}\rangle}, \qquad \langle a\mathbf {u}+ b\mathbf {w}, \mathbf {v}\rangle = a\langle \mathbf {u},\mathbf {v}\rangle + b\langle \mathbf {w},\mathbf {v}\rangle , \\ \langle \mathbf {u}, \mathbf {u}\rangle \geq 0, \qquad \langle \mathbf {u}, \mathbf {u}\rangle = 0 \iff \mathbf {u}=\mathbf{0} \end{aligned} $$

(1)

for any $\mathbf {u}, \mathbf {v}, \mathbf {w}\in V$ and all $a,b \in \mathbb{C}$. From the very definition of inner product, it follows that

$$ \langle \mathbf {u}, a\mathbf {v}\rangle = \overline{a}\langle \mathbf {u},\mathbf {v}\rangle . $$

The inner product is linear in the first argument and conjugate linear in the second.² We call vectors the elements of $V$. The induced norm is the nonnegative real number $|\mathbf {u}| := \sqrt{\langle \mathbf {u}, \mathbf {u}\rangle}$. We say two vectors are orthogonal if and only if $\langle \mathbf {u},\mathbf {v}\rangle = 0$ (and we write $\mathbf {u}\perp \mathbf {v}$). We call unit vector a vector of norm 1.

We denote by $\mathcal{L}(V)$ the set of all linear operators from $V$ to $V$. We define the identity of $\mathcal{L}(V)$ as the unique operator $\mathbf {I}$ such that $\mathbf {I}\mathbf {u}= \mathbf {u}$ for all $\mathbf {u}\in V$. The product of two operators $\mathbf {A}, \mathbf {B}\in \mathcal{L}(V)$ is defined by the relation

$$ (\mathbf {A}\mathbf {B})\mathbf {u}= \mathbf {A}(\mathbf {B}\mathbf {u}) \qquad \forall \mathbf {u}\in V. $$

Given $\mathbf {A}\in \mathcal{L}(V)$, the adjoint operator $\mathbf {A}^{*}\in \mathcal{L}(V)$ is the unique linear operator verifying the identity

$$ \langle \mathbf {A}\mathbf {u}, \mathbf {v}\rangle = \langle \mathbf {u}, \mathbf {A}^{*}\mathbf {v}\rangle \qquad \forall \mathbf {u}, \mathbf {v}\in V.$$

It is easy to verify that the adjoint operator enjoys the following properties:

$$ (\mathbf {A}+ \mathbf {B})^{*} = \mathbf {A}^{*} + \mathbf {B}^{*}, \qquad (a\mathbf {A})^{*} = \overline{a}\mathbf {A}^{*}, \qquad (\mathbf {A}\mathbf {B})^{*} = \mathbf {B}^{*}\mathbf {A}^{*} \qquad \forall \mathbf {A},\mathbf {B}\in \mathcal{L}(V), \ a \in \mathbb{C}. $$

We say an operator $\mathbf {A}\in \mathcal{L}(V)$ is

• normal if and only if $\mathbf {A}\mathbf {A}^{*} = \mathbf {A}^{*} \mathbf {A}$;

• unitary if and only if $\mathbf {A}^{-1} = \mathbf {A}^{*}$; if moreover $\det \mathbf {A}=1$ it is called special unitary. It is the counterpart of orthogonal and special orthogonal operators in the classical real setting.

• Hermitian (or self-adjoint) if and only if $\mathbf {A}= \mathbf {A}^{*}$.

The spectral theorem guarantees that any normal operator $\mathbf {A}$ is diagonalizable over the complex field. There exist complex numbers $\{\lambda _{i}\}$ (the spectrum) and a family of orthogonal projections $\{\mathbf{U}_{i}\}$ ($\mathbf {U}^{2} = \mathbf {U}$, $\mathbf {U}^{*} = \mathbf {U}$), such that $\sum _{i=1}^{n} \mathbf{U}_{i} = \mathbf {I}$ (orthogonal resolution of the identity of $\mathcal{L}(V)$), $\langle \mathbf {U}_{i}, \mathbf {U}_{j} \rangle = 0$ if $i \neq j$, $\langle \mathbf {U}_{i}, \mathbf {U}_{i} \rangle = 1 \ \forall i$, such that

$$ \mathbf {A}= \sum _{i=1}^{n} \lambda _{i} \mathbf {U}_{i}.$$

where $n$ is the dimension of $V$. It happens that unitary and Hermitian operators are normal.

The dyad between any $\mathbf {u},\mathbf {v}\in V$ is the linear operator $\mathbf {u}\otimes \mathbf {v}\in \mathcal{L}(V)$ defined by the following action on $V$:

$$ \mathbf {u}\otimes \mathbf {v}(\mathbf {w}) = \langle \mathbf {w},\mathbf {v}\rangle \mathbf {u}\qquad \forall \mathbf {w}\in V.$$

(2)

We will call such operators second order tensors. The operation ⊗ denotes the dyadic (or tensor) product (see the Appendix). The symmetric dyadic product is indicated by ⊙, so that $\mathbf {u}\odot \mathbf {v}:= \frac{1}{2}(\mathbf {u}\otimes \mathbf {v}+ \mathbf {v}\otimes \mathbf {u})$. By (2) and the sesquilinearity of the inner product, it follows that

$$ (a\mathbf {u}) \otimes \mathbf {v}= a(\mathbf {u}\otimes \mathbf {v}), \qquad \mathbf {u}\otimes (a \mathbf {v}) = \overline{a}(\mathbf {u}\otimes \mathbf {v}) \qquad \forall a\in \mathbb{C}.$$

We have also

$$ \begin{aligned} (\mathbf {u}\otimes \mathbf {v})(\mathbf {w}\otimes \mathbf {x}) &= \langle \mathbf {w}, \mathbf {v}\rangle (\mathbf {u}\otimes \mathbf {x}) \qquad &&\forall \mathbf {u}, \mathbf {v}, \mathbf {w}, \mathbf {x}\in V, \\ (\mathbf {u}\otimes \mathbf {v})^{*} &= \mathbf {v}\otimes \mathbf {u}\qquad &&\forall \mathbf {u}, \mathbf {v}\in V, \\ \mathbf {A}(\mathbf {u}\otimes \mathbf {v}) &= (\mathbf {A}\mathbf {u})\otimes \mathbf {v}\qquad && \forall \mathbf {A}\in \mathcal{L}(V), \ \mathbf {u},\mathbf {v}\in V. \end{aligned} $$

The space $\mathcal{L}(V)$ is itself a finite dimensional complex vector space under the usual notions of sum and multiplication by a scalar. It can be shown that if $\{\mathbf {u}_{i}\}$ is a basis for $V$, then $\{\mathbf {u}_{i}\otimes \mathbf {u}_{j}\}$ is a basis for $\mathcal{L}(V)$. We shall regard $\mathcal {L}(V)$ and $V\otimes V$ as equivalent spaces.

We define the trace operator (contraction) $\operatorname{tr\,}: \mathcal{L}(V) \to \mathbb{C}$ as the linear operator for which

$$ \operatorname{tr\,}(\mathbf {u}\otimes \mathbf {v}) = \langle \mathbf {u}, \mathbf {v}\rangle \qquad \forall \mathbf {u},\mathbf {v}\in V.$$

The trace enjoys the following property:

$$ \operatorname{tr\,}(\mathbf {A}\mathbf {B}) = \operatorname{tr\,}(\mathbf {B}\mathbf {A}) \qquad \forall \mathbf {A},\mathbf {B}\in \mathcal{L}(V).$$

We endow $\mathcal{L}(V)$ with the Hermitian inner product $\langle \cdot , \cdot \rangle : \mathcal{L}(V) \times \mathcal{L}(V) \to \mathbb{C}$ induced by the trace

$$ \langle \mathbf {A}, \mathbf {B}\rangle := \operatorname{tr\,}(\mathbf {A}\mathbf {B}^{*}) .$$

By this definition, we have

$$ \langle \mathbf {u}\otimes \mathbf {v}, \mathbf {w}\otimes \mathbf {x}\rangle = \operatorname{tr\,}((\mathbf {u}\otimes \mathbf {v})(\mathbf {x}\otimes \mathbf {w})) = \langle \mathbf {x}, \mathbf {v}\rangle \langle \mathbf {u}, \mathbf {w}\rangle . $$

(3)

The induced norm is the nonnegative real $|\mathbf {A}| := \sqrt{\langle \mathbf {A}, \mathbf {A}\rangle}$.

