Tensor Algebra

This section introduces certain key topics related to the algebra of tensors. The choice of topics is not exhaustive, but provides a good starting point for the study of Euclidean tensor analysis.

Multilinear functions and tensors

Let us extend the definition of bilinear functions introduced in the preceding section to multilinear functions. Given inner product spaces $V_1, \ldots, V_n$, consider a map $\mathsf{T}:V_1 \times \ldots \times V_n \to \mathbb{R}$ such that for any $\mathsf{u}_i, \mathsf{v}_i \in V_i$, $1 \le i \le n$, and $a \in \mathbb{R}$, $$\mathsf{T}(\mathsf{v}_1, \ldots, \mathsf{u}_i + a\mathsf{v}_i, \ldots, \mathsf{v}_n) = \mathsf{T}(\mathsf{v}_1, \ldots, \mathsf{u}_i, \ldots, \mathsf{v}_n) + a\,\mathsf{T}(\mathsf{v}_1, \ldots, \mathsf{v}_i, \ldots, \mathsf{v}_n).$$ Such functions, which are linear separately in each of their arguments, are said to be multilinear. We will denote the set of all multilinear functions of this form as $\mathcal{T}(V_1 \times \ldots \times V_n,\mathbb{R})$.

Note that multilinearity and linearity are distinct concepts, as the following example illustrates.

Example

Consider the functions $\mathsf{S}, \mathsf{T}:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined as follows: for any $x, y \in \mathbb{R}$, $$\mathsf{S}(x, y) = x + y, \qquad \mathsf{T}(x, y) = xy.$$ It follows that given $a, u, v \in \mathbb{R}$, $$\begin{split} \mathsf{S}((x,y) + c(u,v)) &= (x + y) + c(u + v) = \mathsf{S}(x,y) + a\mathsf{S}(u,v),\\ \mathsf{T}((x,y) + a(u,v)) &= (x + au)(y + av) = xy + a^2uv + axv + ayu \neq \mathsf{T}(x,y) + a\mathsf{T}(u,v). \end{split}$$ We thus see that $\mathsf{S}$ is linear, but $\mathsf{T}$ is not. On the other hand, note that $$\begin{split} \mathsf{S}(x + au, y) &= x + y + au \neq \mathsf{S}(x,y) + a\mathsf{S}(u,y),\\ \mathsf{T}(x + au, y) &= (x + au)y = xy + auy = \mathsf{T}(x,y) + a\mathsf{T}(u,y). \end{split}$$ We thus see that $\mathsf{S}$ is not multilinear, even though it is linear, and $\mathsf{T}$ is multilinear and not linear.

Let us now focus attention on a single inner product space $V$ of dimension $n$ and multilinear functions defined on finite Cartesian products of $V$. A tensor of order $k$ on $V$ is defined as multilinear function of the form $$\mathsf{A}:\underbrace{V \times \ldots \times V}_{k \text{ terms}} \to \mathbb{R},$$ The set of all multilinear maps of the form $\mathsf{A}:\times^k V \to \mathbb{R}$ is denoted by $\mathcal{T}^k(V)$. Thus $\mathcal{T}^k(V)$ denotes the set of all tensors of order $k$ on $V$. Defining maps $+:\mathcal{T}^k(V) \times \mathcal{T}^k(V) \to \mathcal{T}^k(V)$ and $\cdot:\mathbb{R} \times \mathcal{T}^k(V) \to \mathcal{T}^k(V)$ as $$\begin{split} (\mathsf{A} + \mathsf{B})(\mathsf{u}_1, \ldots, \mathsf{u}_k) &= \mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_k) + \mathsf{B}(\mathsf{u}_1, \ldots, \mathsf{u}_k),\\ (a\mathsf{A})(\mathsf{u}_1, \ldots, \mathsf{u}_k) &= a \, \mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_k), \end{split}$$ for any $\mathsf{A},\mathsf{B} \in \mathcal{T}^k(V)$, $a \in \mathbb{R}$ and $\mathsf{u}_1, \ldots, \mathsf{u}_k \in V$, it is easy to verify that $\mathcal{T}^k(V)$ is a real linear space.

Remark

The set $\mathcal{T}^1(V) = \{\mathsf{T}:V \to \mathbb{R} \,|\, \mathsf{T} \text{ is linear}\}$, defined as the set of all linear functions on $V$, is called the (algebraic) dual space of $V$, and is often written as $V^*$. It turns out that if $V$ is a finite dimensional inner product space, then $V^*$ is canonically isomorphic to $V$. This means that $V$ and $V^*$ are identical for all practical purposes. For this reason, a vector is often called a tensor of order $1$. Further, in this case, the action of any $\mathsf{v} \in \mathcal{T}^1(V)$ on any $\mathsf{u} \in V$ is defined as follows: $$\mathsf{v}(\mathsf{u}) = \mathsf{v} \cdot \mathsf{u}.$$ This is a simple instance of what is known as the Riesz representation theorem. It is conventional to define $\mathcal{T}^0(V) = \mathbb{R}$. Thus, tensors of order $0$ are scalars, tensors of order $1$ are vectors, and tensors of order $2$ can be identified with linear maps. The present definition thus unifies the various kinds of linear spaces on $V$ studied earlier. The set of all $\mathcal{T}^k(V)$ is said to constitute a tensor algebra on $V$.

Example

As a simple but important example of tensors, consider the second order tensor $\hat{\mathsf{I}} \in \mathcal{T}^2(V)$ on $V$ defined as follows: for any $\mathsf{u},\mathsf{v} \in V$, $$\hat{\mathsf{I}}(\mathsf{u},\mathsf{v}) = \mathsf{u} \cdot \mathsf{v}.$$ The multilinearity of $\hat{\mathsf{I}}$, in this case just its bilinearity, follows from the bilinearity of the inner product.

More generally, given any linear map $\mathsf{T} \in L(V,V)$, we can define a second order tensor $\hat{\mathsf{T}} \in \mathcal{T}^2(V)$ as follows: for any $\mathsf{u},\mathsf{v} \in V$, $$\hat{\mathsf{T}}(\mathsf{u},\mathsf{v}) = \mathsf{u} \cdot \mathsf{T}\mathsf{v}.$$ It is left as a simple exercise to verify that the map $\hat{\mathsf{T}}$ is a second order tensor on $V$.

Remark

Defining the identity map on $V$ as the map $\mathsf{I} \in L(V,V)$ such that $\mathsf{I}\mathsf{v} = \mathsf{v}$ for any $\mathsf{v} \in V$, note that the corresponding second order tensor is precisely the tensor $\hat{\mathsf{I}} \in \mathcal{T}^2(V)$ introduced earlier.

