Is this a type-o or intentional?

[tolove]

I've been staring at this confused for a while now, and I've just realized that this might be a type-o. Should I assume that the Levi-Civita symbol is only defined on j,k, and that the i is a type-o, or is there an unwritten rule with this notation that gives the i a meaning?

$\sum$_j$\sum$_k ε_i,j,k ∂_j,k

 

[D H]

Nominally that kind of sum would represent i^th component of a vector. However, since δ_jk is zero if j≠k and ε_ijk is zero if j=k, this is a complicated way of writing the zero vector.

 

[Mark44]

Minor pedantic point: The word is "typo", which is short for typographical error.

 

[tolove]

Nominally that kind of sum would represent i^th component of a vector. However, since δ_jk is zero if j≠k and ε_ijk is zero if j=k, this is a complicated way of writing the zero vector.

Thank you very much! The i isn't being defined because it doesn't matter what i is in this situation.

And to mark... I didn't know that, thanks

 

[Ray Vickson]

I've been staring at this confused for a while now, and I've just realized that this might be a type-o. Should I assume that the Levi-Civita symbol is only defined on j,k, and that the i is a type-o, or is there an unwritten rule with this notation that gives the i a meaning?

$\sum$_j$\sum$_k ε_i,j,k ∂_j,k

No typo: the standard ε-symbol, used, eg., in writing vector cross-products in 3 dimensions, is:
$$\epsilon_{i j k} = \left\{ \begin{array}{rccl}<br /> 1 &\text{ if }& ijk &\text{ is an even permutation of 123}\\<br /> -1&\text{ if }&ijk & \text{ is an odd permutation of 123}\\<br /> 0 &&&\text{ otherwise }<br /> \end{array} \right.$$

So, for example, the ith component of $\vec{C} = \vec{A} \times \vec{B}$ is $C_i = \sum_{j,k} \epsilon_{ijk} A_j B_k$.

BTW: 'type-o' is a category of blood (for blood donations); what you probably mean is 'typo'.