Benedict Irwin edited Sketchy Generation.tex  over 9 years ago

Commit id: 304e8b3bbf8a7d9b3403065177103faba3d5c83b

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\epsilon_{ij}=\begin{vmatrix} \textbf{i} & \textbf{j} \\ \textbf{i} & \textbf{j} \end{vmatrix}=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}  \end{equation}  we can see this makes up the symmetry of the $\textbf{E}$ entries in the $F$ tensor above. However, by a Laplace expansion we have a dimensional change from 3 to 2 which can be expressed by \begin{equation}  \epsilon_{ijk}=\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ \textbf{i} & \textbf{j} & \textbf{k} \\ \textbf{i} & \textbf{j} & \textbf{k} \end{vmatrix} =   \textbf{i}\begin{vmatrix} \textbf{j} & \textbf{k} \\ \textbf{j} & \textbf{k} \end{vmatrix}-\textbf{j}\begin{vmatrix} \textbf{i} & \textbf{k} \\ \textbf{i} & \textbf{k} \end{vmatrix}+\textbf{k}\begin{vmatrix} \textbf{i} & \textbf{j} \\ \textbf{i} & \textbf{j} \end{vmatrix}  \end{equation}  But this means the rank 3 tensor which has depth slices ( the $k^th$ matrix $ij$ where $k$ is depth, and $i$ and $j$ are still row and column respectively. We could say the $k^th$ shelf in analogy.) \begin{equation}  \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}_1  \begin{bmatrix}0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}_2  \begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}_3  \end{equation}