Tensor products

Given two tensors $\mathsf{A} \in \mathcal{T}^k(V)$ and $\mathsf{B} \in \mathcal{T}^l(V)$, it is possible to combine them to obtain a tensor of higher order. Specifically, the tensor product of $\mathsf{A}$ and $\mathsf{B}$ is defined as the tensor $\mathsf{A}\otimes\mathsf{B} \in \mathcal{T}^{k + l}(V)$ such that for any $\mathsf{u}_1, \ldots, \mathsf{u}_k, \mathsf{u}_{k+1}, \ldots, \mathsf{u}_{k + l} \in V$, $$\mathsf{A} \otimes \mathsf{B}(\mathsf{u}_1, \ldots, \mathsf{u}_k, \mathsf{u}_{k+1}, \ldots, \mathsf{u}_{k + l}) = \mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_k)\mathsf{B}(\mathsf{u}_{k+1}, \ldots, \mathsf{u}_{k + l}).$$ As a special case given vectors $\mathsf{v}, \mathsf{w} \in V$, their tensor product yields a second order tensor $\mathsf{v} \otimes \mathsf{w} \in \mathcal{T}^2(V)$: for any $\mathsf{u}_1, \mathsf{u}_2 \in V$, $$\mathsf{v} \otimes \mathsf{w} (\mathsf{u}_1, \mathsf{u}_2) = (\mathsf{v} \cdot \mathsf{u}_1)(\mathsf{w} \cdot \mathsf{u}_2).$$ The foregoing definition can be extended to define the tensor product of a finite number of tensors. Suppose that $\mathsf{u}_1, \ldots, \mathsf{u}_k$ are $k$ vectors in $V$. The tensor product of these vectors is defined as the multilinear map $$\mathsf{u}_1 \otimes \ldots \otimes \mathsf{u}_k: V^k \to \mathbb{R},$$ such that for any set of $k$ vectors $\mathsf{v}_1, \ldots, \mathsf{v}_k$ in $V$, $$\mathsf{u}_1 \otimes \ldots \otimes \mathsf{u}_k(\mathsf{v}_1, \ldots, \mathsf{v}_k) = (\mathsf{u}_1\cdot \mathsf{v}_1) \ldots (\mathsf{u}_k\cdot \mathsf{v}_k).$$ Thus, the tensor product allows us to construct higher order tensors from lower order tensors. A higher order tensor that is constructed from vectors, like the one shown above, is called a rank-$1$ tensor.

Remark

There is an ambiguity in the tensor product notation that warrants clarification. Given vectors $\mathsf{v}, \mathsf{w} \in V$, the quantity $\mathsf{v}\otimes\mathsf{w}$ can stand for either a linear map, i.e. an element of $L(V,V)$, or a tensor of order $2$, i.e. an element of $\mathcal{T}^2(V)$. The vector spaces $L(V,V)$ and $\mathcal{T}^2(V)$ are isomorphic - this means that they can be naturally identified with each other. Indeed, given any $\mathsf{T} \in L(V,V)$, the bilinear function $\hat{\mathsf{T}} \in \mathcal{T}^2(V)$ defined as $$\hat{\mathsf{T}}(\mathsf{u}_1,\mathsf{u}_2) = \mathsf{u}_1\cdot\mathsf{T}\mathsf{u}_2,$$ for any $\mathsf{u}_1,\mathsf{u}_2 \in V$, can be used to uniquely associate an element of $L(V,V)$ with $\mathcal{T}^2(V)$, and vice versa. This isomorphism is often abused to represent both $\mathsf{T} \in L(V,V)$ and $\hat{\mathsf{T}} \in \mathcal{T}^2(V)$ using the same symbol. In practice, such an ambiguity is averted by specifying the domain and codomain of every function that is used. A similar ambiguity arises in the case of tensors of higher order - the appropriate meaning is to be inferred from the context.

Basis representation

Recall that the set $\mathcal{T}^k(V)$ of all tensors of order $k$ on $V$ is a linear space. This means that we can legitimately ask how we can construct a basis for it, and represent any $k^{\text{th}}$ order tensor in terms of this basis. This question will be explored in this section.

Let us first consider the special case when an orthonormal basis $(\mathsf{e}_i)$ of $V$ is provided. Then, the representation of a $k^{\text{th}}$ order tensor $\mathsf{A} \in \mathcal{T}^k(V)$ is easily computed by considering the action of $\mathsf{A}$ on $k$ vectors $\mathsf{v}_i = \sum v_{ij} \mathsf{e}_j$ in $V$, where $1 \le i \le k$: $$\begin{split} \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_k) &= \mathsf{A}\left(\sum v_{1i_1} \mathsf{e}_{i_1}, \ldots, \sum v_{ki_k} \mathsf{e}_{i_k}\right)\\ &= \sum \mathsf{A}_{i_1\ldots i_k} v_{1i_1} \ldots v_{ki_k}, \end{split}$$ where $$\mathsf{A}_{i_1\ldots i_k} = \mathsf{A}(\mathsf{e}_{i_1}, \ldots, \mathsf{e}_{i_k}).$$ It follows from the orthonormality of $(\mathsf{e}_i)$ that $$\mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}(\mathsf{v}_1, \ldots, \mathsf{v}_k) = v_{1i_1} \ldots v_{ki_k}.$$ Putting these together, we get $$\mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_k) = \sum A_{i_1\ldots i_k} \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}(\mathsf{v}_1, \ldots, \mathsf{v}_k).$$ Since this is true for any choice of vectors $\mathsf{v}_1, \ldots, \mathsf{v}_k \in V$, we get the representation of $\mathsf{A}$ with respect to the orthonormal basis $(\mathsf{e}_i)$ of $V$ as $$\mathsf{A} = \sum A_{i_1\ldots i_k} \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}.$$ The constants $(A_{i_1\ldots i_k})$ are called the components of $\mathsf{A}$ with respect to the basis $(\mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k})$ of $V$.

The representation of $\mathsf{A}$ just derived also informs us that the set of $n^k$ multilinear functions $\{\mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k} \}$ spans $\mathcal{T}^k(V)$. By studying the actin of $\mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}$ on a set of $k$ basis vectors, it is easy to show that these $n^k$ multilinear functions are also linearly independent. This shows us that the set of $n^k$ multilinear maps $\{\mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k} \}$ constitute a basis of $\mathcal{T}(V^k,\mathbb{R})$, and that $\text{dim}(\mathcal{T}(V^k,\mathbb{R})) = (\text{dim}(V))^k$.

Example

Let us consider the second order tensor $\hat{I}\in\mathcal{T}^2(V)$ defined earlier. Suppose that $(\mathsf{e}_i)$ is an orthonormal basis of $V$. The representation of $\hat{\mathsf{I}}$ can be computed easily as follows: $$\hat{\mathsf{I}} = \sum \hat{I}_{ij} \; \mathsf{e}_i \otimes \mathsf{e}_j,$$ where $$\hat{I}_{ij} = \hat{\mathsf{I}}(\mathsf{e}_i,\mathsf{e}_j) = \mathsf{e}_i\cdot\mathsf{e}_j = \delta_{ij}.$$ We thus see that the second order tensor $\hat{\mathsf{I}} \in \mathcal{T}^2(V)$ has the following representation: $$\hat{\mathsf{I}} = \sum \delta_{ij} \mathsf{e}_i \otimes \mathsf{e}_j.$$ Notice how the Krönecker delta symbols defined earlier naturally figure as the components of the tensor $\hat{\mathsf{I}}$.

It is convenient to represent the components of second order tensor as a matrix. For instance, the components of $\hat{\mathsf{I}}$ just computed can be arranged as follows: $$= \begin{bmatrix} 1 & 0 & \ldots & 0\\ 0 & 1 & \ldots & 0\\ \vdots & & \ddots & \vdots\\ 0 & \ldots & \ldots & 1 \end{bmatrix}.$$ Note that any second order tensor can be represented as a matrix, but such a representation is not possible for higher order tensors.

