Raising and Lowering Indices of Levi-Civita Symbols (+---) metric?

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.

Interesting question. The correct expression for the general case would be

$$ c^i = g_{jj'} g_{kk'} \; \frac{\varepsilon^{ijk}}{\sqrt{\left| \det g \right|}} \; a^{j'} b^{k'}, $$

or

$$ c^{i} = g^{i i'} \sqrt{\left| \det g \right|} \; \varepsilon_{i' j k} a^{j} a^{k} $$

if you wish to use the Levi-Civita symbol with lower indices.

where $g_{ij}$ is the pull-back of the spacetime metric on the 3-dimensional spatial slice (or you can start with the 3d space from the beginning), and $\det g$ is the determinant of the covariant metric tensor matrix. This expression can be proven to be a tensor, i.e. $c^i$ is a true tensor in the most general meaning of this word.

For special relativity in Minkowski coordinates, you have

$$ g_{ij} = \pm \delta_{ij}, $$

where the $\pm$ stands for the two possible conventions (-+++ and +---).

It is easy to see that in both cases you end up with

$$ c^i = \varepsilon^{ijk} a^j b^k. $$