# Transformation law of vector fields on $\mathbb{R}^n$

So suppose we have a function $$F$$ from $$\mathbb R^2$$ to $$\mathbb R^2$$ 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 section of the projection map 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 $$\cal M$$ (so think of a Riemannian 4-manifold for instance) and a function $$F$$ from $$\cal M$$ to $$\mathbb R^n$$, 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?

• The tangent space of a point on manifold needs to be defined before you can talk about vectors. This can be done with directional derivatives. This is also why the dimension of the vector space is the dimension of the manifold. The pressure and temperature have no relationship to tangent space. You could define these on a 1D manifold, but the vector space of a 1D manifold must have dimension 1. Mar 31, 2021 at 2:37
• The transformation law of a vector field on a manifold is that of a contravariant vector (field). Mar 31, 2021 at 2:37
• @Jbag1212 what do you mean the tangent space needs to be defined? It is defined in the obvious way for $R^2$. And I am aware of all that, what i'm saying is this: I have a function as defined above and it can be easily shown to be a section of the projection map from $TR^2$ to $R^2$, so mathematically it satisfies the definition of a vector field (which is just a section of the projection map), so give me a reason why it shouldn't obey the vector transformation law. Mar 31, 2021 at 2:43
• Things obey the vector transformation law if they are vectors which are elements of the tangent space at each point. In $\mathbb{R}^2$ suppose we have a basis for the tangent space, and physically we call it "north-south" and "east-west." Physically speaking, the temperature and pressure do not depend the direction of "north-south" and "east-west." So this vector of temperature and pressure is not in the tangent space. Mar 31, 2021 at 2:52
• In physics we look for physical quantities which transform like tensors. If you were just given a function that you don't actually know what it physically represents, then of course you wouldn't be able to proceed. Just because something has multiple indices doesn't mean that it is a tensor. Spinors are one example, And in answer to your question, yes. Mar 31, 2021 at 3:07

TL;DR - I suspect your confusion lies in the Physics 101 example that e.g. the ordered pair ("temperature","pressure") does not define a vector because when we change our coordinates, temperature and pressure don't transform. However, if we are working in cartesian coordinates, the object (temperature)$$\hat x$$ + (pressure)$$\hat y$$ is a linear combination of our basis vectors, and therefore does transform appropriately. This is just as well-defined as the vector $$\mathbf V = 3\hat x + 4\hat y$$.

In other words, if you write down a physically meaningless (but perfectly well-defined) vector field in some coordinate system, then it will change in the usual way when you move to a different coordinate system.

So suppose we have a function $$F$$ from $$\mathbb R^2$$ to $$\mathbb R^2$$ defined by $$F(x,y)=(g(x,y),h(x,y))$$ where $$g$$ and $$h$$ represent temperature and pressure respectively (the point is, they are both scalar fields).

Okay, subtlety number one: Is the domain of $$F$$ the manifold $$\mathcal M =\mathbb R^2$$, or the image of $$\mathcal M$$ under a cartesian coordinate chart? $$x :\mathcal M \rightarrow \mathbb R^2$$ $$(a,b) \mapsto (a,b)$$

Mathematical Interlude

Despite being one of the simplest possible manifolds, $$\mathbb R^2$$ is actually terrible from a pedagogical point of view precisely because it's so easy to get confused on this issue. The manifold $$\mathcal M = \mathbb R^2$$ is abstract; points $$p\in \mathbb R^2$$ consist of ordered pairs of real numbers $$(a,b)$$, but those numbers are not coordinates for $$p$$. We can introduce coordinates by defining a coordinate chart on some open neighborhood of $$p$$. For example, we might coordinatize the upper half-plane via the polar coordinate chart: $$\pi : \mathbb R_+^2 \rightarrow \mathbb R \times(0,\pi)$$ $$(a,b) \mapsto \left(\sqrt{a^2+b^2},\sin^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)\right)$$ where the first coordinate is interpreted as the radial coordinate and the second as the angular coordinate. Any function which is defined at the manifold level - e.g. some $$f:\mathcal M \rightarrow \mathbb R$$ - has a corresponding expression in each coordinate chart. For example, let $$f:\mathcal M \rightarrow \mathbb R$$ be defined by $$(a,b)\mapsto a$$. If we descend into the polar coordinate chart, we could consider the function $$f_\pi: \mathbb R\times (0,\pi)$$ $$(r,\theta) \mapsto (f\circ \pi^{-1})(r,\theta) = f\big(r\cos(\theta),r\sin(\theta)\big) = r\cos(\theta)$$ $$f_\pi$$ is the expression of the (manifold-level) function $$f$$ in the $$\pi$$ coordinate chart. Changing to a different chart entails mapping points back to $$\mathcal M$$ via $$\pi^{-1}$$, then applying the new coordinate chart. For example, if we wanted to use the cartesian chart defined above, we would have $$f_x = f\circ x^{-1} = f\circ \pi^{-1} \circ \pi \circ x^{-1} = f_\pi \circ (\pi\circ x^{-1})$$ The map $$\pi \circ x^{-1}$$ is called the chart transition map between the cartesian chart $$x$$ and the polar chart $$\pi$$; it is easily seen to be $$\pi \circ x^{-1}: \mathbb R_+^2 \rightarrow \mathbb R\times(0,\pi)$$ $$(a,b) \mapsto \left(\sqrt{a^2+b^2},\sin^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)\right)$$ and so $$f_x : (r,\theta)\in \mathbb R\times(0,\pi) \mapsto r\cos(\theta)$$ as anticipated.