Let us quickly consider the representation of $\mathsf{A} \in \mathcal{T}^k(V)$ with respect to a general basis $(\mathsf{g}_i)$ of $V$. As before, the multilinearity of $\mathsf{A}$ can be used to express it in terms of the basis $(\mathsf{g}_i)$ as follows: for any $\mathsf{u}_1, \ldots, \mathsf{u}_k \in V$, $$\begin{split} \mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_k) &= \mathsf{A}\left(\sum u_{1i_1}\mathsf{g}_{i_1}, \ldots, \sum u_{ki_k}\mathsf{g}_{i_k}\right)\\ &= \sum \mathsf{A}(\mathsf{g}_{i_1}, \ldots, \mathsf{g}_{i_k}) u_{1i_1} \ldots u_{ki_k},\\ &= \sum A^*_{i_1 \ldots i_k} (\mathsf{g}^{i_1} \otimes \ldots \otimes \mathsf{g}^{i_k})(\mathsf{u}_1, \ldots, \mathsf{u}_k), \end{split}$$ where $A^*_{i_1\ldots i_k} = \mathsf{A}(\mathsf{g}_{i_1}, \ldots, \mathsf{g}_{i_k})$. Since this holds for any choice of $\mathsf{u}_1, \ldots, \mathsf{u}_k \in V$, it follows that $$\mathsf{A} = \sum A^*_{i_1 \ldots i_k} \mathsf{g}^{i_1} \otimes \ldots \otimes \mathsf{g}^{i_k}.$$ It is straightforward to check that the set of $n^k$ tensors $(\mathsf{g}^{i_1} \otimes \ldots \otimes \mathsf{g}^{i_k})$ form a basis of $\mathcal{T}^k(V)$. This also shows that the dimension of $\mathcal{T}^k(V)$ is $n^k$. The coefficients $A^*_{i_1\ldots i_k}$ are called the covariant components of $\mathsf{A} \in \mathcal{T}^k(V)$ with respect to the basis $(\mathsf{g}^{i_1} \otimes \ldots \mathsf{g}^{i_k})$ of $\mathcal{T}^k(V)$.

An alternative means to represent the basis representation of the tensor $\mathsf{A} \in \mathcal{T}^k(V)$ can be obtained using the relations $\mathsf{g}^i = \sum g^{ij}\mathsf{g}_j$ in the basis expansion derived earlier. This yields the following representation: $$\mathsf{A} = \sum A_{i_1\ldots i_k} \mathsf{g}_{i_1} \otimes \ldots \otimes \mathsf{g}_{i_k},$$ where $$A_{i_1\ldots i_k} = \sum g^{i_1j_1}\ldots g^{i_kj_k}A^*_{j_1\ldots j_k}.$$ It is left as an easy exercise to prove that the set of $n^k$ multilinear maps $(\mathsf{g}_{i_1} \otimes \ldots \otimes \mathsf{g}_{i_k})$ also form a basis of $\mathcal{T}^k(V)$. The corresponding components of the tensor $\mathsf{A} \in \mathcal{T}^k(V)$ are the constants $A_{i_1 \ldots i_k}$, and are called the contravariant components of $\mathsf{A}$.

Remark

It is also possible to defined mixed components of a given tensor by having few of the indices as covariant and the remaining as contravariant, but such generalizations are not considered here in the interest of simplicity.

Change of basis

Note that the representation of a tensor on $V$ is always with respect to some choice of basis on $V$. Let us now study how this representation changes when we change the basis of $V$. As before, let us first consider the simple case of orthonormal bases. Let $(\mathsf{e}_i)$ and $(\mathsf{f}_i)$ be two orthonormal bases of $V$. Let the representation of a $k^{\text{th}}$ order tensor $\mathsf{A} \in \mathcal{T}^k(V)$ with respect to these bases be given by $$\begin{split} \mathsf{A} &= \sum A_{i_1\ldots i_k} \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}, \quad A_{i_1\ldots i_k} = \mathsf{A}(\mathsf{e}_{i_1}, \ldots, \mathsf{e}_{i_k}),\\ \mathsf{A} &= \sum \tilde{A}_{i_1\ldots i_k} \mathsf{f}_{i_1} \otimes \ldots \otimes \mathsf{f}_{i_k}, \quad \tilde{A}_{i_1\ldots i_k} = \mathsf{A}(\mathsf{f}_{i_1}, \ldots, \mathsf{f}_{i_k}). \end{split}$$ If $\mathsf{f}_i = \sum Q_{ji}\mathsf{e}_j$, then we see that $$\begin{split} \sum A_{i_1\ldots i_k} \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k} &= \sum \tilde{A}_{i_1\ldots i_k} \mathsf{f}_{i_1} \otimes \ldots \otimes \mathsf{f}_{i_k}\\ &= \sum \tilde{A}_{i_1\ldots i_k} \left(\sum Q_{j_1i_1}\mathsf{e}_{j_1}\right) \otimes \ldots \otimes \left(\sum Q_{j_ki_k}\mathsf{e}_{j_k}\right)\\ &= \sum Q_{j_1i_1} \ldots Q_{j_ki_k} \tilde{A}_{i_1\ldots i_k} \mathsf{e}_{j_1} \otimes \ldots \otimes \mathsf{e}_{j_k}, \end{split}$$ which shows that $$\begin{split} A_{j_1\ldots j_k} &= \sum Q_{j_1i_1} \ldots Q_{j_ki_k} \tilde{A}_{i_1\ldots i_k}\\ \Rightarrow \tilde{A}_{i_1\ldots i_k} &= \sum Q^{-1}_{i_1j_1} \ldots Q^{-1}_{i_kj_k} A_{j_1\ldots j_k}\\ &= \sum Q_{j_1i_1} \ldots Q_{j_ki_k} A_{j_1\ldots j_k}. \end{split}$$

Remark

We can equivalently derive this transformation rule as follows $$\begin{split} \tilde{A}_{i_1\ldots i_k} &= \mathsf{A}(\mathsf{f}_{i_1}, \ldots, \mathsf{f}_{i_k})\\ &= \mathsf{A}\left(\sum Q_{j_1i_1}\mathsf{e}_{j_1}, \ldots, \sum Q_{j_ki_k}\mathsf{e}_{j_k}\right)\\ &= \sum Q_{j_1i_1}\ldots Q_{j_1i_1} A_{j_1 \ldots j_k}. \end{split}$$ Notice how the orthonormality of the two bases $(\mathsf{e}_i)$ and $(\mathsf{f}_i)$ of $V$ are implicitly used in this derivation.