We say a second order tensor is harmonic if it is Hermitian and traceless. The set of all harmonic second order tensors will be denoted by

$$ \mathrm{Harm}(V):=\{\mathbf {A}\in \mathcal{L}(V) \operatorname{\,:\,}\mathbf {A}= \mathbf {A}^{*}, \ \operatorname{tr\,}\mathbf {A}= 0\}. $$

The linear operators from $\mathcal{L}(V)$ to $\mathcal{L}(V)$ are themselves tensors. We define $\mathcal{L}(V\otimes V)$ as the set of all such operators. Its elements will be also referenced to as fourth order tensors, whose action is

$$ \mathbf {A}\otimes \mathbf {B}(\mathbf {X}) = \langle \mathbf {X},\mathbf {B}\rangle \mathbf {A}\qquad \forall \mathbf {X}\in V\otimes V.$$

(4)

Given $\mathbf {A}, \mathbf {B}\in \mathcal{L}(V)$ we define the conjugator $\mathbf {A}\boxtimes \mathbf {B}\in \mathcal{L}(V\otimes V)$ as the tensor verifying

$$ \mathbf {A}\boxtimes \mathbf {B}(\mathbf {u}\otimes \mathbf {v}) = \mathbf {A}\mathbf {u}\otimes \mathbf {B}\mathbf {v}\qquad \forall \mathbf {u}, \mathbf {v}\in V.$$

It is easy to see that $(\mathbf {A}\boxtimes \mathbf {B})^{*} = \mathbf {A}^{*} \boxtimes \mathbf {B}^{*}$ and that

$$ (a\mathbf {A}) \boxtimes \mathbf {B}= a(\mathbf {A}\boxtimes \mathbf {B}), \qquad \mathbf {A}\boxtimes (a\mathbf {B}) = \overline{a}(\mathbf {A}\boxtimes \mathbf {B}) \qquad \forall a\in \mathbb{C}.$$

Moreover

$$ (\mathbf {A}\boxtimes \mathbf {B})(\mathbf{C}\boxtimes \mathbf{D}) = \mathbf {A}\mathbf{C} \boxtimes \mathbf {B}\mathbf{D}.$$

(5)

The trace operator on $\mathcal{L}(V\otimes V)$ is defined by

$$ \operatorname{tr\,}(\mathbf {A}\otimes \mathbf {B}) = \langle \mathbf {A}, \mathbf {B}\rangle \qquad \forall \mathbf {A},\mathbf {B}\in \mathcal{L}(V). $$

The counterpart identity of (3) holds also for second order tensors. We define the identity on $\mathcal{L}(V\otimes V)$ as $\mathbb{I}:= \mathbf {I}\boxtimes \mathbf {I}$ such that $\mathbb{I}\mathbf {A}= \mathbf {A}$ for all $\mathbf {A}\in \mathcal{L}(V)$. A fourth order tensor $\mathbb{A}$ has minor symmetries if $\mathbb{A}(\mathbf {A}^{*}) = \mathbb{A}\mathbf {A}$ and $(\mathbb{A}\mathbf {A})^{*} = \mathbb{A}\mathbf {A}$ for all $\mathbf {A}\in \mathcal{L}(V)$. A fourth order tensor $\mathbb{A}$ has the major symmetry if $\mathbb{A}= \mathbb{A}^{*}$.

We conclude by recalling the Cayley-Hamilton theorem for $\mathcal{L}(V)$ when the dimension of $V$ is 2:

$$ \mathbf {A}^{2} -(\operatorname{tr\,}\mathbf {A})\mathbf {A}+ (\det \mathbf {A}) \mathbf {I}= \mathbf{0} \qquad \forall \mathbf {A}\in \mathcal{L}(V); $$

taking traces, we get also the useful scalar equation

$$ |\mathbf {A}|^{2}-(\operatorname{tr\,}\mathbf {A})^{2}+2\det \mathbf {A}=0 \qquad \forall \mathbf {A}\in \mathcal{L}(V). $$

(6)

3 Polar Representation of Plane Tensors

We are mostly interested in the case $V$, endowed with a Hermitian inner product introduced in Sect. 2, is the complexification of some real vector space $V_{\mathbb{R}}$ of dimension 2. Every element $\mathbf {z}$ of $V$ can be put uniquely in the form $\Re (\mathbf {z}) + \mathrm {i}\mkern 1mu\Im (\mathbf {z})$ with $\Re (\mathbf {z}), \Im (\mathbf {z}) \in V_{\mathbb{R}}$. We define also $\overline{\mathbf {z}} := \Re (\mathbf {z}) - \mathrm {i}\mkern 1mu\Im (\mathbf {z})$. By using this notion, it is immediate to verify the following very useful property:

$$ \overline{\langle \mathbf {v},\mathbf {u}\rangle} = \langle \overline{\mathbf {v}}, \overline{\mathbf {u}}\rangle \qquad \forall \mathbf {u},\mathbf {v}\in V. $$

(7)

The key observation (trivial, perhaps) is that special orthogonal tensors (rotations) $\mathbf {Q}\in \operatorname{SO(2)}$ are unitary tensors of $\mathcal{L}(V)$, and as such they can be diagonalized over the complex field. We have denoted by $\operatorname{SO(2)}$ the set of special orthogonal tensors in dimension two.

Remark 1

It is convenient to regard real vectors as complex ones, by the trivial extension $\mathbf {u}\mapsto \mathbf {u}+ \mathrm {i}\mkern 1mu\mathbf{0}$. Every linear operator $\mathbf {A}\in \mathcal{L}( V_{\mathbb{R}})$ admits a natural extension on $\mathcal{L}(V)$, still denoted by $\mathbf {A}$, given by $\mathbf {A}\mathbf {z}= \mathbf {A}\Re (\mathbf {z}) + \mathrm{i}\mathbf {A}\Im (\mathbf {z})$ for all $\mathbf {z}\in V$. The same holds for higher order tensors.

By the spectral theorem, there exist $r\in \mathbb{C}$, $|r|=1$, and a unit vector $\mathbf {p}\in V$, $\mathbf {p}\perp \overline{\mathbf {p}}$, such that the spectral decomposition of $\mathbf {Q}$ reads

$$ \mathbf {Q}= r \mathbf {p}\otimes \mathbf {p}+ \overline{r}\,\overline{\mathbf {p}}\otimes \overline{\mathbf {p}}. $$

Let us pose for later convenience

$$ \mathbf {P}:= \mathbf {p}\otimes \mathbf {p}, \qquad \overline{\mathbf {P}} := \overline{\mathbf {p}} \otimes \overline{\mathbf {p}}, \qquad \mathbf{L}:= \mathbf {p}\otimes \overline{\mathbf {p}}, \qquad \overline{\mathbf {L}} := \overline{\mathbf {p}}\otimes \mathbf {p}. $$

(8)

Tensors $\mathbf {P}, \overline{\mathbf {P}} \in \mathcal{L}(V)$ are Hermitian, mutually orthogonal, unit tensors .³ In particular, they are orthogonal projectors along $\mathbf {p}, \overline{\mathbf {p}}$, respectively. Tensors $\mathbf {L}, \overline{\mathbf {L}} \in \mathcal{L}(V)$ are unit, traceless, mutually orthogonal tensors. Note also that

$$ \mathbf {P}\perp \overline{\mathbf {P}}\perp \mathbf {L}\perp \overline{\mathbf {L}} \perp \mathbf {P}. $$

Let $\mathbf{W}:= \mathbf {P}- \overline{\mathbf {P}}$. Resolving the identity of $\mathcal{L}(V)$ as $\mathbf {I}= \mathbf {P}+ \overline{\mathbf {P}}$, we have immediately

$$ 2\mathbf {P}= \mathbf {I}+ \mathbf {W}, \qquad 2\overline{\mathbf {P}} = \mathbf {I}- \mathbf {W}. $$

(9)

Note that for any $\mathbf {x}\in V_{\mathbb{R}}$ we have

$$ \begin{aligned}[b] \langle \mathbf {x}\otimes \mathbf {x}, \mathbf {W}\rangle &= \langle \mathbf {x}\otimes \mathbf {x}, \mathbf {p}\otimes \mathbf {p}- \overline{\mathbf {p}}\otimes \overline{\mathbf {p}} \rangle \\ &= \langle \mathbf {p}, \mathbf {x}\rangle \langle \mathbf {x}, \mathbf {p}\rangle - \langle \overline{\mathbf {p}}, \mathbf {x}\rangle \langle \mathbf {x}, \overline{\mathbf {p}} \rangle \\ &=\langle \mathbf {p}, \mathbf {x}\rangle \langle \mathbf {x}, \mathbf {p}\rangle - \langle \mathbf {x}, \mathbf {p}\rangle \langle \mathbf {p}, \mathbf {x}\rangle = 0. \end{aligned} $$

(10)

The following identities will be extremely useful.

Lemma 2

Let $\mathbf {P}, \mathbf {L}$ be defined as in (8). Then

$$ \mathbf {P}\boxtimes \mathbf {P}= \mathbf {P}\otimes \mathbf {P}, \qquad \mathbf {P}\boxtimes \overline{\mathbf {P}}=\mathbf {L}\otimes \mathbf {L}, \qquad \mathbf {L}\boxtimes \mathbf {L}= \mathbf {P}\otimes \overline{\mathbf {P}}, \qquad \mathbf {L}\boxtimes \overline{\mathbf {L}} = \mathbf {L}\otimes \overline{\mathbf {L}}, $$

$$ \mathbf {L}\boxtimes \overline{\mathbf {P}} = \mathbf {L}\otimes \overline{\mathbf {P}}, \qquad \mathbf {L}\boxtimes \mathbf {P}= \mathbf {P}\otimes \overline{\mathbf {L}}, \qquad \mathbf {P}\boxtimes \overline{\mathbf {L}} = \mathbf {L}\otimes \mathbf {P}, \qquad \mathbf {P}\boxtimes \mathbf {L}= \mathbf {P}\otimes \mathbf {L}. $$

Proof

Fix generic $\mathbf {u}, \mathbf {v}, \mathbf {w}, \mathbf {x}\in V$. All the identities in the statement of the lemma follow from the fundamental identity

$$ \begin{aligned} (\mathbf {u}\otimes \mathbf {v})\boxtimes (\mathbf {x}\otimes \mathbf {w})\mathbf {a}_{1}\otimes \mathbf {a}_{2} &= (\mathbf {u}\otimes \mathbf {v})\mathbf {a}_{1}\otimes (\mathbf {x}\otimes \mathbf {w}) \mathbf {a}_{2} \\ &= \langle \mathbf {a}_{1},\mathbf {v}\rangle \langle \mathbf {w},\mathbf {a}_{2}\rangle \mathbf {u}\otimes \mathbf {x}\\ &= \langle \mathbf {a}_{1}\otimes \mathbf {a}_{2}, \mathbf {v}\otimes \mathbf {w}\rangle \mathbf {u}\otimes \mathbf {x}\\ &= (\mathbf {u}\otimes \mathbf {x})\otimes (\mathbf {v}\otimes \mathbf {w}) \mathbf {a}_{1}\otimes \mathbf {a}_{2} \qquad \forall \mathbf {a}_{1}, \mathbf {a}_{2}\in V. \end{aligned} $$

 □

By using Lemma 2, we deduce that the identity of $\mathcal{L}(V\otimes V)$ can be rewritten as

$$ \begin{aligned}[b] \mathbb{I}&:= \mathbf {I}\boxtimes \mathbf {I}\\ &= \mathbf {P}\boxtimes \mathbf {P}+ \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}} + \mathbf {P}\boxtimes \overline{\mathbf {P}} + \overline{\mathbf {P}}\boxtimes \mathbf {P}\\ &=\mathbf {P}\otimes \mathbf {P}+ \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} + \mathbf {L}\otimes \mathbf {L}+ \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}. \end{aligned} $$

(11)

3.1 Vectors

We start our investigation from vectors. It is worth starting from vectors because it clarifies the modus operandi that we will adopt later on for second and fourth order tensors.

