Phase choice for 6j symbols

Question

If you define 6j symbols completely formally via trivalent graphs (take http://math.ucr.edu/home/baez/qg-fall2000/qg10.2.html for a start, but be careful - looks like Racah coefficients to me...well, he already mentions "fudge factors" :-) I see no -1^whatever. Now in any standard work (e.g. M. Rotenberg et al.) you see a $(-1)^{S+k}$ in Biedenharn-Elliott, you see a $(-1)^{j1+j2+j3}$ when evaluating a 6j symbol containing a zero... which looks like a completely unnatural phase choice to me, especially given that outside $SU_2$, irreps are not simple integers and you can't take -1 to the power of that anyway.
What looks possible to me is that you could fix a gauge by setting the theta graph colored with $j1,j2,j3$ to $(-1)^{j1+j2+j3}$, and analogous that of $R_1,R_2,R_3$ to, eh, whatever is sensible, and thus have the 3-node 6j symbols concide with the standard ones (or so I think). Still, I find it much more natural to just set all thetas to 1.
Is there a "natural" phase choice? (AFAIK, my spectrocopist colleagues invented the 6j symbols and might had had other things in mind. :-) For $SU_2$? For another Lie group? (Please argue why it's more natural than the one I like best :-)

Answer 1

There are two real forms of $SL(2,C)$, the compact real form $SU(2)$ and the split real form $SL(2,R)$. Mathematicians prefer the split real form and physicists the compact real form. The two cases have different signs. There is also a choice of sign conventions involved even once you have decided which real form you want.

Edit: There is another choice of sign. The two dimensional representation has an invariant symplectic form. Alternatively you can take this to be a super vector space with no even part in which case it has an invariant symmetric inner product. You need the latter convention for diagrams with unoriented edges.