End Interlude

The point of that somewhat length example is that when you say $$F:\mathbb R^2\rightarrow \mathbb R^2$$, its not clear whether you are defining an express at the manifold level - in which case there are no coordinates being used at all, and no transformations to consider - or at the level of (presumably cartesian) coordinates, in which case your $$F$$ is really $$F_x \equiv F \circ x^{-1}$$, and a change of chart is effected by simply inserting a chart transition map, e.g. $$F_\pi \equiv F_x \circ (x\circ \pi^{-1})$$.

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)

Okay. Based on this, I will assume we are working in cartesian coordinates. You are defining a vector field $$\mathbf V$$ on $$\mathbb R^2$$ whose $$x$$-component is the temperature and whose $$y$$-component is the pressure. That defines a little directional derivative which sits at each point, with the vector field being

$$\mathbf V = g(x,y)\frac{\partial}{\partial x} + h(x,y) \frac{\partial}{\partial y}$$

Mathematically, since this is a section of the projection map 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 don't know what you mean here. It is a perfectly well-defined vector field. It doesn't have any physical significance, as far as I can tell, but that doesn't mean it isn't a vector field.

A change of coordinates induces a change of basis, so you're asking whether the components of $$\mathbf V$$ change when we go from the cartesian basis $$\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right\}$$ to e.g. the polar basis $$\left\{\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}\right\}$$, and the answer is obviously yes - the polar coordinate unit vectors generally point in different directions than the cartesian unit vectors, after all. If you replace $$\frac{\partial}{\partial x}\mapsto \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}$$ $$\frac{\partial}{\partial y}\mapsto \frac{\partial r}{\partial y} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}$$ in our original expression for $$\mathbf V$$, then when the dust settles you'll have something of the form

$$\mathbf V = V^r \frac{\partial}{\partial r} + V^\theta \frac{\partial}{\partial \theta}$$

where $$V^{r}$$ and $$V^\theta$$ are some (position-dependent) linear combinations of $$g$$ and $$h$$. That's all the vector transformation rule is - expressing the same vector field using different basis vectors requires different components, which should be fairly obvious.

Actually, this isn't quite right - your functions $$g$$ and $$h$$ are really $$g_x=g\circ x^{-1}$$ and $$h_x = h\circ x^{-1}$$, so under change of chart they would also be replaced by $$g_\pi = g_x \circ (x\circ \pi^{-1})$$ and $$h_\pi = h_x \circ (x\circ \pi^{-1})$$.

More generally if we have an $$n$$-dimensional smooth manifold $$\mathcal M$$ (so think of a Riemannian 4-manifold for instance) and a function $$F$$ from $$M$$ to $$\mathbb R^n$$, will the physical nature of this function (i.e. depending on what physical quantity it represents) govern its transformation behavior?

No. As long as you have a well-defined function at the manifold level, that immediately translates into an expression in whatever coordinate chart you wish to work in. There is nothing transformative about this idea - if you have a point $$p$$ which is being mapped to some other space and you label $$p$$ by some coordinates, then you get a function which eats those coordinates. If you change coordinates, you change the function.

In this case, that other space was the tangent bundle, and we performed a corresponding chart transformation on that (i.e. a change of basis) which was induced by the chart transformation on the manifold $$\mathcal M=\mathbb R^2$$, which added a layer of complexity.

• I posted 4 pages from Chapter 4 of "Flanders" on imgur.com/gallery/CZSR28w . Every point is attached to a local orthogonal frame $\mathbf{e}_i$ where for each $i=1,2,3$ is a smooth vector field. Flanders writes: "What we shall do is express everything in terms of the $\mathbf {e}_i$, ... apply$d$, and $d\mathbf{x}=\sigma_1 \mathbf{e}_1+\sigma_2 \mathbf{e}_2+\sigma_3 \mathbf{e}_3$" Mar 31, 2021 at 13:49
• How can we apply "$d$" to a vector field when it is defined in Chapter 3 for forms but not tangent vectors? What sort of creature $d\mathbf{x}$ is: a covariant or a contravariant vector? Mar 31, 2021 at 13:56
• @hyportnex Flanders does the following. The underlying set for Euclidean space is $\mathbb R^3$; consider a cartesian chart $x$ as per my answer, which eat a point $p$ and spit out its cartesian coordinates, $x(p)=(x^1(p),x^2(p),x^3(p))$. Each $x^i$ is a function from $\mathcal M\rightarrow \mathbb R$, so we can define cooresponding 1-forms $\mathrm dx^i$; we then throw these together into a one-form valued vector $\mathrm d\mathbf x= dx^i \frac{\partial}{\partial x^i}$. Mar 31, 2021 at 14:10
• @hyportnex More formally, $\mathrm d\mathbf x$ is a $(1,1)$-tensor whose cartesian coordinate form is $\mathrm d\mathbf x = \mathrm dx^i \otimes \frac{\partial}{\partial x^i}$, i.e. its components are $(\mathrm d\mathbf x)^i_{\ \ j}= \delta^i_j$. In a different coordinate system, its components will of course be different. Mar 31, 2021 at 14:14
• Thank you for clearing this up. Would you also say that what Flanders does is a bit of a stealthy underhanded way by which he introduces (mixed) tensors? Mar 31, 2021 at 14:24