Let $\mathbf {x}\in V_{\mathbb{R}}$ be a generic real vector. In general we cannot expect $\mathbf {x}$ to be equal to $\mathbf {Q}\mathbf {x}$ for all $\mathbf {Q}\in \operatorname{SO(2)}$. In fact

$$ \mathbf {Q}\mathbf {x}= r \langle \mathbf {x},\mathbf {p}\rangle \mathbf {p}+ \overline{r} \langle \mathbf {x},\overline{\mathbf {p}}\rangle \overline{\mathbf {p}}. $$

However, one shall observe that

$$ r\langle \mathbf {x},\mathbf {p}\rangle \overline{r}\langle \mathbf {x},\overline{\mathbf {p}} \rangle = \langle \mathbf {x},\mathbf {p}\rangle \langle \mathbf {x},\overline{\mathbf {p}} \rangle . $$

In plane language, the product of the projections of $\mathbf {x}$ along $\mathbf {p}$ and $\overline{\mathbf {p}}$ is equal to the product of projections of $\mathbf {Q}\mathbf {x}$ along $\mathbf {p}$ and $\overline{\mathbf {p}}$, for every $\mathbf {Q}\in \operatorname{SO(2)}$. Accordingly, let us pose

$$ \iota (\mathbf {x}):= \langle \mathbf {x}, \mathbf {p}\rangle \langle \mathbf {x}, \overline{\mathbf {p}}\rangle = \langle \mathbf {x}, \mathbf {p}\rangle \overline{\langle \mathbf {x}, \mathbf {p}\rangle} = |\langle \mathbf {x},\mathbf {p}\rangle |^{2}. $$

Proposition 3

Let $\mathbf {x}\in V_{\mathbb{R}}$. Then $\iota (\mathbf {x})$ is invariant, in the sense that

$$ \iota (\mathbf {x}) = \iota (\mathbf {Q}\mathbf {x}) \qquad \forall \mathbf {Q}\in \operatorname{SO(2)}.$$

Moreover

$$ 2\iota (\mathbf {x}) = |\mathbf {x}|^{2}.$$

Proof

We have, by using the spectral decomposition of $\mathbf {Q}$,

$$ \begin{aligned} I(\mathbf {Q}\mathbf {x}) &= \langle \mathbf {Q}\mathbf {x}, \mathbf {p}\rangle \langle \mathbf {Q}\mathbf {x}, \overline{\mathbf {p}}\rangle \\ &=\langle \mathbf {x}, \mathbf {Q}^{*}\mathbf {p}\rangle \langle \mathbf {x}, \mathbf {Q}^{*} \overline{\mathbf {p}}\rangle \\ &=\langle \mathbf {x}, \frac{1}{r}\mathbf {p}\rangle \langle \mathbf {x}, \frac{1}{\overline{r}}\overline{\mathbf {p}}\rangle \\ &=\langle \mathbf {x}, \mathbf {p}\rangle \langle \mathbf {x}, \overline{\mathbf {p}}\rangle = \iota (\mathbf {x}) \qquad \forall \mathbf {Q}. \end{aligned} $$

For the characterization of $\iota (\mathbf {x})$, using (9) and (10), we obtain

$$ \begin{aligned} \iota (\mathbf {x}) &= \langle \mathbf {x}, \mathbf {p}\rangle \langle \mathbf {x}, \overline{\mathbf {p}}\rangle \\ &=\langle \mathbf {x}\otimes \mathbf {x}, \mathbf {P}\rangle \\ &=\frac{1}{2}\langle \mathbf {x}\otimes \mathbf {x}, \mathbf {I}\rangle \\ &=\frac{1}{2} \operatorname{tr\,}(\mathbf {x}\otimes \mathbf {x}) = \frac{1}{2} |\mathbf {x}|^{2}. \end{aligned} $$

 □

By Proposition 3, we have recovered the well-known fact that the norm of a vector is invariant with respect to arbitrary rotations. We can state the following representation theorem for vectors.

Theorem 4

There exist $r \in \mathbb{R}$, $r \geq 0$, and a unit vector $\mathbf {h}\in V_{\mathbb{R}}$ such that any vector $\mathbf{x} \in V_{\mathbb{R}}$ admits the following representation:

$$ \mathbf {x}= r \mathbf {h}.$$

Proof

Let $\mathbf {x}\in V_{\mathbb{R}}$ be a generic vector. By projecting it on the eigenbasis of $\mathbf {Q}\in \operatorname{SO(2)}$, we have $\mathbf {x}= \langle \mathbf {x},\mathbf {p}\rangle \mathbf {p}+ \langle \mathbf {x}, \overline{\mathbf {p}}\rangle \overline{\mathbf {p}}$. Let $r := \sqrt{2\iota (\mathbf {x})}$ and $\mathbf {h}:= \frac{1}{r}(\langle \mathbf {x},\mathbf {p}\rangle \mathbf {p}+ \langle \mathbf {x}, \overline{\mathbf {p}}\rangle \overline{\mathbf {p}})$. We trivially have $\mathbf {x}= r \mathbf {h}$. To conclude, it is obvious that $r \geq 0$ and that $\mathbf {h}\in V_{\mathbb{R}}$, since it is the sum of two complex conjugate terms. Moreover $|\mathbf {h}|^{2} = \langle \mathbf {h}, \mathbf {h}\rangle = \frac{1}{r^{2}} 2 | \langle \mathbf {x}, \mathbf {p}\rangle |^{2} = \frac{2 \iota}{2 \iota}=1$. □

3.2 Second Order Tensors

Let $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ be a real second order tensor. Let $\mathbb{Q}:= \mathbf {Q}\boxtimes \mathbf {Q}$ for $\mathbf {Q}\in \operatorname{SO(2)}$. ℚ is a fourth order orthogonal tensor that rotates second order tensors. A direct computation reveals that its spectral representation reads

$$ \mathbb{Q}= \mathbf {P}\boxtimes \mathbf {P}+ \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}+r^{2} \mathbf {P}\boxtimes \overline{\mathbf {P}} + \overline{r}^{2} \overline{\mathbf {P}} \boxtimes \mathbf {P}, $$

(12)

and that

$$ (1, \mathbf {P}), \qquad (1,\overline{\mathbf {P}}), \qquad (r^{2}, \mathbf {L}), \qquad (\overline{r}^{2}, \overline{\mathbf {L}})$$

(13)

are the eigencouples eigenvalue/eigentensor of ℚ. A scrutiny of (13) suggests defining, in the same spirit of the observations raised in Sect. 3.1,

$$ \iota _{1}(\mathbf {X}) := \langle \mathbf {X}, \mathbf {P}\rangle , \qquad \iota _{2}( \mathbf {X}) := \overline{\iota _{1}(\mathbf {X})}, \qquad \iota _{3}(\mathbf {X}) := \langle \mathbf {X}, \overline{\mathbf {L}}\rangle \langle \mathbf {X}, \mathbf {L}\rangle = | \langle \mathbf {X}, \mathbf {L}\rangle |^{2}.$$

Proposition 5

Let $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ be a real second order tensor. Then $\iota _{1}(\mathbf {X})$, $\iota _{2}(\mathbf {X})$, $\iota _{3}(\mathbf {X})$ are invariants, in the sense that

$$ \iota _{i}(\mathbf {X}) = \iota _{i}(\mathbb{Q}\mathbf {X}) \qquad \forall \mathbb{Q}, \ i=1,2,3.$$

Moreover

$$ \iota _{1}(\mathbf {X}) = \overline{\iota _{2}(\mathbf {X})} = \frac{1}{2} \operatorname{tr\,}\mathbf {X}+ \frac{1}{2} \operatorname{tr\,}(\mathbf {X}\mathbf{W}^{*}), \qquad \iota _{3}(\mathbf {X}) = \frac{1}{4} (|\mathbf {X}|^{2} - 2\det \mathbf {X}).$$

If in addition $\mathbf {X}$ is symmetric, $\iota _{1}(\mathbf {X}) = \iota _{2}(\mathbf {X}) = \frac{1}{2} \operatorname{tr\,}\mathbf {X}$.

