anonanon 152k1212 gold badges233233 silver badges398398 bronze badges $endgroup$ 4 $begingroup$ That is best many thanks a bunch! I also posted a similar query listed here: math.stackexchange.com/inquiries/2135276/… that You could have some Perception on? $endgroup$

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ε i j k ε l m n =

I'm utilizing the Einstein notation, so multiple indices are summed) or they offer two 3-indices symbols with diverse indicator

These Qualities are important for being familiar with the conduct from the tensor in different coordinate units.

From this actions, we can certainly deduce what $epsilon^ ij4k $ can be devoid of experiencing the computations once again: it is simply the result of exchanging the last two indices of $epsilon^ ijk4 $,

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The Levi-Civita image will allow the determinant of a square matrix, along with the cross products of two vectors in three-dimensional Euclidean Room, to be expressed in Einstein index notation.

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The covariant Levi-Civita tensor (also referred to as the Riemannian quantity variety) in any coordinate process that matches the selected orientation is

The crucial element defining house on the image is full antisymmetry within the indices. When any two indices are interchanged, equivalent or not, the image is negated:

Do you see how this could impact the value of $epsilon^ alphabetagammadelta epsilon_ alpha'beta'gamma'delta' $? $endgroup$