Example

Let us revisit the second order tensor $\hat{\mathsf{I}} \in \mathcal{T}^2(V)$ that we studied earlier. Suppose that $(\mathsf{e}_i)$ and $(\mathsf{f}_i)$ are orthonormal bases of $V$ such that $\mathsf{f}_i = \sum Q_{ji}\mathsf{e}_j$. Let the representation of $\hat{\mathsf{I}}$ with respect to these bases be as follows: $$\begin{split} \hat{\mathsf{I}} &= \sum \hat{I}_{ij} \; \mathsf{e}_i \otimes \mathsf{e}_j\\ &= \sum \hat{I}'_{ij} \; \mathsf{e}_i \otimes \mathsf{e}_j. \end{split}$$ The components $\hat{I}_{ij}$ and $\hat{I}'_{ij}$ of $\hat{\mathsf{I}}$ with respect to the orthonormal bases $(\mathsf{e}_i)$ and $(\mathsf{f}_i)$, respectively, can be computed as follows: $$\begin{split} \hat{I}_{ij} &= \hat{\mathsf{I}}(\mathsf{e}_i, \mathsf{e}_j) = \mathsf{e}_i \cdot \mathsf{e}_j = \delta_{ij},\\ \hat{I}'_{ij} &= \hat{\mathsf{I}}(\mathsf{f}_i, \mathsf{f}_j) = \mathsf{f}_i \cdot \mathsf{f}_j = \delta_{ij}. \end{split}$$ We thus see that, in this case, $\hat{I}_{ij} = \hat{I}'_{ij} = \delta_{ij}$.

Alternatively, we can compute the components $\hat{I}'_{ij}$ in terms of the components $\hat{I}_{ij}$ as follows: $$\hat{I}'_{ij} = \sum Q_{ki}Q_{lj}\hat{I}_{kl} = \sum Q_{ki}Q_{lj}\delta_{kl} = \sum Q_{ki}Q_{kj} = \delta_{ij}.$$ The last step follows from the fact that $\mathsf{Q}$ is an orthogonal matrix.

Let us briefly consider the case when general bases are employed. Suppose that the tensor $\mathsf{A} \in \mathcal{T}^k(V)$ has components $A^*_{i_1 \ldots i_k}$ and $\tilde{A}^*_{i_1 \ldots i_k}$ with respect to bases $(\mathsf{g}^{i_1} \otimes \ldots \otimes \mathsf{g}^{i_k})$ and $(\tilde{\mathsf{g}}^{i_1} \otimes \ldots \otimes \tilde{\mathsf{g}}^{i_k})$, respectively. The relationship between these components is easily computed as follows: $$\begin{split} \tilde{A}^*_{i_1 \ldots i_k} &= \mathsf{A}(\tilde{\mathsf{g}}_{i_1}, \ldots, \tilde{\mathsf{g}}_{i_k})\\ &= \mathsf{A}\left(\sum (\tilde{\mathsf{g}}_{i_1}\cdot\mathsf{g}^{j_1})\mathsf{g}_{j_1}, \ldots, \sum (\tilde{\mathsf{g}}_{i_k}\cdot\mathsf{g}^{j_k})\mathsf{g}_{j_k}\right)\\ &= \sum (\tilde{\mathsf{g}}_{i_1}\cdot\mathsf{g}^{j_1})\ldots(\tilde{\mathsf{g}}_{i_k}\cdot\mathsf{g}^{j_k}) A^*_{j_1 \ldots j_k}. \end{split}$$ The inverse of this relation can also be computed similarly.

Contraction

Let $V$ be a finite dimensional inner product space, and let $(\mathsf{e}_i)$ be an orthonormal basis of $V$. Given a tensor $\mathsf{A} \in \mathcal{T}^k(V)$ of order $k$, where $k \ge 2$, the $(i,j)$-contraction of $\mathsf{A}$ is the tensor $\mathcal{C}_{i,j}\mathsf{A} \in \mathcal{T}^{k-2}(V)$ of order $(k - 2)$ defined as follows: for any $\mathsf{v}_1, \ldots, \mathsf{v}_k \in V$, $$\begin{split} & \mathcal{C}_{i,j}\mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i-1}, \mathsf{v}_{i+1}, \ldots, \mathsf{v}_{j-1},\mathsf{v}_{j+1}, \ldots, \mathsf{v}_k)\\ = & \sum_a \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i-1}, \mathsf{e}_a, \mathsf{v}_{i+1}, \ldots, \mathsf{v}_{j-1}, \mathsf{e}_a, \mathsf{v}_{j+1}, \ldots, \mathsf{v}_k). \end{split}$$ Choosing $(\mathsf{v}_i)_{i=1}^k$ to be the set $\mathsf{e}_{i_j})_{j=1}^k$, we see that $$[\mathcal{C}_{i,j}\mathsf{A}]_{i_1\ldots i_{i-1}i_{i+1} \ldots i_{j-1}i_{j+1}\ldots i_k} = \sum_a A_{i_1\ldots i_{i-1}a i_{i+1} \ldots i_{j-1}a i_{j+1}\ldots i_k}.$$ The definition of the contraction is best illustrated with the help of some simple examples.

Example

Suppose that $\mathsf{A} \in \mathcal{T}^4(V)$ is a fourth order tensor on $V$, and $(\mathsf{e}_i)$ is an orthonormal basis of $V$. The $(2,3)$-contraction of $\mathsf{A}$ is the second order tensor $\mathcal{C}_{2,3}\mathsf{A} \in \mathcal{T}^2(V)$ defined as follows: for any $\mathsf{u},\mathsf{v} \in V$, $$\mathcal{C}_{2,3}\mathsf{A}(\mathsf{u},\mathsf{v}) = \sum_a \mathsf{A}(\mathsf{u},\mathsf{e}_a,\mathsf{e}_a,\mathsf{v}).$$ Expressing both sides in terms of the orthonormal basis $(\mathsf{e}_i)$, we get $$[\mathcal{C}_{2,3}\mathsf{A}]_{ij} u_i v_j = \sum_a A_{iaaj} u_i v_j.$$ Since this is true for any choice of $\mathsf{u}, \mathsf{v} \in V$, we get the following result: $$[\mathcal{C}_{2,3}\mathsf{A}]_{ij} = \sum_a A_{iaaj}.$$ Thus, we can express the second order tensor $\mathcal{C}_{2,3}\mathsf{A}$ in terms of the orthonormal basis $(\mathsf{e})_i$ of $V$ as $$\mathcal{C}_{2,3}\mathsf{A} = \sum_a A_{iaaj} \mathsf{e}_i \otimes \mathsf{e}_j.$$ Other possible contractions of $\mathsf{A}$ can be expressed similarly. For instance, $$\begin{split} \mathcal{C}_{1,2}\mathsf{A} &= \sum_a A_{aaij} \mathsf{e}_i \otimes \mathsf{e}_j,\\ \mathcal{C}_{1,4}\mathsf{A} &= \sum_a A_{aija} \mathsf{e}_i \otimes \mathsf{e}_j, \end{split}$$ and so on; other possible contractions can be computed similarly.

Example

Suppose that $\mathsf{A} \in \mathcal{T}^2(V)$ is a second order tensor. In this case, there is only one possible contraction of the tensor $\mathsf{A}$, namely $\mathcal{C}_{1,2}\mathsf{A} \in \mathbb{R}$. This is easily computed as follows: if $(\mathsf{e}_i)$ is an orthonormal basis of $V$, then $$\mathcal{C}_{1,2}\mathsf{A} = \sum_a \mathsf{A}(\mathsf{e}_a,\mathsf{e}_a) = \sum A_{aa}.$$ In this special case, the contraction $\mathcal{C}_{1,2}\mathsf{A}$ is called the trace of $\mathsf{A}$, and is often written as $\text{tr}(\mathsf{A})$. The trace of the linear map associated with $\mathsf{A}$ will be defined in a later section.

