Omega rework

[thurinj]

This is a very long overdue PR for the omega angle calculation and confidence curve plot.

I just finished doing a large swath of validation test to make sure everything is right and follows the Tape and Tape 2016 and Tape and Tape 2017 papers.

Here's an example that reproduces the 2017's Figure 3, where we move the considered point vertically along the center of the lune (full moment tensor space for reference in dotted line):

and the Figure 2 (moving along the equator of the eigenvalue lune):

Before the rework

The _calculate_omega appeared to yield very inconsistent result when compared with the expected analytical solution (the vertical lines are the critical ω_i for the derivative of the fractional volume curve):

After the rework

The issue came from the fact that the original method tried to use the moment tensor as a 6D vector, and some of the computation were not accounting for the off-diag terms of the tensor properly. The new formulation explicitly builds the 3x3 tensor to compute the inter-tensor distance.

This pull request also adds the confidence curve figure, with average confidence score plotted over the shaded area like in Tape and Tape 2016's Figure 1.

To this day, we had made some checks with @SeismoFelix, but only on the DC and FMT space. We now have also the checks on arbitrary coordinates on the lune.

@rmodrak, @SeismoFelix or @ammcpherson, if you want to have a look before I merge that'd be awesome.

ps: @carltape turns out there is another method to compute the critical angle, based on the angular distance between the 6 permutations around the eigensphere. Here an example, say we have $Λ^{DC}$:
$$\Lambda^{DC} = \left(\tfrac{1}{\sqrt{2}},\ 0,\ -\tfrac{1}{\sqrt{2}}\right)$$
if we compute the dot product between the 6 permutations:
(0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0)
we get the following angles:

Pair	Dot product	$\omega$
$(0,1,2)\cdot(0,1,2)$	$+1$	$0°$
$(0,1,2)\cdot(0,2,1)$	$+\tfrac{1}{2}$	$60°$
$(0,1,2)\cdot(1,0,2)$	$+\tfrac{1}{2}$	$60°$
$(0,1,2)\cdot(1,2,0)$	$-\tfrac{1}{2}$	$120°$
$(0,1,2)\cdot(2,0,1)$	$-\tfrac{1}{2}$	$120°$
$(0,1,2)\cdot(2,1,0)$	$-1$	$180°$

[thurinj]

Not sure why the check is failing after the hotfix in commit 53e2091. I am going to try out to close and re-open the PR to check if it solves the issue.