Del operator - order of operations Reading through an old scientific paper, I've come across some potentially ambiguous notation using the del operator.  In this example, $f$, $g$, and $x$ are scalars, and $\textbf{V}$ is a vector. The equation is:
$$x = \nabla \cdot f \textbf{V} g.$$
Should this be interpreted to indicate:
$$x = \nabla \cdot \left(f g\textbf{V} \right),$$
or does it possibly indicate:
$$x =g \nabla \cdot \left(f \textbf{V}\right)?$$
Is the answer indicated by a strict order of operations argument, or is the first form some sort of commonly known shorthand notation? I'm not sure why the two scalars would be separated if they were both intended to be part of the dot product.
 A: See the discussion on the Del operator in Wikipedia: it helps to disambiguate notation involving the $\nabla$ operator. See in particular the section entitled "Precautions". Then you might want to scroll to the top of the webpage to read the entire entry.
See also the Product Rules when using the $\nabla$ operator.
The use of $\nabla$ and how it operates depends to some degree on the context in which it is used. The "three possible meanings—gradient, divergence, and curl—can be formally viewed as the product of scalars, dot product, and cross product, respectively, of the del "operator" with the field. These formal products do not necessarily commute with other operators or products."
A: Are these things fields on a manifold? I so, the paper may be following the common convention in differential geometry that "vector field" on a manifold means a linear operator that takes scalar fields to scalar fields and satisfies certain conditions. Morally you can think of it as
$$ \mathbf V g \equiv \mathbf V \cdot (\nabla g) $$
which means that $\mathbf V g$ is a scalar field and $f \mathbf V g = f(\mathbf Vg) = (f\mathbf V)g$ is a different scalar field (made by pointwise multiplication of $f$ with $\mathbf V g$.
On the other hand, if that is the meaning, then the dot after the explicit $\nabla$ in your expression is a bit mysterious, and the result of the entire expression ought to be a (co)vector rather than a scalar. So you'll have to decide for yourself from the context whether this suggestion makes sense at all.
