How to write $R_{ij}$ as a matrix? Suppose that $V$ is a vector space of dimension $n$ and $R: V \otimes V \to V \otimes V$ a linear map. Then we can write $R$ as a $n^2 \times n^2$ matrix. Let $R_{ij}: V^{\otimes m} \to V^{\otimes m}$ be the linear map which acts on the $i$-th and $j$-th components of $V^{\otimes m}$ as $R$ and acts on the other components by identity. How to write $R_{ij}$ as a matrix? For example, let $m=3$ and $n=2$. Suppose that $R: V \otimes V \to V \otimes V$ is given by $$ \left(\begin{array}{cccc} -2 & 0 & 0 & 0\\ 0 & -1 & -2 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -2 \end{array}\right). $$ I think that the matrices of $R_{12}, R_{13}, R_{23}$ are $8 \times 8$ matrices. How to write down the matrices of $R_{12}, R_{13}, R_{23}$? Thank you very much.
 A: One way, although I'm not sure if it is the easiest or most efficient way, would be use braiding maps. If $i$ and $j$ were adjacent, then the problem would be easily solved by repeatedly making use of the Kronecker product. The fact that $i$ and $j$ may be separated becomes trickier, but nevertheless doable.
First, we calculate $R_{12}$, which is just $R\otimes I$. As a matrix, this is 
$$R_{12} = \left(\begin{array}{cccc} -2I & 0 & 0 & 0\\ 0 & -I & -2I & 0\\ 0 & 0 & -I & 0\\ 0 & 0 & 0 & -2I \end{array}\right),$$
where $I$ are $2\times 2$ identity matrices, and the $0$s are $2\times 2$ zero matrices. Next, we need a way to move the action of $R$ onto different components. And this is best accomplished using a braiding map which permutes the tensor factors.
Let $T_\sigma$ denote the map with action
$$T_\sigma\left(\otimes_i \mathbf{v}_i\right) = \otimes_i \mathbf{v}_{\sigma(i)},$$
extended linearly. Then $R_{ij}$ would just be
$$R_{ij} = T_{(1i)(2j)}^{-1}\circ R_{12}\circ T_{(1i)(2j)},$$
where $(1i)(2j)$ is the permutation which exchanges $1\leftrightarrow i$ and $2\leftrightarrow j$. 
Now, suppose $V$ is $n$-dimensional with basis given by $\{\mathbf{v}_i\}_{i=0}^{n-1}$. Consider the basis $\{\mathbf{v}_{i_1}\otimes \cdots \otimes \mathbf{v}_{i_m}\}$ of $V^{\otimes m}$. Written in terms of column vectors, we will find that 
$$\mathbf{v}_{i_1}\otimes \cdots \otimes \mathbf{v}_{i_m} = \mathbf{e}_{[i_m \cdots i_1]_n},$$
where $\mathbf{e}_j$ denotes the $j$th standard basis vector, and where $[i_m \cdots i_1]_n$ denotes the number given by $i_m \cdots i_1$ in base $n$:
$$[i_m \cdots i_1]_n = \sum_{k=1}^m i_k\cdot n^{k-1}.$$
The action of the braiding map $T_{(1i)(2j)}$ is then
$$T_{(1i)(2j)}\left(\mathbf{e}_{[i_m \cdots i_1]_n}\right) = \mathbf{e}_{[i_m \cdots i_{j+1}i_2i_{j-1}\cdots i_{i+1}i_1i_{i-1}\cdots i_ji_i]_n},$$
which is a handful to write, but should be conceptually clear. For example, to find $R_{23}$, we would need to write down the braiding map $T_{(213)}$, which as a matrix looks like
$$\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}.$$
Conjugating the $R_{12}$ we found previously by this large permutation matrix will give us $R_{23}$. I will leave this calculation, and the finding of $R_{13}$, to you. Hopefully the method is clear.
