I read some older posts about this question, but I don't know if I'm getting it. I'm working with a Lagrangian involving some Levi Civita symbols, and when I calculate a term containing $\epsilon^{ijk}$ I obtain the contrary sign using $\epsilon_{ijk}$. I always apply the normal rules: $\epsilon_ {ijk}=\epsilon^{ijk}=1$; $\epsilon_ {jik}=\epsilon^{jik}=-1$ etc. I believed that there is no difference between covariant and contravariant Levi-Civita symbol. What do you know about this?
[Physics] Levi Civita covariance and contravariance
covariancetensor-calculus
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This thread on physicsforums elaborates a bit on the difference between Levi-Civita symbols and tensors. Based on that, I conclude...
1) Your index notation formula for the magnetic field should use the Levi-Civita tensor, then. The "symbol" is a convenient thing, but this expression must be written with tensors.
2) Carroll likely made a mistake and meant to talk about the Levi-Civita tensor's transformation properties.
3) Any expression where the same index appears on the bottom twice (or the top twice) is just laziness on the part of the author. It's a common laziness, especially in contexts where one doesn't discriminate between covariant and contravariant components, however.
Actually, I'm not sure what your question is here.
4) Again, I would say that this expression should use the tensor, not the symbol.
5) I see no reason why you wouldn't.
Just some general remarks on the Levi-Civita tensor/symbol and what they represent: a flat space has a unique "volume form" or "pseudoscalar". There is a unit volume element that all other volumes are scalar multiples of. This volume has an orientation (think of a volume spanned by vectors according to the right-hand rule vs. a left-hand rule).
The Levi-Civita tensor and symbol are related to this notion. The tensor represents the components of the unit volume element with respect to volume elements built by combinations of basis vectors.
The volume element can be used to perform duality operations. This is the foundation of the Hodge star operator. Using the $N$-dimensional Levi-Civita tensor on a tensor object with $k$ free indices yields a new object with $N-k$ free indices. This can be described geometrically, too. In 3d, scalars would go to volumes, vectors to planes, planes to vectors, and volumes to scalars under this duality operation. It is the explicit mapping of planes to vectors that is so often performed with duality--it allows us to cheat in 3d by dealing with only scalars and vectors. Planes and volumes can then be mapped back to vectors and scalars by duality. This is exactly what is done with the magnetic field and angular momentum. You should see clearly that both of these vectors, if not for the use of the Levi-Civita tensor, would be expressions with 2 free indices, and antisymmetric on those indices. These objects are called bivectors.
The Levi-Civita tensor and symbol are often used in physics and math to treat expressions through duality rather than directly--even when, in my opinion, this obscures the real physics of the problem or covers up for a shortcoming of the notation. Just the other day around here we had a question about building up 4-volumes from a single plane. Geometrically, this is obvious--you can't build a 4-volume from a single plane. But in index notation, it was cumbersome at best, involving finding the dual plane through use of the Levi-Civita tensor and taking traces.
Overall, the Levi-Civita tensor and its many indices can be difficult to work with, especially in arbitrary coordinate systems. I once heard a professor bemoan that all the identities another professor had taught students with Levi-Civita had only used the symbol--i.e. the tensor in cartesian coordinates--and so they weren't valid in arbitrary coordinate systems. The solution suggested was to teach students about tensor densities, which was met with skepticism at best, since there were only three professors in the whole department that, in the other professor's view, either cared about or even knew about tensor densities. I think part of this view is why the Levi-Civita symbol is often used instead; it's just easier to prove some things in cartesian coordinates, even if the resulting expression is not really correct (not really a tensor because the metric determinant has been ignored, etc.).
Ok. I'm going to put my response as an answer to my question since it involves some new information I've found and want to document here. Part of my confusion has stemmed from the fact that different authors use different notation regarding the transition from Levi-Civita symbols or tensor densities ($\tilde{\epsilon}$) and Levi-Civita tensors ($\epsilon$). Here are the two clearest references I have found on the subject and they illustrate the two possible conventional choices. Sean Carroll's lecture notes on geometry and spacetime ch. 2 and Christopher Pope's Electrodynamics lecture notes. I find Pope's notes to give a little bit more detail which made the difference for me.
The point is the Levi-Civita symbol with the lower indices*, $\tilde{\epsilon}_{ijk}$ is defined as an object which is anti-symmetric in its indices. i.e. $\tilde{\epsilon}_{123}=+1$ whereas $\tilde{\epsilon}_{132}=-1$. Furthermore, (as explained better in the Pope notes) the symbol takes the same value in all coordinate frames. That is if we transform from one set of coordinates $\{x^i\}$ to another set of coordinates $\{x^{'i}\}$ we have transform $\tilde{\epsilon}_{ijk}$ transforms to $\tilde{\epsilon}^{'}_{ijk}$ with the relation that $\tilde{\epsilon}^{'}_{ijk}=\tilde{\epsilon}_{ijk}$
It is then possible to prove, using some facts about determinants, that $$\tilde{\epsilon}_{ijk}=\tilde{\epsilon}^{'}_{ijk} = \left|\frac{\partial x'}{\partial{x}}\right|\frac{\partial x^a}{\partial x^{'i}}\frac{\partial x^b}{\partial x^{'j}}\frac{\partial x^c}{\partial x^{'k}}\tilde{\epsilon}_{abc}$$
That is to say $\tilde{\epsilon}_{ijk}$ transforms like a tensor density of weight +1. It turns out $\sqrt{|\text{det}(g_{ij})|}=\sqrt{|\text{det }g|}$ is a tensor density of weight -1 (as proven in the Pope notes) so by multiplying these two tensor densities you get a new tensor density of weight 0, i.e. a regular tensor (indicated by the absence of the tilde).