Note that the definition of the contraction operation uses the orthonormal basis $(\mathsf{e}_i)$ of $V$. It turns out that if we use a different orthonormal basis, the definition remains unaltered. To see this, suppose that $(\mathsf{f}_i)$ is another orthonormal basis of $V$ that is related to the orthonormal basis $(\mathsf{e}_i)$ through the relations $\mathsf{f}_i = \sum Q_{ji}\mathsf{e}_j$. Then, for any tensor $\mathsf{A} \in \mathcal{T}^k(V)$ of order $k \ge 2$, it follows that $$\begin{split} & \mathcal{C}_{i,j}\mathsf{A}(\mathsf{v}_{i_1}, \ldots, \mathsf{v}_{i_{i-1}}, \mathsf{v}_{i_{i+1}}, \ldots, \mathsf{v}_{i_{j-1}}, \mathsf{v}_{i_{j + 1}}, \ldots, \mathsf{v}_k)\\ =& \sum \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i_{i-1}}, \mathsf{f}_a, \mathsf{v}_{i_{i+1}}, \ldots, \mathsf{v}_{i_{j-1}}, \mathsf{f}_a, \mathsf{v}_{i_{j+1}}, \ldots, \mathsf{v}_k)\\ =& \sum Q_{ba}Q_{ca}\mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i_{i-1}}, \mathsf{e}_b, \mathsf{v}_{i_{i+1}}, \ldots, \mathsf{v}_{i_{j-1}}, \mathsf{e}_c, \mathsf{v}_{i_{j+1}}, \ldots, \mathsf{v}_k)\\ =& \sum \delta_{bc} \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i_{i-1}}, \mathsf{e}_b, \mathsf{v}_{i_{i+1}}, \ldots, \mathsf{v}_{i_{j-1}}, \mathsf{e}_c, \mathsf{v}_{i_{j+1}}, \ldots, \mathsf{v}_k)\\ =& \sum \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i_{i-1}}, \mathsf{e}_a, \mathsf{v}_{i_{i+1}}, \ldots, \mathsf{v}_{i_{j-1}}, \mathsf{e}_a, \mathsf{v}_{i_{j+1}}, \ldots, \mathsf{v}_k). \end{split}$$ Here $(\mathsf{v}_i)_{i=1}^k$ are any $k$ vectors in $V$. We thus see that the contraction operation does not depend on whether we choose the basis $(\mathsf{e}_i)$ or the basis $(\mathsf{f}_i)$. In this sense, the definition of the contraction operation is said to be well-defined since we get the same tensor irrespective of the choice of orthnormal basis. This remains true even if use a general basis, but this requires a more careful treatment, as will be discussed next.

Let $(\mathsf{g}_i)$ be a general basis of $V$. Given a tensor $\mathsf{A} \in \mathcal{T}^k(V)$ of order $k$, where $k \ge 2$, the $(i,j)$-contraction of $\mathsf{A}$ can be defined as follows: for any $\mathsf{v}_1, \ldots, \mathsf{v}_k \in V$, $$\begin{split} & \mathcal{C}_{i,j}\mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i-1}, \mathsf{v}_{i+1}, \ldots, \mathsf{v}_{j-1},\mathsf{v}_{j+1}, \ldots, \mathsf{v}_k)\\ = & \sum_a \mathsf{A}(\mathsf{v}_1, \ldots, \mathsf{v}_{i-1}, \mathsf{g}_a, \mathsf{v}_{i+1}, \ldots, \mathsf{v}_{j-1}, \mathsf{g}^a, \mathsf{v}_{j+1}, \ldots, \mathsf{v}_k). \end{split}$$ Notice how both the basis $\mathsf{g}_i)$ and its reciprocal basis $(\mathsf{g}^i)$ figure in this equation. It can be verified with a simple calculation that the order in which the bases $\mathsf{g}_i)$ and $(\mathsf{g}^i)$ appear in this expression is not important. Thus, $$\sum \mathsf{A}(\ldots, \mathsf{g}_a, \ldots, \mathsf{g}^a, \ldots) = \sum \mathsf{A}(\ldots, \mathsf{g}^a, \ldots, \mathsf{g}_a, \ldots).$$ Further, a calculation analogous to the one presented earlier can be used to establish the fact the contraction operation is well defined in the sense that using a different (general) basis $\mathsf{f}_i$ of $V$ results in the same tensor as when the basis $\mathsf{g}_i)$ is used.

The basis representation of $\mathcal{C}_{i,j}\mathsf{A}$ with respect to the general basis $(\mathsf{g}_i)$ of $V$ can be computed using the foregoing definitions. Recall that $\mathsf{A} \in \mathcal{T}^k(V)$ has the following basis representation: $$\mathsf{A} = \sum A_{a_1 \ldots a_k} \mathsf{g}_{a_1} \otimes \ldots \otimes \mathsf{g}_{a_k}.$$ It is left as an exercise to verify that the $(i,j)$ contraction of $\mathsf{A}$ is given by $$\begin{split}\mathcal{C}_{i,j}\mathsf{A} &= \sum g_{a_ia_j}A_{a_1\ldots a_{i-1} a_i a_{i+1} \ldots a_{j-1} a_j a_{j+1}\ldots a_k}\\ & \qquad\quad\mathsf{g}_{a_1} \otimes \mathsf{g}_{a_{i-1}} \otimes \mathsf{g}_{a_{i+1}} \otimes \ldots \otimes \mathsf{g}_{a_{j-1}}\otimes\mathsf{g}_{a_{j+1}}\otimes \ldots \otimes \mathsf{g}_{a_k}. \end{split}$$ Notice the extra factor $g_{a_ia_j}$, unlike the representation with the choice of an orthonormal basis.

Example

In the special case of a second order tensor $\mathsf{A} \in \mathcal{T}^2(V)$, there is only one possible contraction, $\mathcal{C}_{1,2}\mathsf{B} \in \mathbb{R}$, which we introduced earlier as the trace of $\mathsf{A}$. With respect to a general basis $(\mathsf{g}_i)$ of $V$, it can be checked that $$\mathsf{A} = \sum A_{ij} \, \mathsf{g}_i \otimes \mathsf{g}_j \quad\Rightarrow\quad \text{tr}(\mathsf{A}) = \sum g_{ab}A_{ab}.$$ Note that in the special case when $(\mathsf{g}_i)$ is an orthonormal basis of $V$, the expression for the trace of $\mathsf{A}$ reduces to the familiar form $\text{tr}(\mathsf{A}) = \sum A_{aa}$.

Generalized dot product of tensors

Note!

This terminology is not standard, but it is adequate for the purposes of this course.

Certain special operations called generalized dot products, or simply dot products, are now introduced between tensors of different orders. These are introduced on account of their prevalence in the continuum mechanics literature. Throughout this section, $V$ denotes a finite dimensional inner product space.

