Hyper-determinant

Hyper-determinant


With the Levi-Civita symbol we can define a determinant:

ε^{a1,a2,a3,...} = 0 if any 2 of the indices the same. (1 otherwise) (Also, antisymmetric)

Determinant of a 2-tensor is:

det_2(A) = ε^{a}ε^{b}A^{a1,b1}A^{a2,b2}A^{a3,b3} ....

We can define

f^{a1,a2,a3,...} = 0 if any 3 of the indices the same.

Determinant of a 3-tensor is:

 det_3(A) = f^{a}f^{b}f^{c}A^{a1,b1,c1}A^{a2,b2,c2}A^{a3,b3,c3} ....

In the case where it is a 2x2x2 symmetric 3-tensor this should give the discriminant of a cubic.

Let
g_n{a1,a2,a3...} =0 if any no n of the indices are the same.

Determinant of a n-tensor is:
det_n(A) = g_n^{a}g_n^{b}g_n^{c}...A^{a1,b1,c1,...}A^{a2,b2,c2,...}A^{a3,b3,c3,...} ....
Properties
Assuming all components of A are independent (not true for symmetric tensors!).

det_1(A) = 0 by definition.

d/dAin d/dAim det_2(A) = 0

d/dAin d/dAim d/Aio det_3(A) = 0

Inverses
We can generalise the definition of an inverse for a 2-tensor:
(A^-1)ij  =   d/dAij log( det_2(A) )
Bijk = d/dAijk log( det_3(A) ) == ???
Cijkl = d/dAijkl log( det_4(A) ) == ???
What properties do these have?





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