$$\epsilon_{ijk} = \sqrt{|\text{det }g|}\tilde{\epsilon}_{ijk}$$
Both authors agree on this much. And so far none of this would have been changed by choosing the (-+++) metric as opposed to the (+---) metric**. However, the next step is finding the upper index object. Since $\epsilon_{ijk}$ is a tensor we can raise its indices:
$$\epsilon^{ijk} = g^{ii'}g^{jj'}g^{kk'}\epsilon_{i'j'k'} = g^{ii'}g^{jj'}g^{kk'}\tilde{\epsilon}_{i'j'k'}\sqrt{|\text{det }g|} \\ =\text{det}(g^{-1}) \sqrt{|\text{det }g|} \tilde{\epsilon}_{ijk} = \frac{\text{sgn}(g)}{\sqrt{|g|}}\tilde{\epsilon}_{ijk}$$
At this point we have
$$\epsilon^{ijk} = \frac{\text{sgn}(g)}{\sqrt{|\text{det }g|}}\tilde{\epsilon}_{ijk}$$
This is where people make a convention choice. Carroll (and many others that I have seen), for example, makes the chioce that $\tilde{\epsilon}^{ijk} = \tilde{\epsilon}_{ijk}$ so that we get
$$\epsilon^{ijk} = \frac{\text{sgn}(g)}{\sqrt{|\text{det }g|}}\tilde{\epsilon}^{ijk}$$
Pope takes the convention that $\tilde{\epsilon}^{ijk} = \text{sgn}(g)\tilde{\epsilon}_{ijk}$ so that
$$\epsilon^{ijk} = \frac{1}{\sqrt{|\text{det }g|}}\tilde{\epsilon}^{ijk}$$
So we've already made at least two convention chioces. The first is how $\epsilon_{ijk}$ relates to $\tilde{\epsilon}_{ijk}$ and the second is how $\tilde{\epsilon}_{ijk}$ relates to $\tilde{\epsilon}^{ijk}$. I think we have yet another choice now of how to define the cross product. I think the responsible thing to do is to keep in mind all of the machinery up to this point and make a manifestly covariant definition of the cross product. This would look like: $$(A\times B)^i = \epsilon^{i}{}_{jk}A^jB^k=g^{ii'}\epsilon_{i'jk}A^j B^k = \pm \epsilon_{ijk}A^j B^k=\pm\tilde{\epsilon}_{ijk}A^j B^k$$
Where $\pm$ indicates the next convention choice of mostly positive or mostly negative metric signature. I think this has the disadvantage that depending on whether you take the mostly positive or mostly negative metric you get a relative minus sign in the definition of the cross product, but I think this is actually to be expected since it moves the spatial part from being a right handed to left handed coordinate system. It also has the disadvantage that you really have to remember quite a few places for negative signs. It has the advantage that is covariant so that if you do follow the rules you can do the manipulations more easily I guess.
For example @Solenodon Paradoxus gives a formula for the "correct expression" for the cross product but I'm not sure on what grounds that is the "correct expression". It follows proper Einstein convention but the symbol used is the Levi-Civita symol which is not a tensor so there's not actually a rule saying the expression SHOULD follow the Einstein convention which leaves confusion as to the motivation behind any particular definitions when there are so many choices.
*I believe this is a matter of convention. We could alternatively take the Levi-Civita symbol with upper indices to be the object which is anti-symmetric in its indices.
**but the story would maybe be different if we had chosen the upper indexed Levi-Civita symbol to be the usual anti-symmetric object instead.
edit1: for less awkward notation we could define $(A \times B)^i = \epsilon^{ijk}A_jB_k$ which would be equivalent to the definition given. For my problem I'm working on I'm thinking of contravariant components of vectors as the "physical" quantities so I'm prefering to write expression in terms of those.
Best Answer
Comment to the question (v2): Apart from the issue of various overall sign conventions found in the literature, note that:
On one hand, there is the Levi-Civita symbol with upper (lower) indices, whose entries are only $0$s and $\pm 1$s; it is a contravariant (covariant) pseudotensor density, respectively.
On the other hand, there is the Levi-Civita tensor with upper (lower) indices, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$; it is a contravariant (covariant) pseudotensor, respectively.