Proof

We shall prove that $\iota _{i}(\mathbf {X}) = \iota _{i}(\mathbb{Q}\mathbf {X})$. We have, for $i=1$

$$ \begin{aligned} \iota _{1}(\mathbb{Q}\mathbf {X}) = \langle \mathbb{Q}\mathbf {X}, \mathbf {P}\rangle = \langle \mathbf {X}, \mathbb{Q}^{*}\mathbf {P}\rangle = \langle \mathbf {X}, \mathbf {P}\rangle = \iota _{1}( \mathbf {X}). \end{aligned} $$

For $i=2$ the proof is completely analogous. For $i=3$ we have

$$ \begin{aligned} \iota _{3}(\mathbb{Q}\mathbf {X}) &= \langle \mathbb{Q}\mathbf {X}, \overline{\mathbf {L}} \rangle \langle \mathbb{Q}\mathbf {X}, \mathbf {L}\rangle \\ &= \langle \mathbf {X}, \mathbb{Q}^{*}\overline{\mathbf {L}} \rangle \langle \mathbf {X}, \mathbb{Q}^{*}\mathbf {L}\rangle \\ &= r^{2}\overline{r}^{2}\langle \mathbf {X}, \overline{\mathbf {L}} \rangle \langle \mathbf {X}, \mathbf {L}\rangle \\ &= \iota _{3}(\mathbf {X}) \qquad \forall \mathbb{Q}. \end{aligned} $$

For the characterization of the invariants, we have by using (9)

$$ \iota _{1}(\mathbf {X}) = \langle \mathbf {X}, \mathbf {P}\rangle = \frac{1}{2}\langle \mathbf {X}, \mathbf {I}\rangle + \frac{1}{2} \langle \mathbf {X}, \mathbf {W}\rangle = \frac{1}{2} \operatorname{tr\,}\mathbf {X}+ \frac{1}{2} \langle \mathbf {X}, \mathbf {W}\rangle .$$

If $\mathbf {X}$ is actually symmetric, $\langle \mathbf {X}, \mathbf {W}\rangle =0$ by (10). For $i=3$ we deduce, by using (11),

$$ \begin{aligned} 2\iota _{3} &= \langle \mathbf {X}, \overline{\mathbf {L}}\rangle \langle \mathbf {X}, \mathbf {L}\rangle + \langle \mathbf {X}, \mathbf {L}\rangle \langle \mathbf {X}, \overline{\mathbf {L}}\rangle \\ &= \langle \mathbf {X}\otimes \mathbf {X}, \overline{\mathbf {L}}\otimes \overline{\mathbf {L}} + \mathbf {L}\otimes \mathbf {L}\rangle \\ &=\langle \mathbf {X}\otimes \mathbf {X}, \mathbb{I}- \mathbf {P}\otimes \mathbf {P}- \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}\rangle \\ &= \operatorname{tr\,}\mathbf {X}\otimes \mathbf {X}- 2 |\iota _{1}(\mathbf {X})|^{2} \\ &= |\mathbf {X}|^{2} - \frac{1}{2} (\operatorname{tr\,}\mathbf {X})^{2}. \end{aligned} $$

By the Cayley-Hamilton theorem (6), substituting for $(\operatorname{tr\,}\mathbf {X})^{2}$ in the previous expression, we find the required form of $\iota _{3}(\mathbf {X})$. □

We can now prove the following representation theorem for symmetric, 2D second order real tensors.

Theorem 6

There exist $T, R\in \mathbb{R}$, $R\geq 0$, and a tensor $\mathbf {H}\in \mathrm{Harm}( V_{\mathbb{R}})$, $|\mathbf {H}| = \sqrt{2}$, such that any symmetric tensor $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ admits the following representation:

$$ \mathbf {X}= T\mathbf {I}+ R \, \mathbf {H}.$$

Proof

Let $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ be a generic symmetric, real tensor. It has the following decomposition along the eigenbasis of ℚ:

$$ \begin{aligned} \mathbf {X}&= \langle \mathbf {X},\mathbf {P}\rangle \mathbf {P}+ \langle \mathbf {X}, \overline{\mathbf {P}} \rangle \overline{\mathbf {P}} + \langle \mathbf {X},\mathbf {L}\rangle \mathbf {L}+ \langle \mathbf {X},\overline{\mathbf {L}} \rangle \overline{\mathbf {L}} \\ &=\iota _{1}(\mathbf {X})(\mathbf {P}+ \overline{\mathbf {P}}) + \langle \mathbf {X},\mathbf {L}\rangle \mathbf {L}+ \langle \mathbf {X},\overline{\mathbf {L}} \rangle \overline{\mathbf {L}} \\ &=T\mathbf {I}+ \langle \mathbf {X},\mathbf {L}\rangle \mathbf {L}+ \langle \mathbf {X}, \overline{\mathbf {L}} \rangle \overline{\mathbf {L}}, \end{aligned} $$

where we posed $T:=\iota _{1}(\mathbf {X})$. Define now $R:=\sqrt{\iota _{3}(\mathbf {X})}$ and $\mathbf {H}:= \frac{1}{R}(\langle \mathbf {X},\mathbf {L}\rangle \mathbf {L}+ \langle \mathbf {X}, \overline{\mathbf {L}} \rangle \overline{\mathbf {L}})$. It is clear that $\mathbf {H}$ is real and that $\mathbf {H}= \mathbf {H}^{*}$. Moreover, since $\operatorname{tr\,}\mathbf {L}= \langle \mathbf {p},\overline{\mathbf {p}}\rangle = 0$, we conclude that $\mathbf {H}$ is harmonic. Finally,

$$ \begin{aligned} |\mathbf {H}|^{2} = \frac{2 |\langle \mathbf{X},\mathbf{L}\rangle |^{2}}{\iota _{3}} = 2. \end{aligned} $$

 □

Remark 7

• i) The tensor $\mathbf {H}$ appearing in Theorem 6 has the following spectral decomposition:

  $$ \mathbf {H}= \mathbf {h}\otimes \mathbf {h}- \mathbf {h}^{\perp}\otimes \mathbf {h}^{\perp}, $$

  for some unit vector $\mathbf {h}\in V_{\mathbb{R}}$. It follows directly from the fact that $\operatorname{tr\,}\mathbf {H}= 0$ and $|\mathbf {H}| = \sqrt{2}$. It represents a mirror reflection across $\mathbf {h}$.

• ii) Note that $\mathbf {H}$ is invariant under rotations of $k\pi$, $k \in \mathbb{Z}$. Such transformations send $\mathbf {h}$ (resp. $\mathbf {h}^{\perp}$) into $\pm \mathbf {h}$ (resp. $\pm \mathbf {h}^{\perp}$), depending on the oddity of $k$. Therefore, $\mathbb{Q}(k\pi ) \mathbf {H}= \mathbf {H}$.

• iii) In Theorem 6 only $T$ and $R$ are tensor invariants. Unit vectors in $V_{\mathbb{R}}$ belongs to the unit circumference, so that one scalar parameter is needed to fully describe $\mathbf {h}$. If one fixes a basis, such parameter can be chosen to be an angle, measured from some reference axis. This angle is usually denoted by $\Phi$ in the polar representation nomenclature. $T$, $R$, and $\Phi$ are the Verchery polar parameters for 2D symmetric second order tensors.

3.3 Fourth Order Tensors

We first prove a couple of useful identities.

Lemma 8

Let $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ be a real tensor. For every complex tensor $\mathbf {A}, \mathbf {B}\in \mathcal{L}(V)$

$$ \langle \mathbf {X}\otimes \mathbf {X}, \mathbf {A}\otimes \mathbf {B}\rangle = \langle \mathbf {X}\otimes \mathbf {X}, \overline{\mathbf {B}}\otimes \overline{\mathbf {A}}\rangle . $$

Proof

We have

$$ \begin{aligned} \langle \mathbf {X}\otimes \mathbf {X}, \mathbf {A}\otimes \mathbf {B}\rangle &= \operatorname{tr\,}((\mathbf {X}\otimes \mathbf {X})(\mathbf {B}\otimes \mathbf {A})) \\ &= \langle \mathbf {B}, \mathbf {X}\rangle \langle \mathbf {X}, \mathbf {A}\rangle \\ &= \langle \mathbf {X}, \overline{\mathbf {B}} \rangle \langle \overline{\mathbf {A}}, \mathbf {X}\rangle \\ &= \operatorname{tr\,}((\mathbf {X}\otimes \mathbf {X})(\overline{\mathbf {A}}\otimes \overline{\mathbf {B}})) \\ &= \langle \mathbf {X}\otimes \mathbf {X}, \overline{\mathbf {B}}\otimes \overline{\mathbf {A}} \rangle . \end{aligned} $$

 □

Let us introduce the following fourth order tensor:

$$ \begin{aligned} \mathbf {I}\otimes \mathbf {I}&:= \mathbf {P}\otimes \mathbf {P}+ \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} + \overline{\mathbf {P}}\otimes \mathbf {P}+ \mathbf {P}\otimes \overline{\mathbf {P}}. \end{aligned} $$

(14)

Lemma 9

Let $\mathbf {X}\in \mathcal{L}( V_{\mathbb{R}})$ be a symmetric real tensor. Then

$$ \langle \mathbf {X}\otimes \mathbf {X}, \mathbf {P}\otimes \mathbf {P}\rangle = \langle \mathbf {X}\otimes \mathbf {X}, \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}\rangle = \langle \mathbf {X}\otimes \mathbf {X}, \mathbf {P}\otimes \overline{\mathbf {P}}\rangle = \langle \mathbf {X}\otimes \mathbf {X}, \overline{\mathbf {P}}\otimes \mathbf {P}\rangle = \frac{1}{4}\langle \mathbf {X}\otimes \mathbf {X}, \mathbf {I}\otimes \mathbf {I}\rangle . $$

Proof

By recalling (9) and (10), we deduce

$$ \langle \mathbf {X}\otimes \mathbf {X}, \mathbf {P}\otimes \overline{\mathbf {P}}\rangle = \operatorname{tr\,}((\mathbf {X}\otimes \mathbf {X})(\overline{\mathbf {P}}\otimes \mathbf {P})) = \langle \mathbf {X}, \mathbf {P}\rangle \langle \overline{\mathbf {P}}, \mathbf {X}\rangle = \frac{1}{4}\langle \mathbf {X}, \mathbf {I}\rangle \langle \mathbf {I}, \mathbf {X}\rangle = \frac{1}{4}\langle \mathbf {X}\otimes \mathbf {X}, \mathbf {I}\otimes \mathbf {I}\rangle . $$

