TITLE: Determinant of a tensor
QUESTION [14 upvotes]: Is there such a thing as the determinant of a tensor of rank $\gt 2$?  I am tried to think how it might be defined -- potentially like, the determinant of the tensor $A=a_{ijk}$ is $\det(A)=\epsilon^{ijk}\epsilon^{lmn}a_{1il}a_{2jm}a_{3kn}$.  But I don't know if this would even mean anything.  Does the idea of determinants generalize to higher rank tensors?

REPLY [5 votes]: There are of course extensions to Determinants for Tensors of Higher Order. In General, the determinant for a rank $(0,\gamma)$ covariant tensor of order $\Omega$ follows the following convention. In order to preserve covariance, the Levi-Civita Symbols are used. We will need $\gamma$ index labels ie. $\{a,b,c,d,...\}$ where each index label can represent multiple index values ie. $\{a_1,a_2,..,a_\gamma\}$:
$$\text{det}\left(T\right)=\frac{1}{\Omega!}\epsilon^{a_\omega b_\omega c_...}\epsilon^{a_2b_2c_2...}...\epsilon^{a_\gamma b_\gamma c_\gamma}T_{a_1a_2..a_\gamma}T_{b_1b_2..b_\gamma}T_{c_1c_2..c_\gamma}$$
For example if we are dealing with a rank 3 tensor of order 3, $\tilde{\textbf{C}}=C_{ijk}\phantom{.}\textbf{Z}^i\otimes\textbf{Z}^j\otimes\textbf{Z}^k$ where $\textbf{Z}^i$ is the contravariant basis of the tensor space, then we see that:
$$\det(\tilde{\textbf{C}})=\frac{1}{3!}\epsilon^{ijk}\epsilon^{mnl}\epsilon^{pqr}C_{imp}C_{jnq}C_{klr}$$
Which will preserve the tensorial nature of the space.