Does the relationship beteeen the Levi-Civita symbol and the Kronecker Delta hold for more than two products?

Question

I know the following relationship between Levi-Civita and the Kronecker Delta, $\text{sgn}_{ijk} ~ \text{sgn}_{imn}= \delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km}. ~~~$But i don not know about such a relation between more than two products. Do such a formula exist?

Answer 1

One generalization are the Cauchy-Binet formulas $$ \det(AB^T)=\sum_{I:1\le i_1<i_2<..<i_m\le n}\det(A_I)\det(B_I) $$ where $A,B$ are $m\times n$ matrices with $m<n$ and $A_I$, $B_I$ are the quadratic sub-matrices consisting of the columns with index $i_1,...,i_m$, so that $(A_I)_{kl}=A_{k,i_l}$.

In the case of $m=2$ you get your original identity $$ \det\pmatrix{⟨A_{1,.},B_{1,.}⟩&⟨A_{1,.},B_{2,.}⟩\\ ⟨A_{2,.},B_{1,.}⟩&⟨A_{2,.},B_{2,.}⟩} =\sum_{i_1<i_2}\det\pmatrix{A_{1,i_1}&A_{1,i_2}\\A_{2,i_1}&A_{2,i_2}}\det\pmatrix{B_{1,i_1}&B_{1,i_2}\\B_{2,i_1}&B_{2,i_2}} $$ where on the left side the of scalar products corresponds to the Kronecker deltas. The sum of determinant products on the right of the general formula can only expressed using epsilon tensors if $m=n-1$, here thus for $n=3$, where the determinants of the pair $i_1,i_2$ can be enumerated by the missing index.