The other cases are proved similarly. □

Let $\mathbb{X}\in \mathcal{L}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ be a generic fourth order real tensor. Define $\mathtt{Q} := \mathbb{Q}\boxtimes \mathbb{Q}$ as the eighth order tensor that rotates fourth order tensors. A direct computation reveals that

$$ \begin{aligned}[b] \mathtt{Q} = & \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}} + r^{4} \mathbf {P}\boxtimes \overline{\mathbf {P}} \boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}+ r^{2} \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \mathbf {P}+ r^{2} \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}} \\ &+\overline{r}^{4} \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}} + \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}+ \overline{r}^{2} \overline{\mathbf {P}} \boxtimes \mathbf {P}\boxtimes \mathbf {P}\boxtimes \mathbf {P}+ \overline{r}^{2} \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}} \\ & + \overline{r}^{2}\mathbf {P}\boxtimes \mathbf {P}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}} + r^{2}\mathbf {P}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}} \boxtimes \mathbf {P}+ \mathbf {P}\boxtimes \mathbf {P}\boxtimes \mathbf {P}\boxtimes \mathbf {P}+ \mathbf {P}\boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}} \\ & + \overline{r}^{2}\overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}} \boxtimes \mathbf {P}\boxtimes \overline{\mathbf {P}} + r^{2}\overline{\mathbf {P}} \boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}+ \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}\boxtimes \mathbf {P}\boxtimes \mathbf {P}+ \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}\boxtimes \overline{\mathbf {P}}. \end{aligned} $$

(15)

It is the spectral decomposition of $\mathtt{Q}$. The sixteen eigencouples (eight pairs of complex conjugate eigencouples) are given by

$$ (1, \mathbf {L}\otimes \mathbf {L}), \qquad (1,\overline{\mathbf {L}}\otimes \overline{\mathbf {L}}), $$

$$ (1, \mathbf {P}\otimes \mathbf {P}), \qquad (1, \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}), $$

$$ (1,\mathbf {P}\otimes \overline{\mathbf {P}}), \qquad (1,\overline{\mathbf {P}}\otimes \mathbf {P}), $$

$$ (r^{4}, \mathbf {L}\otimes \overline{\mathbf {L}}), \qquad (\overline{r}^{4}, \overline{\mathbf {L}}\otimes \mathbf {L}), $$

$$ (r^{2}, \mathbf {L}\otimes \mathbf {P}), \qquad (\overline{r}^{2}, \overline{\mathbf {L}}\otimes \overline{\mathbf {P}}), $$

$$ (r^{2}, \mathbf {L}\otimes \overline{\mathbf {P}}), \qquad (\overline{r}^{2}, \overline{\mathbf {L}}\otimes \mathbf {P}), $$

$$ (r^{2},\mathbf {P}\otimes \overline{\mathbf {L}}), \qquad (\overline{r}^{2}, \overline{\mathbf {P}}\otimes \mathbf {L}), $$

$$ (r^{2}, \overline{\mathbf {P}}\otimes \overline{\mathbf {L}}), \qquad (\overline{r}^{2}, \mathbf {P}\otimes \mathbf {L}). $$

Proceeding in analogy to the previous subsections, a scrutiny of the eigencouples of $\mathtt{Q}$ suggests introducing the following 43 scalar candidate invariants. Six are linear (three pairs of complex conjugate scalars):

$$ \iota _{1}(\mathbb{X}) = \overline{\iota _{2}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {L}\rangle , $$

$$ \iota _{3}(\mathbb{X}) = \overline{\iota _{4}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {P}\rangle ,$$

$$ \iota _{5}(\mathbb{X}) = \overline{\iota _{6}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {P}\otimes \overline{\mathbf {P}}\rangle ;$$

seventeen are quadratic (five positive real and six pairs of complex conjugate scalars)

$$ \iota _{7}(\mathbb{X}) := |\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}} \rangle |^{2},$$

$$ \iota _{8}(\mathbb{X}) := |\langle \mathbb{X}, \mathbf {L}\otimes \mathbf {P}\rangle |^{2},$$

$$ \iota _{9}(\mathbb{X}) := |\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}} \rangle |^{2},$$

$$ \iota _{10}(\mathbb{X}) := |\langle \mathbb{X}, \mathbf {P}\otimes \overline{\mathbf {L}} \rangle |^{2},$$

$$ \iota _{11}(\mathbb{X}) := |\langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle |^{2},$$

$$ \iota _{12}(\mathbb{X}) = \overline{\iota _{13}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \mathbf {P}\rangle , $$

$$ \iota _{14}(\mathbb{X}) = \overline{\iota _{15}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {P}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle , $$

$$ \iota _{16}(\mathbb{X}) = \overline{\iota _{17}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {P}}\otimes \mathbf {L}\rangle , $$

$$ \iota _{18}(\mathbb{X}) = \overline{\iota _{19}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle , $$

$$ \iota _{20}(\mathbb{X}) = \overline{\iota _{21}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \overline{\mathbf {P}} \otimes \mathbf {L}\rangle , $$

$$ \iota _{22}(\mathbb{X}) = \overline{\iota _{23}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle \langle \mathbb{X},\mathbf {P}\otimes \overline{\mathbf {L}} \rangle ; $$

and twenty are cubic (ten complex conjugate pairs)

$$ \iota _{24}(\mathbb{X}) = \overline{\iota _{25}(\mathbb{X})} :=\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \mathbf {P}\rangle ,$$

$$ \iota _{26}(\mathbb{X}) = \overline{\iota _{27}(\mathbb{X})} :=\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \overline{\mathbf {P}}\rangle ,$$

$$ \iota _{28}(\mathbb{X}) = \overline{\iota _{29}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{30}(\mathbb{X}) = \overline{\iota _{31}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {P}} \otimes \mathbf {L}\rangle \langle \mathbb{X}, \overline{\mathbf {P}}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{32}(\mathbb{X}) = \overline{\iota _{33}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X},\overline{\mathbf {L}} \otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \overline{\mathbf {P}}\rangle ,$$

$$ \iota _{34}(\mathbb{X}) = \overline{\iota _{35}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \mathbf {P}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{36}(\mathbb{X}) = \overline{\iota _{37}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {P}}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{38}(\mathbb{X}) = \overline{\iota _{39}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{40}(\mathbb{X}) = \overline{\iota _{41}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \overline{\mathbf {P}}\otimes \mathbf {L}\rangle ,$$

$$ \iota _{42}(\mathbb{X}) = \overline{\iota _{43}(\mathbb{X})} := \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \mathbf {P}\otimes \mathbf {L}\rangle \langle \mathbb{X}, \overline{\mathbf {P}}\otimes \mathbf {L}\rangle .$$

Remark 10

Not all forty-three invariants are independent. There must exist twenty-eight relations between them such that only fifteen invariants are genuinely independent. For instance, it can be readily seen that $\iota _{12}(\mathbb{X})\iota _{13}(\mathbb{X}) = \iota _{8}(\mathbb{X}) \iota _{9}( \mathbb{X})$ or that $\iota _{24}(\mathbb{X})\iota _{25}(\mathbb{X}) = \iota _{7}(\mathbb{X}) (\iota _{8}( \mathbb{X}))^{2}$. The characterization of all of the forty-three relations is still open [29].

We are mostly interested in hyperelasticity-type tensors. Linear elasticity tensors have both minor symmetries, since they map symmetric second order tensors (linearized strain tensor) into symmetric second order tensors (Cauchy stress tensor). In the case of hyperelasticity, they have also the major symmetry. We denote by $\mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ the subset of $\mathcal{L}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ of 2D, fourth order tensors having major and minor symmetries. It is known that the spectral theorem for any $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ guarantees the existence of real scalars $\{\lambda _{i}\}$ and symmetric, real second order tensors $\{\mathbf{X}_{i}\}$ such that

$$ \mathbb{X}= \sum _{i=1}^{3} \lambda _{i} \mathbf{X}_{i}\otimes \mathbf{X}_{i}. $$

(16)

Restricting our investigation to symmetric fourth order tensors deeply simplifies the scenario, as illustrated by the following proposition.

Proposition 11

Let $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$. Then $\iota _{i}(\mathbb{X})$, $i=1,\dots ,43$ are invariants in the sense that

$$ \iota _{i}(\mathbb{X}) = \iota _{i}(\mathtt{Q}\mathbb{X}), \qquad \forall \mathtt{Q}, \ i=1,\dots 43. $$

Moreover

$$ \iota _{1}(\mathbb{X})=\iota _{2}(\mathbb{X}) = \frac{1}{2}\langle \mathbb{X}, \mathbb{I}- \frac{1}{2}\mathbf {I}\otimes \mathbf {I}\rangle , \qquad \iota _{3}(\mathbb{X})= \cdots =\iota _{6}(\mathbb{X}) = \frac{1}{4} \langle \mathbb{X}, \mathbf {I}\otimes \mathbf {I}\rangle , $$

$$ \iota _{7}(\mathbb{X}) = |\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}} \rangle |^{2}, \qquad \iota _{8}(\mathbb{X})=\cdots =\iota _{23} = | \frac{1}{2}\langle \mathbb{X}, \mathbf {I}\otimes \mathbf {L}\rangle |^{2}, $$

$$ \iota _{24}(\mathbb{X})=\cdots =\iota _{43}(\mathbb{X}) = \frac{1}{4}\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle (\langle \mathbb{X}, \mathbf {I}\otimes \mathbf {L}\rangle )^{2}, \qquad |\iota _{24}(\mathbb{X})| = \sqrt{ \iota _{7}(\mathbb{X})} \iota _{8}(\mathbb{X}). $$

