pulliing out eps

Note on "pulling out ε" PhL 3.25.12 Suppose you have something like this equation X'n = g'-1/2ε'nac g'cb (g'-1/2ε'bdeB'e;d);a where both sides are vectors in x'-space. An array of questions now arises. Question 1: Can you "pull out" the ε'bde object from the cov deriv? The answer is that you can only "pull out" a scalar constant, not a tensor constant. The reason is that the object ε'bde;a ≠ 0, see tensor doc example. Here are the details right here: Consider (ε'bde Qde);a = (ε'bde);a Qde + ε'bde (Qde);a  But I don't think (ε'bde);a = 0 even through ε' is a constant! Let's see, first do this by hand Babc;α ≡ ∂α Babc + ΓanαBnbc + ΓbnαBanc + ΓcnαBabn Then apply εabc;α ≡ ∂α εabc + Γanαεnbc + Γbnαεanc + Γcnαεabn = Γanαεnbc + Γbnαεanc + Γcnαεabn Ouch. So it only a scalar constant you can "pull out" of a covariant derivative. Question 2: How do you "write out" the residual object (g'-1/2B'e;d);a ? Question 3: Does this residual object even exist? After updating tensor doc Appendix D, these questions now have answers. The object (g'-1/2B'e;d) is a rank-2 tensor density of weight +1 (assuming B has weight 0). Here is what the covariant derivative looks like for a rank-2 covariant tensor density Ced of weight W (from Examples) Ced;a ≡ ∂a Ced – ΓneαCnd – ΓndαCen + W Γκκα Ced Now set Ced = (g'-1/2B'e;d) to get (g'-1/2B'e;d);a =(g'-1/2B'e;d),a – Γ'neα(g'-1/2B'n;d) – Γ'ndα(g'-1/2B'e;n) + W Γ'κκα(g'-1/2B'e;d) and in our case W = 1. So there it is, and it ain't purdy! At least it does exist! Question 4: Can you take the covariant derivative of a tensor density? I learned how to do this and updated Appendix D of tensor doc accordingly on 3.26.12.