To motivate the definition of the generalized dot product of tensors on $V$, it is helpful to first consider a few important special cases. In the simplest case, given two vectors $\mathsf{u}, \mathsf{v} \in V$, note that the inner product of these two vectors can be expressed in terms of the contraction operation as follows: $$\mathsf{u} \cdot \mathsf{v} = \mathcal{C}_{1,2}(\mathsf{u} \otimes \mathsf{v}).$$ This restatement of the inner product in terms of the contraction operation will serve as the starting point for its generalization to the dot product of two arbitrary tensors.

Suppose that $\mathsf{u}, \mathsf{v}, \mathsf{w} \in V$ are any three vectors in $V$. Then the vector $(\mathsf{u} \otimes \mathsf{v}) \cdot \mathsf{w} \in V$ is defined as follows: $$(\mathsf{u} \otimes \mathsf{v}) \cdot \mathsf{w} = \mathcal{C}_{2,3}(\mathsf{u} \otimes \mathsf{v} \otimes \mathsf{w}).$$ To understand what this means, consider the action of the vector $\mathcal{C}_{2,3}(\mathsf{u} \otimes \mathsf{v} \otimes \mathsf{w})$ on an arbitrary vector $\mathsf{z} \in V$: if $(\mathsf{g}_i)$ is a general basis of $V$, then $$\begin{split} \mathcal{C}_{2,3}(\mathsf{u} \otimes \mathsf{v} \otimes \mathsf{w})(\mathsf{z}) &= \sum (\mathsf{u} \otimes \mathsf{v} \otimes \mathsf{w})(\mathsf{z}, \mathsf{g}_a, \mathsf{g}^a)\\ &= \sum u_i v_j w_k (\mathsf{g}_i \otimes \mathsf{g}_j \otimes \mathsf{g}_k)\left(\sum z_b \mathsf{g}_b, \mathsf{g}_a, \mathsf{g}^a\right)\\ &= \sum g_{ib} u_i v_j w_k z_b g_{ja} \delta_{ka}\\ &= \sum g_{jk} v_j w_k \, \sum g_{ia} u_a z_a\\ &= (\mathsf{v} \cdot \mathsf{w})(\mathsf{u} \cdot \mathsf{z})\\ &= (\mathsf{v} \cdot \mathsf{w})\mathsf{u}(\mathsf{z}). \end{split}$$ Since this is true for any $\mathsf{z} \in V$, it follows that $$(\mathsf{u} \otimes \mathsf{v}) \cdot \mathsf{w} = (\mathsf{v} \cdot \mathsf{w}) \mathsf{u}.$$ This result also provides a means to compute the dot product $\mathsf{A} \cdot \mathsf{v}$ of a second order tensor $\mathsf{A} \in \mathcal{T}^2(V)$ and a vector $\mathsf{v} \in V$: if $(\mathsf{g}_i)$ is any basis of $V$, then $$\begin{split} \mathsf{A} \cdot \mathsf{v} &= \left(\sum A_{ij} \mathsf{g}_i \otimes \mathsf{g}_j\right)\cdot\left(\sum v_k \mathsf{g}_k\right)\\ &= \sum A_{ij} v_k g_{jk} \mathsf{g}_i \in V. \end{split}$$ Note that $\tilde{\mathsf{A}} \in L(V,V)$ denote the linear map corresponding to the second order tensor $\mathsf{A} \in \mathcal{T}^2(V)$, then, for any $\mathsf{v} \in V$, $\mathsf{A} \cdot \mathsf{v} = \tilde{\mathsf{A}}\mathsf{v}$. This is most easily seen by representing both sides this equation with respect to an orthonormal basis of $V$. This shows that the dot product defined here is consistent with the theory of linear maps developed earlier.

Remark

Note that given an arbitrary second order tensor $\mathsf{A} \in \mathcal{T}^2(V)$ and an arbitrary vector $\mathsf{v} \in V$, $\mathsf{A} \cdot \mathsf{v} \in V$ and $\mathsf{v} \cdot \mathsf{A} \in V$ are, in general, different vectors. The generalized dot product of tensors is thus not necessarily symmetric.

As an extension of the foregoing ideas, we now define dot products for tensors of second order. Given vectors $\mathsf{u}, \mathsf{v}, \mathsf{w}, \mathsf{z} \in V$, the dot product $(\mathsf{u} \otimes \mathsf{v}) \cdot (\mathsf{w} \otimes \mathsf{z}) \in \mathbb{R}$ between the second order tensors $\mathsf{u} \otimes \mathsf{v} \in \mathcal{T}^2(V)$ and $\mathsf{w} \otimes \mathsf{z} \in \mathcal{T}^2(V)$ is defined as follows: $$(\mathsf{u} \otimes \mathsf{v}) \cdot (\mathsf{w} \otimes \mathsf{z}) = \mathcal{C}_{1,2}\left(\mathcal{C}_{1,3}\left(\mathsf{u} \otimes \mathsf{v} \otimes \mathsf{w} \otimes \mathsf{z}\right)\right) = (\mathsf{u} \cdot \mathsf{w})(\mathsf{v} \cdot \mathsf{z}).$$ The simplest means to understand this by considering the dot products of two second order tensors $\mathsf{A}, \mathsf{B} \in \mathcal{T}^2(V)$. If $(\mathsf{g}_i)$ denotes an orthonormal basis of $V$, and $\mathsf{A} = \sum A_{ij} \mathsf{g}_i \otimes \mathsf{g}_j$ and $\mathsf{B} = \sum B_{ij} \mathsf{g}_i \otimes \mathsf{g}_j$ are the representations of $\mathsf{A}$ and $\mathsf{B}$, respectively, with respect to this basis, then note that $$\mathsf{A} \cdot \mathsf{B} = \sum A_{ij} B_{ij}.$$

Remark

It is important to note that different authors follow different conventions regarding this. For instance, it common in the Continuum Mechanics literature to use the following notations: for any two second order tensors $\mathsf{A},\mathsf{B} \in \mathcal{T}^2(V)$, $$\begin{split} \mathsf{A} \cdot \mathsf{B} &= \sum A_{ij}B_{ji},\\ \mathsf{A} : \mathsf{B} &= \sum A_{ij}B_{ij}, \end{split}$$ where $(A_{ij})$ and $(B_{ij})$ are the components of $\mathsf{A}$ and $\mathsf{B}$, respectively, with respect to some orthonormal basis of $V$. Note that the dot product $\mathsf{A} \cdot \mathsf{B}$, according to our definition, is $\mathsf{A} : \mathsf{B}$ according to this definition. Care must be therefore exercised when reading the literature to understand the appropriate meaning of a quantity like $\mathsf{A} \cdot \mathsf{B}$.

The reason for defining the generalized dot product the way we have done is to ensure a uniform and simple notation for many differential and integral identities that we will encounter later on. It is important, however, to keep in mind that this is essentially a matter of convention.

As a final and useful example, note that if $\mathsf{A} \in \mathcal{T}^k(V)$ is a tensor of order $k$, where $k > 2$, and $\mathsf{B} \in \mathcal{T}^2(V)$ is a second order tensor, then, with respect to an orthonormal basis $(\mathsf{g}_i)$ of $V$, $$\mathsf{A} \cdot \mathsf{B} = \sum A_{i_1 \ldots i_{k-2} a b} B_{ab}.$$ The extension of this notion of dot products can be similarly extended to tensors of higher orders.

