Tensor operations

The previous section introduced operations between tensors. In this section, we will focus on operations on single tensors. We will use $a$ for a scalar (0th order tensor), $\underline{\boldsymbol{ u}}$ for a vector (1st order tensor), $\boldsymbol{ a}$ for a 2nd order tensor, and $\textbf{\textsf{ A}}$ for a 4th order tensor. The following operations are discussed:

Transposition, "$\bullet^{\mathrm{T}}$"

For zero and first order tensors, transposition is not defined or has no effect (depending on your point of view). I.e. $a^{\mathrm{T}}=a$ and $\underline{\boldsymbol{ u}}^{\mathrm{T}}=\underline{\boldsymbol{ u}}$. Note that, in contrary to linear algebra, we do not distinguish between row and column vectors. A vector is just a 1st order tensor.

Second-order tensors

When we take the transpose of a 2nd order tensor, the basis change order, but the indices remain the same. If both basis systems, $\underline{\boldsymbol{ e}}_{ i}$ and $\underline{\boldsymbol{ e}}_{ j}$ are the same, transposition is equivalent to to switching the indices on $a_{ij}$ to $a_{ji}$. When doing this in index notation, we implicitly assume that the basis vector order remains unchanged.

Using the index notation, we can derive some important rules for transposed tensor expressions

We can also write the transposition directly on the index symbol, $a^{\mathrm{T}}_{ij}$. For a tensor with equal basis this is the same as transposition of the tensor. However, it might also be used in expressions when the indices are not contracted with basis vectors. Consider the expression $a_{ij} b_i = a^{\mathrm{T}}_{ji} b_i$, where in the first we contract $b$ with $a$'s first index, and in the second we contract with $a^{\mathrm{T}}$'s second index (the same as $a$'s first index). Hence, we have that $a_{ij} = a^{\mathrm{T}}_{ji}$

Fourth order tensors

When taking the major transpose of a 4th order tensor, the basis change order. The first two bases exchange places with the two last while maintaining their internal order. For equal base vectors, transposition is equivalent to switching the indices $\textsf{ A}_{ \textcolor{blue}{ ij}\textcolor{red}{ kl}}$ to $\textsf{ A}_{ \textcolor{red}{ kl}\textcolor{blue}{ ij}}$. In this case, we implicitly assume the same basis vector order.

Usually, the transpose of a 4th order tensor implies the major transpose. However, we could also consider minor transposition, i.e. going from $\textsf{ A}_{ ijkl}$ to $\textsf{ A}_{ jikl}$, $\textsf{ A}_{ ijlk}$, or $\textsf{ A}_{ jilk}$. These would all be minor transpositions but have no clear symbol such as $\bullet^{\mathrm{T}}$.

Trace, $\mathrm{tr}(\bullet)$

The trace is an operation normally associated with second order tensors, and is the sum of the diagonal entries

The trace measures the spherical (volumetric) part of a tensor. For example, the pressure, $p$, is defined as $p=-\mathrm{tr}(\boldsymbol{ \sigma})/3$ where $\boldsymbol{ \sigma}$ is the stress tensor.

Norm, "$\left\vert\left\vert \bullet\right\vert\right\vert$"

The norm of a tensor gives a scalar measure of its magnitude.

If we break this down to common orders, we have

• Scalar (0th order): $\left\vert\left\vert a\right\vert\right\vert=\sqrt{a^2}=\left\vert a\right\vert$ (absolute value)

• Vector (1st order): $\left\vert\left\vert \underline{\boldsymbol{ u}}\right\vert\right\vert=\sqrt{\underline{\boldsymbol{ u}}\cdot\underline{\boldsymbol{ u}}}=\sqrt{u_i u_i}$ (eucledian length of vector)

• 2nd order tensor: $\left\vert\left\vert \boldsymbol{ a}\right\vert\right\vert=\sqrt{\boldsymbol{ a}:\boldsymbol{ a}}=\sqrt{a_{ij} a_{ij}}$

