transformation law of vector fields in $R^n$ I asked this on the physics forum but I also want to get a mathematician's perspective if that's okay.
So suppose we have a function $F$ from $R^2$ to $R^2$ defined by $F(x,y) = (g(x,y),h(x,y))$ where $g$ and $h$ represent temperate and pressure respectively (the point is, they are both scalar fields). From the viewpoint of differential geometry the above function can be seen as a (coordinate representation of a) vector field (i.e. a map from the manifold to the tangent bundle); and here's where my confusion lies: Mathematically, since this is a vector field it is forced to obey the vector transformation law, yet physically it's intuitively clear that this is not a vector field but rather just a bunch of scalars so do we implement a vector transformation law or not (under a change of coordinates); does it even make sense to talk about? I'm having trouble putting the rigorous definitions of differential geometry into physical context.
More generally if we have an n-dimensional smooth manifold $M$ (so think of a Riemannian 4-manifold for instance) and a function $F$ from $M$ to $R^n$, will the physical nature of this function (i.e. depending on what physical quantity it represents) govern its transformation behavior? If yes, where is this (if at all) taken into account in the mathematical framework of differential geometry?
 A: From the point of view of differential geometry functions $\mathbb R^2\to\mathbb R^2$, vector fields on $\mathbb R^2$ and one-forms on $\mathbb R^2$ are different types of geometric objects. However but (I would say unfortunately), they can all be identified with functions $\mathbb R^2\to\mathbb R^2$ provided that you make some choices. In the case of $\mathbb R^2$ there are very natural choices and from classical analysis one is used to express everything in terms of these choices.
Now it you take an object that is natural in one of these interpretations and "delibarately" interpret it in a wrong picture, then it becomes much less natural or even meaningless. What you bring up exactly is an exmple for this. The values of $F$ are just pairs of numbers whereas for a vector field, the natural interpretation would be that $F(x,y)$ is a vector (based at $(x,y)$ and hence describing a "direction" in that point). So while you can interpret $F$ as a vector field, this interpretation has no real meaning, because it depends on your specific choice of coordinates.
There is a similar story between one-forms and vector fields. For a real valued function $f:\mathbb R^2\to \mathbb R$, the natural interpretation of the derivative is as a one-form $df$. In this interpretation, the map $f\mapsto df$ is compatible with arbitrary diffeomorphismcs (i.e. changes of coordinates). But it is possible (and common) to convert this into a vector field (using the inner product on $\mathbb R^2$) which leads to the gradient of $f$. But in this interpretation, that map that sends $f$ to its gradient is only compatibie with rigid motions and not with diffeomorphisms.
