Consider two vectors (i.e. first-order tensors) $\boldsymbol{a}$ and $\boldsymbol{b}$ which can be expressed in index notation as $a_{i}\,\boldsymbol{e}_{i}$ and $b_{i}\,\boldsymbol{e}_{i}$ respectively. These vectors have a scalar product given by

\begin{equation} \boldsymbol{a}\cdot\boldsymbol{b}=a_{i}b_{i} \qquad\boldsymbol{a},\,\boldsymbol{b}\in\mathcal{R}^{3}\,, \end{equation}

and an outer product, denoted by $\otimes$, that yields a second-order tensor $\boldsymbol{C}$ given by

\begin{align} \boldsymbol{C}&=\boldsymbol{a}\otimes\boldsymbol{b}\\ &=a_{i}b_{j}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j} \\ &= C_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j} \qquad \boldsymbol{C}\in\mathcal{R}^{3}\times\mathcal{R}^{3}\,. \end{align}

Similarly, the second-order tensors $\boldsymbol{A}$ and $\boldsymbol{B}$, or $A_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}$ and $B_{ij}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}$ respectively, have a scalar product given by

\begin{equation} \boldsymbol{A}:\boldsymbol{B}=A_{ij}B_{ij} \,, \end{equation}

an inner product given by

\begin{align} \boldsymbol{A}\boldsymbol{B}&=A_{ij}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j})B_{k\ell}(\boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell}) \\ &=A_{ij}B_{j\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{\ell})\,, \end{align}

and an outer product, also denoted by $\otimes$, that yields a fourth-order tensor $\mathbb{C}$ given by

\begin{align} \mathbb{C}&=\boldsymbol{A}\otimes\boldsymbol{B}\\ &=A_{ij}B_{k\ell}\,\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell}\\ & = \mathbb{C}_{ijk\ell} \, \boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell} \qquad \mathbb{C}\in\mathcal{R}^{3}\times\mathcal{R}^{3}\times\mathcal{R}^{3}\times\mathcal{R}^{3} \,. \end{align}

Finally, the product of a fourth-order tensor $\mathbb{A}$ and a second-order tensor $\boldsymbol{B}$ is defined as

\begin{align} \mathbb{A}\boldsymbol{B}&=\mathbb{A}_{ijk\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j}\otimes \boldsymbol{e}_{k}\otimes\boldsymbol{e}_{\ell})B_{mn}(\boldsymbol{e}_{m}\otimes\boldsymbol{e}_{n})\\ &=\mathbb{A}_{ijk\ell}B_{k\ell}(\boldsymbol{e}_{i}\otimes\boldsymbol{e}_{j})\,, \end{align}

The question is. If there is another tensor product, denoted by $\boxtimes$, and defined by

\begin{align} (\boldsymbol{A}\boxtimes\boldsymbol{B})(\boldsymbol{a}\otimes\boldsymbol{b}) &= \boldsymbol{A}\boldsymbol{a}\otimes\boldsymbol{B}\boldsymbol{b} \,\text{, or} \\ (\boldsymbol{A}\boxtimes\boldsymbol{B})\boldsymbol{C} &= \boldsymbol{A}\boldsymbol{C}\boldsymbol{B}^{T} \end{align}

how do the products $\boldsymbol{A}\otimes\boldsymbol{B}$ and $\boldsymbol{A}\boxtimes\boldsymbol{B}$ differ from each other? What do they represent physically? And, how would the product $\boldsymbol{A}\boxtimes\boldsymbol{B}$ be expressed in index notation?

  • $\begingroup$ "correctly presented in index notation" There are several conventions on this. I prefer the convention with upper and lower indices, in which case $a_ib_i$ is nonsensical, and you instead want $g^{ij}a_ib_j$. $\endgroup$
    – Arthur
    Commented Jul 26, 2020 at 9:30
  • $\begingroup$ Regarding your last equation $\begin{equation} (\boldsymbol{A}\boxtimes\boldsymbol{B})(\boldsymbol{a}\otimes\boldsymbol{b}) = \boldsymbol{A}\boldsymbol{a}\otimes\boldsymbol{B}\boldsymbol{b} \end{equation}$, I think it's more of a matter of how you define the missing "$\cdot$" between those two. For instance, $(A\otimes B) \cdot (u\otimes v) = Au \otimes Bv$ when "$\otimes$" is the kronecker product and "$\cdot$" is the regular matrix product. When instead you want "$\otimes$" to mean the regular tensor product, then "$\cdot$" needs to be the appropriate tensor contraction, and the eq. holds. $\endgroup$
    – Hyperplane
    Commented Jul 28, 2020 at 11:56
  • $\begingroup$ @Hyperplane I have edited the post to hopefully clear this uncertainty up. $\endgroup$ Commented Jul 28, 2020 at 12:24

1 Answer 1


I think there is some confusion about the distinction between second order tensors and linear maps. On the one hand, your definition of $A\otimes B$ seems to imply that $A=A_{ij}e_i\otimes e_j$, which would imply that $A\in V\otimes V$, $V$ being the vector space in which the vectors live. On the other hand, in your definition of $A\boxtimes B$ you seem to treat $A$ as a linear mapping, in which $A$ would instead be an element of $V\otimes V^*\approx Hom(V,V)$. Of course if $V$ is finite dimensional, then there is an isomorphism between $V$ and $V^*$ anyway, but it can make things conceptually clearer to distinguish vectors from dual vectors.

With that said, let us consider $A$ and $B$ to be linear mappings $V\to V$, in which case $A\boxtimes B$ is an element of $Hom(V\otimes V, V\otimes V)$ and we have a sequence of isomorphisms:

\begin{eqnarray*} Hom(V\otimes V, V\otimes V) &\approx & (V\otimes V)\otimes (V\otimes V)^*\\ &\approx & (V\otimes V)\otimes (V^*\otimes V^*)\\ &\approx & (V\otimes V^*)\otimes (V\otimes V^*)\\ &\approx Hom(V,V)\otimes Hom(V,V)\\ \end{eqnarray*} and it is straightofrward to see that the image of $A\boxtimes B$ under this sequence is isomorphisms is precisely $A\otimes B\in Hom(V,V)\otimes Hom(V,V)$.

To see this more explicitly, let $e_i$ denote a basis of $V$, with $e_i^*$ denoting the dual basis. Then the element of $V\otimes V^*$ corresponding to $A$ is given by $A_{ij} e_i\otimes e_j^*$. Note that here and in the rest of this answer, we will omit the explicit summation symbol over repeated indices.

On the other hand, we have

$$(A\boxtimes B) (e_k\otimes e_l)=Ae_k\otimes Be_l=A_{ik}e_i\otimes B_{jl}e_j=A_{ik}B_{jl} e_i\otimes e_j$$

Therefore we can trace through the image of $A\boxtimes B$ under the sequence of isomorphisms above as follows:

\begin{eqnarray*} A\boxtimes B&\mapsto &A_{ik}B_{jl}(e_i\otimes e_j)\otimes(e_k\otimes e_l)^*\\ &\mapsto &A_{ik}B_{jl}(e_i\otimes e_j)\otimes(e_k^*\otimes e_l^*)\\ &\mapsto & A_{ik} (e_i\otimes e_k^*)\otimes B_{jl} (e_j\otimes e_l^*)\\ &\mapsto & A\otimes B \end{eqnarray*}

  • $\begingroup$ How then would one write $\boldsymbol{A}\boxtimes\boldsymbol{B}$ in index notation? $\endgroup$ Commented Aug 3, 2020 at 14:10

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