Proof

Let $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$. Then, by (16), it can be written as $\mathbb{X}= \sum _{i=1}^{3} \lambda _{i} \mathbf{X}_{i}\otimes \mathbf{X}_{i}$ for some $(\lambda _{i},\mathbf {X}_{i}) \in \mathbb{R}\times \mathcal{L}( V_{\mathbb{R}})$. By Lemma 8 we deduce immediately $\iota _{1}(\mathbb{X}) = \iota _{2}(\mathbb{X})$ (by linearity, it is sufficient to apply the Lemma on each basis $\mathbf{X}_{i}\otimes \mathbf{X}_{i}$). Moreover, we have

$$ \begin{aligned} 2\iota _{1}(\mathbb{X}) &= \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {L}+ \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}\rangle \\ &= \langle \mathbb{X}, \mathbb{I}- \mathbf {P}\otimes \mathbf {P}- \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}\rangle \\ &=\langle \mathbb{X}, \mathbb{I}\rangle - 2\langle \mathbb{X},\mathbf {P}\otimes \mathbf {P}\rangle \\ &=\langle \mathbb{X}, \mathbb{I}\rangle - \frac{1}{2}\langle \mathbb{X},\mathbf {I}\otimes \mathbf {I}\rangle \\ &=\langle \mathbb{X}, \mathbb{I}- \frac{1}{2}\mathbf {I}\otimes \mathbf {I}\rangle \end{aligned} $$

where we have used (11), Lemma 8, (9) and (10). Similarly, by Lemma 9 we deduce immediately $\iota _{3}(\mathbb{X})=\cdots =\iota _{6}(\mathbb{X}) = \frac{1}{4} \langle \mathbb{X}, \mathbf {I}\otimes \mathbf {I}\rangle$.

To verify that $\iota _{8}(\mathbb{X}) = \cdots =\iota _{23}(\mathbb{X})$, note that by (9), (10) and (16) we can essentially replace $\mathbf {P}, \overline{\mathbf {P}}$ with $\frac{1}{2}\mathbf {I}$. Moreover, by recalling (16) and Lemma 8, $\iota _{8}(\mathbb{X}),\dots , \iota _{23}(\mathbb{X})$ can be put in the form $\frac{1}{4}\langle \mathbb{X}, \mathbf {L}\otimes \mathbf {I}\rangle \langle \mathbb{X}, \mathbf {I}\otimes \mathbf {L}\rangle$. Again using Lemma 8 (applied to every term of the spectral basis of $\mathbb{X}$), we deduce the final expression as in the statement of the proposition.

With the same strategy, one can verify that all $\iota _{24}(\mathbb{X}),\dots , \iota _{43}(\mathbb{X})$ are equal to $\frac{1}{4}\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle ( \langle \mathbb{X}, \mathbf {I}\otimes \mathbf {L}\rangle )^{2}$. That $|\iota _{24}(\mathbb{X})| = \sqrt{\iota _{7}(\mathbb{X})}\iota _{8}(\mathbb{X})$ is straightforward.

In fact, we have

$$ \begin{aligned} |\iota _{24}(\mathbb{X})|^{2} &= \iota _{24}(\mathbb{X}) \overline{\iota _{24}(\mathbb{X})} \\ &=\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {L}} \otimes \mathbf {P}\rangle \langle \mathbb{X}, \overline{\mathbf {L}}\otimes \mathbf {L}\rangle \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle \\ &=|\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle |^{2} \, | \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle |^{4} \\ &=|\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle |^{2} \, | \langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {P}}\rangle |^{4} \\ &= \iota _{7}(\mathbb{X}) |\iota _{9}(\mathbb{X})|^{2} \\ &= \iota _{7}(\mathbb{X}) |\iota _{8}(\mathbb{X})|^{2}, \end{aligned} $$

since we have shown that $\iota _{9}(\mathbb{X}) = \iota _{8}(\mathbb{X})$. □

Remark 12

In Proposition 11 we have found out that there are two linear invariants, i.e., $\iota _{1}$, $\iota _{3}$, two quadratic invariants, i.e., $\iota _{7}$, $\iota _{8}$, and two cubic invariants corresponding to the real and imaginary parts of $\iota _{24}$. However, the latter are not independent, since we also found out that a constraint exists between the absolute value of $\iota _{24}$ and the quadratic invariants.

We are now in a position to prove the representation theorem for plane elasticity-like fourth order tensors.

Theorem 13

There exist $T_{0}, T_{1}, R_{0}, R_{1} \in \mathbb{R}$, $R_{0}, R_{1} \geq 0$, and two real tensors $\mathbf {H}_{0}, \mathbf {H}_{1} \in \mathrm{Harm}( V_{\mathbb{R}})$, $|\mathbf {H}_{0}| = |\mathbf {H}_{1}| = \sqrt{2}$, such that any real tensor $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ admits the following representation:

$$ \mathbb{X}= 2T_{0} \mathbb{I}+ (2T_{1} - T_{0}) \mathbf {I}\otimes \mathbf {I}+ 2R_{0} \left (\mathbf {H}_{0}\otimes \mathbf {H}_{0} - (\mathbb{I}- \frac{1}{2}\mathbf {I}\otimes \mathbf {I})\right ) + 4R_{1}\mathbf {I}\odot \mathbf {H}_{1}.$$

(17)

Proof

Let $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$ be a generic fourth order tensor with major and minor symmetries. Expressing $\mathbb{X}$ in the eigenbasis of ℚ, we obtain after using Proposition 11

$$ \mathbb{X}= \iota _{1}(\mathbb{X}) \mathbb{I}+ (\iota _{3}(\mathbb{X})-\frac{1}{2}\iota _{1}( \mathbb{X}))\mathbf {I}\otimes \mathbf {I}+ \sqrt{\iota _{7}(\mathbb{X})}\mathbb{H}_{0} + \sqrt{4 \iota _{8}(\mathbb{X})}\mathbb{H}_{1}, $$

where we have simply posed

$$\begin{aligned} \mathbb{H}_{0} := {}&\frac{1}{\sqrt{\iota _{7}(\mathbb{X})}} \left (\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \mathbf {L}\otimes \overline{\mathbf {L}} + \langle \mathbb{X},\overline{\mathbf {L}}\otimes \mathbf {L}\rangle \overline{\mathbf {L}} \otimes \mathbf {L}\right ), \\ \mathbb{H}_{1} :={}& \frac{1}{\sqrt{4\iota _{8}(\mathbb{X})}} ( \langle \mathbb{X},\mathbf {L}\otimes \mathbf {P}\rangle \mathbf {L}\otimes \mathbf {P}+ \langle \mathbb{X},\overline{\mathbf {L}}\otimes \overline{\mathbf {P}} \rangle \overline{\mathbf {L}}\otimes \overline{\mathbf {P}}+ \langle \mathbb{X},\mathbf {L}\otimes \overline{\mathbf {P}}\rangle \mathbf {L}\otimes \overline{\mathbf {P}} \\ &{}+ \langle \mathbb{X},\overline{\mathbf {L}}\otimes \mathbf {P}\rangle \overline{\mathbf {L}} \otimes \mathbf {P}+ \langle \mathbb{X},\overline{\mathbf {P}}\otimes \mathbf {L}\rangle \overline{\mathbf {P}}\otimes \mathbf {L}+ \langle \mathbb{X},\mathbf {P}\otimes \overline{\mathbf {L}}\rangle \mathbf {P}\otimes \overline{\mathbf {L}} \\ &{}+ \langle \mathbb{X},\mathbf {P}\otimes \mathbf {L}\rangle \mathbf {P}\otimes \mathbf {L}+ \langle \mathbb{X},\overline{\mathbf {P}}\otimes \overline{\mathbf {L}}\rangle \overline{\mathbf {P}}\otimes \overline{\mathbf {L}}) \\ ={}&\frac{1}{2\sqrt{4\iota _{8}(\mathbb{X})}} ( \langle \mathbb{X},\mathbf {L}\otimes \mathbf {I}\rangle \mathbf {L}\otimes \mathbf {I}+ \langle \mathbb{X},\overline{\mathbf {L}}\otimes \mathbf {I}\rangle \overline{\mathbf {L}}\otimes \mathbf {I}+ \langle \mathbb{X},\mathbf {I}\otimes \mathbf {L}\rangle \mathbf {I}\otimes \mathbf {L}\\ &{}+ \langle \mathbb{X},\mathbf {I}\otimes \overline{\mathbf {L}}\rangle \mathbf {I}\otimes \overline{\mathbf {L}}) \\ ={}&\frac{1}{\sqrt{4\iota _{8}(\mathbb{X})}} \mathbf {I}\odot ( \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {I}\rangle \mathbf {L}+ \langle \mathbb{X},\overline{\mathbf {L}} \otimes \mathbf {I}\rangle \overline{\mathbf {L}}). \end{aligned}$$

It is easy to check that $\mathbb{H}_{0}, \mathbb{H}_{1}$ are real Hermitian and traceless. For $\mathbb{H}_{0}$ it is sufficient to show that, using (11),

$$ \begin{aligned} \langle \mathbb{I}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle &= \langle \mathbf {P}\otimes \mathbf {P}+ \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} + \mathbf {L}\otimes \mathbf {L}+ \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle \\ &= \langle \mathbf {P}\otimes \mathbf {P}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle + \langle \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle + \langle \mathbf {L}\otimes \mathbf {L}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle + \langle \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle = 0. \end{aligned} $$

For $\mathbb{H}_{1}$, similarly, one can check that the tensors spanning $\mathbb{H}_{1}$ are orthogonal to $\mathbb{I}$. For instance, the first terms gives

$$ \begin{aligned} \langle \mathbb{I}, \mathbf {L}\otimes \mathbf {P}\rangle &= \langle \mathbf {P}\otimes \mathbf {P}+ \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} + \mathbf {L}\otimes \mathbf {L}+ \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}, \mathbf {L}\otimes \mathbf {P}\rangle \rangle \\ &= \langle \mathbf {P}\otimes \mathbf {P},\mathbf {L}\otimes \mathbf {P}\rangle + \langle \overline{\mathbf {P}}\otimes \overline{\mathbf {P}}, \mathbf {L}\otimes \mathbf {P}\rangle + \langle \mathbf {L}\otimes \mathbf {L}, \mathbf {L}\otimes \mathbf {P}\rangle + \langle \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}, \mathbf {L}\otimes \mathbf {P}\rangle = 0. \end{aligned} $$

In addition, $\langle \mathbb{H}_{0}, \mathbb{H}_{1}\rangle =0$: one can observe that the tensors spanning both $\mathbb{H}_{0}, \mathbb{H}_{1}$ are the eigentensors of $\mathtt{Q}$, and hence are all mutually orthogonal.