Example

Suppose that $\mathsf{A}, \mathsf{B} \in \mathcal{T}^2(V)$ are second order tensors on $V$. Then, it is true that $$\mathsf{A} \cdot \mathsf{B} = \mathsf{B} \cdot \mathsf{A}.$$ This is most easily seen with the help of the basis representation with respect to an orthonormal basis of $V$: in this case, $$\mathsf{A} \cdot \mathsf{B} = \sum A_{ij} B_{ij} = \sum B_{ij} A_{ij} = \mathsf{B} \cdot \mathsf{A},$$ thereby establishing the claim.

Example

Suppose that $\mathsf{A} \in \mathcal{T}^2(V)$ is a second order tensor on $V$ and $\mathsf{u},\mathsf{v} \in V$ are two vectors in $V$, then $$\mathsf{A} \cdot (\mathsf{u} \otimes \mathsf{v}) = \mathsf{u} \cdot \mathsf{A}\mathsf{v}.$$ This is also easily established by choosing an orthonormal $(\mathsf{g}_i)$ basis of $V$. The component form of $\mathsf{A} \cdot (\mathsf{u} \otimes \mathsf{v})$ then reads $$\begin{split}\mathsf{A} \cdot (\mathsf{u} \otimes \mathsf{v}) &= \sum A_{ij} u_i v_j\\ &= \left(\sum u_k \mathsf{g}_k\right) \cdot \left(\sum A_{ij} v_j \mathsf{g}_i\right)\\ &= \mathsf{u} \cdot \mathsf{A}\mathsf{v}.\end{split}$$ It can be shown similarly that $(\mathsf{u} \otimes \mathsf{v})\cdot\mathsf{A} = \mathsf{A}^T\mathsf{u} \cdot \mathsf{v}$.

Volume forms

Let $V$ be an inner product space of dimension $n$. A tensor $\mathsf{A} \in \mathcal{T}^k(V)$ is said to be symmetric if, for any $\mathsf{u}_1, \ldots, \mathsf{u}_k \in V$, $$\mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_i, \ldots, \mathsf{u}_j, \ldots, \mathsf{u}_n) = \mathsf{A}(\mathsf{u}_1, \ldots, \mathsf{u}_j, \ldots, \mathsf{u}_i, \ldots, \mathsf{u}_n),$$ where $1 \le i < j \le n$. If the sign reverses when any two arguments are interchanged, then the tensor is said to be skew-symmetric.

Remark

It turns out that the set of all antisymmetric tensors of a given order have a rich algebraic structure, called exterior algebra.

A volume form on $V$ is a skew-symmetric tensor $\mathsf{\epsilon}:\times^n V \to \mathbb{R}$ of order $n$. It is customary to denote the set of all volume forms on $V$ using the notation $\Omega^n(V)$: $$\Omega^n(V) = \{ \mathsf\epsilon \in \mathcal{T}^n(V) \,|\, \mathsf\epsilon \text{ is skew-symmetric}\}.$$ It is important to note that the trivial volume form $\mathsf{0} \in \mathcal{T}^n(V)$ that maps every set of $n$ vectors in $V$ to zero is excluded from this discussion. A volume form that is not trivial is said to be non-trivial. All volume forms considered here are assumed to be non-trivial.

Given two volume forms $\mathsf{\epsilon}, \mathsf{\omega} \in \Omega^n(V)$, it can be shown that there exists a scalar $a \in \mathbb{R}$ such that $\mathsf\omega = a\mathsf{\epsilon}$. An equivalent way of stating this is that the set of all antisymmetric tensors of order $n$ over an $n$-dimensional vector space is a vector space of dimension $1$: $\text{dim}(\Omega^n(V)) = 1$.

Proof

Let $\mathsf{e}_1,\ldots,\mathsf{e}_n$ be an orthonormal basis of $V$, and let $\omega \in \Omega^n(V)$ be a volume form. The component representation of the volume form $\omega$ with respect to this basis $(\mathsf{e}_1, \ldots, \mathsf{e}_n$ is $$\omega = \sum \omega_{i_1\ldots i_n} \mathsf{e}_{i_1} \otimes \ldots \mathsf{e}_{i_n}.$$ It is important to note that the components $\omega_{i_1\ldots i_n}$ are skew-symmetric. It is convenient to introduce the Levi-Civita symbol $$\epsilon_{i_1\ldots i_n} = \begin{cases} 1, & \{i_1, \ldots, i_n\} \text{ is an even permutation of } \{1, \ldots, n\},\\ -1, & \{i_1, \ldots, i_n\} \text{ is an odd permutation of } \{1, \ldots, n\},\\ 0, & \text{otherwise}. \end{cases}$$ Using this, we see at once that $$\omega = \left(\sum \epsilon_{i_1\ldots i_n}\omega_{i_1\ldots i_n}\right) \mathsf{e}_1 \otimes \ldots \otimes \mathsf{e}_n.$$ If $\eta \in \Omega^n(V)$ is another volume form, it admits the representation $$\eta = \left(\sum \epsilon_{i_1\ldots i_n}\eta_{i_1\ldots i_n}\right) \mathsf{e}_1 \otimes \ldots \otimes \mathsf{e}_n.$$ It follows at once that $\omega = a\eta$, where $$a = \frac{\sum \epsilon_{i_1\ldots i_n}\omega_{i_1\ldots i_n}}{\sum \epsilon_{i_1\ldots i_n}\eta_{i_1\ldots i_n}}.$$ Note that this fraction is well defined since we work only with non-trivial volume forms. This shows that $\text{dim}(\Omega^n(V)) = 1$.

It is conventional to choose the special volume form $\epsilon \in \Omega^n(V)$ such that $$\epsilon(e_1, \ldots, e_n) = 1.$$ We thus see that the Levi-Civita symbols are just the components of the tensor $\epsilon$ with respect to the orthonormal basis $(\mathsf{e}_1, \ldots, \mathsf{e}_n)$. The tensor $\epsilon$ is called the Levi-Civita tensor.

Picking a particular volume form $\mathsf{\epsilon} \in \Omega^n(V)$, the set of all volume forms $\mathsf{\omega} \in \Omega^n(V)$ such that $\mathsf\omega = a\mathsf{\epsilon}$ for some $a > 0$ are said to constitute an orientation of $V$. Given this orientation on $V$, a basis $(\mathsf{g}_i)$ of $V$ is said to be right-handed if $\epsilon(\mathsf{g}_1, \ldots, \mathsf{g}_n) > 0$, and left-handed otherwise.

Remark

It is emphasized that the choice of orientation is arbitrary. Typically, a special volume form that has mathematical or physical relevance is chosen and the positive orientation is defined with respect to this choice. With a particular choice of orientation, it is customary to focus only on only right-handed bases.

