# Tensor ordering in index lowering operation

If we take two vectors and want to contract them with the metric tensor to find some frame invariant quantity:

$$A^a B^b g_{ab}=\vec A\cdot \vec B$$

is there a convention on where the metric tensor should be placed in this equation to emphasise that we are contracting one tensor or the other? Because I have seen this equation equivalantly written:

$$g_{ab}A^a B^b=\vec A \cdot \vec B$$

Are there cases where the placement is important or is there some convention on the placement of the metric tensor in equations like this (and in more complicated examples where the result is not just a scalar)?.

• Most people write the $g$ first. It’s just a convention, but a very wide-spread one. – G. Smith Feb 23 at 17:34

There is no difference in that $$g_{ab}A^aB^b = A^ag_{ab}B^B = A^aB^bg_{ab}$$. The only time there would be a difference in an index expression like this is if $$g_{ab}$$ has elements not commuting with what it's been contracted with, but this would not be the case if for example your context is ordinary general relativity.
To give an example of readability, consider the contraction $$g_{ab}R^b_{cde}$$; in this case I've put the indices being contracted as close as possible in the expression, which draws your eye to $$b$$ contracted with $$b$$ faster as opposed to writing say, $$R^{b}_{cde}g_{ab}$$. It's a very minute difference, but this is to illustrate what I mean by readability.
By the Einstein convention there is an implicit sum $$\sum_{a,b=1}^d g_{ab}A^aB^b\,.$$ Clearly you will agree that for any term of the sum $$g_{ab}$$, $$A^a$$ and $$A^b$$ are just real numbers. Therefore they commute as always.