Let $z:=\sqrt{\langle \mathbb{X}, \mathbf {L}\otimes \overline{\mathbf {L}}\rangle}$, and hence $\overline{z} = \sqrt{\langle \mathbb{X}, \overline{\mathbf {L}}\otimes \mathbf {L}\rangle}$ .⁴ Define $\mathbf {H}_{0} := \frac{1}{|z|}(z\mathbf {L}+ \overline{z}\overline{\mathbf {L}})$. It is clear that it is harmonic (compare it with $\mathbf {H}$ in Theorem 6) and that $|\mathbf {H}_{0}|=\sqrt{2}$. Hence

$$ \begin{aligned} \mathbf {H}_{0}\otimes \mathbf {H}_{0} &= \frac{1}{|z|^{2}}\left ( |z|^{2} \mathbf {L}\otimes \mathbf {L}+ |\overline{z}|^{2} \overline{\mathbf {L}}\otimes \overline{\mathbf {L}} + z^{2} \mathbf {L}\otimes \overline{\mathbf {L}} + \overline{z}^{2} \overline{\mathbf {L}}\otimes \mathbf {L}\right ) \\ &= \left ( \mathbb{I}- \mathbf {P}\otimes \mathbf {P}- \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} \right ) + \frac{1}{|z|^{2}}\left (z^{2} \mathbf {L}\otimes \overline{\mathbf {L}} + \overline{z}^{2} \overline{\mathbf {L}}\otimes \mathbf {L}\right ) \\ &= \left ( \mathbb{I}- \mathbf {P}\otimes \mathbf {P}- \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} \right ) + \frac{\sqrt{\iota _{7}(\mathbb{X})}}{|z|^{2}} \mathbb{H}_{0}, \end{aligned} $$

from which $\mathbb{H}_{0} = \mathbf {H}_{0}\otimes \mathbf {H}_{0} - (\mathbb{I}- \frac{1}{2}\mathbf {I}\otimes \mathbf {I})$ taking into account symmetries. Define now

$$ \mathbf {H}_{1} := \frac{1}{\sqrt{4\iota _{8}(\mathbb{X})}} ( \langle \mathbb{X}, \mathbf {L}\otimes \mathbf {I}\rangle \mathbf {L}+ \langle \mathbb{X},\overline{\mathbf {L}} \otimes \mathbf {I}\rangle \overline{\mathbf {L}}). $$

By looking at the definition of $\mathbb{H}_{1}$, it is clear that $\mathbb{H}_{1} = \mathbf {I}\odot \mathbf {H}_{1}$. We conclude by posing $2T_{0} := \iota _{1}(\mathbb{X})$, $T_{1} := \iota _{3}(\mathbb{X})$, $2R_{0} := \sqrt{\iota _{7}(\mathbb{X})}$, $4R_{1}=\sqrt{4\iota _{8}(\mathbb{X})}$. □

Remark 14

Arguing as in Remark 7, we find that $\mathbf {H}_{0}, \mathbf {H}_{1}$ have the following spectral decomposition

$$ \mathbf {H}_{i} = \mathbf {h}_{i}\otimes \mathbf {h}_{i} - \mathbf {h}_{i}^{\perp}\otimes \mathbf {h}_{i}^{\perp}, \qquad i=0,1 $$

for some unit vector $\mathbf {h}_{i}\in V_{\mathbb{R}}$. Geometrically, $\mathbf {H}_{i}$ represents a mirror symmetry across $\mathbf {h}_{i}$. Moreover, $\mathbf {H}_{i}$ are invariant under rotations of $k\pi$, $k \in \mathbb{Z}$. Again, if one fixes a basis, each $\mathbf {h}_{i}$ can be parameterized by an angle, $\Phi _{i}$ in the polar representation nomenclature, measured from a reference axis. Note that we have also $\mathbf {H}_{i}\mathbf {H}_{i} = \mathbf {I}$, so that $(\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}) \mathbf {I}= \mathbf {I}$.

In deriving the representation Theorem 13 we did not make use of the invariant $\iota _{24}(\mathbb{X})$, defined in Proposition 11. By plugging (17) into the expression of $\iota _{24}(\mathbb{X})$ as in the statement of Proposition 11, we find that

$$ \iota _{24}(\mathbb{X}) = 8 R_{0} R_{1}^{2} (\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle )^{2}. $$

The next proposition is useful to further characterize such an invariant.

Proposition 15

There exists a real scalar $\Delta$ such that

$$ (\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle )^{2} = \mathrm{e}^{\mathrm {i}\mkern 1mu4\Delta}. $$

Proof

We want to determine $a\geq 0$ and $b \in \mathbb{R}$ such that $\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle = a \mathrm{e}^{\mathrm {i}\mkern 1mub}$. As for $a$, we have

$$ \begin{aligned} a^{2} = |\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle |^{2} &= |\langle \mathbf {H}_{0}, \mathbf {L}\rangle |^{2} \, |\langle \mathbf {H}_{1}, \overline{\mathbf {L}}\rangle |^{2} \\ &= \frac{z \, \overline{z}}{|z|^{2}} \, \frac{4 |\langle \mathbb{X}, \mathbf {L}\otimes \mathbf {I}\rangle |^{2}}{16 \iota _{8}(\mathbb{X})} \\ & = 1, \end{aligned} $$

where $z$ has been defined in the proof of Theorem 13 and where we have used again the definition of $\mathbf {H}_{0}$, $\mathbf {H}_{1}$ given in the same proof.

The real part of $\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle$ can be easily characterized. In fact

$$ \begin{aligned} \Re (\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle ) &= \frac{1}{2}\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}+ \overline{\mathbf {L}}\otimes \overline{\mathbf {L}}\rangle \\ &=\frac{1}{2}\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbb{I}- \mathbf {P}\otimes \mathbf {P}- \overline{\mathbf {P}}\otimes \overline{\mathbf {P}} \rangle \\ &=\frac{1}{2}\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbb{I}\rangle \\ &=\frac{1}{2}\langle \mathbf {H}_{0}, \mathbf {H}_{1} \rangle , \end{aligned} $$

where we have used (11) and the very definition of $\mathbf {H}_{i}$ (as in the proof of Theorem 13) to pass from the second to the third line. An easy computation reveals that

$$ \frac{1}{2}\langle \mathbf {H}_{0}, \mathbf {H}_{1}\rangle = \langle \mathbf {h}_{0}, \mathbf {h}_{1}\rangle ^{2} - \langle \mathbf {h}_{0}^{\perp}, \mathbf {h}_{1}\rangle ^{2} = \langle \mathbf {h}_{1}, \mathbf {H}_{0}\mathbf {h}_{1}\rangle . $$

From a geometric viewpoint, the latter quantity is the cosine of twice the angle between $\mathbf {h}_{0}$ and $\mathbf {h}_{1}$. In fact, as already pointed out, $\mathbf {H}_{0}\mathbf {h}_{1}$ is easily seen to be the mirror image of $\mathbf {h}_{1}$ across $\mathbf {h}_{0}$. If we denote by $\Delta$ such angle, we conclude that $\Re (\langle \mathbf {H}_{0}\otimes \mathbf {H}_{1}, \mathbf {L}\otimes \mathbf {L}\rangle ) = \cos (2\Delta )$ and that $b=2\Delta$.

Taking square, the claim follows. □

Remark 16

• i) The representation result of Theorem 13 is the so-called harmonic decomposition of $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$. This form was already found in [4], where the derivation crucially relies on the Cartesian representation of $\mathbb{X}$ in terms of Verchery polar parameters.

• ii) Note that if $R_{0}=R_{1}=0$, $\mathbb{X}= 2T_{0} \mathbb{I}+ (2T_{1}-T_{0})\mathbf {I}\otimes \mathbf {I}$ is the “standard” form of the isotropic linear hyperelasticity tensor. Thus, if $\mu ,\lambda$ are the usual Lamé constants of isotropic linear hyperelasticity, we get $\iota _{1}(\mathbb{X}) = 2T_{0} = 2\mu$, $2T_{1}-T_{0} = \lambda$, so that $\iota _{3}(\mathbb{X}) = 2T_{1} = \mu + \lambda$ is the bulk modulus $\kappa$. Hence the linear invariants fully describe the isotropic part of $\mathbb{X}$, while the anisotropy is described by $R_{0}, R_{1}$ and $\mathbf {H}_{0}, \mathbf {H}_{1}$.

• iii) From Proposition 15, we deduce that $\iota _{24}(\mathbb{X}) = 8 R_{0} R_{1}^{2} \mathrm{e}^{\mathrm {i}\mkern 1mu4 \Delta}$. Since both $R_{0}$, $R_{1}$ are invariants, we conclude that $\Delta$ is an invariant as well, although it does not appear directly into the representation Theorem 13. Note that $\Delta$, being the angle between $\mathbf {h}_{0}$ and $\mathbf {h}_{1}$, is well-defined in our coordinate-free setting. If one fixes a basis, it is equal to $\Phi _{0}- \Phi _{1}$ (see Remark 14), modulo the sign.