• 4th order tensor: $\left\vert\left\vert \textbf{\textsf{ A}}\right\vert\right\vert=\sqrt{\textbf{\textsf{ A}}::\textbf{\textsf{ A}}}=\sqrt{\textsf{ A}_{ ijkl}\textsf{ A}_{ ijkl}}$

Inverse, "$\bullet^{-1}$"

The inverse is only defined for even order tensors. Instead of defining it generally, we will here only consider 0th, 2nd and 4th order tensors

Scalars

The inverse, $a^{-1}$, of a scalar, $a$, is such that $a a^{-1} = 1$. In this case, we can solve it directly as $a^{-1} = 1/a$.

2nd order tensors

Here the 2nd order identity tensor, $\boldsymbol{ I}$ was used, see Special Tensors.

Similar to the transposition there are relations for taking the inverse of a dot product between two tensors

4th order tensors

Algorithms for determining the inverse of 4th order tensors are virtually non-existent. However, we may represent the entities in a 4th order tensor as a matrix using the Voigt representation, $\underline{\underline{\mathsf{ A}}}$. In this notation, the double contraction reduces to a regular matrix-matrix product: $\textbf{\textsf{ A}}:\textbf{\textsf{ A}}$ is equivalent to $\underline{\underline{\mathsf{ A}}}\cdot\underline{\underline{\mathsf{ A}}}$. Hence, to calculate the inverse of a 4th order tensor, it is usually necessary to convert to the Voigt representation, calculate the inverse, and then convert back.

Determinant, "$\mathrm{det}(\bullet)$"

The standard determinant is only defined for 2nd order tensors. However, it is possible to generalize to higher dimensions. For a 2nd order tensor, it is straight-forward to calculate as the determinant of a matrix by filling the matrix by the tensor coefficients, e.g. in 2d:

$$\begin{aligned} \boldsymbol{ a}=a_{ij} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j}, \quad \left[a_{ij}\right] = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \\ \mathrm{det}(\boldsymbol{ a}) = a_{11} a_{22} - a_{21} a_{12} \end{aligned}$$

We can also define this by using the Levi-Civita symbol,

$$\begin{aligned} \text{2d: }\mathrm{det}(\boldsymbol{ a}) &= \frac{1}{2!} \varepsilon_{i_1 j_1} \varepsilon_{i_2 j_2} a_{i_1 i_2} a_{j_1 j_2} = \varepsilon_{ij} a_{i1} a_{j2} \\ \text{3d: }\mathrm{det}(\boldsymbol{ a}) &= \frac{1}{3!} \varepsilon_{i_1 j_1 k_1} \varepsilon_{i_2 j_2 k_2} a_{i_1 i_2} a_{j_1 j_2} a_{k_1 k_2} = \varepsilon_{ijk} a_{i1} a_{j2} a_{k3} \end{aligned}$$

which allows us to generalize to higher tensor orders, such as for a 4th order tensor in 3d:

$$\begin{aligned} \mathrm{det}(\boldsymbol{ A}) &= \frac{1}{3!} \varepsilon_{i_1 j_1 k_1} \varepsilon_{i_2 j_2 k_2} \varepsilon_{i_3 j_3 k_3} \varepsilon_{i_4 j_4 k_4} A_{i_1 i_2 i_3 i_4} A_{j_1 j_2 j_3 j_3} A_{k_1 k_2 k_3 k_4} \end{aligned}$$

Exponentiation, "$\bullet^n$"

Sometimes, it is convenient to write a series of dot products between the same tensor. We can do this as we would for scalars, by using exponentiation:

Similarly, we can define this for a fourth order tensor.

Exponential, "$\exp(\bullet)$"

The exponential function have some very nice properties for differentiation. It also occurs frequently as a solution to differential equations. It is therefore interesting to define the exponential of a tensor. For a 2nd order tensor, we can define the exponential as