Rather than discussing the general theory of volume forms, it is useful to look at volume forms on the three dimensional Euclidean space $\mathbb{R}^3$. The standard volume form on $\mathbb{R}^3$, written $\mathsf{\epsilon} \in \Omega^3(\mathbb{R}^3)$, sometimes called the Levi-Civita tensor, is defined as follows: if $(\mathsf{e}_i)$ denotes the standard basis of $\mathbb{R}^3$, $$\mathsf{\epsilon} = \sum \epsilon_{ijk} \mathsf{e}_i \otimes \mathsf{e}_j \otimes \mathsf{e}_k,$$ where $\epsilon_{ijk}$ is the Levi-Civita symbol, defined as follows: $$\epsilon_{ijk} = \begin{cases} 1, & \{i,j,k\}\text{ is an even permutation of $\{1,2,3\}$},\\ -1, & \{i,j,k\}\text{ is an odd permutation of $\{1,2,3\}$},\\ 0, & \text{otherwise}, \end{cases}$$ Note that $\mathsf\epsilon(\mathsf{e}_1, \mathsf{e}_2, \mathsf{e}_3) = \epsilon_{123} = 1$ - the standard basis of $\mathbb{R}^3$ is thus right handed with respect to the standard volume form of $\mathbb{R}^3$.

Remark

A permutation of order $n$ is formally defined as a bijection of the form $\pi:I_n \to I_n$, where $I_n = \{1, \ldots, n\} \subseteq \mathbb{N}$. A special kind permuatation is the swap permutation $\pi_{i,j}:I_n \to I_n$ that interchanges the $i^{\text{th}}$ and $j^{\text{th}}$ elements in $I_n$. It can be shown that any arbitrary permutation of $I_n$ can be obtained as a succession of swaps of $I_n$. Further, the number of such swaps, for a given permutation, is either always even or always odd. This can be used to assign a parity, or a sign, to the permuation. The notation $(-1)^\pi$ is often used to denote the sign of the permutation $\pi:I_n \to I_n$. If $(-1)^\pi = 1$, the permutation is said to be even, while if $(-1)^\pi = -1$, the permutation is said to be odd.

Remark

Recall that the three dimensional Euclidean space $\mathbb{R}^3$ admits a special algebraic operation called the cross product: given any two vectors $\mathsf{u} = \sum u_i \mathsf{e}_i \in \mathbb{R}^3$ and $\mathsf{v} = \sum v_i \mathsf{e}_i \in \mathbb{R}^3$, where $(\mathsf{e}_i)$ is the standard basis of $\mathbb{R}^3$, their cross product is defined as the vector $\mathsf{u} \times \mathsf{v} \in \mathbb{R}^3$ given by $$\mathsf{u} \times \mathsf{v} = (u_2 v_3 - u_3 v_2)\mathsf{e}_1 + (u_3 v_1 - u_1 v_3)\mathsf{e}_2 + (u_1 v_2 - u_2 v_1)\mathsf{e}_3.$$ This can be written compactly using the Levi-Civita symbols $\epsilon_{ijk}$, as follows: $$\mathsf{u} \times \mathsf{v} = \sum \epsilon_{ijk} u_j v_k \mathsf{e}_i,$$ as can be easily checked.

An alternative definition of the cross product of two vectors in $\mathbb{R}^3$ can be given in terms of the standard volume form of $\mathbb{R}^3$. Given vectors $\mathsf{v}, \mathsf{w} \in \mathbb{R}^3$, their cross product $\mathsf{v} \times \mathsf{w} \in \mathbb{R}^3$ satisfies the following relation: for any $\mathsf{u} \in \mathbb{R}^3$, $$\mathsf{u} \cdot (\mathsf{v} \times \mathsf{w}) = \mathsf\epsilon(\mathsf{u}, \mathsf{v}, \mathsf{w}).$$ This expression can be used to define the cross product in terms of the volume form. It is left as an exercise to verify this. (Hint. Choose $\mathsf{u}$ to be the standard basis $\mathsf{e}_i$. This immediately yields: $(\mathsf{v}\times\mathsf{w})_i = \sum \epsilon_{ijk} v_j w_k$.)

The advantange in defining the cross product in terms of the volume form is that this allows us to extend the notion of a cross product to higher (and lower!) dimensional analogues called wedge products. We will not be studying wedge products here, but they play an important role in exterior algebra and the theory of differential forms.

Example

We will now introduce a few important identities relating the Levi-Civita symbols $\epsilon_{ijk}$ and the Kronecker delta symbols $\delta_{ij}$, called the $\epsilon-\delta$ identities. These can be written succinctly as follows: $$\begin{split} \epsilon_{ijk}\epsilon_{lmn} &= \text{det} \begin{bmatrix} \delta_{il} & \delta_{im} & \delta_{in}\\ \delta_{jl} & \delta_{jm} & \delta_{jn}\\ \delta_{kl} & \delta_{km} & \delta_{kn} \end{bmatrix},\\ \sum \epsilon_{ijk}\epsilon_{imn} &= \text{det} \begin{bmatrix} \delta_{jm} & \delta_{jn}\\ \delta_{km} & \delta_{kn} \end{bmatrix},\\ \sum \epsilon_{ijk}\epsilon_{ijn} &= 2\delta_{kn},\\ \sum \epsilon_{ijk}\epsilon_{ijk} &= 6. \end{split}$$ These relations are easily proved by direct inspection. For instance, to prove the first identity, note that $\epsilon_{ijk}\epsilon_{klm}$ is $0$ if any of the pair of indices $(i,j,k)$, or $(l,m,n)$ has repeated indices, owing to the skew-symmetry of the Levi-Civita symbol. In this case, the right hand side is also zero since this would correspond to the determinant of a matrix with either two identical rows, or two identical columns. Suppose then that the indices do not repeat. Then there are only two possibilities: either $(l,m,n)$ is an even permutation of $(i,j,k)$, or not. In the former case, $\epsilon_{ijk}\epsilon_{lmn} = 1$. Note that if $(l,m,n)$ is an even permutation of $(i,j,k)$, then exactly one of the three possibilities are true: $(l,m,n) = (i,j,k)$, or $(n,l,m) = (i,j,k)$, or $(m,n,l) = (i,j,k)$. In each of these cases, it can be verified by direct substitution that the identity is true. The case when $(l,m,n)$ is an odd permutation of $(i,j,k)$ is computed similarly, thereby estabilishing the validity of the first identity. It is left as a simple exercise to prove the remaining three identities based on the first by direct substitution.

Example

As an illustration of the use of the $\epsilon-\delta$ identities, let us prove the following identity: given any three vectors $\mathsf{u},\mathsf{v},\mathsf{w} \in \mathbb{R}^3$, it is true that $$\mathsf{u} \times (\mathsf{v} \times \mathsf{w}) = (\mathsf{u}\cdot\mathsf{w})\mathsf{v} - (\mathsf{u}\cdot\mathsf{v})\mathsf{w}.$$ Using the definition of the cross product, we see that $$\begin{split} \mathsf{u}\times(\mathsf{v} \times \mathsf{w}) &= \sum \epsilon_{ijk} u_j (\mathsf{v} \times \mathsf{w})_k \mathsf{e}_i\\ &= \sum \epsilon_{ijk}\epsilon_{klm} u_j v_l w_m \mathsf{e}_i\\ &= \sum (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) u_j v_l w_m \mathsf{e}_i\\ &= \left(\sum u_m w_m\right)\sum v_l \mathsf{e}_l - \left(\sum u_j v_j\right)\sum w_m \mathsf{e}_m\\ &= (\mathsf{u}\cdot\mathsf{w})\mathsf{v} - (\mathsf{u}\cdot\mathsf{v})\mathsf{w}. \end{split}$$ The idenity is thus proved.