• iv) $T_{0}$, $T_{1}$, $R_{0}$, $R_{1}$, $\Phi _{0}$, $\Phi _{1}$ are the polar parameters of Verchery. However, only $T_{0}$, $T_{1}$, $R_{0}$, $R_{1}$, $\Delta$ (the angle between $\mathbf {h}_{0}$ and $\mathbf {h}_{1}$) are tensor invariants. Note that the coordinate-free approach makes it clear that neither $\Phi _{0}$ nor $\Phi _{1}$ are invariants, since they are not even well defined in this framework.

• v) When $\mathbf {H}_{0}$ and $\mathbf {H}_{1}$ are coaxial (so that $\mathbf {H}_{0}\mathbf {H}_{1} = \mathbf {H}_{1}\mathbf {H}_{0}$), the two harmonics are exactly in phase (or in counterphase). Such a condition characterizes the ordinary orthotropy (geometric orthotropy) of the elastic tensor. Assume in fact $\mathbf {H}_{1}$ is coaxial with $\mathbf {H}_{0}$ and take $\mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{\mathbb{R}})$. If we consider the transformation $\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i} ( \mathbb{X})$, recalling the properties shown at the end of Remark 14, we have

  $$ \begin{aligned} (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}) \mathbb{I}&= (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\boxtimes \mathbf {I}\\ &= (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\boxtimes (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\\ &= \mathbf {I}\boxtimes \mathbf {I}= \mathbb{I}, \end{aligned} $$
  $$ \begin{aligned} (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}) \mathbf {I}\otimes \mathbf {I}&= (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\otimes ( \mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\\ &=\mathbf {I}\otimes \mathbf {I}, \end{aligned} $$
  $$ \begin{aligned} (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}) \mathbf {H}_{0}\otimes \mathbf {H}_{0} &= (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {H}_{0} \otimes (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {H}_{0} \\ &=\mathbf {H}_{i}\mathbf {H}_{0}\mathbf {H}_{i}\otimes \mathbf {H}_{i}\mathbf {H}_{0}\mathbf {H}_{i} \\ &=(\mathbf {H}_{0}\otimes \mathbf {H}_{0}), \end{aligned} $$
  $$ \begin{aligned} (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}) \mathbf {I}\otimes \mathbf {H}_{1} &= (\mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {I}\otimes ( \mathbf {H}_{i}\boxtimes \mathbf {H}_{i})\mathbf {H}_{1} \\ &=(\mathbf {I}\otimes \mathbf {H}_{1}), \end{aligned} $$

  and we finally deduce

  $$ \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i}\boxtimes \mathbf {H}_{i} ( \mathbb{X}) = \mathbb{X}, \qquad \forall \mathbb{X}\in \mathcal {E}( V_{\mathbb{R}}\otimes V_{ \mathbb{R}}), \ i=0,1. $$

  In plane language, $\mathbb{X}$ is invariant under a mirror reflection across $\mathbf {h}_{i}$, i.e., $\mathbb{X}$ is orthotropic.

• vi) For the other kinds of anisotropy, if $R_{0}=0$ we have the so called $R_{0}$-orthotropy [27]. If $R_{1}=0$ we have square symmetry. If $T_{0}=T_{1}$ we have a rari-constant material [30]. We refer to [29, Ch. 4] and references therein for an exhaustive presentation.

• vii) In principle, such an approach extends to 3D tensors.

Appendix:  A Note on Tensors

Tensors admit several interpretations. Here we elaborate on the interpretation underlying (2). The reader is addressed to [1, 2, 25] for more details. A map (a functional) $V_{1}\times V_{2}\to \mathbb{C}$ is called bilinear if it is linear with respect to each of its arguments (separately). The tensor product space $V_{1} \otimes V_{2}$ is defined as $\mathcal{L}(V_{1}^{*}\times V_{2}^{*}; \mathbb{C})$, i.e., the space of all bilinear functionals on $V^{*}_{1} \times V^{*}_{2}$ with complex scalar values, where $V^{*}$ is the dual space of $V$. A tensor of order two $\mathbf {v}_{1} \otimes \mathbf {v}_{2} :V_{1}^{*}\times V_{2}^{*} \to \mathbb{C}$ is an element of $V_{1} \otimes V_{2}$, defined by

$$ \mathbf {v}_{1} \otimes \mathbf {v}_{2} (\mathbf {x}_{1}^{*}, \mathbf {x}_{2}^{*}) = \mathbf {x}_{1}^{*}( \mathbf {v}_{1})\mathbf {x}_{2}^{*}(\mathbf {v}_{2}) \qquad \forall (\mathbf {x}_{1}^{*}, \mathbf {x}_{2}^{*}) \in V_{1}^{*} \times V_{2}^{*}. $$

The spaces $V_{i}$ being endowed with Hermitian inner products $\langle \cdot , \cdot \rangle _{V_{i}}$, by the Riesz representation theorem the previous action rewrites as

$$ \mathbf {v}_{1} \otimes \mathbf {v}_{2} (\mathbf {x}_{1}^{*}, \mathbf {x}_{2}^{*}) = \langle \mathbf {v}_{1}, \mathbf {x}_{1} \rangle _{V_{1}} \langle \mathbf {v}_{2}, \mathbf {x}_{2} \rangle _{V_{2}}, $$

where $\mathbf {x}_{i}$ is the Riesz representation of $\mathbf {x}^{*}_{i}$ in $V$.

We are mostly interested in the case the tensor product space is of the form $V \otimes V^{*}$. By using Riesz representation theorem, we get

$$ \begin{aligned} \mathbf {v}_{1} \otimes \mathbf {v}_{2}^{*} (\mathbf {x}_{1}^{*}, \mathbf {x}_{2}^{**}) &= \mathbf {x}_{1}^{*}(\mathbf {v}_{1}) \mathbf {x}_{2}^{**}(\mathbf {v}_{2}^{*}) \\ &=\langle \mathbf {v}_{1}, \mathbf {x}_{1} \rangle _{V}\langle \mathbf {v}_{2}^{*}, \mathbf {x}_{2}^{*} \rangle _{V^{*}} \\ &=\langle \mathbf {v}_{1}, \mathbf {x}_{1} \rangle _{V}\langle \mathbf {x}_{2}, \mathbf {v}_{2} \rangle _{V} \end{aligned} $$

where $\mathbf {x}_{1}$ is the Riesz representation of $\mathbf {x}_{1}^{*}$ in $V$ and $\mathbf {x}_{2}^{*}$ is the Riesz representation of $\mathbf {x}_{2}^{**}$ in $V^{*}$. To deduce the last equality, we shall apply again the Riesz representation theorem to $\mathbf {v}_{2}^{*}$ and $\mathbf {x}_{2}^{*}$. Hence, denoting by $\mathbf {v}_{2}, \mathbf {x}_{2}$ the Riesz representations of $\mathbf {v}_{2}^{*}, \mathbf {x}_{2}^{*}$ in $V$, respectively, we have

$$ \langle \mathbf {v}_{2}^{*}, \mathbf {x}_{2}^{*} \rangle _{V^{*}} = \langle \langle \cdot , \mathbf {v}_{2} \rangle _{V}, \langle \cdot , \mathbf {x}_{2} \rangle _{V} \rangle _{V^{*}} = \langle \mathbf {x}_{2}, \mathbf {v}_{2} \rangle _{V}. $$

We clarify that the Hermitian inner product on $V^{*}$ is a sesquilinear operator $\langle \cdot ,\cdot \rangle _{V^{*}}:V^{*}\times V^{*} \to \mathbb{C}$ with analogous properties as in (1). By using the Riesz representation theorem, we have found out the relationship with the inner product on the predual space $V$.

Thus, denoting the inner product on $V$ simply by $\langle \cdot , \cdot \rangle$, we finally have

$$ \begin{aligned} \mathbf {v}_{1}\otimes \mathbf {v}_{2}^{*}(\mathbf {x}^{*}_{1}, \mathbf {x}^{**}_{2}) &= \langle \mathbf {v}_{1}, \mathbf {x}_{1} \rangle \langle \mathbf {x}_{2}, \mathbf {v}_{2} \rangle \\ &= \langle \mathbf {v}_{1} \langle \mathbf {x}_{2}, \mathbf {v}_{2}\rangle , \mathbf {x}_{1} \rangle \\ &:= \langle \mathbf {v}_{1} \otimes \mathbf {v}_{2}(\mathbf {x}_{2}), \mathbf {x}_{1} \rangle \end{aligned} $$

(18)

which establishes the isomorphism between $\mathcal{L}(V)$ and $V \otimes V^{*}$. In fact $\mathbf {x}_{2} \mapsto \mathbf {v}_{1} \langle \mathbf {x}_{2}, \mathbf {v}_{2}\rangle$ is linear from $V$ to $V$, and hence there exists a unique linear operator $\mathbf {A}$, that we reasonably baptize $\mathbf {v}_{1} \otimes \mathbf {v}_{2}$, such that $\mathbf {A}\mathbf {x}_{2} = \mathbf {v}_{1} \langle \mathbf {x}_{2}, \mathbf {v}_{1}\rangle$ for every $\mathbf {x}_{2}$. Note also that in the last line of (18) $\mathbf {v}_{1} \otimes \mathbf {v}_{2}$ belongs to $\mathcal {L}(V)$ while in the first line $\mathbf {v}_{1} \otimes \mathbf {v}^{*}_{2}$ is an element of $V \otimes V^{*}$. By the definiteness of the inner product, the last two lines define the action of dyads on $V$, i.e., (2). The same argument extends to multilinear maps, providing higher order tensors.