What is the coordinate-free definition of the Levi-Civita Symbol?

Question

For clarity I mean the symbol $\epsilon_{ijk}=e_{ijk}/|g|$ where $g$ is the determinant of the metric tensor and $e_{ijk}$ is anti-symmetric in all its indices and $e_{123}=1$. Is there a purely geometrical interpretation that doesn't rely on defining each component as I have done here? For instance the tensor $\delta^i_j$ can be defined in terms of its components (1 when $i=j$ otherwise 0) or by saying its the identity transform on the tangent space.

An example of the usage is to define a dual vector to a given tensor via $a_i=\epsilon_{ijk}A^{jk}$, so I would also appreciate answers that give a coordinate independent way of understanding this statement.

Answer 1

Ok, I can't quite figure this out till the end, but maybe with the help of comments we can finish it off.

Let's assume for simplicity that $A^{ij}$ is a real matrix, as complex matrices and geometry simultaneously are a bit past my ability to imagine. Anyway, any real matrix can be decomposed into symmetric and antisymmetric part

$$A = \frac{A+A^T}{2} + \frac{A-A^T}{2} = B + C$$

It should be easy to see that Levi-Civita operation on the symmetric part is zero $\epsilon_{ijk}B^{jk} = 0$. So only the anti-symmetric part is left.

In 3D, an arbitrary anti-symmetric matrix can be written with the help of Levi-Civita symbol as

$$C_{ij} = \epsilon_{ijk}c^k$$ where $c^k$ is an arbitrary 3D vector. Thus, the full operation OP has requested can be written as

$$\epsilon_{ijk}A^{jk} = \epsilon_{ijk}C^{jk} = \epsilon_{ijk}\epsilon^{jkl}c_l = 2\delta_i^l c_l = 2c_i$$

Since $C_{ij}x^j = -\vec{c} \times \vec{x}$, the original operation $\epsilon_{ijk}A^{jk} = 2c_i$ can be interpreted as

"For the part of the mapping $A$ that acts as a cross-product, extract negative twice the vector that would be cross-producted with".

I think my wording is a bit clumsy. Perhaps colleagues that are more fresh on fluid dynamics would be able to rewrite the above statement in terms of something like vorticity

Answer 2

Maybe is the evaluation of a basic k-form in a set of $k$ basic vectors:
$$\varepsilon^1\wedge\varepsilon^2\wedge...\wedge\ \varepsilon^k \ (e_{i_1},e_{i_2},...,e_{i_k})=\epsilon_{i_1i_2...i_k}$$ where $\varepsilon^r(e_s)=\delta^r{}_s$, here $\varepsilon^r$ being its dual basis.

Consider the next instance: if we have a base change $$b_1=Ae_1+Be_2$$ and $$b_2=Ce_1+De_2$$ then $$\varepsilon^1\wedge\varepsilon^2(b_1,b_2) =\varepsilon^1\wedge\varepsilon^2(Ae_1+Be_2,Ce_1+De_2).$$ So $$\varepsilon^1\wedge\varepsilon^2(b_1,b_2)=AD-BC$$ and this is geometrical since $AD-BC$ is the area spawned by the new base $b_1,b_2$.